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https://mathoverflow.net/questions/15209 | 17 | Let $f: \mathbb{Z}\_p \rightarrow \mathbb{C}\_p$ be any continuous function. Then Mahler showed there are coefficients $a\_n \in \mathbb{C}\_p$ with
$$
f(x) = \sum^{\infty}\_{n=0} a\_n {x \choose n}.
$$
This is known as the Mahler expansion of $f$. Here, to make sense of $x \choose n$ for $x \not\in \mathbb{Z}$ define
$$
{x\choose n} = \frac{x(x-1)\ldots(x-n+1)}{n!}.
$$
There's an important application to number theory: expressing a function in terms of its Mahler expansion is one step in translating the old fashioned interpolation-based language of $p$-adic $L$-functions into the more modern measure-theoretic language. This application is explained in Coates and Sujatha's book *Cyclotomic Fields and Zeta Values*.
However, when the $p$-adic $L$-function is more complicated than Kubota-Leopoldt's, it seems to me that this "translation" really requires one to be able to write down a Mahler expansion of a function $f$ with much larger domain, e.g. the ring of integers of the completion of the maximal unramified extension of $\mathbb{Q}\_p$ or some finitely ramified extension. (See, for instance, line (8), p. 19 of de Shalit's *Iwasawa Theory of Elliptic Curves with Complex Multiplication*).
I can't find a published reference that these functions really have Mahler expansions. Mahler's paper uses in an essential way that the positive integers are dense in $\mathbb{Z}\_p$, so it doesn't instantly generalize.
So is it true or false that for a ring of integers $\mathcal{O}$ in a finitely ramified complete extension of $\mathbb{Q}\_p$, and a function $f: \mathcal{O} \rightarrow \mathbb{C}\_p$, there is a Mahler expansion as above?
| https://mathoverflow.net/users/1018 | When does a p-adic function have a Mahler expansion? | It is false for the valuation ring in any nontrivial finite extension of $\mathbb{Q}\_p$. The coefficients of the Mahler expansion of a continuous function $\mathcal{O} \to \mathbb{C}\_p$ are determined by its restriction to $\mathbb{Z}\_p$ (they are given as $n$-th differences of the sequence of values on nonnegative integers, in fact). But there are different continuous functions $\mathcal{O} \to \mathbb{C}\_p$ with the same restriction to $\mathbb{Z}\_p$.
Even worse, the Mahler expansions need not even converge because if $x$ is not in $\mathbb{Z}\_p$, the binomial coefficient values may have negative valuation.
**EDIT:** As Kevin Buzzard and dke suggest, one can give a positive answer if your question is interpreted differently. The point of this edit is to make a few explicit remarks in these two directions.
1) If it is known in advance that $f \colon \mathcal{O} \to \mathbb{C}\_p$ is represented by a single convergent power series, then the Mahler expansion of $f|\_{\mathbb{Z}\_p}$ converges to $f$ on all of $\mathcal{O}$. This can be deduced from the theorem that a continuous function $\mathbb{Z}\_p \to \mathbb{C}\_p$ is analytic if and only if the Mahler expansion coefficients $a\_n$ satisfy $a\_n/n! \to 0$ (see Theorem 54.4 in *Ultrametric calculus: an introduction to $p$-adic analysis* by W. H. Schikhof).
2) If one chooses a $\mathbb{Z}\_p$-basis of $\mathcal{O}$, then $f$ can be interpreted as a continuous function $\mathbb{Z}\_p^r \to \mathbb{C}\_p$, and any such function has a multivariable Mahler expansion
$$\sum a\_n \binom{x\_1}{n\_1} \cdots \binom{x\_r}{n\_r},$$
where the sum is over tuples $n=(n\_1,\ldots,n\_r)$ with $n\_i \in \mathbb{Z}\_{\ge 0}$, and $a\_n \to 0$ $p$-adically.
| 19 | https://mathoverflow.net/users/2757 | 15212 | 10,194 |
https://mathoverflow.net/questions/15220 | 49 | Let $K$ be a Galois extension of the rationals with degree $n$. The [Chebotarev Density Theorem](http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem) guarantees that the rational primes that split completely in $K$ have density $1/n$ and thus there are infinitely many such primes. As Kevin Buzzard pointed out to me in a [comment](https://mathoverflow.net/questions/15151/product-of-all-fp-p-prime/15157#15157), there is a simpler way to see that there are infinitely many rational primes that split completely in $K$, namely that the [Dedekind zeta-function](http://en.wikipedia.org/wiki/Dedekind_zeta_function) $\zeta\_K(s)$ has a simple pole at $s = 1$. While this result is certainly much easier to prove than Chebotarev's Theorem, it is still not an elementary proof.
>
> Is there a known elementary proof of the fact that there are infinitely many rational primes that split completely in $K$?
>
>
>
Selberg's elementary proof of Dirichlet's Theorem for primes in arithmetic progressions handles the case where $\text{Gal}(K/\mathbb{Q})$ is Abelian. I don't know anything about the general case. Since Dirichlet's Theorem is stronger than required, it is possible that an simpler proof exists even in the Abelian case.
*Remarks on the meaning of elementary*. I am aware that there is no uniformly recognized definition of "elementary proof" in number theory. While I am not opposed to alternate definitions, my personal definition is a proof which can be carried out in first-order arithmetic, i.e. without quantification over real numbers or higher-type objects. Obviously, I don't require it to be explicitly formulated in that way — even logicians don't do that! Odds are that whatever you believe is elementary is also elementary in my sense.
Kurt Gödel [observed](http://plato.stanford.edu/entries/goedel/#SpeUpThe) that proofs of (first-order) arithmetical facts can be much, much shorter in second-order arithmetic than in first-order arithmetic. This observation explains some of the effectiveness of analytic number theory, which is implicitly second-order. In view of Gödel's observation, it is possible that we have encountered arithmetical facts with a reasonably short second-order proof (i.e. could be found in an analytic number theory textbook) but no reasonable first-order proof (i.e. the production of any such proof would necessarily exhaust all of our natural resources). The above is unlikely to be such, but it is interesting to know that beasts of this type could exist...
| https://mathoverflow.net/users/2000 | Is there an "elementary" proof of the infinitude of completely split primes? | By the primitive element theorem, $K=\mathbb{Q}(\alpha)$ for some nonzero $\alpha \in K$, and we may assume that the minimal polynomial $f(x)$ of $\alpha$ has integer coefficients. Let $\Delta$ be the discriminant of $f$. Since $K/\mathbb{Q}$ is Galois, a prime $p \nmid \Delta$ splits completely in $K$ if and only if there is a degree $1$ prime above $p$, which is if and only if $p | f(n)$ for some $n \in \mathbb{Z}$. Suppose that the set $P$ of such primes is finite. Enlarge $P$ to include the primes dividing $\Delta$. Let $t$ be a positive integer such that $\operatorname{ord}\_p t> \operatorname{ord}\_p f(0)$ for all $p \in P$. For any integer $m$, we have $f(mt) \equiv f(0) \;(\bmod \; t)$, so $\operatorname{ord}\_p f(mt) = \operatorname{ord}\_p f(0)$ for all $p \in P$. But $f(mt) \to \infty$ as $m \to \infty$, so eventually it must have a prime factor outside $P$, contradicting the definition of $P$.
| 71 | https://mathoverflow.net/users/2757 | 15221 | 10,199 |
https://mathoverflow.net/questions/15235 | 4 | More precisely, is there a map of schemes $X$ --> $Y$ such that $f$ gives a homeomorphism between $X$ and a closed subset of $Y$, but the corresponding map on sheaves is not surjective?
| https://mathoverflow.net/users/7 | Homeomorphism onto a closed subset of a scheme that isn't a closed immersion | Yes, for example if $K \subset L$ is an inclusion fields, then the induced map
Spec $L \to $ Spec $K$ is a homeomorphism (both source and target are single points),
but the induced map on sheaves is the given inclusion of $K$ into $L$, which is
surjective only if $K = L$.
For another example, let $X'\to Y$ be a closed immersion of schemes over ${\bar{\mathbb F}}\\_p$, and let $X \to X'$ be the relative Frobenius morphism.
Then $X\to X'$ is a homeomorphism on underlying topological spaces but is not an isomorphism of schemes, and so the composite $X\to Y$ is a closed embedding on underlying spaces but not a closed immersion of schemes.
As one last example, let $X'$ be the cuspidal cubic given by $y^2 = x^3$ in the affine
plane $Y$ (over $\mathbb C$, say), and let $X$ be the normalization of $X'$ (which is just
the affine line). Then $X \to X'$ is a homeomorphism on underlying spaces, but is not
an isomorphism of schemes. The composite $X \to Y$ is thus not a closed immersion,
but induces a closed embedding of underlying topological spaces.
| 8 | https://mathoverflow.net/users/2874 | 15236 | 10,208 |
https://mathoverflow.net/questions/15238 | 5 | If we have a linear recurrence sequence where each term depends on all previous terms, say
$a\_n = \sum\_{k=0}^{n-1} \binom{n}{k} a\_k, \quad a\_0 = 1$
is there any way to estimate the growth of a\_n in terms of a Big-O notation?
I suppose the growth must be super-exponential, because if $a\_1, \ldots, a\_{n-1}$ grows exponentially, say, $q^i$, then we have $a\_n = (q+1)^n - q^n$. Hence The exponent grows from $q$ to $q+1$. But I am not sure if this serves as an argument.
Thanks!
| https://mathoverflow.net/users/3736 | How to estimate the growth of a recurrence sequence | A very powerful way to estimate the growth of a recurrence is to look at the analytic properties of the generating function that it implies. In this case we should take the exponential generating function $f(x) = \sum\_{n \ge 0} \frac{a\_n}{n!} x^n$, giving the identity
$$2f(x) = e^x f(x) + 1$$
hence $f(x) = \frac{1}{2 - e^x}$ (one can also deduce this by a purely combinatorial argument). This function is meromorphic, so the growth rate of $a\_n$ is dictated by the position of its poles. The pole closest to the origin is at $x = \ln 2$, which gives $a\_n \sim \frac{n!}{(\ln 2)^n}$. The other poles contribute similar terms.
The wonderful thing about this technique is that it can work **even if you can't solve for the generating function** because recurrences that imply certain types of identities for the generating function can still control its analytic properties. The best reference I know for this kind of stuff is Flajolet and Sedgewick's *Analytic Combinatorics*, which is available free online.
| 16 | https://mathoverflow.net/users/290 | 15239 | 10,210 |
https://mathoverflow.net/questions/15251 | 5 | Motivation/example. Consider $K = \mathbb{Q}(\sqrt[3]{2})$. This is a number field with ring of integers $O\_K = \mathbb{Z}[\sqrt[3]{2}]$. We have a norm map $N\_{K/\mathbb{Q}}$ which maps $x + y\sqrt[3]{2} + z\sqrt[3]{4}$ to $x^3 + 2y^3 + 4z^6 - 6xyz$; restricting to $\mathcal{O}\_K$ gives of course the same form. Using standard results about factorization of prime ideals, it is not too hard to see which integers are norms of elements in $\mathcal{O}\_K$. Therefore we can get the values of $n$ (with some work...) for which $x^3 + 2y^3 + 4z^3 - 6xyz = n$ has integral solutions, and I guess it is also possible to say something about the solutions for a fixed $n$ - although it is not obvious to me how to do this in general.
The same is of course true for many interesting quadratic forms - to cite a famous example: we can get all positive integers $n$ which can be written as the sum of two perfect squares, or more generally as $x^2 + \alpha y^2$ (for some interesting values of $\alpha$).
Questions. Is this a fruitful method to study diophantine equations? Are there interesting "large" classes of higher degree polynomials/diophantine equations which can be treated by this sort of argument? What is known in general about such "norm forms"? How to decide whether a polynomial is a norm form? Et cetera :)
(I know that it is a bit vague... I didn't find any useful references.)
| https://mathoverflow.net/users/1107 | homogeneous forms as norms | Just a remark : It is certainly interesting to study the equation $N\_{L|K}(x)=y$, where $L|K$ is an extension of number fields, and there must be an extensive literature on the subject. Hasse proved that if such an equation is solvable everywhere locally (at every place of $K$), and if $L|K$ is *cyclic*, then there is a global solution. This is sometimes called the Hasse Norm Principle.
Hasse also found examples where $L|K$ is galoisian and the equation is solvable everywhere locally, but not globally. Even for $K=\mathbb{Q}$, there are infinitely many biquadratic extensions $L$ which provide counter-examples.
| 4 | https://mathoverflow.net/users/2821 | 15253 | 10,218 |
https://mathoverflow.net/questions/15265 | 3 | As someone who mostly does symbolic computation, I've always been puzzled by the fascination mathematicians seem to have with Lp(R) (for p<∞)? To be more precise, there are **no** non-trivial polynomials in that space and, to me, polynomials are not only the simplest functions, they are the building blocks of most everything which can be (easily) manipulated algorithmically. And *restricting to a compact support* is really a non-answer, since one of the great things about polynomials is that they are global, analytic functions.
To ask a more precise question: are there some spaces of (total, real-valued) functions which are both nice from a functional analysis point of view, and contain all the polynomials?
| https://mathoverflow.net/users/3993 | Polynomials and L^p(R) | You are referring to $L^p(\mathbb{R}, \mathcal{B}, \mu)$ in the case that $\mathbb{R}$ is endowed with Lebesgue measure $\mu$. Consider instead the measure $\nu$ given by $d\nu = f d\mu$, where $f$ is in the [Schwartz space](http://en.wikipedia.org/wiki/Schwartz_space) and $f$ does not take the value zero. Because the product of a polynomial with $f$ is also in the Schwartz space, and the Schwartz space is contained in $L^p(\mathbb{R}, \mathcal{B}, \mu)$, it follows that polynomials are in $L^p(\mathbb{R}, \mathcal{B}, \nu)$.
| 13 | https://mathoverflow.net/users/1847 | 15267 | 10,229 |
https://mathoverflow.net/questions/15082 | 22 | Artin has a theorem (10.1 in Laumon, Moret-Bailly) that if $X$ is a stack which has separated, quasi-compact, representable diagonal and an fppf cover by a scheme, then $X$ is algebraic. Is there a version of this theorem that holds for fpqc covers?
| https://mathoverflow.net/users/28 | fpqc covers of stacks | It is false. I'm not sure what the comment about algebraic spaces has to do with the question, since algebraic spaces do admit an fpqc (even \'etale) cover by a scheme. This is analogous to the fact that the failure of smoothness for automorphism schemes of geometric points is not an obstruction to being an Artin stack. For example, $B\mu\_n$ is an Artin stack over $\mathbf{Z}$ even though $\mu\_n$ is not smooth over $\mathbf{Z}$ when $n > 1$. Undeterred by this, we'll make a counterexample using $BG$ for an affine group scheme that is fpqc but not fppf.
First, we set up the framework for the counterexample in some generality before we make a specific counterexample. Let $S$ be a scheme and $G \rightarrow S$ a $S$-group whose structural morphism is affine. (For example, if $S = {\rm{Spec}}(k)$ for a field $k$ then $G$ is just an affine $k$-group scheme.) If $X$ is any $G$-torsor for the fpqc topology over an $S$-scheme $T$ then the structural map $X \rightarrow T$ is affine (since it becomes so over a cover of $T$ that splits the torsor). Hence, the fibered category $BG$ of $G$-torsors for the fpqc topology (on the category of schemes over $S$) satisfies effective descent for the fpqc topology, due to the affineness requirement.
The diagonal $BG \rightarrow BG \times\_S BG$ is represented by affine morphisms since for any pair of $G$-torsors $X$ and $Y$ (for the fpqc topology) over an $S$-scheme $T$, the functor ${\rm{Isom}}(X,Y)$ on $T$-schemes is represented by a scheme affine over $T$. Indeed, this functor is visibly an fpqc sheaf, so to check the claim we can work locally and thereby reduced to the case $X = Y = G\_T$ which is clear.
Now impose the assumption (not yet used above) that $G \rightarrow S$ is fpqc. In this case I claim that the map $S \rightarrow BG$ corresponding to the trivial torsor is an fpqc cover. For any $S$-scheme $T$ and $G$-torsor $X$ over $T$ for the fpqc topology, the functor
$$S \times\_{BG} T = {\rm{Isom}}(G\_T,X)$$
on $T$-schemes is not only represented by a scheme affine over $T$ (namely, $X$) but actually one that is an fpqc cover of $T$. Indeed, to check this we can work locally over $T$, so passing to a cover that splits the torsor reduces us to the case of the trivial $G$-torsor over the base (still denoted $T$), for which the representing object is $G\_T$.
So far so good: such examples satisfy all of the hypotheses, and we just have to prove in some example that it violates the conclusion, which is to say that it does not admit a smooth cover by a scheme. Take $S = {\rm{Spec}}(k)$ for a field $k$, and let $k\_s/k$ be a separable closure and $\Gamma = {\rm{Gal}}(k\_s/k)^{\rm{opp}}$. (The "opposite" is due to my implicit convention to use left torsors on the geometric side.) Let $G$ be the affine $k$-group that "corresponds" to the profinite group $\Gamma$ (i.e., it is the inverse limit of the finite constant $k$-groups $\Gamma/N$ for open normal $N$ in $\Gamma$). To get a handle on $G$-torsors, the key point is to give a more concrete description of the ``points'' of $BG$.
Claim: If $A$ is a $k$-algebra and $B$ is an $A$-algebra, then to give a $G$-torsor structure to ${\rm{Spec}}(B)$ over ${\rm{Spec}}(A)$ is the same as to give a right $\Gamma$-action on the $A$-algebra $B$ that is continuous for the discrete topology such that for each open normal subgroup $N \subseteq \Gamma$ the $A$-subalgebra $B^N$ is a right $\Gamma/N$-torsor (for the fpqc topology, and then equivalently the \'etale topology).
Proof: Descent theory and a calculation for the trivial torsor. QED Claim
Example: $A = k$, $B = k\_s$, and the usual (right) action by $\Gamma$.
Corollary: If $A$ is a strictly henselian local ring then every $G$-torsor over $A$ for the fpqc topology is trivial.
Proof: Let ${\rm{Spec}}(B)$ be such a torsor. By the Claim, for each open normal subgroup $N$ in $\Gamma$, $B^N$ is a $\Gamma/N$-torsor over $A$. Since $A$ is strictly henselian, this latter torsor is trivial for each $N$. That is, there is a $\Gamma/N$-invariant section $B^N \rightarrow A$. The non-empty set of these is finite for each $N$, so by set theory nonsense with inverse limits of finite sets (ultimately not so fancy if we take $k$ for which there are only countably many open subgroups of $\Gamma$) we get a $\Gamma$-invariant section $B \rightarrow A$. QED Corollary
Now suppose there is a smooth cover $Y \rightarrow BG$ by a scheme. In particular, $Y$ is non-empty, so we may choose an open affine $U$ in $Y$. I claim that $U \rightarrow BG$ is also surjective. To see this, pick any $y \in U$ and consider the resulting composite map
$${\rm{Spec}} \mathcal{O}\_{Y,y}^{\rm{sh}} \rightarrow BG$$
over $k$. By the Corollary, this corresponds to a trivial $G$-torsor, so it factors through the canonical map ${\rm{Spec}}(k) \rightarrow BG$ corresponding to the trivial $G$-torsor. This latter map is surjective, so the assertion follows. Hence, we may replace $Y$ with $U$ to arrange that $Y$ is affine. (All we just showed is that $BG$ is a quasi-compact Artin stack, if it is an Artin stack at all, hardly a surprise in view of the (fpqc!) cover by ${\rm{Spec}}(k)$.)
OK, so with a smooth cover $Y \rightarrow BG$ by an affine scheme, the fiber product
$$Y' = {\rm{Spec}}(k) \times\_{BG} Y$$
(using the canonical covering map for the first factor) is an affine scheme since we saw that $BG$ has affine diagonal. Let $A$ and $B$ be the respective coordinate rings of $Y$ and $Y'$, so by the Claim there is a natural $\Gamma$-action on $B$ over $A$ such that the $A$-subalgebras $B^N$ for open normal subgroups $N \subseteq \Gamma$ exhaust $B$ and each $B^N$ is a $\Gamma/N$-torsor over $A$. But $Y' \rightarrow {\rm{Spec}}(k)$ is smooth, and in particular locally of finite type, so $B$ is finitely generated as a $k$-algebra. Since the $B^N$'s are $k$-subalgebras of $B$ which exhaust it, we conclude that $B = B^N$ for sufficiently small $N$. This forces such $N$ to equal $\Gamma$, which is to say that $\Gamma$ is finite.
Thus, any $k$ with infinite Galois group does the job. (In other words, if $k$ is neither separably closed nor real closed, then $BG$ is a counterexample.)
| 38 | https://mathoverflow.net/users/3927 | 15269 | 10,230 |
https://mathoverflow.net/questions/15261 | 4 | Given a locally free sheaf $M$ on $\mathbb{P}^2$ with $h^0(M)=1$.
Is it true that we have $h^2(M)=0$ in this case?
I got this idea from Friedman's book "Algebraic Surfaces and holomorphic vector bundles".
In Chapter 4, p.109, Ex. 4 he wrote: $h^0(Hom(V,V))=1$, by Serre duality $h^2(Hom(V,V))=h^0(Hom(V,V)\otimes K)=0$ since $K=O(-3)\subset O$.
But I don't see why $h^0(Hom(V,V)\otimes K)=0$ follows from $K=O(-3)\subset O$. I mean it could stil have dimension one or is it because $O(-3)$ has no global sections? Would $h^0(Hom(V,V)\otimes O(-i))=0$ still be true for $i=1,2$? Can one generalize to arbitrary locally free sheaves or is this only correct in this special case, i.e. $V$ stable?
${\bf Edit:}$ Given a simple sheaf $V$ on $\mathbb{P}^2$, Bjorn's answer shows $H^0(Hom(V,V)\otimes O(-i))=0$ for $i>0$ which can be written as $Hom(V,V(-i))=0$ for $i>0$. Can this be generalized?
For example given a sheaf of rings or algebras $R$ and a simple left $R$-module M, do we always have $Hom\_R(M,M(-i))=0$ for $i>0$? Or do I need to be more careful in this case?
I remembered this question reading arxiv.org/abs/0810.0067, page 8, where such an equality shows up, without further explanation, so i thought the argumentation should carry over to this more general case.
| https://mathoverflow.net/users/3233 | Serre duality and low dimensional cohomology groups | 1) A locally free sheaf $M$ on $\mathbb{P}^2$ with $h^0(M)=1$ need not satisfy $h^2(M)=0$. For example, if $M=\mathcal{O} \oplus \mathcal{O}(-n)$ for $n>3$, then $h^0(M) = 1 + 0 = 0$, but $M^\vee = \mathcal{O} \oplus \mathcal{O}(n)$ and $K=\mathcal{O}(-3)$, so Serre duality shows that $h^2(M) = h^0(M^\vee \otimes K) = h^0(\mathcal{O}(-3)) + h^0(\mathcal{O}(n-3)) > 0$.
2) On the other hand, the statement is true when $M = \mathbf{Hom}(V,V)$ and $h^0(M)=1$. Here is why. First, $M$ is self-dual, so in this case Serre duality says that $h^2(M) = h^0(M \otimes K)$. If $V$ were a nonzero vector space over a field $k$, there would be a natural map $k \hookrightarrow \operatorname{Hom}(V,V)$ whose image consists of the multiplications by scalars; similarly, for a nonzero vector bundle $V$, there is a natural map $\mathcal{O} \hookrightarrow M = \mathbf{Hom}(V,V)$, and the global section $1$ of $\mathcal{O}$ maps to an everywhere nonvanishing global section of $M$. In your setting, $h^0(M)=1$, so this global section spans $H^0(M)$. Because this global section is everywhere nonvanishing, it cannot be in the image of $H^0(M \otimes \mathcal{O}(-i)) \hookrightarrow H^0(M)$ for any $i>0$. In particular, $H^0(M \otimes K)=0$, so $h^2(M)=0$ by the Serre duality argument.
3) The conclusion $h^2(M)=0$ can fail if you replace $V$ by a nonzero vector bundle for which $h^0(\mathbf{Hom}(V,V)) \ne 1$. For instance, if $V=\mathcal{O} \oplus \mathcal{O}(n)$ for $n \ge 3$, then $\mathbf{Hom}(V,V) = \mathcal{O} \oplus \mathcal{O}(n) \oplus \mathcal{O}(-n) \oplus \mathcal{O}$, which after twisting by $\mathcal{O}(-3)$ still has nonzero global sections.
| 15 | https://mathoverflow.net/users/2757 | 15274 | 10,232 |
https://mathoverflow.net/questions/15273 | 10 | Hi,
I came across the space $BU\_\otimes$ when struggling with twisted K-theory. Segal proved that this is an H-space, right? I have read a dozen times by now that the group $[X, BU\_\otimes]$ consists of the vector bundles of virtual dimension 1. I can clearly see the semi-group structure there, but what does the inverse of such a bundle look like?
Best regards,
Ulrich Pennig
| https://mathoverflow.net/users/3995 | BU with tensor product H-space structure | I'll write $U(X)=[X,BU\_\otimes]$. So $U(X)\subset K(X)=[X,Z\times BU]$ is the multiplicatively closed subset corresponding to "virtual bundles of rank 1".
Likewise, I'll write $I(X)=[X,BU]\subset K(X)$ for the ideal of "virtual bundles of rank 0".
There's a bijection $a\mapsto 1+a$ from $I(X)$ to $U(X)$.
Here's the claim: if $X$ is compact (let's say it's a finite CW-complex of dimension $n$), then $I(X)$ is a nilpotent ideal (in fact, $I(X)^{n+1}=0$). This is a generic fact about multiplicative cohomology theories, and it can be proved in a number of ways; you can think about it in terms of the multiplicative properties of the Atiyah-Hirzebruch spectral sequence, for instance.
For such $X$, it is clear that elements of $1+a\in U(X)$ are invertible, given by the series $(1+a)^{-1}=1+a+a^2+\cdots$ which terminates since $a\in I(X)$ and $I(X)$ is nilpotent.
For infinite dimensional $X$, you need to argue a little harder. If $X=\lim\_{\to} X\_i$ where the $X\_i$ are finite CW-complexes, then there is a surjection $K(X)\to \lim\_{\leftarrow} K(X\_i)$, which restricts to surjections for $I(X)$ and $U(X)$.
The kernel is a $\lim^1$-term. If the $\lim^1$-term vanishes, then it's clear that elements of $U(X)$ are invertible, since their images in the $U(X\_i)$ are invertible. It's enough to check that $\lim^1$-vanishes in the case that $X=BU\_\otimes$, which is a standard calculation.
| 14 | https://mathoverflow.net/users/437 | 15275 | 10,233 |
https://mathoverflow.net/questions/15271 | 0 | (I removed my motivation because it may be misleading :) )
Let $A$ be a noetherian commutative ring and let $M \neq 0$ be a finitely generated zero-dimensional (i.e. $\mathrm{dim} \ \mathrm{Supp}(M) = 0$) $A$-module. Then the submodule $0 < M$ has primary decomposition $0 = \bigcap\_{\mathfrak{p} \in \mathrm{Supp}(M)} M(\mathfrak{p})$, where $M(\mathfrak{p})$ is the $\mathfrak{p}$-primary component of $M$, i.e. the kernel of the canonical morphism $M \rightarrow M\_{\mathfrak{p}}$. I have proven (it's hopefully correct) that the canonical morphism $\mathbf{j}:M \rightarrow \bigoplus\_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$ ist an isomorphism and that $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) \leq M/M(\mathfrak{q}) \leq \bigoplus\_{\mathfrak{p} \in \mathrm{Supp}(M)} M/M(\mathfrak{p})$, where $M \lbrack \mathfrak{q}^\infty \rbrack$ is the $\mathfrak{q}$-torsion part of $M$ ($0$-th local cohomology with support $\mathfrak{q}$). Now, my question is if in general $\mathbf{j}(M \lbrack \mathfrak{q}^\infty \rbrack) = M/M(\mathfrak{q})$ so that $M = \bigoplus\_{\mathfrak{p} \in \mathrm{Supp}(M)} M \lbrack \mathfrak{p}^\infty \rbrack$, or do I need to assume that $A$ is a Dedekind ring (and perhaps also that $M$ is a torsion module) and if so, how can I prove this?
| https://mathoverflow.net/users/717 | Primary decomposition of zero-dimensional modules | If $M$ is finitely generated and has $0$-dimensional support, then $M$ is in fact supported at finitely many maximal ideals (a $0$-dimensional closed subset of the Spec of a Noetherian ring is just a finite union of maximal ideals), and one has the isomorphism
$M = \oplus\_{\mathfrak p} M\_{\mathfrak p} = \oplus\_{\mathfrak p} M[\mathfrak p^{\infty}].$
The first equality just specifies the fact that, since the quasi-coherent sheaf on Spec $A$
attached to $M$ is supported at finitely many closed points, it is a sky-scraper sheaf at these points, and its global sections are just the sum of its stalks at those finitely many points. (In particular, at all ${\mathfrak p}$ not in the support of $M$, the localization
$M\_{\mathfrak p}$ vanishes, and so does not contribute to the direct sum, so really the direct sum is just over the finitely many points in the support.)
To see the second equality (which is the crux of the question as far as I can tell), note
that we reduce to the local case: $M\_{\mathfrak p}$ is finitely generated over $A\_{\mathfrak p}$, and has support equal to the closed point ${\mathfrak p}$ of Spec $A\_{\mathfrak p}$.
A consideration of the very definition of support will now show that each element of $M\_{\mathfrak p}$ is
annihilated by some power of ${\mathfrak p}$, and hence that $M\_{\mathfrak p} =
M[\mathfrak p^{\infty}]$.
From this decomposition everything else is easy to work out: for example,
$M(\mathfrak p)$ (which if I understand correctly is defined to be the kernel of
the map to $M\_{\mathfrak p}$) is precisely $\oplus\_{\mathfrak q \neq \mathfrak p}
M\_{\mathfrak q} = \oplus\_{\mathfrak q \neq \mathfrak p} M[\mathfrak q^{\infty}].$
In particular, if we want to isolate $M[\mathfrak q^{\infty}],$ we just intersect
the $M(\mathfrak p)$ for all $\mathfrak p \neq \mathfrak q$. This explains the
formula in Arminius's answer.
In short: rather than having to memorize or quote technical results from Eisenbud (or elsewhere),
one can use geometric reasoning on Spec $A$ to answer such questions. (All the technical reasoning has been embedded in one step: the proof of the correspondence between $A$-modules and quasi-coherent sheaves on Spec $A$.)
| 6 | https://mathoverflow.net/users/2874 | 15283 | 10,239 |
https://mathoverflow.net/questions/15297 | 0 | In more rigorous language:
" **V**: a vector space having an uncountable base
**S**: The set of subspaces of **V** that have countable dimension.
Can we construct explicitly a chain in the poset **S** (ordered by inclusion), such that this chain has NO upper bound in **S**? "
Apparently, this chain must have uncountable terms. Also,because S doesn't satisfy Zorn's lemma, we know such chain must exist in **S**.
But how do we construct it?
| https://mathoverflow.net/users/4000 | How do we construct (in a vector space) a chain of countable dimensional subspaces that can only be bounded by an subspace of uncountable dimension? | The other answers asked you first to well-order the whole vector space (or a basis for it), and those answers are perfectly correct, but perhaps you don't like well order the whole space. So let me describe a construction that appeals directly to the Axiom of Choice.
Let V be your favorite vector space having uncountable dimension. For each countable dimension subpace W, let aW be an element of V that is not in W. Such a vector exists, since W is countable dimensional and V is not, and we choose such elements by the Axiom of Choice.
Having made these choices, the rest of the construction is now completely determined. Namely, we construct a linearly ordered chain of countable dimensional spaces, whose union is uncountable dimension. Let V0 be the trivial subspace. If Vα is defined and countable dimensional, let Vα+1 be the space spanned by Vα and the element aVα. If λ is a limit ordinal, let Vλ be the union of all earlier Vα. It is easy to see that { aVβ | β < α} is a basis for Vα. Thus, the dimension of each Vα is exactly the cardinality of α. In particular, if ω1 is the first uncountable ordinal, then Vω1 will have uncountable dimension, yet be the union of all Vα for α < ω1, which all have countable dimension, as desired.
If you forbid one to use the Axiom of Choice, then it is no longer true that every vector space has a basis (since it is consistent with ZF that some vector spaces have no basis), and the concept of dimension suffers in this case. But some interesting things happen. For example, it is consistent with the failure of AC that the reals are a countable union of countable sets. R = U An, where each An is countable. (The irritating difficulty is that although each An is countable, one cannot choose the functions witnessing this uniformly, since of course R is uncountable.) But in any case, we may regard R as a vector space over Q, and if we let Vn be the space spanned by A1 U ... U An, then we can still in each case make finitely many choices to witness the countability and conclude that each Vn is countable dimensional, but the union of all Vn is all of R, which is not countable dimensional.
| 7 | https://mathoverflow.net/users/1946 | 15302 | 10,254 |
https://mathoverflow.net/questions/15309 | 28 | [Qiaochu Yuan](https://mathoverflow.net/users/290/qiaochu-yuan) in his [answer](https://mathoverflow.net/questions/15292/why-cant-there-be-a-general-theory-of-nonlinear-pde/15298#15298) to [this question](https://mathoverflow.net/questions/15292/why-cant-there-be-a-general-theory-of-nonlinear-pde) recalls a [blog post](http://terrytao.wordpress.com/2009/11/05/the-no-self-defeating-object-argument/) (specifically, comment 16 therein) by Terry Tao:
>
> For instance, one cannot hope to find an algorithm to determine the existence of smooth solutions to arbitrary nonlinear partial differential equations, because it is possible to simulate a Turing machine using the laws of classical physics, which in turn can be modeled using (a moderately complicated system of) nonlinear PDE
>
>
>
* Is this "it is possible" an application of a *Newton thesis*, much as the usual [Church-Turing thesis](http://en.wikipedia.org/wiki/Church%E2%80%93Turing_thesis), going along the lines of «if something is imaginably doable in real life, one can simulate it with the usual equations of physics», or has the simulation actually been actually carried out?
* Does one need PDEs to simulate Turing machines, or are ODEs good enough?
| https://mathoverflow.net/users/1409 | Simulating Turing machines with {O,P}DEs. | [ODEs are good enough.](https://doi.org/10.1007/11494645_21) Your comment got me started digging.
| 28 | https://mathoverflow.net/users/1847 | 15310 | 10,261 |
https://mathoverflow.net/questions/15031 | 10 | ### Background
The (rational) Calogero-Moser system is the dynamical system which describes the evolution of $n$ particles on the line $\mathbb{C}$ which repel each other with force proportional to the cube of their distance. If the particles have (distinct!) position $q\_i$ and momentum $p\_i$, then the Hamiltonian which describes this system is
$$ H=\sum\_i p\_i^2+\sum\_{i\neq k}\frac{1}{(x\_i-x\_k)^2} $$
There are many interesting properties of this system, but one of the first interesting properties is that it is `completely integrable'. This means that solving it explicitly amounts to solving a series of straight-forward integrals.
The integrability can most easily be shown by showing that the phase space for this system includes into a symplectic reduction of a certain matrix space, and then noticing that the above Hamiltonian is a restriction of a integrable Hamiltonian on the whole space. This is done by assigning to any ensemble of points $q\_i$ and momenta $q\_i$ a pair of $n\times n$ matrices $X$ and $Y$, where $X$ is the diagonal matrix with $q\_i$ on the diagonal entries, while $Y$ is given by
$$ Y\_{ii}=p\_i, \; Y\_{ik}=(x\_i-x\_k)^{-1}, \; i\neq k $$
This matrix assignment defines a map from the configuration space $CM\_n$ of the CM system to the space of pairs of matrices. The space of pairs of matrices $(X,Y)$ is naturally a symplectic space from the bilinear form $(X,Y)\cdot (X',Y')=Tr(XY')-Tr(X'Y)$, and the action of $GL\_n$ by simultaneous conjugation naturally has a moment map. Therefore, we sympletically reduce the space of pairs of matrices at a specific coadjoint orbit (not the origin) and get a new symplectic space $\overline{CM}\_n$.
Composing the above matrix assignment with symplectic reduction, we get a map $CM\_n\rightarrow \overline{CM}\_n$. This map turns out to be a symplectic inclusion which has dense image. We also notice that the functions $Tr(Y^i)$, as $i$ goes from $1$ to $n$, descend to a Poisson-commuting family of functions on $\overline{CM}\_n$, and because $\overline{CM}\_n$ is $2n$ dimensional, each of the functions $Tr(Y^i)$ gives an integrable flow on $\overline{CM}\_n$. Finally, we notice that $Tr(Y^2)$ restricts to $H$ on $CM\_n$.
### The Massive Version of the CM System
Now, make the following change to the system. To every particle, assign a number $m\_i$ (the *mass*), which can be in $\mathbb{C}$, but I am interested in the case where the $m\_i$ are positive integers. Define a the massive CM Hamiltonian as
$$H\_m=\sum\_i\frac{p\_i^2}{m\_i}+\sum\_{i\neq k}\frac{m\_im\_k}{(x\_i-x\_k)^2} $$
The physical meaning of this equation is that particles still have force proportional to the inverse of the cube of their distance, but the force is proportional to the *mass* of that particle; also, particles resist acceleration proportional to their mass. If the force were to drop off proportional to the inverse square of their distance, and attract instead of repel, this would model how massive particles move under the influence of gravity.
### Questions
1. Is this system integrable?
2. Can it be realized in a similar matrix form?
3. Does it have any interesting or new behavior than the usual CM system?
### An Idea
It is *almost* possible to realize this Hamiltonian in a simple modification of the previous approach. Let $M$ denote the diagonal matrix with the $m\_i$s on the diagonal. Then
$$Tr(MYMY)=\sum\_im\_i^2p\_i^2+\sum\_{i\neq k}\frac{m\_im\_k}{(x\_i-x\_k)^2}$$
The functions $Tr( (MY)^i)$ should again be a Poisson commutative family. Rescaling the $p\_i$ by $m\_i^{3/2}$ gives the massive Hamiltonian $H\_m$; **however**, this rescaling is not symplectic, and so it won't preserve the flows.
### Another Idea
In the case of integer $m\_i$, one possibility is to work with $N\times N$ matrices rather than $n\times n$ matrices, where $N=\sum m\_i$. Then it is possible to construct a matrix $X$ with eigenvalues $q\_i$, each occuring with multiplicity $m\_i$, as well as a matrix $Y$ such that $(X,Y)$ defines a point in $\overline{CM}\_N$. The Hamiltonian $Tr(Y^2)$ even restricts to the correct 'massive' Hamiltonian $H\_m$. **However**, the flow described by this Hamiltonian on $\overline{CM}\_n$ will in almost all cases immediately separate eigenvalues that started together, which we don't want. If we restrict the Hamiltonian to the closed subspace where the eigenvalues are required to stay together, then this gives the desired flow. Unfortunately, restricting to a closed subvariety doesn't preserve a Hamiltonian being integrable.
| https://mathoverflow.net/users/750 | Is the 'massive' Calogero-Moser system still integrable? | The paper "Meromorphic Parametric Non-Integrability, the Inverse Square Potential" by E. J. Tosel, proves almost what was claimed in the comments. Except for Jacobi's theorem:
>
> The 3-body problem on a line with *arbitrary masses* and
> inverse square potential is completely integrable with rational first integrals.
>
>
>
and Calogero-Moser’s Theorem:
>
> The n-body problem with *equal masses* on a line with an inverse square potential is completely integrable. More precisely, there exists a complete family of commuting first integrals which are rational in $(Q,P)$.
>
>
>
all other cases are non-integrable. The main theorem is:
>
> **Theorem 3** (Non-integrability meromorphic in linear momenta and masses, rational
> in positions).
>
>
> (i) For $n = 4$, the $n$-body problem on a line with an inverse square potential does
> not have a complete system of generically independent first integrals which are
> rational with respect to $Q$ and meromorphic with respect to $P$ and $(m \_i) \_{1\le i\le n}$
>
>
> (ii) For $n = 3$ and $p \geq 2$, the $n$-body problem in $\mathbb{R} ^p$ with an inverse square potential does not have a complete system of generically independent first integrals
> which are rational with respect to $Q$ and meromorphic with respect to $P$ and
> $(m \_i) \_{1\le i\le n}$
>
>
> A corollary of this theorem is that Calogero-Moser’s theorem deals with an
> exceptional case: there cannot exist a Lax pair $(L,B)$ which would depend meromorphically
> on the masses for $n$ bodies on a line ($n \geq 4$).
>
>
>
There is a remark there that the rationality condition in the case of the line needs only be checked in $n-4$ positions.
| 7 | https://mathoverflow.net/users/2384 | 15315 | 10,265 |
https://mathoverflow.net/questions/15322 | 8 | I'm reading J.P. May's [Concise Course in Algebraic Topology](http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf "Concise Course in Algebraic Topology"), and I'm having a lot of trouble visualizing how things work in Chapter 8, "Based cofiber and fiber sequences". Of course this is pretty basic stuff, but it's really cool to me that there are such clear topological analogues to the usual exact sequences in homological algebra. Still, I can't even get a clear picture of what a smash product looks like for any but the most basic of spaces, and based cones/suspensions/loopspaces make my head hurt.
a) Will I be alright if in my head I just sort think of a smash product as a usual product (for example), with the understanding that I need to tack on an extra condition that I really shouldn't think too hard about?
b) Why all the fuss about based homotopy theory, anyways?
c) While I'm at it, can anyone suggest a book that is less terse? I feel like this one rarely gives the motivation and visual intuition that I'd like...
| https://mathoverflow.net/users/303 | visualizing what's going on in based homotopy theory, et al. | Another book with pictures of reduced suspension etc. is Ronnie Brown's Topology and Groupoids.
see <http://www.bangor.ac.uk/~mas010/topgpds.html>. Which is also excellent for non-based stuff.
Don't believe all you hear about the unbased case being grotty! It is beautiful, but it is possibly easier to learn Alg. Top. in the based situation first, especially if it is that the someone else has decided you should do. :-)
I have put some material that might help (Abstract Homotopy...) on my n-Lab home page (follow the links from Tim Porter (found by a search)).
For a) I had something like your problem when I started, but then thought of the based cylinder as a cylinder with a long base point! Of course, you really need to squidge that line to a point. It is safe when mapping out of a smash to retain the subspace that is to be squidged just always mapping all of it to the base. (In other words, don't agonise about the smash at this stage. Use it as a device for the moment and after you learn to use it and see how it behaves its strangeness will probably have dissapated.)
| 6 | https://mathoverflow.net/users/3502 | 15324 | 10,269 |
https://mathoverflow.net/questions/15314 | 3 | Sorry for the impreciseness of the title. It is merely meant for an analogy.
Exchange of limiting operations and integrations are basically derived from Lebesgue's dominated convergence theorem. For instance, let $f: \mathbb{R}^2 \to \mathbb{R}$ be Borel measuable. Let $f(\cdot, u) \in C^k(I)$ for some open set $I$ and for all $u$ in a Borel set $D$. Let
$g = \int\_D f(x,u) {\rm d} u$.
Then a sufficient condition for $g \in C^k(I)$ is that $f^{(k)}(x, \cdot)$ is dominated by an integrable function on $D$, i.e., $\sup\_{x \in I} |f^{(k)}(x, \cdot)| \in L^1(D)$, and $g^{(k)}(x) = \int\_D f^{(k)}(x,u) {\rm d} u$ holds in $I$.
My question is about **when is real-analyticity preserved under integration**, say, if $f$ is **real-analytic** in $I$ for each $u$, i.e., $f(\cdot, u) \in C^{\omega}(I)$ for all $u \in D$, what will be a sufficient condition for $g \in C^{\omega}(I)$?
Following the above rationale, we will obtain the following condition: for each $x\_0 \in I$,
1) the radius of convergence of $f(x, u) = \sum\_k a\_k(u) (x-x\_0)^k$ is bounded away from zero for all $u \in D$.
2) integrability condition: $\int\_D \sum\_k a\_k(u) (x-x\_0)^k {\rm d} u < \infty$.
Then the analyticity of $g$ follows from Fubini's theorem.
Questions:
1) Is there other sufficient condition different from the above 'superficial' generalization, maybe exploring other characterization of real analyticitiy? The absolute integrability might not be easy to check.
2) Is there a more local version, which might give the radius of convergence of $g$.
Thanks!
| https://mathoverflow.net/users/3736 | "exchange" of real analyticity and integration | It is rather hard to work with the coefficients of the Taylor expansion directly but, fortunately, one does not have to. Real analyticity is actually analiticity in a strip (of varying width) around the real line and it is quite hard to imagine the situation when $f(x)$ can be computed/estimated on the line but not nearby. So, for all "practical" purposes, it is enough to check that for every point $x\in\mathbb R$, there is a disk $B\subset\mathbb C$ centered at $x$ such that the integrals $\int\_D|f(z,u)|\,du$ are uniformly bounded on $B$. Then the analyticity follows from Fubini and Cauchy formula.
| 6 | https://mathoverflow.net/users/1131 | 15329 | 10,271 |
https://mathoverflow.net/questions/15327 | 23 | There are already a lot of discussion about the motivation for prime spectrum of commutative ring. In my perspective(highly non original), there are following reasons for the importance of prime spectrum.
1. $\text{Spec}(R)$ is the minimal spectrum containing $\text{Spec}\_{\rm max}(R)$ which has good functoriality which means the preimage of a prime ideal is still a prime ideal.
2. if $p\in \text{Spec}(R)$, then $S\_{p} = R-p$ is multiplicative set. Then one can localize.
3. $S\_{p}^{-1}R$ for $p\in \text{Spec}(R)$ is a local ring (has unique maximal ideal which is equivalent to have unique isomorphism class of simple modules). Local ring is easy to deal with and the maximal ideal can be described in explicitly, i.e $m=S\_p^{-1}p$
(My advisor told me P.Cartier pushed Grothendieck to built commutative algebraic geometry machinery based on prime spectrum and the reasons mentioned above are the reasons they used prime spectrum)
Addtional reason: one can have good definitions of topological space and a structure sheaf on it so that one can recover this commutative ring back as its global section
Now, my question is for the people coming from commutative world, what **other properties** do you expect the spectrum of a noncommutative ring should have?
I am aware that people are coming from different branches, there might be various kinds of noncommutative ring arising in your study. Therefore, the question for people coming from different branches of mathematics is that which kind of noncommutative ring do you meet and what properties do you feel that the spectrum of noncommutative ring should have to satisfy your need?
The main motivation for me to ask this question is I am learning noncommutative algebraic geometry. In the existence work by Rosenberg, there are several kinds of spectrum(at least six different spectrum) for different purposes and they satisfy the analogue properties(noncommutative version)I mentioned above and coincide with prime spectrum when one impose the condition of commutativity. **I wonder check whether these spectrum satisfied the other reasonable demand**
| https://mathoverflow.net/users/1851 | What properties "should" spectrum of noncommutative ring have? | I know almost nothing about noncommutative rings, but I have thought a bit about what the general concept of spectra might or should be, so I'll venture an answer.
One other property you might ask for is that it has a good categorical description. I'll explain what I mean.
The spectrum of a commutative ring can be described as follows. (I'll just describe its underlying set, not its topology or structure sheaf.) We have the category **CRing** of commutative rings, and the full subcategory **Field** of fields. Given a commutative ring $A$, we get a new category $A/$**Field**: an object is a field $k$ together with a homomorphism $A \to k$, and a morphism is a commutative triangle. The set of connected-components of this category $A/$**Field** is $\mathrm{Spec} A$.
There's a conceptual story here. Suppose we think instead about algebraic topology. Topologists (except "general" or "point-set" topologists) are keen on looking at spaces from the point of view of Euclidean space. For example, a basic thought of homotopy theory is that you probe a space by looking at the paths in it, i.e. the maps from $[0, 1]$ to it. We have the category **Top** of all topological spaces, and the subcategory **Δ** consisting of the standard topological simplices $\Delta^n$ and the various face and degeneracy maps between them. For each topological space $A$ we get a new category **Δ**$/A$, in which an object is a simplex in $A$ (that is, an object $\Delta^n$ of **Δ** together with a map $\Delta^n \to A$) and a morphism is a commutative triangle. This new category is basically the singular simplicial set of $A$, lightly disguised.
There are some differences between the two situations: the directions have been reversed (for the usual algebra/geometry duality reasons), and in the topological case, taking the set of connected-components of the category wouldn't be a vastly interesting thing to do. But the point is this: in the topological case, the category **Δ**$/A$ encapsulates
>
> how $A$ looks from the point of view of simplices.
>
>
>
In the algebraic case, the category $A/$**Field** encapsulates
>
> how $A$ looks from the point of view of fields.
>
>
>
$\mathrm{Spec} A$ is the set of connected-components of this category, and so gives partial information about how $A$ looks from the point of view of fields.
| 23 | https://mathoverflow.net/users/586 | 15330 | 10,272 |
https://mathoverflow.net/questions/15333 | 20 | I was wondering: is there a good place to find exercises in Hodge theory? Mostly computations and proving small (preferably nifty) theorems, is what I have in mind. Something roughly like the Problems in Group Theory book by Dixon, or many other such problem books.
The biggest difficulty I'm having is that I don't have an adequate number of problems to do, as Griffiths & Harris's Principles has none, and Voisin's otherwise excellent books have only a couple of exercises each chapter, and I've not been able to find a book or paper on the arxiv that does.
| https://mathoverflow.net/users/622 | Exercises in Hodge Theory | One suggestion: "Period mappings and Period Domains", by Carlson, Muller-Stach, and Peters, in the Cambridge studies in advanced mathematics series. It's a very nice read, and each chapter comes with examples and problems.
| 6 | https://mathoverflow.net/users/3545 | 15338 | 10,277 |
https://mathoverflow.net/questions/12322 | 9 | An ideal $\mathfrak{a}$ is called irreducible if $\mathfrak{a} = \mathfrak{b} \cap \mathfrak{c}$ implies $\mathfrak{a} = \mathfrak{b}$ or $\mathfrak{a} = \mathfrak{c}$. Atiyah-MacDonald Lemma 7.11 says that in a Noetherian ring, every ideal is a finite intersection of irreducible ideals. Exercise 7.19 is about the uniqueness of such a decomposition.
7.19. Let $\mathfrak{a}$ be an ideal in a noetherian ring. Let
$$\mathfrak{a} = \cap\_{i=1}^r \mathfrak{b}\_i = \cap\_{j=1}^s \mathfrak{c}\_j$$ be two minimal decompositions of $\mathfrak{a}$ as an intersection of irreducible ideals. [I assume minimal means that none of the ideals can be omitted from the intersection.] Prove that $r = s$ and that (possibly after reindexing) $\sqrt{\mathfrak{b}\_i} = \sqrt{\mathfrak{c}\_i}$ for all $i$.
Comments: It's true that every irreducible ideal in a Noetherian ring is primary (Lemma 7.12), but I don't think our result follows from the analogous statement about primary decomposition. For example, here is Example 8.6 from Hassett's $\textit{Introduction to Algebraic Geometry}$.
8.6 Consider $I = (x^2, xy, y^2) \subset k[x,y]$. We have $$I = (y, x^2) \cap (y^2, x) = (y+x, x^2) \cap (x, (y+x)^2),$$ and all these ideals (other than $I$) are irreducible.
If my interpretation of "minimal" is correct, then this is a minimal decomposition using irreducible ideals, but it is not a minimal primary decomposition, because the radicals are not distinct: they all equal $(x,y)$.
There is a hint in the textbook: Show that for each $i = 1, \ldots, r$, there exists $j$ such that $$\mathfrak{a} = \mathfrak{b}\_1 \cap \cdots \cap \mathfrak{b}\_{i-1} \cap \mathfrak{c}\_j \cap \mathfrak{b}\_{i+1} \cap \cdots \cap \mathfrak{b}\_r.$$ I was not able to prove the hint.
I promise this exercise is not from my homework.
**Update.** There doesn't seem to be much interest in my exercise. I've looked at various solution sets on the internet, and I believe they all make the mistake of assuming that a minimal irreducible decomposition is a minimal primary decomposition. Does anyone know of a reference which discusses irreducible ideals? Some google searches have produced Hassett's book that I mention above and not much else.
| https://mathoverflow.net/users/71 | Atiyah-MacDonald, exercise 7.19 - "decomposition using irreducible ideals" | Here is a solution to the hint:
First of all, note that since all the ideals in question contain $\mathfrak a$, we may replace
$A$ by $A/\mathfrak a$, and so assume that $\mathfrak a = 0$; this simplifies the notation somewhat.
Next, the condition that $\mathfrak b\_1 \cap \cdots \cap \mathfrak b\_r = 0$ is equivalent
to the requirement that the natural map $A \to A/\mathfrak b\_1 \times \cdot \times
A/\mathfrak b\_r$ (the product ot the natural quotient maps) is injective, while the
condition that $\mathfrak b\_i$ is irreducible is equivalent to the statement that
if $I$ and $J$ are non-zero ideals in $A/\mathfrak b\_i$, then $I \cap J \neq 0$ also.
Now suppose given our two irreducible decompositions of $0$. Choose $i$ as in the hint,
and set $I\_j := \mathfrak b\_1 \cap \cdots \cap \mathfrak b\_{i-1} \cap \mathfrak c\_j\cap \mathfrak b\_{i+1}
\cap \cdots \cap \mathfrak b\_r,$ for each $j =1,\ldots,s$.
Then $I\_1\cap \cdots \cap I\_s = 0$ (since it is contained in the intersection
of the $\mathfrak c\_j$, which already vanishes).
Now we recall that $A$ embeds into the product of the $A/\mathfrak b\_{i'}$.
Note that $I\_j$ is contained in $\mathfrak b\_{i'}$ for $i'\neq i$. Thus,
if we let $J\_j$ denote the image of $I\_j$ in $A/\mathfrak b\_i$, then
we see that the image of $I\_j$ under the embedding
$A \hookrightarrow A/\mathfrak b\_1\times\cdots\times A/\mathfrak b\_i \times \cdots \times A/\mathfrak b\_r$
is equal to $0 \times \cdots \times J\_j \times\cdots \times 0$.
Thus the intersection of the images of the $I\_j$, which is the image
of the intersection of the $I\_j$ (since we looking at images under an embedding)
is equal to $0\times \cdots \times (\bigcap J\_j) \times \cdots \times 0.$
Thus, since the intersection of the $I\_j$ is equal to $0$, we see
that $\bigcap J\_j = 0.$ But $\mathfrak b\_i$ is irreducible, and so
one of the $J\_j = 0$. Equivalently, the corresponding $I\_j = 0.$
This proves the hint.
(I think the exercise should be a fairly easy deduction from the hint. The
statement that $r = s$ at least follows directly, using the hint together with
minimality of the two decompositions.)
| 14 | https://mathoverflow.net/users/2874 | 15340 | 10,279 |
https://mathoverflow.net/questions/15336 | 9 | **Background/motivation**
It is a classical fact that we have a natural isomorphism $Sym^n (V^\*) \cong Sym^n (V) ^\ast$ for vector spaces $V$ over a field $k$ of characteristic 0. One way to see this is the following.
On the one hand elements of $Sym^n (V^\*)$ are symmetric powers of degree n of linear forms on $V$, so they can be identified with homogeneous polynomials of degree n on $V$. On the other hand elements of $Sym^n (V) ^\ast$ are linear functionals on $Sym^n V$; by the universal property of $Sym^n V$ these correspond to n-multilinear symmetric forms on $V$. The isomorphism is then as follows.
An n-multilinear symmetric form $\phi$ corresponds to the homogeneous polynomial $p(v) = \phi(v, \dots, v)$. In the other direction to a polynomial $p(v)$ we attach the multinear form obtained by polarization $\phi(v\_1, \dots, v\_n) = \frac{1}{n!}\sum\_{I \subset [n]} (-1)^{n - \sharp I} p(\sum\_{i \in I} v\_i)$. Here $[n]$ is the set $\lbrace 1, \dots, n \rbrace$.
**Problem**
Of course this will not work for $n$ greater than the characteristic of $k$ if the latter is positive.
One can expect that an isomorphism $Sym^n (V^\*) \cong Sym^n (V) ^\ast$ holds also in positive characteristic, and that this should be trivially true by using the universal properties of the symmetric powers. The problem is that if I try to define a natural map between the two spaces using the universal properties I have at some point to divide by $n!$ anyway.
Still there may be some natural isomorphism that I cannot see. Or maybe there is not a natural isomorphism, but I don't know how to prove this.
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> Is there a natural isomoprhism $Sym^n (V^\*) \cong Sym^n (V)^\ast$ in positive characteristic?
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| https://mathoverflow.net/users/828 | Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic? | The answer is no (and well-known to people working in the representation theory of algebraic groups in positive characteristic). In fact for $V$ finite dimensional and of dimension $>1$ the two vector spaces are not
isomorphic as $GL(V)$-modules ($GL(V)$ is either considered naively as an abstract group when the field $k$
is infinite or as an algebraic group in the general case) for $n=p$ equal to the characteristic.
Under the assumption that $V$ is finite dimensional we may instead formulate the problem as the impossibility
of having a $GL(V)$-isomorphism $Sym^n(V) \cong Sym^n(V^\ast)^\ast$. Now, we have an injective $GL(V)$-map $V^{(p)} \to Sym^p(V)$ given by $v \mapsto v^p$, where $V^{(p)}=k\bigotimes\_kV$, where $k$ acts on the left hand side through the $p$'th power (concretely if we choose a basis for $V$ then the action on $V=k^m$ is given by the group homomorphism $GL\_m(k) \to GL\_m(k)$ which takes $(a\_{ij})$ to $(a^p\_{ij})$). As $\dim V > 1$ we have $\dim V^{(p)}=\dim V < \dim Sym^p(V)$ so that the inclusion is proper.
It is easily verified that $V^{(p)}$ is irreducible and it is in fact the unique irreducible submodule of $S^p(V)$. This can be seen by starting with an arbitrary non-zero element $f$ of $S^p(V)$ and then acting on it by suitable linear combinations of the action of elementary matrices of $GL(V)$ until one arrives at a non-zero element of (the image of) $V^{(p)}$. This is more easily understood if one uses the fact that we have an action
of an algebraic and consider the induced action by its Lie algebra. Choosing again a basis of $V$ we have elements $x\_i\partial/\partial x\_j$ whose action on a monomial are very visible. In this way it is clear that
starting with any monomial of degree $p$ one may apply a sequence of such operators to obtain a non-zero
multiple of a monomial of the form $x\_k^p$. This plus some thought shows the statement.
Assume now that we have a $GL(V)$-isomorphism $Sym^p(V) \cong Sym^p(V^\ast)^\ast$. Dualising the inclusion $V^{\ast(p)} \hookrightarrow Sym^p(V^\ast)$ and composing with the isomorphism we got a quotient map $Sym^p(V) \to V^{(p)}$. It is easy to see that in Jordan-Hölder sequence of $Sym^p(V)$ $V^{(p)}$ so that the composite $V^{(p)} \to Sym^p(V) \to V^{(p)}$ must be an isomorphism and hence the inclusion $V^{(p)} \hookrightarrow Sym^p(V)$ is split, contradicting that $V^{(p)}$ is the unique simple submodule.
| 18 | https://mathoverflow.net/users/4008 | 15344 | 10,282 |
https://mathoverflow.net/questions/15354 | 18 | This questions is inspired by an exercise in Hungerford that I have only partially solved. The exercise reads: "A domain is a UFD if and only if every nonzero prime ideal contains a nonzero principal ideal that is prime." (For Hungerford, 'domain' means commutative ring with $1\neq 0$ and no zero divisors).
One direction is easy: if $R$ is a UFD, and $P$ is a nonzero prime ideal, let $a\in P$, $a\neq 0$. Then factor $a$ into irreducibles, $a = c\_1\cdots c\_m$. Since $P$ is a prime ideal in a commutative ring, it is completely prime so there is an $i$ such that $c\_i\in P$, and therefore, $(c\_i)\subseteq P$. Since $c\_i$ is a prime element (because $R$ is a UFD), the ideal $(c\_i)$ is prime.
I confess I am having trouble with the converse, and will appreciate any hints.
But on that same vein, I started wondering if there was a similar "ideal theoretic" condition that describes Euclidean domains. Other classes of domains have ideal theoretic definitions: PID is obvious, of course, but less obvious perhaps are that GCD domains can be defined by ideal-theoretic conditions (given any two principal ideals $(a)$ and $(b)$, there is a least principal ideal $(d)$ that contains $(a)$ and $(b)$, least among all principal ideals containing $(a)$ and $(b)$), as can Bezout domains (every finitely generated ideal is principal). Does anyone know if there is an ideal theoretic definition for Eucldean domains?
| https://mathoverflow.net/users/3959 | Characterizations of UFD and Euclidean domain by ideal-theoretic conditions | Dear Arturo,
The exercise in question is actually a theorem of Kaplansky. It appears as Theorem 5 on page 4 of his *Commutative Rings*. [I was not able to tell easily whether the result appears for the first time in this book.] The proof is reproduced in Section 10 of an expository article I have written [but probably not yet finished] on factorization in integral domains:
[http://alpha.math.uga.edu/~pete/factorization.pdf](http://alpha.math.uga.edu/%7Epete/factorization.pdf)
Regarding your second question, there has been some work on understanding Euclidean domains from more intrinsic perspectives. Two fundamental articles are:
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> Motzkin, Th. The Euclidean algorithm. Bull. Amer. Math. Soc. 55, (1949). 1142--1146.
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[http://alpha.math.uga.edu/~pete/Motzkin49.pdf](http://alpha.math.uga.edu/%7Epete/Motzkin49.pdf)
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> Samuel, Pierre About Euclidean rings. J. Algebra 19 1971 282--301.
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[http://alpha.math.uga.edu/~pete/Samuel-Euclidean.pdf](http://alpha.math.uga.edu/%7Epete/Samuel-Euclidean.pdf)
I have not had the chance to digest these papers, so I'm not sure if they directly answer your question (maybe not, but I think they will be helpful).
| 19 | https://mathoverflow.net/users/1149 | 15356 | 10,290 |
https://mathoverflow.net/questions/15357 | 8 | The question is self-contained finite group theory but the motivation requires more background.
The finite groups I am interested in are the groups $Sp(n,F\_3)$. For $n$ even these are the usual symplectic groups over the field with three elements. For $n$ odd these are the odd symplectic groups. These are a semi-direct product of a symplectic group with a Heisenberg group and we have a sequence of subgroups $Sp(n,F\_3)\rightarrow Sp(n+1,F\_3)$.
Consider one of these inclusions and look at induction/restriction of irreducible complex representations. My question is: take an irreducible representation of the subgroup and induce. Is this representation a direct sum of pair-wise non-isomorphic irreducible representations (sometimes this is called multiplicity free)?
I can prove this for $n$ even because the irreducible representations of $Sp(2m+1,F\_3)$ can be constructed using Clifford theory (known to physicists as Mackey theory). I can use the computer for small $n$. So my question is really for $n$ odd.
---
I can give more information which hints at my interest. Define a sequence of groups $G(n)$ by a sequence of finite presentations so that we have surjective homomorphisms $B(n)\rightarrow G(n)$ where $B(n)$ is the usual braid group. Take generators $\sigma\_1,\ldots ,\sigma\_{n-1}$ and the Artin relations. In addition take $\sigma\_i^3$ and for $n\ge 5$ take $(\sigma\_1\sigma\_2\sigma\_3\sigma\_4)^{10}=1$. Then we have $G(n)=Sp(n-1,F\_3)$.
Then we also have $G(n)\times G(m)\rightarrow G(n+m)$ compatible with $B(n)\times B(m)\rightarrow B(n+m)$. This is similar to the symmetric groups.
---
I am aware that this question can be generalised. I have deliberately restricted to a simple example as I would like to have one case fully worked out before generalising. If you have a proof of the above and your proof generalises, that's different!
| https://mathoverflow.net/users/3992 | From symmetric groups to symplectic groups? | Without working it out completely, I can't give a yes or no answer.
But here's a start: Let $G = Sp(2n)$ (a group of $2n$ by $2n$ matrices, in the notation I use) and let $K = Sp(2n-2) \ltimes H$, where $H$ is the appropriate Heisenberg group. Let's work over any finite field of odd characteristic -- I'd guess that if what you want is true, then it's true in this generality.
You wish to show (by Frobenius reciprocity) that restriction from $G$ to $K$ is multiplicity-free. A standard way to accomplish this would be to prove the following:
*Claim:* The convolution ring $A = C[K \backslash G / K]$ of $K$-bi-invariant functions on $G$ is commutative.
In other words, try to prove that $(G,K)$ is a Gelfand pair. The standard method involves accomplishing the following:
*Task:* Find an anti-involution $\sigma$ of $G$ (meaning $\sigma^2 = Id$ and $\sigma(gh) = \sigma(h) \sigma(g)$, such that every $K$-double-coset $K g K$ in $G$ is stable under $\sigma$, i.e., if $\sigma(g) \in K g K$ for all $g \in G$.
Such an involution yields an anti-automorphism of $A$, making it commutative -- this is the "Gelfand-Kazhdan method".
My advice: try something like conjugation by a matrix (the matrix $J$ defining the symplectic form, perhaps) followed by the transpose, for the anti-involution. It's up to you to analyze the double cosets, but I bet it's been done, at least in low rank (2n = 4, perhaps). The double cosets can get unwieldy, but in your case it almost suffices to analyze the double cosets for the "Heisenerg parabolic" $P$ containing $K$. The $P$ double cosets in $G$ can be analyzed via the Weyl group.
Final advice: the "odd symplectic groups" that you refer to are often called Jacobi groups in the literature due to their relevance to Jacobi forms. Someone may have worked out some of this already!
| 4 | https://mathoverflow.net/users/3545 | 15358 | 10,291 |
https://mathoverflow.net/questions/15284 | 13 | In my meaning, a **direct sum** in a category should really be called a "biproduct". If $X,Y$ are objects, then a direct sum $X \oplus Y$ is an object $Z$ along with isomorphisms $\hom(Z,A) = \hom(X,A) \times \hom(Y,A)$ and $\hom(A,Z) = \hom(A,X) \times \hom(A,Y)$ for all objects $A$. A direct sum is unique up to canonical isomorphism if it exists, of course. A category **has (finite) direct sums** if it has a zero object (an object that is both initial and terminal; i.e. "the direct sum of zero things") and if $X\oplus Y$ exists for any objects $X,Y$. If a category has direct sums, then it is naturally enriched in abelian monoids. I believe that an **additive category** is a category with direct sums in which all the hom-sets (which are already abelian monoids) are actually abelian *groups*.
There are many times when people say "include all direct sums". For example:
**Example:**
Let $\mathcal C$ be any category (enriched over $\rm SET$). Then I can make it enriched over $\rm ABGP$ by applying the $\rm Free: SET \to ABGP$ functor to each hom-set. So now I have a new catefory ${\rm Free}(\mathcal C)$ in which I can add morphisms. But often I want to add objects, too, so I do something like "take the matrix category" ${\rm Mat}(\mathcal C)$, whose objects are finite sequences of objects in $\mathcal C$ and whose morphism are matrices of morphism in ${\rm Free}(\mathcal C)$. Then it's more or less obvious that ${\rm Mat}(\mathcal C)$ is an additive category. If $\mathcal C$ is freely generated by some (objects and) morphisms, then ${\rm Mat}(\mathcal C)$ is presumably "the free additive category generated by those morphisms".
But often I'm not content with free additive categories. For example, I might want to present a category by generators and relations.
>
> **Question:** Is it clear that when I take the quotient of an additive category by some ideal (as an $\rm ABGP$-enriched category), that it still has direct products?
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Or perhaps I really want the *abelian* category presented by generators and relations. Or maybe I just want every idempotent to split, in which case I might take the [Karoubi envelope](http://ncatlab.org/nlab/show/Karoubi+envelope).
>
> **Question:** If I extend my category to split all idempotents, or to include kernels and cokernels, or ..., is it clear that it still has direct products?
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A very explicit application contained in these constructions is the formation of the exterior tensor product of categories: if $\mathcal B,\mathcal C$ are additive categories, then $\mathcal B \boxtimes \mathcal C$ is the free additive category generated by $\mathcal B \times \mathcal C$ with a bunch of relations.
| https://mathoverflow.net/users/78 | How cavalier can I be when demanding a category have direct sums? | The answer to the first question is yes. If A and B have a direct sum A ⊕ B in C, then there are inclusions iA : A → A ⊕ B, iB : B → A ⊕ B and projections pA : A ⊕ B → A, pB : A ⊕ B → B such that pAiA = 1, pBiB = 1, and iApA + iBpB = 1. Conversely, the existence of such maps in an Ab-enriched category make A ⊕ B a direct sum of A and B, even if we do not asssume a priori that A and B have a direct sum. Now if we form the quotient C/I by an ideal I, and two objects A and B with a direct sum A ⊕ B in C, the image of this system of maps presents the image of A ⊕ B as the direct sum of the images of A and B. In short, direct sums are absolute colimits, and as the quotient functor C → C/I is essentially surjective (indeed, bijective on objects), every pair of objects of C/I inherits a direct sum from C.
| 15 | https://mathoverflow.net/users/126667 | 15364 | 10,295 |
https://mathoverflow.net/questions/15366 | 287 | I wonder if anyone else has noticed that the market for [expository papers](https://www.grammarly.com/blog/expository-writing/#:%7E:text=Expository%20writing%2C%20as%20its%20name,or%20attempting%20to%20persuade%20them.) in mathematics is very narrow (more so than it used to be, perhaps).
Are there any journals which publish expository work, especially at the "intermediate" level? By intermediate, I mean neither (i) aimed at an audience of students, especially undergraduate students (e.g. Mathematics Magazine) nor (ii) surveys of entire fields of mathematics and/or descriptions of spectacular new results written by veteran experts in the field (e.g. the Bulletin, the Notices).
Let me give some examples from my own writing, mostly just to fix ideas. (I do *not* mean to complain.)
1. About six years ago I submitted an expository paper "On the discrete geometry of Chicken McNuggets" to the American Mathematical Monthly. The point of the paper was to illustrate the utility of simple reasoning about lattices in Euclidean space to give a proof of Schur's Theorem on the number of representations of an integer by a linear form in positive integers. The paper was rejected; one reviewer said something like (I paraphrase) "I have the feeling that this would be a rather routine result for someone versed in the geometry of numbers." This shows that the paper was not being viewed as expository -- i.e., a work whose goal is the presentation of a known result in a way which will make it accessible and appealing to a broader audience. I shared the news with my officemate at the time, Dr. Gil Alon, and he found the topic interesting. Together we "researchized" the paper by working a little harder and proving some (apparently) new exact formulas for representation numbers. This new version was accepted by the Journal of Integer Sequences:
<https://cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html>
This is not a sad story for me overall because I learned more about the problem ("The Diophantine Problem of Frobenius") in writing the second version with Gil. But still, something is lost: the first version was a writeup of a talk that I have given to advanced undergraduate / basic graduate audiences at several places. For a long time, this was my "general audience" talk, and *it worked* at getting people involved and interested: people always came up to me afterward with further questions and suggested improvements, much more so than any arithmetic geometry talk I have ever given. The main result in our JIS paper is unfortunately a little technical [not deep, not sophisticated; just technical: lots of gcd's and inverses modulo various things] to state, and although I have recommended to several students to read this paper, so far nothing has come of it.
2. A few years ago I managed to reprove a theorem of Luther Claborn (every abelian group is isomorphic to the class group of some Dedekind domain) by using elliptic curves along the lines of a suggestion by Michael Rosen (who reproved the result in the countable case). I asked around and was advised to submit the paper to *L'Enseignement Mathematique*. In my writeup, I made the conscious decision to write the paper in an expository way: that is, I included a lot of background material and explained connections between the various results, even things which were not directly related to the theorem in question. The paper was accepted; but the referee made it clear that s/he would have preferred a more streamlined, research oriented approach. Thus *EM*, despite its name ("Mathematical Education"), seems to be primarily a research journal (which likes papers taking new looks at old or easily stated problems: it's certainly a good journal and I'm proud to be published in it), and I was able to smuggle in some exposition under the cover of a new research result.
3. I have an expository paper on factorization in integral domains:
[http://alpha.math.uga.edu/~pete/factorization.pdf](http://alpha.math.uga.edu/%7Epete/factorization.pdf)
[**Added**: And a newer version: [http://alpha.math.uga.edu/~pete/factorization2010.pdf](http://alpha.math.uga.edu/%7Epete/factorization2010.pdf). ]
It is not finished and not completely polished, but it has been circulating around the internet for about a year now. Again, this completely expository paper has attracted more attention than most of my research papers. Sometimes people talk about it as though it were a preprint or an actual paper, but it isn't: I do not know of any journal that would publish a 30 page paper giving an intermediate-level exposition of the theory of factorization in integral domains. Is there such a journal?
**Added**: In my factorization paper, I build on similar expositions by the leading algebraists P. Samuel and P.M. Cohn. I think these two papers, published in 1968 and 1973, are both excellent examples of the sort of "intermediate exposition" I have in mind (closer to the high end of the range, but still intermediate: one of the main results Samuel discusses, Nagata's Theorem, was published in 1957 so was not exactly hot off the presses when Samuel wrote his article). Both articles were published by the *American Mathematical Monthly*! I don't think the Monthly would publish either of them nowadays.
**Added**: I have recently submitted a paper to the Monthly:
[http://alpha.math.uga.edu/~pete/coveringnumbersv2.pdf](http://alpha.math.uga.edu/%7Epete/coveringnumbersv2.pdf)
(By another coincidence, this paper is a mildly souped up answer to MO question #26. But I did the "research" on this paper in the lonely pre-MO days of 2008.)
Looking at this paper helps me to see that the line between research and exposition can be blurry. I think it is primarily an expository paper -- in that the emphasis is on the presentation of the results rather than the results themselves -- but I didn't have the guts to submit it anywhere without claiming some small research novelty: "The computation of the irredundant linear covering number appears to be new." I'll let you know what happens to it.
(**Added**: it was accepted by the Monthly.)
| https://mathoverflow.net/users/1149 | Which journals publish expository work? | I'm not too familiar with [Expositiones Mathematicae](https://www.sciencedirect.com/journal/expositiones-mathematicae), but have you given them a look?
**EDIT:** The article I happened to have seen, which made me think that Expo Math might be along the lines Pete Clark was looking for, is [this paper](https://arxiv.org/abs/0811.1480) of T. Bühler - it modestly claims to no originality save for assembling disparate parts of the literature and writing down what's old news to connoisseurs (I'm paraphrasing here!) but of course this is, in a sense, *precisely* its originality & worth.
| 73 | https://mathoverflow.net/users/763 | 15369 | 10,297 |
https://mathoverflow.net/questions/15373 | 4 | Suppose $K$ is a (not necessarily algebraically closed) field, and $G\_1$ and $G\_2$ are *split* semisimple algebraic groups over $K$ which become isomorphic over $\bar{K}$, the algebraic closure of $K$. Are $G\_1$ and $G\_2$ isomorphic over $K$? What about if the $G$s are reductive?
It seems like this should follow (at least in the semisimple case) from Tits' general structure theorem for semisimple groups over a not-necessarily algebraically closed field; as explained in section 35.5 of Humphreys' *Linear Algebraic Groups*, a semisimple algebraic group is determined by its $\bar{K}$ isomorphism class, its anisotropic kernel (which looks like it is trivial for a split group) and its `index' (for which there again only seems to be one choice for a split group). But I am not expert enough to completely trust this argument...
| https://mathoverflow.net/users/3513 | If split algebraic groups are potentially isomorphic, are they isomorphic? | The answer is yes, for arbitrary split connected reductive groups over any field. The main point is that the Existence, Isomorphism, and Isogeny Theorems (relating split connected reductive groups and root data) are valid over any field. One reference is SGA3 near the end (which works over any base scheme), but in Appendix A.4 of the book "Pseudo-reductive groups" there is given a direct proof over fields via faithfully flat descent, taking as input the results over algebraically closed fields (since for some reason the non-SGA3 references always seem to make this restriction).
[Caveat: that A.4 gives a complete treatment for the Isomorphism and Isogeny Theorems over general ground fields, and that is what the question is really about anyway; for the Existence Theorem in the case of exceptional types I don't know a way to "pull it down" from an algebraic closure, instead of having to revisit the constructions to make them work over prime fields or $\mathbf{Z}$.]
| 8 | https://mathoverflow.net/users/3927 | 15378 | 10,302 |
https://mathoverflow.net/questions/12117 | 11 | I apologize if these questions seem naive or loaded.
Is there an analogous theory of highest weights for irreducible finite-dimensional representations of Lie algebras of algebraic group (or perhaps group schemes) over a non-algebraically closed field (resp. a "nice" ring, say a Dedekind domain).
Are there analogous results to Lie's theorems in the case of algebraic groups (perhaps even arbitrary group schemes)? I am aware of Jantzen's book on representations of algebraic groups, but if I remember correctly, he does everything over an algebraically closed base field. I have not studied the book in detail to convince myself that the arguments there will carry over to the non-algebraically closed case.
I suppose the Borel-Bott-Weil-(Schmidt) construction of highest weights using sections of cohomology groups of line bundles may be generalized to a more arithmetic setting (as Jantzen has done in his book). Is there any progress in this direction beyond algebraic groups, say to include a "nice" class of group schemes? I am more curious of the case of classical groups.
Concerning more general group schemes, I have looked up parts of SGA3, but I did not find any clearly stated results connecting the Lie algebra of a group scheme (as defined there using universal properties) to the underlying group scheme.
A more general and more loaded question: to what extent is a smooth scheme determined by its tangent space at a distinguished point. I am aware of the notion of jet schemes, are there some important or at least neat results in this area anyone would like to share?
Thanks in advance.
| https://mathoverflow.net/users/3247 | An arithmetic highest weight theory? | Johnson, you have one of the foremost experts in the world on such matters (over general fields) just upstairs from your office. Make use of that.
Many of the basic constructions work for split groups over fields, but proving good properties (such as irreducibility and classification results) requires being over a field of characteristic 0. (Once constructions are made, to prove things one can extend scalars to an algebraic closure, or even reduce to the familiar case over $\mathbf{C}$ if so inclined, by the "Lefschetz Principle".) The Lie algebra is a good invariant (e.g., faithful!) over fields of characteristic 0, but even then it only retains at best information about groups up to isogeny. Another case where it gives a useful invariant is over $\mathbf{Z}/p\mathbf{Z}$-algebras where, together with the $p$-Lie algebra structure, it gives an equivalence with the category of finite locally free groups $G$ with vanishing relative Frobenius morphism such that the sheaf of invariant 1-forms is locally free over the base (loosely speaking because such vanishing allows one to get by with truncated exponential in degrees $< p$); this is explained in SGA3, VII$\_{\rm{A}}$, 7.2, 7.4
More generally one cannot expect to get mileage out of the Lie algebra alone (but it is still perfectly useful via its role in classification by means of root systems, among other things).
| 6 | https://mathoverflow.net/users/3927 | 15381 | 10,305 |
https://mathoverflow.net/questions/15392 | 6 | I am trying to prove a version of quantum Schur-Weyl duality. I hope to be able to generalize the proof of the Schur-Weyl duality between $U\_q(\mathfrak{gl}\_n)$ and the Hecke algebra $H\_r$. So I am looking for a good reference for this with a careful proof. It would also be nice to see a proof that uses the quantum coordinate ring of $GL\_n$ instead of the enveloping algebra (and therefore is phrased in terms of decomposing $V^{\otimes r}$ as a left-comodule for this coordinate ring and a right $H\_r$-module).
| https://mathoverflow.net/users/3318 | Reference for quantum Schur-Weyl duality | This goes back to Jimbo I think. A reference is:
"A q-difference analogue of $U(\mathfrak g)$, Hecke algebra and the Yang-Baxter equation'', Lett. Math. Phys. 11 (1986).
It has been much studied though, so there are lots of subsequent papers, some of which might be closer to what you are looking for? For example [this](https://arxiv.org/abs/0806.0264) paper studies an analogue of Schur-Weyl duality for "walled Brauer algebras", and [this](https://arxiv.org/abs/math/0108038) paper studies a two-parameter version.
| 7 | https://mathoverflow.net/users/1878 | 15395 | 10,313 |
https://mathoverflow.net/questions/15396 | 4 | Let be $d$ a positive integer, $\Omega=\mathbb{R}^{\mathbb{Z}^d}$ and fix $R\geq 2$. We define weighted Banach spaces
$$ \Omega\_p:=\left\{ x\in \Omega\left| \left[\sum\_{i\in\mathbb{Z}^d}\frac{|x\_i|^R}{(1+|i|)^p}\right]^{\frac{1}{R}} < \infty\right. \right\},\ \ \ \ \ \ \ p>d. $$
My question is why the embeddings
$\Omega\_p \hookrightarrow \Omega\_{p'}$ are compact whenever $p< p ' $ ?
| https://mathoverflow.net/users/2386 | Embeddings of Weighted Banach Spaces | This is a special case of a much more general phenomenon, so I'm writing an answer which deliberately takes a slightly high-level functional-analytic POV; I think (personally) that this makes it easier to see the wood for the trees, even if it might not be the most direct proof. However, depending on your mathematical background it might not be the most helpful; so apologies in advance.
Anyway, start with a very general observation: let $E$ be a (real or complex) Banach space, and for each $n=1,2,\dots$ let $T\_n:E\to E$ be a bounded linear operator which has finite rank. (In particular, each $T\_n$ is a compact operator.)
Lemma: Suppose that the sequence $T\_n$ converges in the operator norm to some linear operator $T:E\to E$. Then $T$ is compact.
(The proof ought to be given in functional-analytic textbooks, so for sake of space I won't repeat the argument here.)
Now we consider these specific spaces $\Omega\_p$. Let $T:\Omega\_p\to\Omega\_{p'}$ be the embedding that you describe.
For each $n$, define $T\_n: \Omega\_p \to \Omega\_{p'}$ by the following rule:
$$ T\_n(x)\_i = x\_i \ {\rm if } \ |i| \leq n \ {\rm and} \ 0 \ {\rm otherwise} $$
Then each $T\_n$ has finite rank (because every vector in the image is supported on the finite set $\{ i \in {\mathbb Z}^d \vert \ \vert i\vert \leq n\}$). I claim that $T\_n$ converges to $T$ in the operator norm, which by our lemma would imply that $T$ is compact, as required.
We can estimate this norm quite easily (and indeed all we need is an upper bound). Let $x\in\Omega\_p$ have norm $\leq 1$; that is,
$$ \sum\_{i\in{\mathbb Z}^d} |x\_i|^R (1+ \vert i\vert)^{-p} \leq 1 $$
Then the norm of $(T-T\_n)(x)$ in $\Omega\_{p'}$ is going to equal $C^{1/R}$, where
$$ \eqalign{
C &:= \sum\_{i\in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p'} \\\\
& = \sum\_{i \in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert i \vert)^{p-p'} \\\\
& \leq \sum\_{i \in {\mathbb Z}^d : \vert i\vert > n} |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'} \\\\
& \leq \sum\_{i \in {\mathbb Z}^d } |x\_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'}
& = (1+\vert n \vert)^{p-p'}
}
$$
This shows that $\Vert T-T\_n\Vert \leq (1+\vert n\vert)^{(p-p')/R}$ and the right hand side can be made arbitrarily small by taking $n$ sufficiently large. That is, $T\_n\to T$ in the operator norm, as claimed, and the argument is complete (provided we take the lemma on trust).
Note that we used very little about the special nature of your weights. Indeed, as Bill Johnson's answer indicates, the only important feature is that in changing weight you in effect multiply your vector $x$ by a "multiplier sequence" which lies in $c\_0({\mathbb Z}^d)$, i.e. the entries "vanish at infinity".
**Edit 17-02-10:** the previous paragraph was perhaps slightly too terse. What I meant was the following: suppose that you have two weights $\omega$ and $\omega'$, such that the ration $\omega/\omega'$ lies in $c\_0({\mathbb Z}^d)$. Then the same argument as above shows that the corresponding embedding will be compact. Really, this is what Bill's answer was driving at: it isn't the weights which are important, it's the fact that the factor involved in *changing* weight is given by something "vanishing at infinity".
| 5 | https://mathoverflow.net/users/763 | 15406 | 10,320 |
https://mathoverflow.net/questions/15400 | 9 | There is an ubiquitous pattern of questions concerning assumedly any kind of mathematical object or structure: groups, graphs, numbers, categories, and so on. It goes like this (informally):
>
> Can a class of objects or structures of a given kind *X* that is characterized by some "external condition" *Y* be defined by a condition *Z* in their respective "internal"
> language, and if so: how?
>
>
>
Well-known examples ("external condition" **=** "internal condition"):
1. groups $G$ isomorphic to a subgroup of the symmetric group on $G$ **=** all groups (*Cayley's theorem*)
2. graphs embeddable in the plane **=** graphs not containing a subgraph that is a subdivision of $K\_5$ or $K\_{3,3}$ (*Kuratowski's theorem*)
3. numbers *n* of trees on *k* labeled vertices **=** numbers *n = k**k*-2 for some *k* > 1 (*Cayley's theorem on trees*)
4. numbers *n* with only one group of order *n* **=** numbers *n = p1 · p2 · ... · pk* for some *k* > 0, where the *pi* are distinct primes and no *pj-1* is divisible by any *pi* (*cyclic numbers*, see [Sloane's A003277](http://www.research.att.com/~njas/sequences/A003277))
Further examples from MO:
* [Which graphs are Cayley graphs?](https://mathoverflow.net/questions/14830/which-graphs-are-cayley-graphs)
* [Can we recognize when a category is
equivalent to the category of models
of a first order theory?](https://mathoverflow.net/questions/13155/can-we-recognize-when-a-category-is-equivalent-to-the-category-of-models-of-a-fir)
* [Can you determine whether a graph is
the 1-skeleton of a polytope?](https://mathoverflow.net/questions/9255/can-you-determine-whether-a-graph-is-the-1-skeleton-of-a-polytope)
>
> **Question #1:** What's the proper way to
> characterize this pattern of
> questions? What's the common
> context / rationale?
>
>
>
---
>
> **Question #2:** How is the introductory question to be
> posed properly?
>
>
>
| https://mathoverflow.net/users/2672 | An ubiquitous pattern of questions | There are three sibling theorems of logic which guarantee that such characterizations are bound to happen.
* [The Craig Interpolation Theorem](http://en.wikipedia.org/wiki/Craig_interpolation)
* [The Robinson Joint Consistency Theorem](http://en.wikipedia.org/wiki/Robinson%27s_joint_consistency_theorem)
* [The Beth Definability Theorem](http://en.wikipedia.org/wiki/Beth_definability)
[The wikipedia entries need some work. I suggest you look up these theorems in a good logic book, for example Hodges's *Model Theory* or his more accessible *Shorter Model Theory*.]
I will focus on the Beth Definability Theorem, though the other two siblings lead to similar conclusions in slightly different contexts.
Suppose you have a first-order language L0 and a larger language L. Let T be a theory in L and let φ(x) be a formula of the larger language L with the following property. Whenever A1 and A2 are two models of T which have the same universe A and the same interpretation for all parts of the small language L0, then A1 ⊧ φ(a) iff A2 ⊧ φ(a) for all a ∈ A. Beth's Definability Theorem says that there must be a formula φ0(x) of the smaller language L0 such that T ⊦ ∀x(φ(x) ↔ φ0(x)).
The connection with your question is as follows. The base language L0 is the 'internal' language of the structures you really care about, while the larger language L has some additional 'external' data. The theory T characterizes the structures with external data that you care about, and φ(x) is a property of such structures that you are interested in. If φ(x) is sufficiently independent of the external data, then φ(x) must be equivalent to an internal formula φ0(x).
Not all of the examples you give are easily cast into this formalism, but the basic flavor is the same. Unfortunately, the Beth Definability Theorem (and its proof) does not say much on how to find the internal formula φ0(x) but, at least, it says that the search will not be in vain.
| 16 | https://mathoverflow.net/users/2000 | 15413 | 10,324 |
https://mathoverflow.net/questions/15424 | 1 | Is the tr.deg of Q\_p over Q 1? and what about C over Q?
| https://mathoverflow.net/users/1238 | What is the transcendence degree of Q_p and C over Q? | In both cases the transcendence degree is the cardinality of the continuum. CH is not needed.
This is a corollary of the following result: let $K$ be any infinite field, and let $L/K$ be any extension. Then
$\# L = \operatorname{max} (\# K, \operatorname{trdeg}\_K L)$.
To prove this, in turn it suffices to establish the following two results (each of which is straightforward):
1) If $K$ is infinite and $L/K$ is algebraic, then $\# L = \# K$.
2) If $K$ is any infinite field, $T = \{t\_i\}\_{i \in I}$ is an arbitrary set of indeterminates and $K(T)$ is a purely transcendental function field in the indeterminates $T$, then $ \# K(T) \leq \# T + \# K$.
| 9 | https://mathoverflow.net/users/1149 | 15428 | 10,334 |
https://mathoverflow.net/questions/15447 | 13 | Is there any standard way to read (in English) the Legendre symbol (a|p) (<http://en.wikipedia.org/wiki/Legendre_symbol>), say, similar to "a choose b"
which is used for the binomial coefficients?
| https://mathoverflow.net/users/3635 | Is there a standard way to read the Legendre symbol? | I say "a on b" for the Legendre/Jacobi/Kronecker symbol. This works because, as an American, I say "a over b" for an ordinary fraction.
| 14 | https://mathoverflow.net/users/1149 | 15449 | 10,343 |
https://mathoverflow.net/questions/15435 | 2 | Just a minor curiosity that's flitted across my mind, but that's (part of) what this site's for, right?:
Is it possible for Hom(a, -) and Hom(b, -) to both be monadic functors from C to Set, for non-isomorphic objects a and b in C? Ideally, the answer would come with either a nice example or an outline of a nice proof of impossibility (i.e., proof that all monadic representable functors on a category are isomorphic).
| https://mathoverflow.net/users/3902 | If a category is "monadic", is it necessarily so in a unique manner? | You can certainly have non-equivalent monadic functors. Here's one example: Let $\mathcal{V}\_k$ be the category of $k$-vector spaces. For a vector space $V$, let $H\_V: \mathcal{V}\_k\to \mathcal{V}\_k$ be the functor
$$
H\_V(W) = hom\_k(V,W).
$$
Such a functor is always monadic, as long as $V$ is non-zero and finite dimensional. The associated monad is
$$
T\_V(W) = hom\_k(V, V\otimes\_k W) = End\_k(V)\otimes\_k W,
$$
so this is presenting a Morita equivalence: $k$-vector spaces are equivalent to modules over the matrix ring $End\_k(V)$.
You wanted functors to set; let $U\_V:\mathcal{V}\_k\to Set$ be given by the same formula as $H\_V$. Then again, this will be monadic, as long as $V$ is non-zero and finite dimensional (and I'm not sure you even really need the finite dimensionality condition for either of these examples; **added:** you certainly don't in the first example, since $H\_V$ is an exact functor, so the hypotheses of the Barr-Beck theorem certainly hold, though $T\_V$ is not tensoring with an endomorphism ring if $V$ is infinite.).
| 6 | https://mathoverflow.net/users/437 | 15454 | 10,348 |
https://mathoverflow.net/questions/15448 | 8 |
>
> What are the derivations of the algebra of *continuous* functions on a topological manifold?
>
>
>
A *supermanifold* is a locally ringed space (X,O) whose underlying space is a *smooth* manifold X, and whose sheaf of functions locally looks like a graded commutative algebra which is an exterior algebra over the sheaf of smooth functions. The [wikipedia article](http://en.wikipedia.org/wiki/Supermanifold) has a detailed description. The n[-lab post](http://ncatlab.org/nlab/show/supermanifold) is a little thinner, but by pointing is out I've probably set in motion events that will alter it to be the superior article. Ahh, Heisenberg Uncertainty on the internet.
What neither of these article tell you is **why must the manifold be smooth**?
Once I thought I understood the reason. Now I'm not so sure. The reason I was brought up with came from looking at some examples of maps of supermanifolds. For simplicity let X be just an ordinary manifold regarded as a (trivial) supermanifold. Let $\mathbb{R}^{0|2}$ be the supermanifold whose underlying manifold is just a point and whose ring of functions is the exterior algebra $A=\mathbb{R}[\theta\_1, \theta\_2]$, where $\theta\_i$ is an odd generator. The algebra A is four dimensional as a real algebra.
A map of supermanifolds $\mathbb{R}^{0|2} \to X$ is the same as a map of algebras $O(X) \to A$ where $O(X)$ is functions on X (of a to-be-determined type). Since X is a trivial supermanifold this is the same as pair of linear maps $a,b$:
$$ f \mapsto a(f) + b(f) \theta\_1 \theta\_2$$
It is easy to see that $a$ all by itself is an algebra map $O(X) \to \mathbb{R}$. This is the same thing as evalutation at a point $x\_0 \in X$. Once I knew a reference for this, but I can't seem to find it. I think it is true as stated, that algebra homomorphisms $O(X) \to \mathbb{R}$ are in bijection with the points of the manifold $X$, where $O(X)$ is either smooth or continuous functions, without taking into account any topology. If this is false, then we should take into account the topology as well. In any event, $a(f) = f(x\_0)$ is evaluation at $x\_0$. What is $b$?
By using the algebra property of the above assignment we see that $b$ must satisfy:
$$b(fg) = f(x\_0) b(g) + b(f) g(x\_0).$$
In other words, $b$ is a derivation of $O(X)$ at the point $x\_0$.
So now the argument goes something like, varying over all the map from $\mathbb{R}^{0|2}$ to $X$ we see that the functions must have all first derivatives at all points and so must be $C^1$-functions. Using more odd manifolds $\mathbb{R}^{0|4}$, $\mathbb{R}^{0|6}$, etc. we see that they must have arbitrarily high derivatives and so must in fact be smooth.
Or do they?
In order to really make this sort of argument hold water I think you need to know something like "there aren't any derivations of the algebra of continuous functions". You need the naive continuous version of the theory to be trivial or obviously useless. Otherwise I don't see a good motivation for why one *must* work smoothly, why it is the only interesting situation. So my question is:
>
> What exactly are the derivations (taking into account topology or not) of the algebra of continous functions on a topological manifold X?
>
>
>
This is surely equivalent to knowing the local question, i.e. when $X = \mathbb{R}^n$. Why is the naive theory of continuous supermanifolds badly behaved?
| https://mathoverflow.net/users/184 | Derivations of C(X)? or Why Must Supermanifolds be Smooth? | Here's a proof from definition. (I don't think it has anything to do with compactness.) Let us show that derivation
$\delta:C(X)\to C(X)$ vanishes for any topological manifold $X$. Indeed, $\delta(1)=0$ as usual, so it suffices to show that whenever $f\in C(X)$ vanishes at $x\in X$, so does
$\delta(f)$. For this, it is enough to write $f$ as a product $f=g\_1g\_2$ with $g\_1(x)=g\_2(x)=0$. Take $g\_1=\sqrt{|f|}$, $g\_2=f/g\_1$, and define $g\_2$ to be zero where it is undefined.
| 14 | https://mathoverflow.net/users/2653 | 15458 | 10,352 |
https://mathoverflow.net/questions/15440 | 32 | There is the following theorem:
"A space $X$ is the inverse limit of a system of discrete finite spaces, if and only if $X$ is totally disconnected, compact and Hausdorff."
A finite discrete space is totally disconnected, compact and Hausdorff and all those properties pass to inverse limits. I guess the other direction might be proved by taking the system of all decompositions of $X$ into disjoint clopen sets. The inverse limit should give $X$ back.
So what happens, if I dismiss the finiteness condition. As mentioned above every inverse limit of discrete spaces is totally disconnected, Hausdorff. So the question is:
"Which totally disconnected Hausdorff spaces are inverse limits of discrete spaces?"
For example I think it is impossible to write $\mathbb{Q}$ as an inverse limit of discrete spaces, but I don't have a proof.
| https://mathoverflow.net/users/3969 | Which spaces are inverse limits of discrete spaces ? | These are the completely ultrametrizable spaces.
Recall that a d:E2→[0,∞) is an ultrametric if
1. d(x,y) = 0 ↔ x = y
2. d(x,y) = d(y,x)
3. d(x,z) ≤ max(d(x,y),d(y,z))
As usual, (E,d) is a complete ultrametric space if every Cauchy sequence converges.
Suppose E∞ is the inverse limit of the sequence En of discrete spaces, with fn:E∞→En being the limit maps. Then
d∞(x,y) = inf { 2-n : fn(x) = fn(y) }
is a complete ultrametric on E∞, which is compatible with the inverse limit topology.
Conversely, given a complete ultrametric space (E,d), the relation x ∼n y defined by d(x,y) ≤ 2-n is an equivalence relation. Let En be the quotient E/∼n, with the discrete topology. These spaces have obvious commuting maps between them, let E∞ be the inverse limit of this system. The map which sends each point of E to the sequence of its ∼n equivalence classes is a continuous map f:E→E∞. Because E is complete, this map f is a bijection. Moreover, a simple computation shows that this bijection is in fact a homeomorphism. Indeed, with d∞ defined as above, we have
d∞(f(x),f(y)) ≥ d(x,y) ≥ d∞(f(x),f(y))/2.
---
As Pete Clark pointed out in the comments, the above is an incomplete answer since the question does not assume that the inverse system is countable. However, the general case does admit a similar characterization in terms of [uniformities](http://en.wikipedia.org/wiki/Uniform_space). For the purposes of this answer, let us say that an *ultrauniformity* is a unformity with a fundamental system of entourages which consists of open (hence clopen) equivalence relations. The spaces in question are precisely the complete Hausdorff ultrauniform spaces.
Suppose E is the inverse limit of the discrete spaces Ei with limit maps fi:E→Ei. Without loss of generality, this is a directed system. Then the sets Ui = {(x,y): fi(x) = fi(y)} form a fundamental system of entourages for the topology on E, each of which is a clopen equivalence relation on E. The universal property of inverse limits guarantees that E is complete and Hausdorff. Indeed, every Cauchy filter on E defines a compatible sequence of points in the spaces Ei, which is the unique limit of this filter.
Conversely, suppose E is a complete Hausdorff ultrauniform space. If U is a fundamental entourage (so U is a clopen equivalence relation on E) then the quotient space E/U is a discrete space since the diagonal is clopen. In fact, E is the inverse limit of this directed system of quotients. It is a good exercise (for Pete's students) to show that completeness and Hausdorffness of E ensure that E satisfies the universal property of inverse limits.
| 42 | https://mathoverflow.net/users/2000 | 15465 | 10,358 |
https://mathoverflow.net/questions/15466 | 10 | If you have a lattice $L \subset \mathbb{C}$, you can compute the following numbers:
$
G\_4(L) = \sum\_{\omega \in L, \omega \neq 0} \frac{1}{\omega^4}, \quad G\_6(L) = \sum\_{\omega \in L, \omega \neq 0} \frac{1}{\omega^6}.
$
By a 'lattice', I just mean a closed discrete additive non-trivial subgroup of $\mathbb{C}$ (so I'm allowing degenerate lattices like $\mathbb{Z}$). Anyhow, these numbers are important invariants of the lattice, because they set up a bijection
$\{ \mbox{Lattices in }\mathbb{C}\}\rightarrow \mathbb{C}^2 \setminus \{0\}$
$L \mapsto (G\_4(L), G\_6(L))$.
But what do these numbers actually measure about the lattice, geometrically? Some kind of combination of angles? Some area? I'm confused. We have these numbers that get used over and over, but what do they actually *measure*?
I guess one possible answer is: consider the Riemann surface (torus) $\mathbb{C} / \Lambda$. Then $G\_4$ and $G\_6$ can be recovered as certain period integrals along the fundamental cycles of the elliptic curve. Is that right? Is there a more direct geometric understanding of $G\_4$ and $G\_6$?
| https://mathoverflow.net/users/401 | What do the numbers G_4 and G_6 of a lattice actually measure? | As far as I know $G\_4$ and $G\_6$ don't have a direct geometric interpretation of the type you are looking for. Rather, they appear as coefficients in the algebraic equation for
$\mathbb C/\Lambda.$ More precisely, the pair
$(\mathbb C/\Lambda, dz)$ consisting of the complex torus $\mathbb C/\Lambda$ and the everywhere-holomorphic differential form $dz$ is isomorphic to the pair
$(E,dx/y),$ where $E$ is the smooth complex projective curve cut out by the (homogeneous equation
associated to) the equation $y^2 = 4 x^3 - 60 G\_4(L) x - 140 G\_6(L).$ (Here the letter $E$
is for "elliptic".)
As you probably know,
this is more or less the content of the theory of
[Weierstrass's elliptic functions](http://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions).
In summary: lattices in $\mathbb C$ are the same thing as elliptic curves over ${\mathbb C}$
equipped with a choice of non-zero holomorphic differential (via $\Lambda \mapsto
(\mathbb C/\Lambda, dz)$, and the quantities $G\_4$ and $G\_6$ give an explicit formula
for this correspondence, by describing the coefficients of the algebraic equation for the corresponding elliptic curve.
| 9 | https://mathoverflow.net/users/2874 | 15468 | 10,360 |
https://mathoverflow.net/questions/15441 | 2 | In the paper hep-th/9712042v2, p. 20, the following setup is given:
A complex manifold M and an n+1-dimensional vector bundle V on it. V has an underlying real bundle $V\_{\mathbb{R}}$ with a flat connection $\nabla$. Also, there is a holomorphic inclusion of a line bundle $L$ in $V$.
Finally, we have the section
$M \to \mathbb{P}[(V\_{\mathbb{R}})\_{\mathbb{C}}],\\
m \mapsto L\_m$
My question is what is meant by the statement that this section is assumed to be "an immersion with respect to $\nabla$"?
I just can't make sense of how the existence of a connection has something to do with an immersion condition for the section.
| https://mathoverflow.net/users/3170 | Immersion with respect to a connection? | Just read further down the same page: "The immersion condition in (b) states that $m \to L\_m$ is an immersion into the 2n + 1 dimensional projective space of a local trivialization of $P(V\_R)\_C$."
You can also say it without reference to the flatness of the connection: the section is transverse to the horizontal distribuition of the connection. Ie, L, thought of as a section of $P(V\_R)\_C$, is "maximally non parallel". Etc etc.
| 3 | https://mathoverflow.net/users/2991 | 15471 | 10,363 |
https://mathoverflow.net/questions/15178 | 3 | Let $K$ be a field, $V$ an $n-$dimensional $K$-vector space and $q: V \to K$ a quadratic form of Witt index $r$. Let $G:=SO(q)$ denote the special orthogonal group associated to $q$.
Then $G$ is an algebraic $K$-group of $K-$rank $r$.
If $q'$ is isometric to $q$, clearly $G':=SO(q')$ is isomorphic to $G$. But this is not necessary for $G$ to be isomorphic to $G'$ (take $n=1$).
My question is: How can I produce a lot of nonisomorphic $SO(q)$'s?
Here "non-isomorphic" should be interpreted either as abstract groups or as algebraic groups.
I'm particularly interested in the case when the Witt rank is 1.
| https://mathoverflow.net/users/3380 | Algebraic groups of relative rank 1 | Fleshing out the Galois cohomology approach suggested by Brian Conrad leads to a clean answer for all $n$, for all fields $K$ of characteristic not $2$, and for all nondegenerate quadratic forms of rank $n$ over $K$. The answer is exactly what moonface claimed:
>
> Given quadratic forms $q$ and $q'$, the algebraic groups $\operatorname{SO}(q)$ and $\operatorname{SO}(q')$ are isomorphic if and only if $q$ and $q'$ are *similar*, i.e., $q$ is equivalent to $\lambda q'$ for some $\lambda \in K^\times$.
>
>
>
The key observation is that the homomorphism from $\operatorname{O}\_n$ to the automorphism group scheme of $\operatorname{SO}\_n$ giving the conjugation action is surjective for all $n$, and the kernel is $\lbrace \pm 1 \rbrace$ for $n>2$. Then for $n>2$, one has the exact sequence of pointed sets
$$ H^1(K,\lbrace \pm 1 \rbrace) \to H^1(K,\operatorname{O}\_n) \to H^1(K,\operatorname{\bf Aut} \operatorname{SO}\_n).$$
The first term is $K^\times/K^{\times 2}$, the second term is the set of equivalence classes of nondegenerate rank $n$ quadratic forms, and the third term is the set of isomorphism classes of $K$-forms of $\operatorname{SO}\_n$. The sequence (and its twists - remember that we are dealing with nonabelian cohomology) shows that two quadratic forms give rise to the same $K$-form of $\operatorname{SO}\_n$ if and only if they are similar.
If $n=2$, a similar argument applies, though one can also see everything explicitly: every rank $2$ quadratic form is similar to $x^2-dy^2$ for some $d \in K^\times$, and the corresponding $\operatorname{SO}(q)$ is the "Pell equation torus" $x^2-dy^2=1$; both depend just on the image of $d$ in $K^\times/K^{\times 2}$. I leave the cases $n=1$ and $n=0$ to those who like to think about such things.
**More details:** When $n$ is odd, we have $\operatorname{O}\_n = \lbrace \pm 1 \rbrace \cdot \operatorname{SO}\_n$, and all automorphisms of $\operatorname{SO}\_n$ are inner.
When $n=2m$ for some $m \ge 2$, we have $-1 \in \operatorname{SO}\_n$, so conjugation by an element of $\operatorname{O}\_n$ outside $\operatorname{SO}\_n$ gives an outer automorphism of $\operatorname{SO}\_n$; correspondingly, the $D\_m$ Dynkin diagram (interpreted appropriately for small $m$) has an involution.
Why does the rotation of the $D\_4$ Dynkin diagram not give an extra outer automorphism when $n=8$? The covering group between the simply connected form $\operatorname{Spin}\_8$ and the adjoint form $\operatorname{PSO}\_8$ is $(\mathbb{Z}/2\mathbb{Z})^2$, so there are three intermediate covers, one of which is $\operatorname{SO}\_8$, but the rotation permutes these three.
(Thanks to my colleague David Vogan for discussing this with me.)
| 7 | https://mathoverflow.net/users/2757 | 15473 | 10,364 |
https://mathoverflow.net/questions/8262 | 18 | Let $L$ be a language on a finite alphabet and let $L\_n$ be the number of words of length $n$. Let $f\_L(x) = \sum\_{n \ge 0} L\_n x^n$. The following are well-known:
* If $L$ is regular, then $f\_L$ is rational.
* If $L$ is unambiguous and context-free, then $f\_L$ is algebraic.
Does there exist a natural family of languages $\mathcal{L}$ containing the context-free languages such that if $L \in \mathcal{L}$, then $f\_L$ is [holonomic](http://en.wikipedia.org/wiki/Holonomic#Holonomic_function)? Is that class of languages also associated to a natural class of automata?
This question is prompted by a remark in Flajolet and Sedgewick where they assert that there is no meaningful generating function formalism associated to context-sensitive languages because of the significant undecidability issues. However, holonomic functions have proven a robust and incredibly useful framework in combinatorics, so I think this is a natural question to ask.
| https://mathoverflow.net/users/290 | Is there a natural family of languages whose generating functions are holonomic (i.e. D-finite)? | I vaguely remember that the paper by Marni Mishna and Mike Zabrocki [Analytic aspects of the shuffle product](http://arxiv.org/abs/0802.2844) sheds some light on the subject.
| 6 | https://mathoverflow.net/users/3032 | 15481 | 10,367 |
https://mathoverflow.net/questions/15219 | 17 | Let $G$ be a permutation group on the finite set $\Omega$. Consider the Markov chain where you start with an element $\alpha \in \Omega$ chosen from some arbitrary starting probability distribution. One step in the Markov chain involves the following:
* Move from $\alpha$ to a random element $g\in G$ that fixes $\alpha$.
* Move from $g$ to a random element $\beta\in \Omega$ that is fixed by $g$.
Since it is possible to move from any $\alpha$ to any $\beta$ in $\Omega$ in a single step (through the identity of $G$) the Markov chain is irreducible and aperiodic. This implies that there is a unique distribution that is approached by iterating the procedure above from any starting distribution. It's not hard to show that the limiting distribution is the one where all orbits are equally likely (i.e. the probability of reaching $\alpha$ is inversely proportional to the size of the orbit containing it).
---
I read about this nice construction in P.J. Cameron's "Permutation Groups", where he brings up what he calls a slogan of modern enumeration theory: "...the ability to count a set is closely related to the ability to pick a random element from that set (with all elements equally likely)."
One special case of this Markov chain is when we let $\Omega=G$ and the action be conjugation, then we get a limiting distribution where all conjugacy classes of $G$ are equally likely. Now, except for this nice result, it would also be interesting to know something about the rate of convergence of this chain. Cameron mentions that it is rapidly mixing (converges exponentially fast to the limiting distribution) in some important cases, but examples where it's not rapidly mixing can also be constructed. My question is:
>
> Question: Can we describe the rate of convergence of the Markov chain described above in terms of group-theoretic concepts (properties of $G$)?
>
>
>
While giving the rate of convergence in terms of the properties of $G$ might be a hard question, answers with sufficient conditions for the chain to be rapidly mixing are also welcome.
| https://mathoverflow.net/users/2384 | Markov chain on groups | This is a classical question which is now well understood, I believe. Start with this famous paper: <http://tinyurl.com/yfyrg63> Although this question is really not about standard r.w. on groups, a survey by Saloff-Coste gives an excellent introduction and literature review: www.math.cornell.edu/~lsc/rwfg.pdf
| 9 | https://mathoverflow.net/users/4040 | 15483 | 10,369 |
https://mathoverflow.net/questions/14863 | 28 | An [alternating permutation](https://arxiv.org/abs/0912.4240) of {1, ..., n} is one were π(1) > π(2) < π(3) > π(4) < ... For example: (24153) is an alternating permutation of length 5.
If $E\_n$ is the number of alternating permutations of length n, the $\sec x + \tan x = \sum\_{n \geq 0} E\_n \frac{x^n}{n!}$ is the exponential generating function.
How does one sample the alternating permutations uniformly at random? Rejection sampling would quickly become inefficient since $\frac{E\_n}{n!} \approx \frac{4}{\pi} \left(\frac{2}{\pi} \right)^n$ decays exponentially in *n*.
| https://mathoverflow.net/users/1358 | Random Alternating Permutations | This is very easy - I often teach this in my combinatorics classes. You start with this Pascal type triangle which you need to precompute:
<http://mathworld.wolfram.com/Seidel-Entringer-ArnoldTriangle.html>
(read Arnold's paper referenced there if the pattern is unclear or my survey paper with Postnikov "Increasing Trees and Alternating Permutations" - see my webpage).
Now note that $a\_{n,i}$, the i-th number in n-th row is the number of alternating permutations with the first letter $\ge i$ (you need to alternate directions). Then you use dynamic programming or simply choose the first letter to be i with probability $a\_{n,i} - a\_{n,i+1}$, which reduces the problem to a alternating permutation of $[n-1]$. I am skipping (easy) details, but I trust you can figure out the rest from here.
UPDATE (May 10, 2015):
Apparently, this algorithm requires $O^\ast(n^3)$ time to compute the numbers $a\_{n,i}$ which is not bad. However, if one only wants to compute the number of alternating permutations, there is a faster way: see [this talk](http://jonfest2011.irmacs.sfu.ca/talk/53) by Richard Brent and the slides [here](http://maths-people.anu.edu.au/~brent/pd/tangent.pdf).
| 21 | https://mathoverflow.net/users/4040 | 15487 | 10,372 |
https://mathoverflow.net/questions/14527 | 15 | The [Robinson-Schensted correspondence](https://en.wikipedia.org/wiki/Robinson%E2%80%93Schensted%E2%80%93Knuth_algorithm) is a bijection between elements of the symmetric group and ordered pairs of standard tableaux of the same shape.
Some simple operations on tableaux correspond to simple operations on the group: switching the tableaux corresponds to inverse on the group.
>
> What about taking the transpose of the tableaux? Does that correspond to something easily described on permutations?
>
>
>
In order to rule out easy guesses, let me describe this on $S\_3$:
* the identity switches with the involution 321.
* the transpositions 213 and 132 switch.
* the 3-cycles 312 and 231 switch.
In general, this operation preserves being order $\leq 2$ (since this is equivalent to the P- and Q-symbols being the same).
| https://mathoverflow.net/users/66 | What bijection on permutations corresponds under RS to transpose? | When you conjugate diagrams and apply RSK or other Young tableau bijections, the answer is typically bad, with some rare exceptions. The right way to think of RSK is to think of row length being continuous while columns still integer (see e.g. my "Geometric proof of the hook-length formula" paper and refs therein).
An exception: there is a "hidden symmetry" for LR-coefficients when you conjugate all three diagrams - see Hanlon-Sundaram paper (1992). Once you know this bijection, there are natural connections to RSK as described in the long Pak-Vallejo paper and a followup by Azenhas-Conflitti-Mamede (search the web for a paper and a ppt presentation). Together, these do give a complete description of your involution, but making sense of it might require quite a bit of work.
| 10 | https://mathoverflow.net/users/4040 | 15490 | 10,375 |
https://mathoverflow.net/questions/15478 | 6 | Can you offer some examples of such rings, other than $\frac{k[x,y]}{(x^{2}, xy)}$. Thanks.
| https://mathoverflow.net/users/3976 | Examples of one-dimensional non-Cohen Macaulay rings | In dimension $1$, Cohen-Macaulay just mean *unmixed*, so all the associated primes have the same dimension. Thus the easiest way to cook up a non-CM ring of dimension $1$ is: Pick your favorite regular ring (say $A=k[x,y,z]$). Take an ideal of dimension $1$, say $I=(x,y)$. Take another ideal of dimension $0$, say $J=(x^3,y^4,z^5)$ such that $J$ is not contained in $I$. Now take $R=A/(I\cap J)$. Geometrically we just throw 2 things of pure dimensions $1$ and $0$ together. In some sense, all non-CM rings of dimension $1$ arise this way.
| 10 | https://mathoverflow.net/users/2083 | 15491 | 10,376 |
https://mathoverflow.net/questions/15486 | 6 | Recall that a *group* is an associative, unital monoid $G$ such that the map $(p\_1,m) : G \times G \to G\times G$ is an isomorphism of sets. Here $p\_1$ is the first projection and $m$ is the multiplication, so the map is $(g\_1,g\_2) \mapsto (g\_1,g\_1g\_2)$.
My question is a basic one concerning the definition of "2-group".
Recall that a *monoidal category* is a category $\mathcal G$ along with a functors $m : \mathcal G \times \mathcal G \to \mathcal G$ and $e: 1 \to \mathcal G$, where $1$ is the category with one object and only identity morphisms, such that certain diagrams commute up to natural isomorphism and those natural isomorphisms satisfy some axioms of their own (the natural isomorphisms are part of the data of the monoidal category).
Then a *2-group* is a monoidal category $\mathcal G$ such that the functor $(p\_1,m): \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ is an equivalence of categories. I.e. there exists a functor $b: \mathcal G \times \mathcal G \to \mathcal G \times \mathcal G$ such that $b\circ (p\_1,m)$ and $(p\_1,m) \circ b$ are naturally isomorphic to the identity. Note that $b$ is determined only up to natural isomorphism of functors.
>
> **Question:** Can I necessarily find such a functor $b$ of the form $b = (p\_1,d)$, where $d : \mathcal G \times \mathcal G \to \mathcal G $ is some functor (called $d$ for "division")? If so, can I necessarily find $d = m\circ(i \times \text{id})$, where $i: \mathcal G \to \mathcal G$ is some functor (called $i$ for "inverse")?
>
>
>
In any case, the natural follow-up question is to ask all these at the level of 3-groups, etc.
| https://mathoverflow.net/users/78 | How strict can I be in the definition of "2-group"? | You can always do this. Take any $b$ and define $d = p\_2 b$. Then $b' = (p\_1, d)$ is equivalent to the original $b$. To see this note that
$$(p\_1, m) \circ b = (p\_1b, m \circ (p\_1 b, d)) \simeq id = (p\_1, p\_2) $$
The first component shows $p\_1 b \simeq p\_1$. We use this transformation $\times id$ to show that
$b \simeq b' = (p\_1, d)$.
Now we consider the equivalence $(p\_1, m) \circ b' \simeq id$. Here we have,
$$(p\_1, m) \circ (p\_1, d) = (p\_1, m \circ (p\_1, d)) \simeq id = (p\_1, p\_2) $$
restricting to $G = G \times \{ 1\} \subseteq G \times G$, this gives a natural isomorphism $m(x, d(x, 1)) \simeq 1$. You can take $i(x) = d(x,1)$, and we have $x i(x) \cong 1$.
We also have
$$(p\_1, d) \circ (p\_1, m) = (p\_1, d \circ (p\_1, m)) \simeq (p\_1, p\_2) $$
which gives a natural isomorphism, $ d(x, xy) \simeq y$ (writing $m(x,y) = xy$).
Thus we have,
$$d(x,y) \simeq d(x,1 y) \simeq d(x, x i(x) y) \simeq i(x) y, $$
which is the formula you were after. So we can replace $d(x,y)$ with $m(i(x), y)$ to get a third inverse functor b''. Note that this doesn't mean that we have a strict 2-group, just that we can define the inverse functors and difference functors you asked about.
Notice also that we didn't really use anything about G being a 1-category as opposed to an n-category (except the associator and unitors) so this argument generalizes to the n-group setting basically verbatim.
| 4 | https://mathoverflow.net/users/184 | 15495 | 10,377 |
https://mathoverflow.net/questions/15493 | 8 | Is a closed morphism with proper fibres proper?
| https://mathoverflow.net/users/nan | morphism closed + fibres proper => proper? | The answer is no. Consider an integral nodal curve $Y$ over an algebraically closed field, normalize the node and remove one of the two points lying over the node. Then you get a morphisme $f : X\to Y$ which is bijective (hence homeomorphic), separated and of finite type, and the fibers are just (even reduced) points. But $f$ is not proper (otherwise it would be finite and birational hence coincides with the normalization map).
In the positive direction, you can look at EGA, IV.15.7.10.
[**Add**] There is an elementary way to see that $f$ is not proper just using the definition. Let $Y'\to Y$ be the normalization of $Y$. So $X$ is $Y'$ minus one closed point $y\_0$. It is enough to show that the base change of $f$ to $X\times Y' \to Y \times Y'$ is not closed. Consider the closed subset
$$\Delta=\left\lbrace (x, x) \mid x\in X \right\rbrace \subset X\times Y'.$$
Its image by $f\_{Y'} : X\times Y' \to Y\times Y'$ is $\left\lbrace (f(x), x) \mid x\in X\right\rbrace$ which is the graph of $Y'\to Y$ minus one point $(f(y\_0), y\_0)$. So $f$ is not universally closed, thus not proper.
| 29 | https://mathoverflow.net/users/3485 | 15503 | 10,382 |
https://mathoverflow.net/questions/15444 | 178 | Define an "eventual counterexample" to be
* $P(a) = T $ for $a < n$
* $P(n) = F$
* $n$ is sufficiently large for $P(a) = T\ \ \forall a \in \mathbb{N}$ to be a 'reasonable' conjecture to make.
where 'reasonable' is open to interpretation, and similar statements for rational, real, or more abstractly ordered sets for $n$ to belong to are acceptable answers.
What are some examples of eventual counterexamples, famous or otherwise, and do different eventual counterexamples share any common features? Could we build an 'early warning system' set of heuristics for seemingly plausible theorems?
edit: The Polya conjecture is a good example of what I was trying to get at, but answers are not restricted to number theory or any one area.
| https://mathoverflow.net/users/3623 | Examples of eventual counterexamples | It was once conjectured that factors of $x^n-1$ over the rationals had no coefficient exceeding 1 in absolute value. The first counterexample comes at $n=105$.
| 161 | https://mathoverflow.net/users/3684 | 15506 | 10,384 |
https://mathoverflow.net/questions/15494 | 16 | The question on games and mathematics that appeared recently on mathoverflow
([Which popular games are the most mathematical?](https://mathoverflow.net/questions/13638/which-popular-games-are-the-most-mathematical))
reminded me of a problem I encountered some time ago : starting with the insane
dream of completely solving the game of bridge with a nice mathematical theory,
I ended up considering extremely simplified versions of bridge. One of them was
as follows : there are only 2 players instead of 4, and instead of the usual
deck there are only 2n cards numbered from 1 to 2n. Each player holds half
of the deck, so this is a "complete information" game : each player knows exactly
what is in his opponent's hand. There are no bids, just a sequence of n moves
where each player drops a card ; as in bridge the strongest card wins the trick
and the winner of the game is the player with the largest number of tricks
in the end (take n odd to avoid draws). Also, the winner of the preceding
trick is the first to play (for the very first move the first player is determined
by some rule, random or other ; this is immaterial to the subsequent discussion).
This looks like a very basic kind of game, especially amenable to
mathematization : for example the set of all initial positions is nicely indexed
by the subsets $I$ of $\lbrace 1,2, \ldots , 2n\rbrace$ whose cardinality is $n$
(say $I$ is the set of cards held by the first player). I was
however unable to answer the following questions :
* Is there an algorithm which, given the initial position, finds out which player
will win if each one plays optimally ? What is the best strategy ?
+ Has this game already been studied by combinatorialists ?
| https://mathoverflow.net/users/2389 | Mathematical solution for a two-player single-suit trick taking game? | Yes, it has been studied by Johan Wästlund in *A solution of two-person single-suit whist*, which gives an efficient algorithm to compute the value of a position in this game (Theorem 10.1). He has also studied the more general situation of multiple suits, but with the restriction that each suit is split evenly between the two players, in *Two-person symmetric whist*. Here some familiar values from combinatorial game theory appear, reflecting the fact that breaking a new suit is generally disadvantageous to the player doing so.
| 21 | https://mathoverflow.net/users/126667 | 15530 | 10,400 |
https://mathoverflow.net/questions/15407 | 5 | Suppose two positive holomorphic line bundles $L\_1 \to X\_1, L\_2\to X\_2$ over two projective complex manifold $X\_1, X\_2$ have isomorphic ring of sections $R=R\_1=R\_2$ where $R\_i=\oplus\_{m=0}^\infty\Gamma(X\_i,mL\_i)$. Isomorphism as graded ${\mathbb C}$- algebras.
Is there any relationship betweeen $X\_1$ and $X\_2$? Eg, some morphism between them? How about relationship to $Proj R$?
Thanks.
| https://mathoverflow.net/users/nan | Relationship between Line Bundles with isomorphic ring of sections | To expand on the answer above: as B. Cais says, if the line bundles are ample (which I think follows from positivity by Kodaira), we have a canonical isomorphism $\mathrm{Proj} R\_i\cong X\_i$. Thus, if the graded rings $R\_i$ are isomorphic, then the induced map of Proj's gives an isomorphism $R\_1\cong R\_2$ carrying one line bundle to the other.
| 4 | https://mathoverflow.net/users/66 | 15537 | 10,406 |
https://mathoverflow.net/questions/15496 | 9 | Let *K* be an ordered field. Define the *n*-sphere:
$$S^n(K) := \{ (x\_1,x\_2,\dots,x\_n+1) \in K^{n+1} \mid \sum\_{i=1}^{n+1} x\_i^2 = 1 \}$$
A set of vectors $v\_1, v\_2, \dots, v\_r \in S^n(K)$ is *orthonormal* if the dot product of any two of them is zero. An *orthonormal basis* is an orthonormal set of cardinality $n + 1$.
1. Is every vector in $S^n(K)$ a member of an orthonormal basis? If not, what is the largest *r* such that very vector is a member of an orthonormal set of size *r*?
2. More generally, given *n* and *s*, what is the largest *r* such that every orthonormal set in $S^n(K)$ of size *s* is contained in an orthonormal set of size *r*?
What's known:
1. For $n = 1$, every vector in $S^n(K)$ is a member of an orthonormal basis, regardless of *K*.
2. If *K* is Pythagorean (i.e., a sum of squares is a square) every orthonormal set completes to an orthonormal basis (use Gram-Schmidt).
Can more be said? I'm most interested in the case of *K* the field of rational numbers or a real number field, and the case $n = 2$.
ADDED LATER: I am assuming that *K* is an ordered field here. Otherwise, we need to modify our definition of orthonormal set to also include the condition that the vectors are linearly independent, which is automatically true for ordered fields. Observations for fields that are not ordered (such as non-real number fields or fields of positive characteristic) would also be much appreciated. MODIFIED: As Bjorn Poonen points out below, linear independence turns out to follow automatically in this case. (Though in general, over non-ordered fields, there can exist orthogonal vectors that are linearly dependent, our condition that the vectors be "normal" rules this out).
| https://mathoverflow.net/users/3040 | Spheres over rational numbers and other fields | Any orthonormal set extends to an orthonormal basis, over any field
of characteristic not $2$. This is a special case of [Witt's theorem](http://en.wikipedia.org/wiki/Witt%27s_theorem).
**EDIT:** In response to Vipul's comment: The proof of Witt's theorem is constructive, and leads to the following recursive algorithm for extending an orthonormal set $\lbrace v\_1,\ldots,v\_r \rbrace$ to an orthonormal basis.
Let $e\_1,\ldots,e\_n$ be the standard basis of $K^n$, where $e\_i$ has $1$ in the $i^{\operatorname{th}}$ coordinate and $0$ elsewhere. It suffices to find a sequence of reflections defined over $K$ whose composition maps $v\_i$ to $e\_i$ for $i=1,\ldots,r$, since then the inverse sequence maps $e\_1,\ldots,e\_n$ to an orthonormal basis extending $v\_1,\ldots,v\_r$. In fact, it suffices to find such a sequence mapping just $v\_1$ to $e\_1$, since after that we are reduced to an $(n-1)$-dimensional problem in $e\_1^\perp$, and can use recursion.
*Case 1:* $q(v\_1-e\_1) \ne 0$, where $q$ is the quadratic form. Then reflection in the hyperplane $(v\_1-e\_1)^\perp$ maps $v\_1$ to $e\_1$.
*Case 2:* $q(v\_1+e\_1) \ne 0$. Then reflection in $(v\_1+e\_1)^\perp$ maps $v\_1$ to $-e\_1$, so follow this with reflection in the coordinate hyperplane $e\_1^\perp$.
*Case 3:* $q(v\_1-e\_1)=q(v\_1+e\_1)=0$. Summing yields $0=2q(v\_1)+2q(e\_1)=2+2=4$, a contradiction, so this case does not actually arise.
| 16 | https://mathoverflow.net/users/2757 | 15540 | 10,408 |
https://mathoverflow.net/questions/12955 | 8 | I'm just looking for an example of an integral, rectifiable varifold, which has no locally bounded first variation.
---
### Recapitulation
for every $m$-rectifiable varifold $\mu$ exists a $m$-rectifiable set $E$ in $\mathbb R^n$, meaning $E=E\_0 \cup \bigcup\_{k\in\mathbb N} E\_k$ with $\mathcal H^m(E\_0)=0$ and $E\_k\subseteq F\_k$ for some $\mathcal C^1$-manifolds $F\_k$ of dimension $m$, and a non-negativ function $\theta\in L^1\_{\text{loc}}(\mathcal H^m|\_E)$ such that $\mu=\theta \mathcal H^m|\_E$. This is a characterisation of $m$-recitifiable varifolds.
The first variation $\delta\mu$ of a varifold $\mu$ is for $\eta\in\mathcal C^1\_c(\mathbb R^n;\mathbb R^n)$ given by
$$\delta\mu(\eta)=\int div\_\mu\eta\,d\mu,$$
where $ div\_\mu(\eta)(x) = \sum\_{i=1}^n \tau\_i^T(x)\cdot D\eta(x)\cdot \tau\_i(x)$ where $\tau\_i(x)$ is a orthogonal basis of the tangentspace of $\mu$ in $x$, which coinsidence $\mu$-almost everywhere with $T\_xF\_i$ for $x\in E\_i\subseteq F\_i$ as above. So $div\_\mu\eta(x)$ is just the divergence in the manifold $F\_i$, with $x\in E\_i\subseteq F\_i$.
We say $\mu$ has an locally bounded first variation, if for all $\Omega'\subseteq \Omega$ there exists $c(\Omega')<\infty$ such that
$$ \delta\mu(\eta) \le C(\Omega',\Omega) \Vert \eta\Vert\_{L^\infty(\Omega)} \qquad\forall\;\eta\in\mathcal C^1\_c(\Omega'). $$
See for more explanation for example <http://eom.springer.de/G/g130040.htm>.
For a $\mathcal C^2$-manifold $M$ in $\mathbb R^n$ with mean curvature $H\_M$ the first variation is
$$ \delta M(\eta)=-\int\_M H\_M \cdot \eta \,dvol\_M -\int\_{\partial M} \tau\_0 \cdot \eta \,dvol\_{\partial M} \qquad\forall\;\eta\in\mathcal C\_c^1(\mathbb R^n)$$
with the inner normal $\tau\_0\in T\_xM\cap(T\_x\partial M)^\bot$ and where the mean curvature is the trace of the second fundamental form $A$ by the meaning of $H\_M(x)=\sum\_{i=1}^m A\_x(\tau\_i,\tau\_i)$ in the normal space of $M$. As obviouse in this case the first variation is locally bounded.
| https://mathoverflow.net/users/3538 | Example for an integral, rectifiable varifold with unbounded first variation | Something is strange here: it seems like for the sawtooth curve (the Lipschitz curve that goes up and down with slope $1$), the first variation is just the sum of $\delta$-measures at the turning points times the unit bisector vectors, so we can have fixed length and arbitrarily large first variation (just make turning points more and more dense), which can be now trivially turned into an example of finite area and infinite first variation: take more and more rigged closed sawtooth curves around infinitely many circles with finite some of radii contained in a compact domain. Am I missing something?
| 6 | https://mathoverflow.net/users/1131 | 15549 | 10,415 |
https://mathoverflow.net/questions/15546 | 2 | I have an inclusion of topological spaces (actually manifolds with corners) $X \to Y$. I can show that for every $x \in X$ there is a neighborhood of $x$ in $Y$ of the form $U \times V$. Also, the intersection of $U \times V$ with $X$ is $U \times V'$.
In fact, $V$ is a neighborhood of the origin in $[0,1)^k$. Also, $V' \subset V$ is given by a set of subsets $\mathcal I$ of $\{1, \ldots, k\}$. It is the union
$$
\bigcup\_{I \in \mathcal I} \{ x\_i = 0 \text{ for all } i \in I \}.
$$
That is, $V'$ is made up of some half-lines, quarter-planes, octants, etc. in $[0,1)^k$.
Question: Is this enough to show that $X \to Y$ is a cofibration of topological spaces?
| https://mathoverflow.net/users/1676 | Simple question of topological cofibration | The inclusion of a CW subcomplex $K$ into a CW complex $L$ is a cofibration. Briefly, we can extend a homotopy from the $n-1$-skeleton to the $n$-skeleton by projecting $e\times I$ to $e\times\{0\}\cup\partial E\times I$ for any $n$-cell $e$ not in $L$.
| 4 | https://mathoverflow.net/users/2349 | 15553 | 10,418 |
https://mathoverflow.net/questions/15438 | 42 | On one of my exams last year, we were given a problem (we chose five or six out of eight problems) on an exam, the goal of which was to prove the Bruhat decomposition for $GL\_n(k)$. I was one of the two people to choose said problem. I gave a very long convoluted argument which although correct was really inelegant. I proved it more than once because I wasn't satisfied with my proof, and I figured out a somewhat slick contradiction argument based on maximizing leading zeroes of rows (number of zeroes before the pivot), but the proof was still a real mess.
Statement of the problem:
Let $G:=GL(V)$ for $V$ a finite dimensional $k$ vector space. Let $B$ be the stabilizer of the standard flag (these will be invertible upper triangular matrices), and let $W$ be the subgroup of permutation matrices.
Show that $G=\coprod\_{w\in W} BwB$, where the $BwB$ are double cosets. That is, show that $G=BWB$.
Question:
Is there a slick proof of this fact maybe using "more machinery"? In particular, is there any sort of "coordinate-free" proof (Can we even define the Borel and Weyl subgroups without coordinates?)?
| https://mathoverflow.net/users/1353 | A slick proof of the Bruhat Decomposition for GL_n(k)? | You are asking to compute the double quotient $B\backslash G /B.$ This is the same as computing $G\backslash (G/B \times G/B)$. A point in $G/B$ is a full flag on $k^n$. So you
are trying to compute the set of pairs $(F\_1,F\_2)$ of flags, modulo the simultaneous action
of $G$.
Another way to think about $G/B$ is that it is the space of Borel subgroups (the coset of
$g$ corresponds to the conjugate $g B g^{-1}$, where $B$ is the upper triangular Borel
that was fixed in the statement of the question).
The passage from flags to Borels is given by mapping $F$ to its stablizer in $G$.
So you can also think that you're trying to describe pairs of Borels $(B\_1,B\_2)$,
modulo simultaneous conjugation by $G$.
Now recall that a torus $T$ in $G$ is a conjugate of the diagonal subgroup. Choosing
a torus in $G$ is the same as choosing a decomposition of $k^n$ as a direct sum of
1-dimensional subspaces (or lines, for short). (These will be the various eigenspaces of the torus acting
on $k^n$.) The diagonal torus corresponds to the standard decomposition of $k^n$
as $n$ copies of $k$.
Now a torus $T$ is contained in a Borel $B$ (let me temporarily use $B$ to
denote any Borel, not just the upper triangular one) if and only if the corresponding
decomposition of $k^n$ into a sum of lines is compatible with the flag that $B$ fixes, i.e. if the flag is given by taking first one line, than the sum of that one with a second, then the sum of those two
with a third, and so on. In particular, choosing a torus $T$ contained in a Borel $B$
determines a "labelled decomposition" of $k^n$, i.e. we may write
$k^n = L\_1 \oplus \ldots \oplus L\_n$, where $L\_i$ is the $i$th line;
just to be clear, the labelling is chosen so that the corresponding flag is just
$L\_1 \subset L\_1\oplus L\_2 \subset \cdots.$ (Again, to be completely clear,
if $T$ is the conjugate by $g \in G$ of the diagonal torus, then $L\_i$ is the
translate by $g$ of the line spanned by the $i$th standard basis vector.)
Note that this labelled decomposition depends not just on $T$ (which only gives an
unlabelled decomposition) but on the Borel $B$ containing $T$ as well. (In more Lie
theoretic language, this is a reflection of that fact that a torus determines
a collection of weights in any representation of $G$,
while a choice of a Borel containing the torus lets you order the weights as well,
by determining a set of positive roots.)
Of course, $B$ will contain more than one torus; or more geometrically,
$k^n$ will admit more than one decomposition into lines adapted to the filtration
$F$ of which $B$ is the stabilizer. But if one thinks about the different possible
lines, you see that $L\_1$ is uniquely determined (it must be the first step in
the flag), $L\_2$ is uniquely determined modulo $L\_1$ (since together with $L\_1$
it spans the second step in the flag), and so on, which shows that any two tori
$T$ in $B$ are necessarily conjugate by an element of $B$, and the same sort
of reasoning shows that the normalizer of $T$ in $B$ is just $T$ (because
if $g \in G$ is going to preserve both the flag and the collection of lines, which
is the same as preserving the ordered collection of lines, all it can do
is act by a scalar on each line, which is to say, it must be an element of $T$).
Now a key fact is that any two Borels, $B\_1$ and $B\_2$, contain a common torus.
In other words, given two filtrations, we can always choose an (unordered)
decomposition of $k^n$ into a direct sum of lines which is adapted to *both*
filtrations. (This is an easy exercise.) Of course the ordering of the
lines will depend which of the two filtrations we use. In other words, we get a set
of $n$ lines in $k^n$ which are ordered one way according to the filtration
$F\_1$ given by $B\_1$, and in a second way according to the filtration
given $F\_2$ by $B\_2$. If we let $w \in S\_n$ be the permutation which takes the first
ordering to the second, then we see that the pair $B\_1$ and $B\_2$ determines
an element $w \in S\_n$. This *is* the Bruhat decomposition.
It wouldn't be hard to continue with this point of view to completely prove
the claimed decomposition, but it will be easier for me (at least notationally)
to switch back to the $B\backslash G/B$ picture.
Thus consider the coset $gB$ in $G/B$, corresponding to the Borel $g B g^{-1}$.
Let me use slightly nonstandard notation, and write $D$ for the diagonal torus;
of course $D \subset B$. We may also find a torus $T \subset B \cap g B g^{-1}$.
Now there is an element $b \in B$, determined modulo $D$, such that $T = b D b^{-1}$.
(This follows from the discussion above about conjugacy properties of tori in
Borel subgroups.) We also have $g D g^{-1} \subset g B g^{-1}$, and there exists
$g b'g^{-1}\in g B g^{-1},$ well defined modulo $g D g^{-1}$, such that
$T = (g b'g^{-1}) g D g^{-1} (g b' g^{-1})^{-1} = g b' D (b')^{-1} g^{-1}.$
We thus find that $b^{-1} g b' \in N(D)/D$, and thus that $g \in B w B$ for
some $w$ in the Weyl group $N(D)/D$. Note that since $b$ and $b'$ are
well defined modulo $D$, the map from $T$ to $w$ is well-defined.
Thus certainly $G$ is the union of the $B w B$. If you consider what I've already
written carefully, you will also see that the different double cosets are disjoint. We can also prove this directly as follows: given $B$ and $g B g^{-1}$, the map
$T \mapsto w$ constructed above is a map from the set of $T$ contained in $B \cap
g B g^{-1}$ to the set $N(D)/D$. Now any two such $T$ are in fact conjugate
by an element of $B \cap g B g^{-1}$. The latter group is connected,
and hence the space of such $T$ is connected.
(These assertions are perhaps most easily seen by thinking in terms of
filtrations and decompositions of $k^n$ into sums
of lines, as above). Since $N(D)/D$ is discrete, we see that $T \mapsto w$ must
in fact be constant, and so $w$ is uniquely determined just by $g B g^{-1}$ alone.
In other words, the various double cosets $B w B$ are disjoint.
The preceding discussion is a litte long, since I've tried to explain (in the
particular special cases under consideration) some general facts about conjugacy
of maximal tori in algebraic groups, using the translation of group theoretic
facts about $G$, $B$, etc., into linear algebraic statements about $k^n$.
Nevertheless, I believe that this is the standard proof of the Bruhat decomposition,
and explains why it is true: the relative position of two flags is described by an element of the Weyl group.
| 54 | https://mathoverflow.net/users/2874 | 15554 | 10,419 |
https://mathoverflow.net/questions/15550 | 8 | I'll explain the problem but what I am looking for is a few suggested methods to approach this problem.
You don't need to know what a microarray but if you are interested look here [link text](http://en.wikipedia.org/wiki/DNA_microarray)
The info below is simplified, not addressed to a micro-biologist (I am not one)
Data:
There are about 250,000 probes on an array that each measure a part of a dna sample.
An array is specific to an individual(germ, person....) If the array is designed for my dna and we put yours on it some probes will match (we are both human I think) and some will not. If they match the measurement is high(the probe glows) as the dna at a probe matched less it glows less. There is a very high variance at each probe across the same dna. So it is very difficult to know if a individual sample matches (same dna) at the probe level.
If the data was nice we could just do a simple Hypothesis test to determine if the sample probe was the same as the dna the array was made for. Because of the high variance this will not work.
Question:
We are looking for a way to consider the measurements at other probes to improve the test at a probe. We kinda know this should work (based on the DNA) but we are not sure how to choose the probes that will best assist for each of the set of 250,000 probes or what statistical method would be best to use the information about the other probes to test an individual probe.
Different example, maybe easier to understand:
A person has 6 characteristics, lets say weight, height, age, income, date of birth (db), education, race. I measure characteristics for myself but my measurement tools are really bad so I make several measurements but I have a very large variance in my measurements. Now I measure another person and want to know if they have a different income than myself. But my measurement tools are so bad almost any other person’s measured income would be within my confidence interval. (I can only measure the other person once, It kills them ☺ ) But I know that other characteristics should be able to help me determine if the income is the same. I don’t know which ones (weight, height, age, db, education, race). I need a method to choose the most useful/helpful characteristics. Now lets say I know that height and age can help determine if our income are the same or different. What would be the statistical method used?
I have what might be called a training set. That is in know what is and is not the same.
@sheldon-cooper answer,
"Note: neither of this will give a "significance score"", A confidence value is good enough. I don't need a statistical "Test"
As an MD I hope that this is an acceptable question for MathOverFlow.net. and I hope you can understand the question as abridged as it is. If some would like to help me rewrite or has suggestion on how I can rewrite the question without all the bio background and as a more specific stats question, let me know.
| https://mathoverflow.net/users/4045 | MicroArray, tesing if a sample is the same with high variance data. | In general, the approach of using additional measurements of other values (not the one you are interested in directly) was useful in many problems in the past, so it sounds like a good idea in your case as well. Here are a few things to look at:
Binary classification
---------------------
If the task is to determine whether a given probe is a 'match', you can just pose it as a binary classification problem. The input features to a classifier will be ALL the probes in the array. The classifier will determine automatically which of these are useful. If the variance at the target probe is actually small, it may decide to just look at that probe only. If other sites (probes) are helpful too, it may decide to look at them as well. Classifiers can automatically construct complex rules that take into account one or many probes, as needed.
This can be simple to try since there are packages readily available and there are not many parameters to tune. I'd try using a classifier called SVM, it works well in many cases, can deal with limited training data (looks like that's an issue for you). Again, there are SVM packages available that you can download and just run (libsvm is one example, but there are many others).
The disadvantage of this is that it may become resource-consuming if you want a classifier for each probe. Your input dimensionality is 250,000. A simple linear classifier needs roughly one number per dimension, so 250,000 numbers. If you need such a classifier for every probe, this will require 250,000^2 numbers total, which is a lot. There are ways to go around this, but that's one concern.
Collaborative filtering
-----------------------
Collaborative filtering basically allows you to predict the value of one item from the values of related items. Say you know that I liked two movies M1, M2, and disliked two others, M3 and M4. Using collaborative filtering, you can predict whether or not I will like a new movie M5. In your case, you are not doing a completely blind prediction for M5; you do have some (although very noisy) observation. So you would combine that observation with the prediction from other features.
The advantage of this is that there is only one model (as opposed to one classifier per probe). So this is easier to scale.
One model would work something like this. It would group probes that are either all "same" or all "different" most of the time. (These groups would be learned by the model from training data.) Then, given a new array, it would use these groups and data from other probes in each group to determine whether a group is, again, "same" or "different". [This](https://mathoverflow.net/questions/14412/matrix-factorization-model) question had a link to what looked like an easy intro to collaborative filtering.
The main disadvantage is, I'm not sure you can find an implementation of the model you need for microarrays online (or anywhere). You'd probably need to derive something suitable for your data, which is not difficult, but requires knowing what to do.
Another disadvantage is, you'll probably need more training data for collaborative filtering to work.
Note: neither of this will give a "significance score" (as in t-test, for example). You'll get a confidence value, but you won't be able to say that the findings are statistically significant with a certain confidence.
| 1 | https://mathoverflow.net/users/3035 | 15557 | 10,421 |
https://mathoverflow.net/questions/15558 | 8 | Do you know any explicit constructions of families of irreducible finite dimensional representations of the three string braid group?
I will describe what I already know and then invite you to suggest alternatives. This is somewhat open-ended so I will also pose some more precise questions.
---
This problem is solved for dimension at most five in
<http://arxiv.org/abs/math/9912013> by Imre Tuba, Hans Wenzl
and published in Pacific Journal of Mathematics. So I am particularly interested in six dimensional representations.
---
The usual presentation is generators $\sigma\_1$, $\sigma\_2$ and the relation $\sigma\_1\sigma\_2\sigma\_1=\sigma\_2\sigma\_1\sigma\_2$. An alternative is generators $s$, $t$ and the relation $s^2=t^3$ where $s=\sigma\_1\sigma\_2\sigma\_1$ and $t=\sigma\_1\sigma\_2$. This shows we have a central extension of the free product of $C\_2$ and $C\_3$.
Given a representation of dimension $N$ take the dimensions of the eigenspaces of $s$ and $t$. Then we have $n\_1+n\_2=N$ and $m\_1+m\_2+m\_3=N$. If the representation is irreducible then $n\_i\ge m\_j$. This data labels the irreducible components of the variety of irreducible representations of dimension $N$. The dimension of the component is $N^2-n\_1^2-n\_2^2-m\_1^2-m\_2^2-m\_3^2+2$. For $N=6$ we have
* (4,2) and (2,2,2) of dimension 6
* (3,3) and (2,2,2) of dimension 8
* (3,3) and (3,2,1) of dimension 6
* (3,3) and (3,3,0) of dimension 2
---
The case (3,3) and (3,2,1) is interesting because these representations do not deform to representations of the symmetric group.
---
Can you give a representation and a braid $\alpha$ such that the trace of $\alpha$ and the trace of the braid given by reading $\alpha$ backwards are different? I have an example of dimension 12.
---
The description as a central extension of the free product means we have a construction which associates a representation to each invertible matrix (for each component).
I would like to improve on this in two ways. First this is unwieldy. Second I would like to be able to specify the eigenvalues of $\sigma\_1$. This means passing to a covering of the moduli space.
| https://mathoverflow.net/users/3992 | Representations of the three string braid group | Bruce,
I can give you representants for an open piece of the moduli space (3,3) resp. (3,2,1). I'll add them in Mathematica-form so that you can plug them into any braid you like to work on. The variables a,b,d,x,y,z are the coordinates of the moduli space and l stands for a third root of unity (i dont know how to tell Mathematica to work with roots of unity, so if you know replace l by that thing).
I obtained these representants from the hexagon of 1-dml simples of the modular group and taking the dimension vector (1,1,1,2,1,0) as required. The corresponding moduli problem is then rational (in x,y,z) over that of pairs of 2x2 matrices of rank one. these are rational (in a,b,d). Hope this helps (and sorry for the lengthy formulas below).
sigma1={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, (-1 + l) z, (b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, (-1 + a) (-1 + l) (l + x) y z, -(-1 + l) (-1 + x) z, -(b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, (a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, -b (-1 + l) (l + x) y z, (-1 + l) (a + b - a d - a x - b x + a d x - a y) z, -b (-1 + l) (-1 + x + y) z}, {-(1 - a - b - d + a d) (-1 + l) (1 + l) x z, (-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, (-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-b (-1 + l) (1 + l) x y z, - b (-1 + l) (-1 + x + y + l y) z, (-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {(-1 + a) (-1 + l) (1 + l) x y z, (-1 + a) (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}
sigma2={{-(1 - a - b - d + a d + x - a x - b x - d x + a d x + l x - a l x - b l x - d l x + a d l x - y + a y) z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, (-1 + l) (1 - a - b - d + a d + l - a l - b l - d l + a d l - y + a y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-b - d + a d + l - a l + b x + d x - a d x - l x + a l x + y - a y + l y - a l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) (l + x) y z, (-1 + l) (-1 + x) z, (b + d - a d) (-1 + l) (-1 + x) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, ( a + b - a d - l + d l - a x - b x + a d x + l x - d l x - a y + l y) z, b (-1 + l) (l + x) y z, -(-1 + l) (a + b - a d - a x - b x + a d x - a y) z, b (-1 + l) (-1 + x + y) z}, {(1 - a - b - d + a d) (-1 + l) (1 + l) x z, -(-1 + l) (b + d - a d - l + a l + b l + d l - a d l) z, -(-1 + l) z, -(1 - a - b - d + a d + l - a l - b l - d l + a d l + x - a x - b x - d x + a d x + l y - a l y) z, -(-1 + l) z, -(b + d - a d) (-1 + l) z}, {b (-1 + l) (1 + l) x y z, b (-1 + l) (-1 + x + y + l y) z, -(-1 + l) (1 - d - x + d x - y) z, b (-1 + l) (l + x) y z, -(-1 + d + a l + b l - a d l + x - d x - a l x - b l x + a d l x + y - a l y) z, b (-1 + l) (-1 + x + y) z}, {-(-1 + a) (-1 + l) (1 + l) x y z, -(-1 + a) (-1 + l) (-1 + x + y + l y) z, (-1 + l) (-1 + x) z, -(-1 + a) (-1 + l) ( l + x) y z, (-1 + l) (-1 + x) z, -(-1 + a + b l + d l - a d l + x - a x - b l x - d l x + a d l x + y - a y) z}}
EDIT March 9th : The component (3,3;3,2,1) is not able to detect braid-reversion. On the other hand, the component of 6-dimensional representations (3,3;2,2,2) can. A minimal braid that can be separated by this family from its reversed braid is s1^-1s2^2s1^-1s2. One can show that every irreducible component of simple B(3)-representations has a Zariski dense family parametrized by a minimal rational variety. For representations of dimension <= 11 explicit such families are given in [this note](http://arxiv.org/abs/1003.1610).
Each of these families can then be turned into a family of 3-string braid invariants over Q(rho) for rho a primitive 3-rd root of unity by specializing the parameters to random integers. For the family (3,3;2,2,2) mentioned above this can be done in SAGE as follows :
```
K.=NumberField(x^2+x+1)
a=randint(1,1000)
b=randint(1,1000)
c=randint(1,1000)
d=randint(1,1000)
e=randint(1,1000)
f=randint(1,1000)
g=randint(1,1000)
h=randint(1,1000)
B=matrix(K,[[1,0,0,a,0,f],[0,1,1,0,1,0],[1,1,0,1,0,0],[0,0,1,0,d,e],[0,1,0,b,c,0],[g,0,1,0,0,1]])
Binv=B.inverse()
mat2=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,1,0,0,0],[0,0,0,-1,0,0],[0,0,0,0,-1,0],[0,0,0,0,0,-1]])
mat3=matrix(K,[[1,0,0,0,0,0],[0,1,0,0,0,0],[0,0,l^2,0,0,0],[0,0,0,l^2,0,0],[0,0,0,0,l,0],[0,0,0,0,0,l]])
s1=h*Binv*mat3*B*mat2
s2=h*mat2*Binv*mat3*B
s1inv=s1.inverse()
s2inv=s2.inverse()
```
One can then check braid-reversion for all knots with at most 8 crossings that are closures of 3-string braids. Here are the tests to perform
```
test41=(s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv).trace()
test52=(s1inv**3*s2inv*s1*s2inv-s2inv*s1*s2inv*s1inv**3).trace()
test62=(s1inv**3*s2*s1inv*s2-s2*s1inv*s2*s1inv**3).trace()
test63=(s1inv**2*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv**2).trace()
test73=(s1**5*s2*s1inv*s2-s2*s1inv*s2*s1**5).trace()
test75=(s1inv**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1inv**4).trace()
test82=(s1inv**5*s2*s1inv*s2-s2*s1inv*s2*s1inv**5).trace()
test85=(s1**3*s2inv*s1**3*s2inv-s2inv*s1**3*s2inv*s1**3).trace()
test87=(s1**4*s2inv*s1*s2inv**2-s2inv**2*s1*s2inv*s1**4).trace()
test89=(s1inv**3*s2*s1inv*s2**3-s2**3*s1inv*s2*s1inv**3).trace()
test810=(s1**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1**3).trace()
test816=(s1inv**2*s2*s1inv**2*s2*s1inv*s2-s2*s1inv*s2*s1inv**2*s2*s1inv**2).trace()
test817=(s1inv**2*s2*s1inv*s2*s1inv*s2**2-s2**2*s1inv*s2*s1inv*s2*s1inv**2).trace()
test818=(s1inv*s2*s1inv*s2*s1inv*s2*s1inv*s2-s2*s1inv*s2*s1inv*s2*s1inv*s2*s1inv).trace()
test819=(s1**3*s2*s1**3*s2-s2*s1**3*s2*s1**3).trace()
test820=(s1**3*s2inv*s1inv**3*s2inv-s2inv*s1inv**3*s2inv*s1**3).trace()
test821=(s1inv**3*s2inv*s1**2*s2inv**2-s2inv**2*s1**2*s2inv*s1inv**3).trace()
```
The following tests should give non-zero invariants : 6.3,7.5,8.7,8.9,8.10 (which are known as 'flypes' in the theory) and 8.17 (the [non-invertible knot](http://en.wikipedia.org/wiki/Invertible_knot) with minimal number of crossings). I tried to include all details in the note [Rationality and dense families of B(3) representations](http://arxiv.org/abs/1003.1610). All comments to this text are welcome. I like to thank Bruce Westbury for providing me with feedback.
| 15 | https://mathoverflow.net/users/2275 | 15563 | 10,424 |
https://mathoverflow.net/questions/15460 | 10 | In his article where he stated what we know as Eisenstein's irreducibility criterion (which actually was first proved by Schönemann, as was Scholz's reciprocity law and Hensel's Lemma), he claimed that the result also holds in the following case: assume that
$$ f(x) = x^m + a\_{m-1}x^{m-1} + \ldots + a\_0, $$
where the $a\_i$ are integers divisible by $p$, and where $p^2 \nmid a\_j$ for some
$0 \le j \le m-1$. There is an obvious counterexample provided by the quadratic polynomial $f(x) = (x-p)^2 = x^2 -2px + p^2$, so that could be the end of that story. But it isn't: I've read somewhere that some form of this criterion holds for polynomials of degree $\ge 3$, and that the degrees of the possible factors of counterexamples can be predicted in terms of this index $j$. Unfortunately, I don't remember the exact statement (those who are familiar with Newton polygons will probably be able to figure out a correct version) or where I've seen this. Can anyone help?
| https://mathoverflow.net/users/3503 | Variants of Eisenstein irreducibility | Basically all such criteria boil down to some argument involving the Newton polygon as Kevin Buzzard mentions in the comments. While something as general as your statement has trivial counter examples the following generalization holds:
>
> Let $R$ be a unique factorization domain and $f(x) =a\_nx^n+\cdots +a\_0\in R[x]$ with $a\_0a\_n\neq 0$. If the Newton polygon of $f$ with
> respect to some prime $p\in R$ consists of the only line segment from $(0,m)$ to $(n, 0)$
> and if $gcd(n,m) = 1$ then $f$ is irreducible in $R[X]$.
>
>
>
I've heard this called the Eisenstein-Dumas criterion of irreducibility (it also proves the example given in the comments). Another generalization of Eisenstein's criterion is the following:
>
> If $p|a\_0,a\_1,\dots,a\_k$ but $p^2\not | a\_0$ then $f(x)$ has an irreducible factor of degree $\geq k+1$
>
>
>
(This is how you prove for example, that a polynomial like $x^n+5x^{n-1}+3$ is irreducible, after checking that it has no linear factors.) If not answering your question, at least I hope that this refreshes your memory of the statement you claim above. :)
| 11 | https://mathoverflow.net/users/2384 | 15564 | 10,425 |
https://mathoverflow.net/questions/15560 | 9 | Hello,
I wonder if the techniques introduced in Neemans paper:
"The Grothendieck duality theorem via Bousfield's techniques and Brown representability "
can be used to establish Verdier duality. More precisely:
Consider the unbounded, derived category $D(M)$ of $\mathbb{Q}$ vector spaces on a compact complex manifold $M$ . I would like to show that $Rf\_!$ has a right adjoint.
In order to use Brown representability one has to show that $D(M)$ is compactly generated.
i.e. there exists a set of objects $c\_i$ that commutes with direct sums:
$$Hom(c\_i,\bigoplus x\_j)=\bigoplus Hom(c\_i,x\_j)$$
and generates $D(M)$:
$$\forall c\_i Hom(c\_i,x)=0 \Rightarrow x=0$$
My problem is that i can't find such a set of generators. I first tried shifts of
$$i\_\*\mathbb{Q}$$ where $i$ is the inclusion of an open subset. However these do neither commute with coproducts nor are they generators (they can not see sheaves without global sections).
My second try was shifts of
$$i\_!\mathbb{Q}$$
these are generators, but again they do not seem to respect coproducts.
Can someone give a set of compact generators? Or is this approach to Verdier duality doomed anyway?
| https://mathoverflow.net/users/2837 | Verdier duality via Brown representability? | The category of sheaves of $\mathbb{Q}$ vector spaces on $M$ is a Grothendieck abelian category. It follows that the derived category of such, $D(M)$ in your notation, is a well generated triangulated category. A proof of this can be found in Neeman's paper "On The Derived Category of Sheaves on a Manifold". In particular $D(M)$ satisfies Brown representability.
I thought I'd also comment on the proof. The point is that by a result of Alonso, Jeremías and
Souto one can extend the Gabriel-Popescu theorem to the level of derived categories. This realizes the derived category of any Grothendieck abelian category as a localization of the derived category of R-modules for some ring R. The kernel of this localization can be generated by a set of objects and so general nonsense gives the desired generating set of $\alpha$-small objects for some regular cardinal $\alpha$ for the localization.
| 9 | https://mathoverflow.net/users/310 | 15567 | 10,428 |
https://mathoverflow.net/questions/14762 | 1 | Hi,
probably this is a fairly newbie question, but is it possible that the a generic Levy process explodes (i.e. tends to infinity for finite time t with positive probability)? If yes, could you please provide an example, if no point me to the proof.
| https://mathoverflow.net/users/3160 | Exploding Levy processes | Hi,
As remarked by Leonid Kovalev a Lévy process doesn't explodes as far as I know, nevertheless as you mention generic Levy Process for which I don't know any definition, you may be thinking of them as diffusions driven by a Lévy process.
Then looking at some SDEs, you can have some cases where explosion time is a.s. finite, and even when the driving procsess is a Brownian Motion, for example I think I can remember that $X\_t$ verifying $d[Ln(X\_t)]=a(b-Ln(X\_t))dt+\sigma.dW\_t$, $X\_0=0$ is of this type.
(for references, google at "Black-Karasinski short rate model")
Regards
| 1 | https://mathoverflow.net/users/2642 | 15575 | 10,432 |
https://mathoverflow.net/questions/14959 | 5 | Let $E/F$ be a quadratic extension of number fields, and let $V$ be an $n$-dimensional Hermitian space over $E$.
Let $\tilde{G} := GU(V)$ and $G := U(V)$. Suppose that $(\pi, V\_{\pi})$ is an irreducible cuspidal representation of $G.$
Is there an irreducible cuspidal representation $(\tilde{\pi}, V\_{\tilde{\pi}})$ of $\tilde{G}$ such that $V\_\pi \subset V\_{\tilde{\pi}}|\_{G}$? Note that here, the restriction is that of cusp forms, not of the representation itself.
| https://mathoverflow.net/users/3186 | extending cusp forms | I believe the answer should be yes, by some version of the following sketch of an argument:
(Note: by restriction of scalars, I regard all groups as being defined over $\mathbb Q$,
and I write ${\mathbb A}$ for the adeles of $\mathbb Q$.)
We are given $V\_{\pi} \subset Cusp(G(F)\backslash G({\mathbb A}\_F)).$
Let $\tilde{C}$ denote the maximal $\mathbb Q$-split torus in the centre of $\tilde{G}$
(this is just a copy of $\mathbb G\_m$), and write $C = \tilde{C}\cap G$. (I guess
this is just $\pm 1$?)
Now $C(\mathbb A)$ acts on $V\_{\pi}$ through some character $\chi$ of $(\mathbb A)/C(\mathbb Q)$. Choose an extension
$\tilde{\chi}$ of $\chi$ to a character of $\tilde{C}(\mathbb A)/\tilde{C}(\mathbb Q)$,
and regard $V\_{\pi}$ as a representation of $\tilde{C} G$ by have $\tilde{C}$ act
through $\tilde{\chi}$. Since $\tilde{C} G$ is normal and Zariksi open in $\tilde{G}$,
we should be able to further extend the $\tilde{C} G(\mathbb A)$-action on $V\_{\pi}$
to an action of $\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A).$
Now if we consider $Ind\_{\tilde{G}(\mathbb Q)\tilde{C} G(\mathbb A)}^{\tilde{G}(\mathbb A)} V\_{\pi},$ we should be able to
find a cupsidal representation $V\_{\tilde{\pi}}$ of the form you want (with $\tilde{C}(A)$ acting via $\tilde{\chi}$).
The intuition is that automorphic forms on $G(\mathbb A)$ are $Ind\_{G(\mathbb Q)}^{G(\mathbb A)} 1,$
and similarly for $\tilde{G}$. We will consider variants of this formula that takes into account central characters, and think about how to compare them for $G$ and $\tilde{G}$.
Inside the automorphic forms, we have those where $C(\mathbb A)$ acts by $\chi$; this we can
write as $Ind\_{G(\mathbb Q)C(\mathbb A)}^{G(\mathbb A)} \chi$, and then rewrite as
$Ind\_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)} \tilde{\chi}.$ This is where $V\_{\pi}$ lives, once we extend it to a repreresentation
of $\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)$ as above.
Now the automorphic forms on $\tilde{G}(\mathbb A)$ with central character $\tilde{\chi}$
are $Ind\_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb A)} \tilde{\chi},$
which we can rewrite as
$Ind\_{\tilde{G}(\mathbb Q) \tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)}
Ind\_{\tilde{G}(\mathbb Q)\tilde{C}(\mathbb A)}^{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)}
\tilde{\chi}.$ This thus contains $Ind\_{\tilde{G}(\mathbb Q)\tilde{C}G(\mathbb A)}^{\tilde{G}(\mathbb A)}V\_{\pi}$ inside it, and so an irreducible constituent of the latter
should be a $V\_{\tilde{\pi}}$ whose restriction (as a space of functions) to $G(\mathbb A)$ contains $V\_{\pi}$.
What I have just discussed is the analogue for $G$ and $\tilde{G}$ of the relation between automorphic forms on $SL\_2$ and $GL\_2$ discussed e.g. in Langlands--Labesse. Hopefully I haven't blundered!
| 3 | https://mathoverflow.net/users/2874 | 15581 | 10,436 |
https://mathoverflow.net/questions/15569 | 14 | How does one show that if $U \subseteq \mathbb{C}^n$ is nonempty and Zariski open then $U$ is also dense in the analytic topology on $\mathbb{C}^n$?
| https://mathoverflow.net/users/4048 | Zariski open sets are dense in analytic topology | It is enough to show that the complement of $U$ has empty interior. Also, that complement is contained in the zero set $Z$ of a non-constant polynomial $f$, so it is enough to show that $Z$ does not contain open sets.
If $z\in Z$ is a point in the interior of $Z$, then the Taylor series of $f$ at $z$ is of course zero. Since $f$ is an entire function, this is absurd.
| 23 | https://mathoverflow.net/users/1409 | 15582 | 10,437 |
https://mathoverflow.net/questions/15514 | 0 | I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them).
These branches meet at certain points (junctions). Each junction is adjacent to a subset of the sectors (at least 3 of them).
By specifying what sectors my junctions are adjacent to, I can completely recover the tree.
This seems like something known, but I would like a reference to it.
The number of trees with n branches is given by
<http://www.oeis.org/A001003>
and this is quite easy to prove.
Furthermore, if I order the sectors in the description of the junctions, I can make this representation unique.
Example:
(0,1,2,3,4,5) represents the tree with only one vertex, and 6 branches connected to this junction.
| https://mathoverflow.net/users/1056 | Describe a tree by junctions | I realized that taking the dual of my trees, I always get an n-gon,
where some chords, the faces in the dual are my junctions. The bijection is now trivial.
| 1 | https://mathoverflow.net/users/1056 | 15599 | 10,446 |
https://mathoverflow.net/questions/15584 | 22 | In the very first chapter Hartshorne proposes the following seemingly trivial exercise (ex. I.2.17(ii)):
>
> Show that a strict complete intersection is a set theoretic complete intersection.
>
>
>
Here are Hartshorne's definitions:
>
> A variety $Y$ of dimension $r$ in $\mathbb{P}^n$ is a (strict) complete intersection if $I(Y)$ can be generated by $n-r$ elements. $Y$ is a set-theoretic complete intersection if $Y$ can be written as the intersection of $n-r$ hypersurfaces.
>
>
>
Here $I(Y)$ is the homogeneous ideal of $Y$. The point is that the first definition seems wrong, since one would naturally require that $I(Y)$ can be generated by $n-r$ *homogeneous* elements (with this definition the exercise becomes trivial).
I have never made my mind if this is a misprint by Hartshorne. So the question is
>
> Is it true that in a polynomial ring any homogeneous ideal generated by $k$ elements is also generated by $k$ homogeneous elements?
>
>
>
If I recall well it is not difficult to find counterexamples in graded rings which are not polynomial rings, so the point of the exercise may be to show that polynomial rings have this special property.
| https://mathoverflow.net/users/828 | Minimal number of generators of a homogeneous ideal (exercise in Hartshorne) | Dear Andrea: Hartshorne was right, but we need to do some work. Let $\mu(I)$ be the minimal number of generators of $I$, and $\mu\_h(I)$ be the minimal number of a homogenous system of generators of $I$. Let $R=k[x\_1,\dots,x\_n]$ and $\mathfrak m=(x\_1,\dots,x\_n)$. Suppose $\mu\_h(I)=m$ and $f\_1,\dots, f\_m$ is a minimal homogenous set of generators. At this point we switch to the local ring $A=R\_{\mathfrak m}$ (the reason: it is easier to do linear algebra over local rings, as anything not in $\mathfrak m$ is now invertible). It will not affect anything since $I\subset\mathfrak m$.
Construct a surjective map $F\_0 = \bigoplus\_{i=1}^m A(-\deg \ f\_i) \to I \to 0$ and let $K$ be the kernel. We claim that $K \subset\mathfrak mF\_0$. If not, then one can find an element $(a\_1,...,a\_m) \in K$ such that $\sum a\_if\_i=0$ and $a\_1$, say, has a degree $0$ term $u\_1\neq 0$. By considering terms of same degree in the sum one sees that there are $b\_i$s such that:$$u\_1f\_1 = \sum\_{i=2}^m b\_if\_i$$
so the system is not minimal, as $u\_1 \in k$, contradiction.
Now tensoring the sequence $$ 0 \to K \to F\_0 \to I \to 0$$ with $k=A/\mathfrak m$. By the claim $K\subset\mathfrak mF\_0$, so $F\otimes k \cong I\otimes k$. It follows that $m= \operatorname{rank} F\_0 = \dim\_k(I\otimes k)$. But over a local ring, the last term is exactly $\mu(I)$, and we are done.
| 21 | https://mathoverflow.net/users/2083 | 15605 | 10,450 |
https://mathoverflow.net/questions/15591 | 13 | Let $X$ be an infinite set, and let $G$ be the symmetric group on $X$. I want to understand $G$ by putting a topology on it, without imposing any more structure on $X$. What 'interesting' possibilities are there and what is known about them?
In particular, I have heard of the pointwise convergence topology (an open neighbourhood of g is a set of permutations that agree on some specified finite set of points), and found some papers on this, but are there any other topologies that have been studied?
What if I take the coarsest topology compatible with the group operations such that either a) the stabiliser of any subset is closed, b) the stabiliser of any partition is closed, or c) both are closed? I think these will be coarser than the pointwise convergence topology, because in the pointwise convergence topology the stabiliser of any first-order structure is closed. Is there a useful characterisation of the open subgroups and/or closed subgroups?
| https://mathoverflow.net/users/4053 | Topologies on an infinite symmetric group | It is known that there are exactly two separable group topologies on $S\_\infty$ (i.e., on the group of permutations of a countable set): one is antidiscrete and the other one is topology of pointwise convergence. This is a statement of Theorem 6.26 in
[here](http://arxiv.org/abs/math/0409567). Hence Polish topology on $S\_\infty$ is unique.
There are some abstract characterizations of closed subgroups, for example they are exactly the Polish groups with a countable basis of identity consisting of open subgroups. Another characterization is: a Polish group G is homeomorphic to a closed subgroup of $S\_\infty$ if and only if it admits a compatible left invariant ultrametric. You may want to look at Section 1.5 of Becker and Kechris "The Descriptive Set Theory of Polish Group Action".
| 9 | https://mathoverflow.net/users/896 | 15616 | 10,456 |
https://mathoverflow.net/questions/15596 | 10 | All the statements of the [Archimedean property](http://en.wikipedia.org/wiki/Archimedean_property) with which I am acquainted fundamentally uses ℕ -- more than as a totally ordered semi-group, really being the 'standard model' of the naturals. It is a fundamental ingredient in showing that the reals are (up to isomorphism) the only complete totally ordered field.
In this particular result, there are two non ingredients which seem to be of a fundamentally different nature: Dedekind completeness (which is about subsets, while all the other axioms are about elements), and the Archimedean property, which pre-supposes the existence of the Naturals. But because being 'Archimedean' already makes sense in (ordered) monoids, that is the one I am most interested in.
Crazy scenario: replace the 'naturals' in the Archimedean property with a non-standard model of the 'naturals' (call this N-Archimedean). Now [the reasoning for the argument that](https://planetmath.org/ArchimedeanOrderedFieldsAreReal) all Archimedean totally ordered fields are sub-fields of ℝ readily lifts, but no longer proves quite the same thing.
In other words, it seems that this *unique position* of ℝ is in part due to ℕ already being baked in to the question. And thus my question: is there a proper generalization of the Archimedean property which is ``properly abstract''?
| https://mathoverflow.net/users/3993 | Is there a version of the Archimedean property which does not presuppose the Naturals? | It is not surprising that some versions of the Archimedean property concern subsets of the order rather than merely elements. The reason is that the Archimedean property is provably not expressible in a first order manner.
This is because the structure of the reals R, as an ordered field, say, (but one can add any structure at all), has elementary extensions to nonstandard models R\* which are non-Archimedean. This means that any statement in the language of ordered fields that is true in the reals R will also be true in the nonstandard reals. To prove that such models exist is an elementary application of the Compactness theorem, and one can also construct them directly via the ultrapower construction. One can also control the cofinality of the nonstandard order. For example, one can arrange that every countable subset of R\* is bounded. Since all these various nonstandard models R\* satisfy exactly the same first order truths as the standard reals R, but are non-Archimedean, it follows that being Archimedean is not first-order expressible.
Being Archimedean is, of course, second-order expressible, and the usual definition is a second order definition. As Neel mentions in the comments, the natural numbers are identifiable as the smallest subset of the ordered field containing 0 and closed under successor n+1.
If one adds the natural numbers N as a predicate to the original model, so that one is looking at R as an ordered field with a unary predicate holding of the natural numbers, then the nonstandard model R\* will include a nonstandard version N\* of the natural numbers. This new field R\*, which is not Archimedean, will nevertheless appear to be Archimedean relative to the nonstandard natural numbers N\*. For example, for any x and y in R\*, there will be a number n in N\* such that nx > y.
Indeed, one can do amazing things along this line. Suppose that V is the entire set-theoretic universe, and let V\* be a nonstandard version of it (such as an ultrapower by a nonprincipal ultrafilter on the natural numbers). Inside V\*, the structure R\* is thought to be the actual real numbers and so V\* thinks R\* is Archimedean, even though back in V we can see that it is mistaken, precisely because V\* is using the wrong set of natural numbers for its conclusion. The model V\* simply cannot see the true set of natural numbers sitting inside R\*, because it does not have that set.
More generally, one can similarly describe what it should mean for any ordered field F to be Archimedean relative to a subring R. Perhaps this simple idea is the generalization for which you are looking? It is mainly amounting to the question of whether the subring is cofinal in the original order.
Thus, it is very natural to look at the possible cofinalities of the orders that arise in ordered fields (or the other types of structures that you consider). For any infinite regular cardinal κ, one may find an elementary extension of the reals R to a nonstandard ordered field R\*, where the order of R\* has a cofinal κ sequence. To do this, just perform a series of κ many extensions, each with new elements on top of the previous model. In κ many steps, the union of the structures you built will have an order with cofinality exactly κ.
If one only uses the ultrapower construction to construct the nonstandard models, however, then there are limits on the resulting cofinality of the order. Understanding these limits is a large part of Shelah's deep work on PCF (= possible cofinality) theory.
| 18 | https://mathoverflow.net/users/1946 | 15620 | 10,460 |
https://mathoverflow.net/questions/15628 | 41 | Let $G$ be a (maybe Lie) group, and $M$ a space (perhaps a manifold). Then a principal $G$-bundle over $M$ is a bundle $P \to M$ on which $G$ acts (by fiber-preserving maps), so that each fiber is a $G$-torsor (a $G$-action isomorphic, although not canonically so, to the action of $G$ on itself by multiplication). A map of $G$-bundles is a bundle map that plays well with the actions.
Then I more-or-less know what the *classifying space* of $G$ is: it's some bundle $EG \to BG$ that's universal in the homotopy category of (principal) $G$-bundles. I.e. any $G$-bundle $P \to M$ has a (unique up to homotopy) map $P\to EG$ and $M \to BG$, and conversely any map $M\to BG$ (up to homotopy) determines a (unique up to isomorphism) bundle $P \to M$ and by pulling back the obvious square.
At least this is how I think it works. [Wikipedia's description of $BG$ is here.](http://en.wikipedia.org/wiki/Classifying_space)
So, let $G$ be a Lie group and $M$ a smooth manifold. On a $G$-bundle $P \to M$ I can think about *connections*. As always, a connection should determine for each smooth path in $M$ a $G$-torsor isomorphism between the fibers over the ends of the path. So in particular, a bundle-with-connection is a (smooth) functor from the path space of $M$ to the category of $G$-torsors. But not all of these are connections: the value of holonomy along a path is an invariant up to "thin homotopy", which is essentailly homotopy that does not push away from the image of the curve. So one could say that a bundle-with-connection is a smooth functor from the thin-homotopy-path-space.
More hands-on, a connection on $P \to G$ is a ${\rm Lie}(G)$-valued one-form on $P$ that is (1) invariant under the $G$ action, and (2) restricts on each fiber to the canonical ${\rm Lie}(G)$-valued one-form on $G$ that takes a tangent vector to its left-invariant field (thought of as an element of ${\rm Lie}(G)$).
Anyway, my question is: is there a "space" (of some sort) that classifies $G$-bundles over $M$ with connections? By which I mean, the data of such a bundle should be the same (up to ...) as a map $M \to $ this space. The category of $G$-torsors is almost right, but then the map comes not from $M$ but from its thin-homotopy path space.
Please re-tag as desired.
| https://mathoverflow.net/users/78 | What is the classifying space of "G-bundles with connections" | There is a stupid answer which is equivalence classes of G-bundles with connection on M are the same as homotopy classes of maps $M \to BG$. That is as long as two G-bundles with connection are considered equivalent if they have the same underlying principal bundle. This isn't meant to be a serious answer, just point out that your question is not exactly well posed.
But more seriously, there is a stack which represents G-principal bundles with connections. It even has a nice form:
$$ Bun\_G^\nabla = [ \Omega^1( - ; \mathfrak{g}) / G]$$
Maps from M to this stack **are** principal G-bundles with connection.
The problem with this stack is that it is not *presentable*. It is not covered by a manifold. It can be describe as a quotient stack, but thing you act on is the sheaf $\Omega^1(-; \mathfrak{g})$ of Lie algebra valued 1-forms. This is a sort of generalized manifold (in a loose sense), but this sheaf is not representable (great exercise!).
If it was a presentable stack, then we could take its classifying space (there are several ways to do this, e.g take the realization of the simplicial manifold obtained by iterated fiber products of the covering manifold). Homotopy classes of maps to this space could then be related to certain isomorphism classes of maps to the stack. But since $Bun^\nabla\_G$ is not presentable we are kinda stuck.
You could ask, well what happens if I replace $\Omega^1(-; \mathfrak{g})$ with an honest topological space that is the best approximation to it (for maps into it). Well it turns out the space which best approximates $\Omega^1(-; \mathfrak{g})$ is the point. So you get the classifying space of the stack $[pt/G]$ which is just the usual BG.
| 18 | https://mathoverflow.net/users/184 | 15633 | 10,469 |
https://mathoverflow.net/questions/15479 | 12 | Constructing the Kummer K3 of an Abelian surface $A$, we have an obvious 22-dimensional collection of classes in $H^2(K3, \mathbb{Z})$ given by the 16 (-2)-curves (which by construction do not intersect each other), and the pushforward-and-pullback of the six classes generating $H^2(A, \mathbb{Z})$. However, this is clearly not all of the classes that I need to find; first of all, the intersection form is wrong---it is certainly not unimodular.
Secondly, there are a few other classes that can be constructed geometrically which are missing---for example, since $\sum\_{i=1}^{16} E\_i$, the sum of the exceptional divisors, is the branch locus of a 2-1 cover of K3, it must be divisible by two, which my naive description misses.
Ultimately what I am hoping to do is to produce a generating function summing over all effective curve classes in the K3 (with coefficients determined by some GW-invariants), but as stated above, I am missing some classes whose description I do not know. I've been looking through Barth, Peters, Van de Ven, and the best statement I can find is Proposition VIII 3.7:
>
> The set of effective classes on a Kahler K3-surface is the semigroup generated by the nodal classes and the integral points in the closure of the positive cone.
>
>
>
That being said, is there a nice concrete description of these somewhere?
| https://mathoverflow.net/users/1703 | What classes am I missing in the Picard lattice of a Kummer K3 surface? | The lattice $L\_{K3}=H^2(K3,\mathbb Z)$ is $2E\_8+3U$, with $E\_8$ negative definite and $U$ the hyperbolic lattice for the bilinear form $xy$. It is unimodular and has signature $(3,19)$.
The 16 (-2)-curves $E\_i$ form a sublattice $16A\_1$ of determinant $2^{16}$. It is not primitive in $L\_{K3}$. The primitive lattice $K$ containing it is computed as follows. Consider a linear combination $F=\frac12\sum a\_i E\_i$ with $a\_i=0,1$. Recall that $E\_i$ are labeled by the 2-torsion points of the torus $A$, i.e. the elements of the group $A[2]$.
Then $F$ is in $K$ $\iff$ the function $a:A[2]\to \mathbb F\_2$,
$i\mapsto a\_i$, is affine-linear. You will find the proof of this statement in Barth-(Hulek-)Peters-van de Ven "Compact complex surfaces", VIII.5.
(The element $\frac12\sum E\_i$ in your example corresponds to the constant function 1, which is affine linear).
Thus, $K$ has index $2^5$ in $16A\_1$ and its determinant is $2^{16}/(2^5)^2=2^6$.
$K$ is called the *Kummer lattice*. By the above, it is a concrete negative-definite lattice of rank 16 with determinant $2^6$. Nikulin proved that a K3 surface is a Kummer surface iff $Pic(X)$ contains $K$.
The orthogonal complement $K^{\perp}$ of $K$ in $L\_{K3}$ is $H^2(A,\mathbb Z)$ but with the intersection form multiplied by 2. As a lattice, it is isomorphic to $3U(2)$. It has determinant $2^6$, the same as $K$. The lattice $L\_{K3}=H^2(K3,\mathbb Z)$ is recovered from the primitive orthogonal summands $K$ and $K^{\perp}$.
However, your question has "Picard lattice" in the title. The Picard group of $X$ is strictly smaller than $H^2(X,\mathbb Z)$. To begin with, it has signature $(1,r-1)$, not $(3,19)$. For a Kummer surface, it contains Kummer lattice $K$ described above, and its intersection with $K^{\perp}$ is the image of the Picard group of $A$. For a Kummer surface one has $r=17,18,19$ or 20.
For the Mori-Kleiman cone of effective curves, which you would need for Gromov-Witten theory, the description you put in a box is already the best possible.
| 13 | https://mathoverflow.net/users/1784 | 15637 | 10,473 |
https://mathoverflow.net/questions/15614 | 6 | Max-flow and linear programming are two big hammers in algorithm design: each are expressive enough to represent many poly-time solvable problems. Some problems are obvious applications of max-flow: like finding a maximum matching in a graph. What I'm looking for are examples of problems that can be solved via clever encodings as flow problems or LP problems -- ones that aren't obvious. I'm looking for questions at a level suitable for a homework problem for an advanced undergraduate or beginning graduate course in algorithms.
Any ideas?
| https://mathoverflow.net/users/3028 | Interesting applications of max-flow and linear programming | Determining whether a sports team has been mathematically eliminated from qualifying for the playoffs is a cute application of max-flow min-cut:
<http://www.cs.princeton.edu/courses/archive/spr03/cs226/assignments/baseball.html>
| 12 | https://mathoverflow.net/users/2233 | 15638 | 10,474 |
https://mathoverflow.net/questions/15641 | 7 | I need a reference (or a short proof) for the following statement:
Suppose a closed manifold $N$ is the result of a surgery (along an embedded sphere) on a closed manifold $M$. Then the difference $\sum dim H\_i(N) - \sum dim H\_i(M)$ (the homology is taken with coefficients in a field) is at most 2.
| https://mathoverflow.net/users/2823 | Surgery and homology: a reference request | To say that a smooth, closed manifold $N$ is obtained by surgery along a (framed) sphere in $M$ is to say that there is a cobordism $P$ from $M$ to $N$ and a Morse function $f\colon P\to [0,1]$, with $f^{-1}(0)=M$, $f^{-1}(1)=N$, and exactly one critical point $c$. By Morse theory, $H\_\*(P,M)$ is then 1-dimensional, generated by the descending disc of $c$. Likewise, $H\_\*(P,N)$ is 1-dimensional, generated by the ascending disc of $c$.
By the homology exact sequence of the pair $(P,M)$, $\dim H\_\*(M)$ differs from $\dim H\_\*(P)$ by $1$. By the homology exact sequence of the pair $(P,N)$, $\dim H\_\*(N)$ also differs from $\dim H\_\*(P)$ by $1$. Hence $|\dim H\_\*(M) - \dim H\_\*(N)|$ is $0$ or $2$.
| 12 | https://mathoverflow.net/users/2356 | 15647 | 10,480 |
https://mathoverflow.net/questions/15643 | 3 | Reading Wojtaszczyk's Banach spaces for analysts, I'm trying to understand his proof that the space of all continuous linear functionals on $(X^\star,\sigma(X^\star, X))$ is $X$.
To show the $ \subseteq$ part, he says let $\varphi$ be any linear functional on $X^\star$ continuous in $\sigma(X^\star, X)$. Then $\{x^\star \in X^\star : |\varphi(x^\star)| < 1\} \supset \{x^\star \in X^\star : |x\_j(x^\star)| < \epsilon, j=1,\ldots,n\}$ for some $\epsilon >0$ and some $x\_1, \ldots, x\_n \in X$. (Isn't this just saying since $\varphi^{-1}((-1,1))$ is open, it contains a neighborhood of 0?)
Then--- this is where I get lost--- he says that the result follows from the fact that if $\varphi\_0, \ldots, \varphi\_n$ are linear forms on a linear space $X$ (without any topology), then $\varphi\_0 \in \text{span}\{\varphi\_j\}\_{j=1}^n$ iff $\text{ker}\varphi\_0 \supset \cap\_{j=1}^n \text{ker} \varphi\_j$.
How is this fact relevant?
| https://mathoverflow.net/users/2586 | characterization of continuous functionals in weak-star topology | Yes to your first question. As for the second, regard the $x\_j$-s as linear functionals on $X^\*$. If you have $x\_j(x^\*)=0$ for all $j$, then every multiple of $x^\*$ is in the the first set you have in your second paragraph; i.e., $|\phi(tx^\*)| <1$ for all $t$ and hence $\phi(x^\*)=0$. Thus $\phi$ is a linear combination of the $x\_j$-s and hence is continuous.
Tell Joel hello for me.
| 7 | https://mathoverflow.net/users/2554 | 15649 | 10,482 |
https://mathoverflow.net/questions/15658 | 41 | I have heard the claim that the derived category of an abelian category is in general additive but not abelian. If this is true there should be some toy example of a (co)kernel that should be there but isn't, or something to that effect (for that matter, I could ask the same question just about the homotopy category).
Unless I'm mistaken, the derived category of a semisimple category is just a ℤ-graded version of the original category, which should still be abelian. So even though I have no reason to doubt that this is a really special case, it would still be nice to have an illustrative counterexample for, say, abelian groups.
| https://mathoverflow.net/users/361 | How do I know the derived category is NOT abelian? | The following nicely does the trick I think...
**Lemma** Every monomorphism in a triangulated category splits.
Proof: Let $T$ be a triangulated category and suppose that $f\colon x\to y$ is a monomorphism. Complete this to a triangle
$x \stackrel{f}{\to} y \stackrel{g}{\to} z \stackrel{h}{\to} \Sigma x$
then $f\circ \Sigma^{-1}h = 0$ as we can rotate backward and maps in triangles compose to zero. Since $f$ is a monomorphism we deduce that $\Sigma^{-1}h$ and hence $h$ are zero. But this implies that $y\cong x\oplus z$ (a proof of this can be found in the first part of my answer [here](https://mathoverflow.net/questions/4782/splitting-in-triangulated-categories/4785#4785) so that $f$ is a split monomorphism. █
Since every kernel is a monomorphism we get the following counterexample. The map
$\mathbb{Z}/p^2\mathbb{Z} \to \mathbb{Z}/p\mathbb{Z}$
does not have a kernel in $D(Ab)$ by virtue of the fact that $\mathbb{Z}/p^2\mathbb{Z}$ is indecomposable. Of course the same thing works in the homotopy category.
| 48 | https://mathoverflow.net/users/310 | 15662 | 10,492 |
https://mathoverflow.net/questions/15666 | 6 | If K is a finite extension of $\mathbb Q\_p$ for some prime number $p$, (possibly need $p \neq 2$), $L\_1$ and $L\_2$ totally ramified abelian extension of $K$, $ \pi\_1, \pi\_2$ are respectively the uniformizer that generates each field. Is it true that $ L\_1 L\_2$ is totally ramified iff $Nm\_{L\_1 / K}(\pi\_1)$, $Nm\_{L\_2 / K}(\pi\_2)$ differs by an element in the intersection of the two norm groups.
Is there a proof of this result (or the correct version of the result) without employing big tools?
Add at 6:49 pm 18th Feb: From Class Field Theory we know that there are one maximal totally ramified abelian extension of a local number field $K$ corresponding to each uniformizer, so I would expect that some version of the above statement is true. At least when $Nm\_{L\_1 / K}(\pi\_1)=Nm\_{L\_2 / K}(\pi\_2)=x$, $ L\_1 L\_2$ is totally ramified, because both are contained in $K^{ram}\_x$.
| https://mathoverflow.net/users/2701 | When is the composition of two totally ramified extension totally ramified? | Let me give an elementary answer in the case of abelian exponent-$p$ extensions of $K$, where $K$ is a finite extension of $\mathbb{Q}\_p$ containing a primitive $p$-th root $\zeta$ of $1$. This is the basic case, and Kummer theory suffices.
Such extensions correspond to sub-$\mathbb{F}\_p$-spaces in $\overline{K^\times} = K^\times/K^{\times p}$ (thought of a vector space over $\mathbb{F}\_p$; not to be confused with the multiplicative group of an algebraic closure of $K$).
It can be shown fairly easily that the unramified degree-$p$ extension of $K$ corresponds to the $\mathbb{F}\_p$-line $\bar U\_{pe\_1}$, where $e\_1$ is the ramification index of $K|\mathbb{Q}\_p(\zeta)$ and $\bar U\_{pe\_1}$ is the image in $\bar K^\times$ of the group of units congruent to $1$ modulo the maximal ideal to the exponent $pe\_1$. This is the "deepest line" in the filtration on $\bar K^\times$. See for example prop. 16 of [arXiv:0711.3878](http://arxiv.org/abs/0711.3878v2).
An abelian extension $L|K$ of exponent $p$ is totally ramified if and only if the subspace $D$ which gives rise to $L$ (in the sense that $L=K(\root p\of D)$) does not contain the line $\bar U\_{pe\_1}$.
Now, if $L\_1$ and $L\_2$ are given by the sub-$\mathbb{F}\_p$-spaces $D\_1$ and $D\_2$, then the compositum $L\_1L\_2$ is given by the subspace $D\_1D\_2$ (the subspace generated by the union of $D\_1$ and $D\_2$). Thus the compositum $L\_1L\_2$ is totally ramified if and only if $D\_1D\_2$ does not contain the deepest line $\bar U\_{pe\_1}$.
**Addendum.** A similar remark can be made when the base field $K$ is a finite extension of $\mathbb{F}\_p((\pi))$. Abelian extensions $L|K$ of exponent $p$ correspond to sub-$\mathbb{F}\_p$-spaces of $\overline{K^+}=K/\wp(K^+)$ (not to be confused with an algebraic closure of $K$), by Artin-Schreier theory. The unramified degree-$p$ extension corresponds to the image of $\mathfrak{o}$ in $\bar K$, which is an $\mathbb{F}\_p$-line $\bar{\mathfrak o}$ (say).
Thus, the compositum of two totally ramified abelian extensions $L\_i|K$ of exponent $p$ is totally ramified precisely when the subspace $D\_1+D\_2$ does not contain the line $\bar{\mathfrak o}$, where $D\_i$ is the subspace giving rise to $L\_i$ in the sense that $L\_i=K(\wp^{-1}(D\_i))$. See Parts 5 and 6 of [arXiv:0909.2541](http://arxiv.org/abs/0909.2541v2).
| 4 | https://mathoverflow.net/users/2821 | 15668 | 10,496 |
https://mathoverflow.net/questions/15611 | 23 | In his answer to a question about simple proofs of the
Nullstellensatz
([Elementary / Interesting proofs of the Nullstellensatz](https://mathoverflow.net/questions/15226/elementary-interesting-proofs-of-the-nullstellensatz)),
Qiaochu Yuan referred to a really simple proof for the case of an
uncountable algebraically closed field.
Googling, I found this construction also in Exercise 10 of a 2008 homework
assignment from a course of J. Bernstein (see the last page of
<http://www.math.tau.ac.il/~bernstei/courses/2008%20spring/D-Modules_and_applications/pr/pr2.pdf>).
Interestingly, this exercise ends with the
following (asterisked, hard) question:
(\*) Reduce the case of arbitrary field $k$ to the case of
an uncountable field.
After some tries to prove it myself, I gave up and returned to googling.
I found several references to the proof provided by Qiaochu Yuan, but no
answer to exercise (\*) above.
So, my question is: To prove the Nullstellensatz, how can the general case
of an arbitrary algebraically closed field be reduced to the
easily-proved case of an uncountable algebraically closed field?
The exercise is from a course of Bernstein called 'D-modules and their
applications.' One possibility is that the answer arises somehow when
learning D-modules, but unfortunately I know nothing of
D-modules. Hence, proofs avoiding D-modules would be particularly
helpful.
| https://mathoverflow.net/users/2734 | To prove the Nullstellensatz, how can the general case of an arbitrary algebraically closed field be reduced to the easily-proved case of an uncountable algebraically closed field? | These logic/ZFC/model theory arguments seem out of proportion to the task at hand. Let $k$ be a field and $A$ a finitely generated $k$-algebra over a field $k$. We want to prove that there is a $k$-algebra map from $A$ to a finite extension of $k$. Pick an algebraically closed extension field $k'/k$ (e.g., algebraic closure of a massive transcendental extension, or whatever), and we want to show that if the result is known in general over $k'$ then it holds over $k$. We just need some very basic commutative algebra, as follows.
Proof: We may replace $k$ with its algebraic closure $\overline{k}$ in $k'$ and $A$ with a quotient $\overline{A}$ of $A \otimes\_k \overline{k}$ by a maximal ideal (since if the latter equals $\overline{k}$ then $A$ maps to an algebraic extension of $k$, with the image in a finite extension of $k$ since $A$ is finitely generated over $k$). All that matters is that now $k$ is perfect and infinite.
By the hypothesis over $k'$, there is a $k'$-algebra homomorphism
$$A' := k' \otimes\_k A \rightarrow k',$$
or equivalently a $k$-algebra homomorphism $A \rightarrow k'$. By expressing $k'$ as a direct limit of finitely generated extension fields of $k$ such an algebra homomorphism lands in such a field (since $A$ is finitely generated over $k$). That is, there is a finitely generated extension field $k'/k$ such that the above kind of map exists. Now since $k$ is perfect, there is a separating transcendence basis $x\_1, \dots, x\_n$, so
$k' = K[t]/(f)$ for a rational function field $K/k$ (in several variables) and a monic (separable) $f \in K[t]$ with positive degree. Considering coefficients of $f$ in $K$ as rational functions over $k$, there is a localization
$$R = k[x\_1,\dots,x\_n][1/h]$$
so that $f \in R[t]$. By expressing $k'$ as the limit of such $R$ we get such an $R$ so that there is a $k$-algebra map
$$A \rightarrow R[t]/(f).$$
But $k$ is infinite, so there are many $c \in k^n$ such that $h(c) \ne 0$. Pass to the quotient by $x\_i \mapsto c\_i$.
QED
I think the main point is twofold: (i) the principle of proving a result over a field by reduction to the case of an extension field with more properties (e.g., algebraically closed), and (ii) spreading out (descending through direct limits) and specialization are very useful for carrying out (i).
| 23 | https://mathoverflow.net/users/3927 | 15672 | 10,499 |
https://mathoverflow.net/questions/15673 | 25 | In his very nice article
>
> Peter Roquette,
> History of valuation theory. I. (English summary) Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 291--355,
> Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002
>
>
>
Roquette states the following result, which he attributes to Kurschak:
Hensel-Kurschak Lemma: Let $(K,|\ |)$ be a complete, non-Archimedean normed field. Let $f(x) = x^n + a\_{n-1} x^{n-1} + \ldots + a\_1 x + a\_0 \in K[x]$ be a polynomial. Assume (i) $f(x)$ is irreducible and (ii) $|a\_0| \leq 1$. Then $|a\_i| \leq 1$ for all $0 < i < n$.
He says that this result is today called Hensel's Lemma and that Hensel's standard proof applies.
This is an interesting result: Roquette explains how it can be used to give a very simple proof of the fact that, with $K$ as above, if $L/K$ is an algebraic field extension, there exists a unique norm on $L$ extending $| \ |$ on $K$. This is in fact the argument I gave in a course on local fields that I am currently teaching.
It was my initial thought that the Hensel-Kurschak Lemma would follow easily from one of the more standard forms of Hensel's Lemma. Indeed, in class last week I claimed that it would follow from
Hensel's Lemma, version 1: Let $(K,| \ |)$ be a complete non-Archimedean normed field with valuation ring $R$, and let $f(x) \in R[x]$ be a polynomial. If there exists $\alpha \in R$ such that $|f(\alpha)| < 1$ and $|f'(\alpha)| = 0$, then there exists $\beta \in R$ with $f(\beta) = 0$ and $|\alpha - \beta| < 1$.
Then in yesterday's class I went back and tried to prove this...without success. (I was not at my sharpest that day, and I don't at all mean to claim that it is not possible to deduce Hensel-Kurschak from HLv1; only that I tried the obvious thing -- rescale $f$ to make it a primitive polynomial -- and that after 5-10 minutes, neither I nor any of the students saw how to proceed.) I am now wondering if maybe I should be trying to deduce it from a different version of Hensel's Lemma (e.g. one of the versions which speaks explicitly about factorizations modulo the maximal ideal).
This brings me to a second question. There are of course many results which go by the name Hensel's Lemma. Nowadays we have the notion of a **Henselian normed field**, i.e., a non-Archimedean normed field in which the exended norm in any finite dimensional extension is unique. (There are many other equivalent conditions; that's rather the point.) Therefore, whenever I state a result -- let us restrict attention to results about univariate polynomials, to fix ideas -- as "Hensel's Lemma", I feel honorbound to inquire as to whether this result holds in a non-Archimedean normed field if and only if the field is Henselian, i.e., that it is equivalent to all the standard Hensel's Lemmata.
Is it true that the conclusion of the Hensel-Kurschak Lemma holds in a non-Archimedean valued field iff the field is Henselian?
More generally, what is a good, reasonably comprehensive reference for the various Hensel's Lemmata and their equivalence in the above sense?
| https://mathoverflow.net/users/1149 | An unfamiliar (to me) form of Hensel's Lemma | What you say at the beginning of your post is right: Hensel-Kurschak's lemma may be deduced from some refined version of Hensel's lemma. Actually, it's what Neukirch does in Algebraic Number Theory (see chapter II, corollary 4.7). His proof relies on the following (see 4.6)
Hensel's lemma: Let $(K,|.|)$ be a complete valued field with valuation ring $R$, maximal ideal $\mathfrak{m}$. Let $f(x) \in R[x]$ be a primitive polynomial (ie $f\ne 0$ mod $\mathfrak{m}$). Suppose $f=\bar{g}\bar{h}$ mod $\mathfrak{m}$, with $\bar{g}$ and $\bar{h}$ relatively prime. Then you can lift $\bar{g}$ and $\bar{h}$ to polynomials $g$ and $h$ in $R[x]$ such that $\textrm{deg}(g)=\textrm{deg}(\bar{g})$ and $f=gh$.
Neukirch goes on with proving that the valuation on $K$ extends uniquely to any algebraic extension (see corollary 4.7), as you say Roquette does.
As regards your last question, you may want to have a look at chapter II, paragraph 6 (appropriately called Henselian Fields) in the book of Neukirch. His definition of Henselian field is that it should satisfy Hensel-Kurschak's lemma. In theorem 6.6, he shows that this property is equivalent to the unique extension of the valuation to algebraic extensions.
| 16 | https://mathoverflow.net/users/4069 | 15679 | 10,503 |
https://mathoverflow.net/questions/15664 | 52 | Here is the criteria for a "perfect" graph editor:
* it should be able to perform an automated, but controllable layout
* one is able to make "manual" enforcements to nodes and edges locations when you need it (or at least such fine automated layout so you don't need "manual" enforcements)
* one could add some math symbols and formulae on a graph
Common vector graphics editors could do the trick, but there is a lot of overhead efforts to draw every node, every edge, every label.
Graphviz is good enough, but sometimes you cannot get needed layout (even if you use several tricks like additional invisible nodes etc) and you should use ladot or dot2tex for math formulae
yEd has nice layouts, but there is a problem with a math text.
This is probably not a math question, but it is common to draw graphs in articles i think.
[Result graph](http://mathtech.ru/0pocket/grf_res.jpg)
(Update: 27.12.2010)
Here is another candidate for the best editor in TeX - Asymptote (asymptote.sf.net).
The very powerful tool at first glance.
(Update: 28.04.2014)
The very tasty semiautomated tool to use with PGF/TiKZ is TiKZEdt. You can extend its palette with your own tools and make the process of diagram creation very simple!
| https://mathoverflow.net/users/3315 | What is the best graph editor to use in your articles? | I already had about 30 pages of graphs typeset with xymatrix for my thesis before discovering tikz; but was so impressed by it that I was happy to rewrite them all. As well as (imho) looking better, it gave me cross-platform compatibility - xypic seems to need pstricks, so on the mac with TeXshop (which uses pdflatex, I assume) the old graphs couldn't even be rendered.
Its ability to construct graphs iteratively can also be a massive timesaver- for instance, I wanted a bunch of otherwise identical rectangles at various positions, so with tikz could just loop over a list of their first coordinate rather than having to tediously cut,paste and modify an appropriate number of copies of the command for a rectangle. Particularly handy when I then decided they all needed to be slightly wider!
There's a gallery of tikz examples [here](http://www.texample.net/tikz/examples/), to give you some idea of what it's capable of (and with the relevant source code- I did find the manual a bit hard to understand and learnt mostly by examples or trial and error).
The vector graphics package inkscape (which I used to use for drawing more complicated graphs for inclusion as eps images) also apparently has a plugin to export as tikz, although I haven't tried that out.
| 30 | https://mathoverflow.net/users/987 | 15690 | 10,508 |
https://mathoverflow.net/questions/14506 | 3 | Are the Pearson and Spearman rank correlation coefficients related in a specific way for uniform random variables? Specifically, is the relationship $\rho\_{spearman} = 2\*\sin(\frac{\pi}{6}\rho\_{pearson})$? If so, why?
| https://mathoverflow.net/users/3875 | Are the Pearson and Spearman rank correlation coefficients related in a specific way for uniform RVs? | This formula is from Pearson 1907, see e.g.
Rank Correlation and Product-Moment Correlation
Author(s): P. A. P. Moran
Source: Biometrika, Vol. 35, No. 1/2 (May, 1948), pp. 203-206
Johan
| 2 | https://mathoverflow.net/users/4070 | 15695 | 10,512 |
https://mathoverflow.net/questions/15696 | 4 | The usual way of getting a category of metric spaces is to take metric spaces as objects, and the *nonexpansive* maps (ie, functions $f : A \to B$ such that $d\_B(f(a), f(a')) \leq d\_A(a, a')$) as morphisms.
However, for my purposes I'd like to use the Banach fixed point theorem to get a category with a trace structure or Conway operators on it, which means I want to consider the contraction mappings on nonempty metric spaces -- that is, there should be $q < 1$ for each mapping $f$ such that $d\_B(f(a), f(a')) \leq q \cdot d\_A(a, a')$.
But nonempty metric spaces and contraction mappings don't form a category, since the identity function is not a contraction map! Is there some way of defining this kind of setup as a category? I'm happy to play games with the metrics (e.g., use ultrametrics, but bounds on them, that sort of thing), if it helps.
| https://mathoverflow.net/users/1610 | "Category" of Nonempty Metric Spaces and Contractive Maps? | Presumably you don't want to allow arbitrary non-expansive maps, otherwise you could simply take that.
One thing that you could do artificially is to take the subcategory of "metric spaces + nonexpansive maps" generated by the strict contractions. This is a bit like adding the unit in to a non-unital ring. That may seem a little forced, though.
This sort of thing is also encountered in two other situations: Hilbert-Schmidt operators on Hilbert spaces, and cobordisms between manifolds (Stolz and Teichner have, at one time, needed something like this, I vaguely recall). One solution, that I think comes from those areas, is to use the idea of a "length" of a morphism. In this case, the length of a morphism would be its contraction factor. Morphisms of length 0 have to be the identity (or an isometric isomorphism, if you don't want to be too [evil](http://ncatlab.org/nlab/show/evil)).
Perhaps you could clarify exactly which nonexpansive maps you wish to disallow?
| 3 | https://mathoverflow.net/users/45 | 15699 | 10,515 |
https://mathoverflow.net/questions/15703 | 29 | What did Newton himself do, so that the "Newton polygon" method is named after him?
| https://mathoverflow.net/users/454 | Newton and Newton polygon | The Newton polygon and [Newton's method](http://en.wikipedia.org/wiki/Newton%27s_method) are closely related. The following theorem was first proven by Puiseux:
>
> if $K$ is an algebraically closed field of characteristic zero, then the field of [Puiseux series](http://en.wikipedia.org/wiki/Puiseux_expansion) over $K$ is the algebraic closure of the field of formal Laurent series over $K$
>
>
>
However according to Wikipedia
>
> This property was implicit in Newton's use of the Newton polygon as early as 1671 and therefore known either as Puiseux's theorem or as the Newton–Puiseux theorem.
>
>
>
A place where this is illustrated in more detail is "A history of algorithms: from the pebble to the microchip" By Jean-Luc Chabert, Évelyne Barbin, page 191. I will quote the first paragraph
>
> Immediately following his description of his numerical method for solving equations, Newton used the same principle to show how to obtain algebraic solutions of equations. He explains how the method of successive linear approximations can be adapted by using a ruler and "small parallelograms", the first version of what is called Newton's polygon. The method was applied in a more general case later, by Puiseux in 1850, both in considering multiple branches and in considering functions of a complex variable
>
>
>
Then it explains Newton's approach in detail. You can follow the references given there. And then we know the story that this nice tool is now used for the understanding of polynomials over local fields even tough originally the local field was the field of formal Laurent series.
| 23 | https://mathoverflow.net/users/2384 | 15708 | 10,520 |
https://mathoverflow.net/questions/15685 | 14 | From [Model Theory](http://en.wikipedia.org/wiki/Model_theory) article from wikipedia : "A theory is satisfiable if it has a model $ M\models T$ i.e. a structure (of the appropriate signature) which satisfies all the sentences in the set $T$". In [structure definition](http://en.wikipedia.org/wiki/Structure_%28mathematical_logic%29) there is also requirement for "container of a structure" to be set.
As we assume then, **every model have to lay inside of some set-container**. This obviously give us in serious trouble, as for set theory, there is no model of this type, and even maybe cannot be. One of possible explanations why set theory cannot be closed inside set ( which will lead us to some well known paradoxes) is opinion that "[there can be no end to the process of set formation](http://en.wikipedia.org/wiki/Absolutely_infinite)" so we have "structure" which cannot be closed inside itself which is obviously rather well state.
As we know that not every theory may have a model (see set-theory) then some question arises:
* What are the coses (other than pure practical - if they are set we know how to work with them) of putting so strong
requirements for **model to be set?**
* **Is there any way to weaken this
requirement?**
* Are there any "explorations" of
possible extension of model theory
with fundamental objects other than
sets?
I presume that some categorical point of view may be useful here, but is there any? I am aware about questions asked before, specially here [8731](https://mathoverflow.net/questions/8731/categorical-foundations-without-set-theory), but it was asked in context of category theory which is of course valid point of view but somehow too fine. I would like to ask in general.
And last one, philosophical question: is then justified, that condition for a theory to have model in set universe is some kind of **anthropomorphic** point of view - just because we cannot build any other structures in effective way we build what is accessible for us way but it has no objective nor universal meaning? Is true that model theory is only a "universal algebra+logic" in universe of the set, or it justifications may be extended to some broader point of view? If yes: which one?
I have hope that this question is good enough for mathoverflow: at least please try to deal wit as **kindly request for references.**
---
Remark: Well formulated point from [n-CathegoryCafe discussion](http://golem.ph.utexas.edu/category/2008/07/category_theory_and_model_theo.html): "In the centre of Model Theory there is " fundamental existence theorem says that the syntactic analysis of a theory [the existence or non-existence of a contradiction] is equivalent to the semantic analysis of a theory [the existence or non-existence of a model]."
**In fact the most important point is: may it be extended on non "set container" universes?**
---
I would like to thank everyone who put here some comments or answers. In is the most interesting that in a light of answer of Joel David Hamkins it seems that for first order theories (FOT) consistency is equivalent to having set model. It is nontrivial, because it is no matter of somehow arbitrary definition of "having model" but it is related to **constructive** proof of Completeness Theorem of Gödel. From ontological point of view it then states that for FOT there is no weaker type of consistency than arising from model theory based on sets, and in some way it is maximal form of consistency simultaneously. Then there is no way to extend for FOT equivalence to non-set containers, which is nontrivial part - the only theories which are consistent in FOT are those which has a set models and this statement is proved not using set theoretical constructions in nonconstructive ways. So it was important to me, and I learn a lot from this even if for specialist it is somehow maybe obvious. I have hope that I understand it right;-)
@Tran Chieu Minh: thank You for pointing to interesting discussion, I will try to understand the meaning of Your remarks here and there.
| https://mathoverflow.net/users/3811 | Is it necessary that model of theory is a set? | You seem to believe that it is somehow contradictory to have a set model of ZFC inside another model of ZFC. But this belief is mistaken.
As Gerald Edgar correctly points out, the Completeness Theorem of first order logic asserts that if a theory is consistent (i.e. proves no contradiction), then it has a countable model. To be sure, the proof of the Completeness Theorem is fairly constructive, for the model is built directly out of the syntactic objects (Henkin constants) in an expanded language. To re-iterate, since you have mentioned several times that you find something problematic with it:
* **Completeness Theorem.** (Goedel 1929) If a theory is consistent, then it has a model that is a set.
The converse is much easier, so in fact we know that a theory is consistent if and only if it has a set model. This is the answer to your question.
More generally, if a theory is consistent, then the upward Lowenheim-Skolem theorem shows that it has models of every larger cardinality, all of which are sets. In particular, since the language of set theory is countable, it follows that if ZFC is consistent, then it has models of any given (set) cardinality.
The heart of your confusion appears to be the mistaken belief that somehow there cannot be a model M of ZFC inside another structure V satisfying ZFC. Perhaps you believe that if M is a model of ZFC, then it must be closed under all the set-building operations that exist in V. For example, consider a set X inside M. For M to satisfy the Power Set axiom, perhaps you might think that M must have the full power set P(X). But this is not so. All that is required is that M have a set P, which contains as members all the subsets of X that exist in M. Thus, M's version of the power set of X may be much smaller than the power set of X as computed in V. In other words, the concept of being the power set of X is not absolute between M and V.
Different models of set theory can disagree about the power set of a set they have in common, and about many other things, such as whether a given set is countable, whether the Continuum Hypothesis holds, and so on. Some of the most interesting arguments in set theory work by analyzing and taking advantage of such absoluteness and non-absoluteness phenomenon.
Perhaps your confusion about models-in-models arises from the belief that if there is a model of ZFC inside another model of ZFC, then this would somehow mean that we've proved that ZFC is consistent. But this also is not right. Perhaps some models of ZFC have models of ZFC inside them, and others think that there is no model of ZFC. If ZFC is consistent, then these latter type of models definitely exist.
* **Incompleteness Theorem.** (Goedel 1931) If a (representable) theory T is consistent (and sufficiently strong to interpret basic arithmetic), then T does not prove the assertion "T is consistent". Thus, there is a model of T which believes T is inconsistent.
In particular, if ZFC is consistent, then there will be models M of ZFC that believe that there is no model of ZFC. In the case that ZFC+Con(ZFC) is consistent, then some models of ZFC will have models of ZFC inside them, and some will believe that there are no such models.
A final subtle point, which I hesitate to mention because it can be confusing even to experts, is that it is a theorem that *every* model M of ZFC has an object m inside it which M believes to be a first order structure in the language of set theory, and if we look at m from outside M, and view it as a structure of its own, then m is a model of full ZFC. This was recently observed by Brice Halmi, but related observations are quite classical. The proof is that if M is an ω-model, then it will have the same ZFC as we do in the meta-theory and the same proofs, and so it will think ZFC is consistent (since we do), and so it will have a model. If M is not an ω-model, then we may look at the fragments of the (nonstandard) ZFC inside M that are true in some Vα of M. Every standard fragment is true in some such set in M by the Reflection Theorem. Thus, by overspill (since M cannot see the standard cut of its ω) there must be some Vα in M that satisfies a nonstandard fragment of its ZFC. Such a model m = VαM will therefore satisfy all of the standard ZFC. But M may not look upon it as a model of ZFC, since M has nonstandard axioms which it thinks may fail in m.
| 36 | https://mathoverflow.net/users/1946 | 15713 | 10,523 |
https://mathoverflow.net/questions/15707 | 8 | I'm trying to obtain a bound for the order of some finite groups, and part of it comes down to the order of a permutation group of degree $n$ that is nilpotent. I imagine these have to be much smaller than the full symmetric group, and a bound that is sub-exponential in $n$ would seem reasonable (given that permutation $p$-groups fall a long way short of having exponential order), but I haven't seen this written down anywhere.
I found one reference that looks promising:
P. Palfy, Estimations for the order of various permutation groups, Contributions to general algebra, 12 (Vienna, 1999), 37-49, Heyn, Klagenfurt, 2000.
However, I can't actually find the article anywhere online. Any suggestions?
| https://mathoverflow.net/users/4053 | Largest possible order of a nilpotent permutation group? | The paper of Vdovin mentioned by Steve shows that the nilpotent subgroups of the symmetric groups of maximal order are either the Sylow 2-subgroups P(n) of Sym(n), or P(n-3) x Alt(3) when n = 2(2k+1)+1.
Vdovin, E. P. "Large nilpotent subgroups of finite simple groups."
Algebra Log. 39 (2000), no. 5, 526-546, 630; translation in Algebra and Logic 39 (2000), no. 5, 301-312.
<http://www.ams.org/mathscinet-getitem?mr=1805754>
<http://dx.doi.org/10.1007/BF02681614>
It looks at the orbits of the center of the nilpotent group, and the action of the quotient on those orbits to give a reasonable bound. Then it shows that all nilpotent subgroups of maximal order are conjugate to the types mentioned.
The paper also handles alternating groups, groups of Lie type, and sporadic groups. For groups of Lie type, the maximal order nilpotent is usually a Sylow p-subgroup for p the characteristic. Sporadics are only handled briefly: the nilpotent subgroups of maximal order are always Sylows and they satisfy the same general bound as the other simple groups.
| 7 | https://mathoverflow.net/users/3710 | 15716 | 10,525 |
https://mathoverflow.net/questions/15592 | 23 | It seems that I have the needed example, but I want it to be simple and self-explaining...
>
> Construct a nontrivial complete metric space $X$ with intrinsic metric which has no nontrivial minimizing geodesics.
>
>
>
**Definitions:**
* A metric $d$ is called *intrinsic* if for any two points $x$, $y$ and any $\epsilon>0$ there is an $\epsilon$-midpoint $z$; i.e. $d(x,z),d(z,y)<\tfrac12 d(x,y)+\epsilon$.
* A minimizing geodesic is *nontrivial* if it connects two distinct points.
* A meric space is *nontrivial* if it contains two distinct points.
**Comments:**
* Clearly, $X$ can not be locally compact.
| https://mathoverflow.net/users/1441 | Intrinsic metric with no geodesics | Well, the unit ball in $c\_0$ is almost what you want (there is no unique shortest curve between points). All we need now is to enhance "bypasses" and to give disadvantage to "straight lines". This can easily be done by taking the distance element to be $(2+\sum\_n 2^{-n}x\_n)^{-1}\|dx\|\_\infty$, which is never less than the usual distance element in $c\_0$ and never greater than 3 times it in the unit ball. Now, if we have any continuous finite length curve $x(t)$ from $y$ to $z$ parametrized by the arclength, we can easily shorten it by replacing the $m$-th position by the maximum of the actual value of $x\_m(t)$ and $y\_m+t(z\_m-y\_m)/d+\frac 12 \min(t,d-t)$, where $d$ is the length of $x(t)$, which will work if $m$ is large enough since $\max\_t|x\_m(t)|\to 0$ as $m\to\infty$ and both functions change slower than the distance along the original curve.
This is certainly self-explaining (the shortest curve escapes from $c\_0$ to $\ell^\infty$) but I do not know if it is simple enough for your purposes.
| 13 | https://mathoverflow.net/users/1131 | 15720 | 10,528 |
https://mathoverflow.net/questions/15687 | 38 | In Toen's and Vezzosi's article *From HAG to DAG: derived moduli stacks* a kind of definition of DAG is given. I am not an expert and can't see what's the relation between DAG and the motivic cohomology ideas of Voevodsky (particularly his category $DG$). It would be nice if somebody could explain that to me.
I addition I am very keen on seeing how these ideas can be used in an explicit example. If I should explain someone why motivic cohomology is a good thing, I would certainly mention the proof of the Milnor conjecture. But can one see the use of derived ideas in a more explicit and down-to-earth example?
| https://mathoverflow.net/users/4011 | What is DAG and what has it to do with the ideas of Voevodsky? | There is a very general nice pattern here:
Let $C$ be a category of test spaces on which we want to model more general spaces. Then
* a "very general space" modeled on $C$ is an object in the [gros](http://ncatlab.org/nlab/show/petit+topos) [∞-topos](http://ncatlab.org/nlab/show/(infinity%2C1)-topos) $Sh\_{(\infty,1)}(C)$ of ∞-stacks on $C$.
Morel-Voevodsky take C to be the [Nisnevich site](http://ncatlab.org/nlab/show/Nisnevich+site). Then $Sh\_{(\infty,1)}(C)$ is the the ∞-topos whose [intrinsic cohomology](http://ncatlab.org/nlab/show/cohomology) is [motivic cohomology](http://ncatlab.org/nlab/show/motivic+cohomology).
Here C happens to be just an ordinary category. More generally, we could take C to be an ∞-category itself. In that case $Sh\_{(\infty,1)}(C)$ could be called the ∞-topos of [derived stack](http://ncatlab.org/nlab/show/derived+stack)s. What Toen Vezzosi do in HAG I and II is to provide a model-cateory theoretic presentation of this. That's why the articles look like they are hard to read: this is a component based way to describe an abstract elegant concept.
Now, the objects in $Sh\_{(\infty,1)}(C)$ are "very general" spaces modeled on $C$. There is a chain of ∞-subcategories of more "tame" spaces inside, though:
* first there are those ∞-stacks on C which are represented by an ∞-stack with a C-valued structure sheaf. This are the [structured ∞-topos](http://ncatlab.org/nlab/show/structured+(infinity%2C1)-topos), that generalize the notion of ringed spaces.
* and then among these are those that are *locally* equivalent to objects in C. These are the [generalized schemes](http://ncatlab.org/nlab/show/generalized+scheme#definition_in_the_variant_of_lurie_5) or "derived scheme" if C is suitably ∞-categorical.
The pattern here is reviewed at [notions of space](http://ncatlab.org/nlab/show/space#NotionsOfSpace).
In principle one could consider such "derived schemes" also in the Morel-Voevodsky gros ∞-topos of ∞-stacks on the Nisnevich site, it just seems that so far nobody looked into this. But there are all kinds of examples of test space categories C that people still have to push through this general nonsense. For instance we can take C to be simply the category of smooth manifolds. Then the above generalities spit out the notion of a [derived smooth manifold](http://ncatlab.org/nlab/show/derived+smooth+manifold).
The punchline being: all these things unify in one grand picture. It is not Toen/Vezzosi/Lurie derived geometry on one hand and Morel/Voevodsky cohomology on the other. Instead all this is parts of one big picture.
And also, if I may say this here, this big picture really is hardly restricted to algebraic geometry. Lurie's [notion of space](http://ncatlab.org/nlab/show/space#NotionsOfSpace) is much, much more general. It describes GEOMETRY. Of whatever sort.
| 25 | https://mathoverflow.net/users/381 | 15721 | 10,529 |
https://mathoverflow.net/questions/15717 | 3 | This is my first question. It appeared while solving a research problem in cryptography. I am computer science student, so I apologize for lack of mathematical rigor in this question. Thanks for any help.
Consider the Riemann zeta function at s = +1. It diverges, but the expression for the function is $\zeta(1) = \lim\_{n \rightarrow \infty} \sum\_{i = 1}^{n} \frac{1}{i}$ , the truncated sum of which are the $n$-th harmonic number, $\mathcal{H}(n)$.
The question is, how about the expression $\zeta(1) = \lim\_{n \rightarrow \infty} \prod\_{\textrm{primes } p\_i \leq n} \frac{1}{1-p\_i^{-1}}$. is the value of the truncated product $\mathcal{H}(n)$ too?
My simulations for large values of $n$ tells me that it is some function of $\log n$ (for example comparing the ratio of the function for $n$ and $n^2$ and $n^3$ etc) How do we prove this?
In summary, what is the value of $\prod\_{\textrm{primes } p\_i \leq n} \frac{1}{1-p\_i^{-1}}$?
Thanks
| https://mathoverflow.net/users/4074 | Truncated product of $\zeta(1)$? | Formula (8) on [this page](http://mathworld.wolfram.com/MertensConstant.html) gives the result
$$\prod\_{p \le n} \frac1{1-p^{-1}} = e^\gamma \log n \,(1 + o(1)).$$
| 11 | https://mathoverflow.net/users/126667 | 15725 | 10,533 |
https://mathoverflow.net/questions/15701 | 15 | Somehow [this question](https://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples) made me think of instances of small exceptions in general, and I remembered the statement I heard once that $S\_5,A\_6,S\_6,A\_7,A\_8,S\_8$ are the only instances of symmetric/alternating groups that are not quotients of $Z/2Z\*Z/3Z=PSL\_2(Z)$ (see [this MathSciNet entry](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=TI&pg7=ALLF&pg8=ET&review_format=html&s4=Wiegold&s5=generators&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=7&mx-pid=280580), which I just found). Does anyone have an idea of a conceptual explanation for this fact?
**Edit:** I also find [this article](http://www.ams.org/bull/1898-04-04/S0002-9904-1898-00473-1/supplementary-information/home.html) which mentions the same result. It's quite interesting that the positive part (for all $n>8$ these groups are quotients) is proved using Bertrand's postulate. I think it's cool that Bertrand's postulate can be used for group theory.
| https://mathoverflow.net/users/1306 | Symmetric groups which are not quotients of Z/2Z*Z/3Z | Yes, this is well known. In fact, there is a name for such groups - this is a (2,3)-generation property. And yes, by now there is a conceptual understanding why all sufficiently large finite simple groups have this property - the basic ideas are outlined in [this](http://www.ams.org/mathscinet/search/publdoc.html?r=1&pg1=CNO&s1=1405944&loc=fromrevtext) helpful MathSciNet review gently explaining the major breakthrough by Liebeck and Shalev (1996). There are more recent developments in the field, both in the asymptotic direction and in the explicit construction, such as figuring out which $PSL(n,q)$ are (2,3)-generated - see papers by Tamburini, etc. - the literature is too big to be reviewed here.
| 10 | https://mathoverflow.net/users/4040 | 15730 | 10,537 |
https://mathoverflow.net/questions/15727 | 7 | When A is an abelian group with trivial G-action (G being a discrete group) we get that Hn(G,A)≅Hn(BG,A). Is there a similar connection between group cohomology and topological cohomology if A is a non-trivial ℤG-module? What *can* we say in that case?
| https://mathoverflow.net/users/3238 | Group cohomology vs. topological cohomology in the case of non-trivial action | This is an example of twisted cohomology. In general, for a generalized cohomology theory (spectrum) E and a space X you can talk about E-twists over X. This is a certain structure on X. Given a particular twist $\tau$, you can then form the $\tau$-twisted cohomology $E^\tau(X)$. This was the subject of a [recent ArXiv paper](http://arxiv.org/abs/1002.3004) by Ando-Blumberg-Gepner. One way to think of this is that E-cohomology of X is naturally graded by the E-twists. (This viewpoint is especially when E is an $E\_\infty$-ring spectrum, in which case twists can be added and there is a cup product-type formula).
In the case of ordinary cohomology, a twist reduces to a local system on X. A *local system* is the same thing as a functor from the fundamental groupoid of X to the category of abelian groups. This can equivalently be thought of as a locally constant sheaf of abelian groups on X, connecting up with Sammy Black's answer. For more general types of cohomology, these don't correspond to sheaves in the usual sense. If you allow "sheaves of spectra" you can get this work, but that is a difficult and long story.
In the case that $X = BG$ (with G discrete), then the fundamental groupoid of X is equivalent to the category G, i.e. the category with a single object with automorphisms the group G. Then a local system, i.e. a functor $G \to Ab$ is exactly the same as a G-module. Then the twisted cohomology of $BG$ with this twist is exactly the group cohomology of G with values in the module. There is also a similar homology story.
| 10 | https://mathoverflow.net/users/184 | 15735 | 10,540 |
https://mathoverflow.net/questions/15731 | 29 | I am curious to collect examples of equivalent axiomatizations of mathematical structures. The two examples that I have in mind are
1. **Topological Spaces.** These can be defined in terms of open sets, closed sets, neighbourhoods, the Kuratowski closure axioms, etc.
2. **Matroids**. These can be defined via independent sets, bases, circuits, rank functions, etc.
Are there are other good examples?
Secondly, what are some advantages of multiple axiomatizations?
Obviously, one advantage is that one can work with the most convenient definition depending on the task at hand. Another is that they allow different generalizations of the object in question. For example, infinite matroids can be axiomatized by adapting the independent set axioms, but it is unknown how to axiomatize them via the circuit axioms. An acceptable answer to the second question would be an example of a proof in one axiom system that doesn't translate easily (not sure how to make this precise) into another axiom system.
| https://mathoverflow.net/users/2233 | Cryptomorphisms | The phenomenon that I think you have in mind has a name: **cryptomorphism**. I learned the name from the writings of Gian-Carlo Rota; Rota's favorite example was indeed matroids. Gerald Edgar informs me that the name is due to Garrett Birkhoff.
I think modern mathematics is replete with cryptomorphisms. In my class today, I presented the "Omnibus Hensel's Lemma". Part a) was: the following five conditions on a valued field are all equivalent. Part b) was: complete fields satisfy these equivalent properties. There are lots more equivalent conditions than the five I listed: see
[An unfamiliar (to me) form of Hensel's Lemma](https://mathoverflow.net/questions/15673/an-unfamiliar-to-me-form-of-hensels-lemma)
and especially Franz Lemmermeyer's answer for further characterizations.
I would say that the existence of cryptomorphisms is a sign of the richness and naturality of a mathematical concept -- it means that it has an existence which is independent of any particular way of thinking about it -- but that on the other hand the existence of not obviously equivalent cryptomorphisms tends to make things more complicated, not easier: you have to learn several different languages at once. For instance, the origin of the question I cited above was the fact that in Tuesday's class I st*p*dly chose the wrong form of Hensel's Lemma to use to try to deduce yet another version of Hensel's Lemma: it didn't work! Since we are finite, temporal beings, we often settle for learning only some of the languages, and this can make it harder for us to understand each other and also steer us away from problems that are more naturally phrased and attacked via the languages in which we are not fluent. Some further examples:
I think that the first (i.e., most elementary) serious instance of cryptomorphism is the determinant. Even the Laplace expansion definition of the determinant gives you something like $n$ double factorial different ways to compute it; the fact that these different computations are not obviously equivalent is certainly a source of consternation for linear algebra students. To say nothing of the various different ways we want students to think about determinants. It is "just" the signed change of volume of a linear transformation in Euclidean space (and the determinant over a general commutative ring can be reduced to this case). And it is "just" the induced scaling factor on the top exterior power. And it is "just" the unique scalar $\alpha(A)$ which makes the adjugate equation $A\*\operatorname{adj}(A) = \alpha(A) I\_n$ hold. And so forth. You have to be fairly mathematically sophisticated to understand all these things.
Other examples:
Nets versus filters for convergence of topological spaces. Most standard texts choose one and briefly allude to the other. As G. Laison has pointed out, this is a disservice to students: if you want to do functional analysis (or read works by American mathematicians), you had better know about nets. If you want to do topological algebra and/or logic (or read works by European mathematicians), you had better know about filters.
There are (at least) three axiomatizations of the concept of **uniform space**: (i) entourages, (ii) uniform covers, (iii) families of pseudometrics. One could develop the full theory using just one, but at various points all three have their advantages. Is there anyone who doesn't wish that there were just one definition that would work equally well in all cases?
| 24 | https://mathoverflow.net/users/1149 | 15754 | 10,552 |
https://mathoverflow.net/questions/15604 | 5 | I am looking for a (already-studied or interesting) class of Matroids such that
- Class of Gammoids are contained in it
One example would be Strongly-base-orderable Matroids. I would also be grateful if someone knows a class of Matroids such that
* Class of Gammoids are contained in it AND
* It is contained in the class of "Strongly-base-orderable" Matroids.
By the way, strongly-base-orderable is a property such that :
GIven any two bases I,J of the Matroid, there exists a bijection \pi between I-J and J-I such that given any subset H of I-J, I- H +\pi(H) and J - \pi(H) + H is a base in the Matroid. (In Oxley's "Matroid theory" pp410 )
Motivation :
Well, I have something I can show for Gammoids but cannot for Strongly-base-orderable Matroids, although computational evidence suggests that it holds for general Matroids.
| https://mathoverflow.net/users/4057 | What interesting class of Matroids are there that contains the class of Gammoids? | The way I understand your question is as follows: you have a property which you believe holds for all matroids (these are rare); you can prove it for gammoids, but your proof does not extend to a certain bigger class of "strongly-base-orderable" (SBO) matroids. I take it you have an example of a SBO matroid where your proof breaks down because some intermediate lemma explicitly fails (even if the main property still holds). Without knowing if this is the right picture and what are all these properties and examples, it's hard to give you a good answer. However, if that is indeed your logic you might as well consider classes of matroids which contain gammoids but are not completely included in the class of SBO matroids. Say, Piff and Welsh proved in 1970 that every gammoid is representable over large enough field. So, how about the class of real-representable matroids? Not all of them are SBO, but so what - perhaps your "bad example" is not there...
| 4 | https://mathoverflow.net/users/4040 | 15756 | 10,553 |
https://mathoverflow.net/questions/15755 | -2 | Let $U$ is a complete lattice with least element 0.
*Weak partitioning* is a collection $S$ of nonempty subsets of $U$ such that $\forall x\in S: x\cap\bigcup(S\setminus\{x\})=0$.
*Strong partitioning* is a collection $S$ of nonempty subsets of $U$ such that $\forall A,B\in PS:(A\cap B=\emptyset \Rightarrow \bigcup A\cap\bigcup B=0)$.
Easy to show that every strong partitioning is weak partitioning.
Is weak and strong partitioning the same?
If not, under which additional conditions these are the same?
| https://mathoverflow.net/users/4086 | Weak partitioning vs. strong partitioning | If the lattice $U$ satisfies the meet distributive law
$$x \wedge \bigvee\_{i \in I} y\_i = \bigvee\_{i \in I} x \wedge y\_i$$
where $(y\_i)\_{i \in I}$ is an arbitrary collection of elements of $U$, then "weak partitioning" implies "strong partitioning." More precisely, you only need the above to hold when the right hand side is $0$.
An example of a complete lattice where weak and strong partitioning are inequivalent is the lattice $U$ consisting of all closed subsets of $\{1,\frac12,\frac13,\ldots,0\}$ (as a subspace of $\mathbb{R}$) and the collection $S = \{\{\frac1n\}: n \geq 1\}$. The weak-partitioning property is easily verified since the points $\frac1n$ are isolated. The strong partitioning property fails for the two sets $A = \{\{\frac1{2n}\}: n \geq 1\}$ and $B = \{\{\frac1{2n+1}\} : n \geq 0\}$, for example, since $\bigvee A = \overline{\bigcup A}$ and $\bigvee B = \overline{\bigcup B}$ both contain the point $0$.
PS: In your formulation of weak and strong partitioning, I interpret $S$ as a collection of *nonzero elements* of $U$, since "nonempty subsets" doesn't make much sense in context.
| 6 | https://mathoverflow.net/users/2000 | 15765 | 10,559 |
https://mathoverflow.net/questions/15751 | 2 | Let A is a complete lattice.
I call a subset $S$ of A *filter-closed* when for every filter base $T$ in $S$ we have $\bigcap T\in S$. (A *filter base* is a nonempty, down directed set.)
I call a subset $S$ of A *chain-closed* when for every non-empty chain $T$ in $S$ we have $\bigcap T\in S$.
**Conjecture** $S$ is filter-closed if and only if $S$ is chain-closed.
| https://mathoverflow.net/users/4086 | Filter-closed vs. chain-closed | Indeed, your conjecture is correct.
**Theorem.** If L is a complete lattice and S is a subset of L, then S is chain-closed iff S is filter-closed.
Proof. Clearly filter-closed implies chain-closed, since every chain is a filter base. Conversely, suppose that S is chain-closed, and that A is a filter base contained in S. Note that S is trivially filter-closed with respect to any finite filter base. So suppose by induction that S is filter-closed with respect to any filter base of size smaller than |A|. Enumerate A = { aα | α < |A| }. Let bβ be the meet of { aα | α < β }. This is the same as the meet of the filter sub-base of A generated by this set. This filter sub-base has size less than |A|, and hence by induction every bβ is in S. Also, the bβ are a descending chain in S, since as we take more aα, the meet gets smaller. Thus, by the chain-closure of S, the meet b of all the bβ is in S. This meet b is the same as the meet of A, and so we have proved that S is filter-closed. QED
This argument is very similar to the following characterization of (downward) complete lattices (which I had posted as my original answer).
**Theorem.** The following are equivalent, for any
lattice L.
* L is complete, in the sense that every subset of L has a
greatest lower bound.
* L is filter complete, meaning that every filter base in
L has a greatest lower bound.
* L is chain complete, meaning that every filter base in L
has a greatest lower bound.
Proof. It is clear that completeness implies filter
completeness, since every filter base is a subset of L, and
filter completeness implies chain completeness, since every
chain is a filter base. For the remaining implication, suppose that L is chain
complete. We want to show that every subset A of L has a
greatest lower bound in L. We can prove this by transfinite
induction on the size of A. Clearly this is true for any
finite set, since L is a lattice. Fix any infinite set A.
Enumerate A as { aα | α < |A| }.
By the induction hypothesis, for each β < |A|, the
set { aα | α < β } has a
greatest lower bound bβ. Note that {
bβ | β < |A| } is a chain, because
as we include more elements into the sets, the greatest
lower bound becomes smaller. Thus, there is an element b in
L that is the greatest lower bound of the
bβ's. It is easy to see that this element b
is also a lower bound of A. QED
One can describe the method as finding a linearly ordered
cofinal sequence through the filter generated by the filter
base. This proof used AC when A was enumerated, and I
believe that this cannot be omitted.
One can modify the argument to show that for every infinite
cardinal κ, then a lattice is κ-complete (every
subset of size less than κ has a glb) iff every
filter base of size less than κ has a glb iff every
chain of size less than κ has a glb.
Note that if the lattice is *bounded* (meaning that it has
a least and greatest element), then having greatest lower
bounds for every set is the same as having least upper
bounds for every set, since the least upper bound of a set
A is the greatest lower bound of the set of upper bounds of
A. Thus, a *complete lattice* is often defined as saying
that every subset has a glb and lub.
There have been a few questions here at MO concerning
complete lattices. See [this one](https://mathoverflow.net/questions/11728) and [this one](https://mathoverflow.net/questions/11435).
Questions about the degree of completeness of a partial
order often arise in connection with forcing arguments, and
when one is speaking of partial completeness and partial
orders (rather than lattices), and the situation is
somewhat more subtle. For example, a partial order P is
said to be κ-closed if every linearly ordered subset
of P of size less than κ has a lower bound. It is
κ-directed closed if every filter base in P of size
less than κ has a lower bound. With these concepts,
it is no longer true that a partial order is
κ-directed closed if and only if it is
κ-closed. One example arising in forcing would be the
forcing to add a slim κ-Kurepa tree, which is
κ-closed but not κ-directed closed. The
difference between these two concepts is related to
questions of large cardinal indestructibility, for Richard
Laver proved that every supercompact cardinal κ can
become indestructible by all κ-directed closed
forcing, but no such cardinal can ever be indestructible by
all κ-closed forcing, precisely because the slim
κ-Kurepa tree forcing destroys the measurability of
κ.
| 6 | https://mathoverflow.net/users/1946 | 15769 | 10,562 |
https://mathoverflow.net/questions/13812 | 1 | $$f(a,x)=\sum\_{\tau=-\infty}^{\infty}\frac{\exp\left(2\pi i\tau x\right)}{(\tau+a)^{p+1}}$$
Can I apply Euler-Maclauren formula to this sum?
where $a\in(0,0.5)$, p is a natural number, and $x$ is a real number
| https://mathoverflow.net/users/3589 | Sum and interpolation of hurwitz zeta functions | Well, if $p$ is an integer, you should realize that $\frac{1}{(\tau+a)^{p+1}}$ can be obtained by integrating $Q(x)e^{-2\pi iax}$ against $e^{-2\pi i\tau x}$ where $Q(x)$ is the (unique) polynomial of degree $p$ satisfying $Q^{(m)}(0)=e^{-2\pi ia}Q^{(m)}(1)$ for $m<p$ and $Q^{(p)}=\frac{(2\pi i)^{p+1}}{e^{-2\pi ia}-1}$.
So, your function is just $Q(x)e^{-2\pi iax}$ on $(0,1)$ extended by periodicity to the entire line. The polynomial $Q$ can be easily found for each particular $p$, so if you need some small range of $p$, you have an exact closed form formula. If you want to consider large $p$, then it is not so useful but the origianal series gives you a high precision approximation if you keep just the first few terms. Either way, you have an "expression one can work with", don't you?
The thing that totally puzzles me is why you think that your series has any relation to the Hurwitz zeta function.
| 1 | https://mathoverflow.net/users/1131 | 15782 | 10,571 |
https://mathoverflow.net/questions/15781 | 14 | Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$? I am going to be on the road soon, so pleas don't be offended if I don't respond quickly to a comment.
| https://mathoverflow.net/users/4092 | Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$? | The genus class field of an extension $K/F$ is defined to be the largest extension $L/K$ with the following properties:
1. $L/K$ is unramified
2. $L$ is the compositum of $K/F$ and an abelian extension $A/F$.
Thus the quick answer to your question is: the Hilbert class field of $K$ is abelian over ${\mathbb Q}$ if and only if the Hilbert class field of $K$ coincides with its genus class field.
The not-so-quick answer would tell you more about the construction of the genus class field. For abelian extensions of the rationals, the construction is easy: everything you'd like to know should be contained in Frölich's book
* *Central extensions, Galois groups, and ideal class groups of number fields*
AMS 1983
Basically you will have to look for the largest abelian extension of ${\mathbb Q}$ with the same conductor as $K/{\mathbb Q}$.
| 18 | https://mathoverflow.net/users/3503 | 15794 | 10,581 |
https://mathoverflow.net/questions/15804 | 9 | Let $\mathbb{P}$ be a probability measure on some probability space $(\Omega,\mathcal{A})$. Are there conditions on the $\sigma$-algebra $\mathcal{A}$ such that for every real number $c\in [0,1]$ we find a set $A\in\mathcal{A}$ with $\mathbb{P}(A)=c$.
It is like the intermediate value theorem for continuous functions.
| https://mathoverflow.net/users/4097 | When does a probability measure take all values in the unit interval? | A measure space $(\mathbb{P},\Omega,\mathcal{A})$ is atomless if for all $A\in\mathcal{A}$ with $\mathbb{P}(A)>0$ there exists $B\subset A, B\in\mathcal{A}$ such that $0<\mathbb{P}(B)<\mathbb{P}(A)$. Now according to a theorem of Sierpinski, the values of an atomless measure space form an interval. In particular, for probability spaces, every value in $[0,1]$ is taken. The original source of the article can be found [here](http://matwbn.icm.edu.pl/ksiazki/fm/fm3/fm3125.pdf) (in french). For a proof in english, you can look at on 215D on page 46 in Fremlin's book [Measure Theory 2](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.116.7816&rep=rep1&type=pdf).
| 17 | https://mathoverflow.net/users/35357 | 15808 | 10,588 |
https://mathoverflow.net/questions/15795 | 17 | This question was inspired by and is somewhat related to [this question](https://mathoverflow.net/questions/14944/have-people-successfully-worked-with-the-full-ring-of-diferential-operators-in-ch/).
In his article "Crystals and the de Rham cohomology of schemes" in the collection "Dix exposes sur la cohomologie des schemas" Grothendieck defines the (small) infinitesimal site of an $S$-scheme $X$ using thickenings of usual opens. He then proceeds to prove that in characteristic $0$ the cohomology with coefficients in $\mathcal{O}\_{X}$ computes the algebraic de Rham cohomology of the underlying scheme. This is remarkable, because the definition of the site does not use differential forms and it is not necessary for $X/S$ to be smooth.
This fails in positive characteristics, and as a remedy, Grothendieck suggests adding the additional data of divided power structures to the site, which he then calls the "crystalline site of $X/S$". This site then has good cohomological behaviour (e.g. if $X$ is liftable to char. $0$, then cohomology computed with the crystalline topos is what it "should be"). The theory of the crystalline topos was of course worked out very successfully by Pierre Berthelot.
> My question is: Even though the infinitesimal site is in some sense not nicely behaved in positive characteristics, have people continued to study it in this context? What kind of results have been obtained, and has it still been useful? I'm particularly interested in results about $D$-modules in positive characteristic (i.e. crystals in the infinitesimal site if $X/S$ is smooth), but I am also curious to see in which other directions progress has been made.
| https://mathoverflow.net/users/259 | The Infinitesimal topos in positive characteristic | There is a paper of Ogus for 1975, "The cohomology of the infinitesimal site",
in which he shows that if $X$ is proper over an algebraically closed field $k$ of char. $p$, and embeds into a smooth
scheme over $k$, then the infinitesimal cohomology of $X$ coincides with etale cohomology with coefficients in $k$ (or more generally $W\_n(k)$ if we work with the infinitesimal site of $X$ over $W\_n(k)$).
Note that, since $k$ has char. $p$, we are talking about etale cohomology with
mod $p^n$ coefficients, so this is "smaller" than the usual etale cohomology;
it just picks up the "unit roots" of Frobenius. So the infinitesimal cohomology
gives the unit root part of the crystalline cohomology.
| 24 | https://mathoverflow.net/users/2874 | 15811 | 10,590 |
https://mathoverflow.net/questions/15812 | 1 | Following the [blow](https://mathoverflow.net/questions/15685/is-it-necessary-that-model-of-theory-is-a-set). I will try to ask question in order to check if I well understand what was pointed. I decide to ask another question, because mathoverflow is not projected to be good environment for discussion, so it would be better to ask another question than to modify previous one.
From wikipedia, we have definition of **[magma](http://en.wikipedia.org/wiki/Magma_%2528algebra%2529)**:
>
> "In abstract algebra, a magma (or
> groupoid; not to be confused with
> groupoids in category theory) is a
> basic kind of algebraic structure.
> Specifically, a magma consists of a
> set M equipped with a single binary
> operation M × M → M. A binary
> operation is closed by definition, but
> no other axioms are imposed on the
> operation."
>
>
>
Axioms of magma structure are [first order theory](http://en.wikipedia.org/wiki/First-order_logic). So lets drop words "set" and "closed" from the definition, and try playing with theory T for which we have:
1. predefined single object E
2. objects which may be whatever You
like to name by letters from
finite alphabet ( not necessary elements of defined set)
3. binary operation defined that for
every pair of objects it sends it into predefined element E.
It has interpretation on the ground of ZFC and then it has finite or infinite models which are sets - in fact it may be considered as trivial magma if we assume that objects are from set $M U \{E\}$. Whilst considered without set background it should also work, and do not be cursed by any obvious paradoxes ( because it has set-models). Then we have an example of theory which, when interpreted in domain of sets, is first order theory and is consistent / has models.
Suppose we **formally** drop requirement that domain of discourse for this theory is set interpretation. So then several questions arises:
1. **Is theory T first order theory?** When interpreted on set domain, it is. But if not in set domain?
2. Is it necessary to interpret it in
set universum, by means of any other arguments than arising from question 1\*?
3. Can we say that T is consistent in domain of set theory ( whilst we do not know in other domains)?
4. **If we drop requirements that universum of objects is in set, will
this theory becomes inconsistent?**
5. Is for any theory obligatory to point
its domain of discourse in a formal
meaning, that is for pure
syntactical definition? Is interpretation necessary?
**By consistent I presume definition that ""Consistent" means that a contradiction cannot be deduced from it."**
Suppose that answer on 4th question is NO, we may say that theory is still consistent even if we drop requirement that it is interpreted on domain of sets. Then we will end with theory which has set-model. From completeness theorem when interpreted on domain of set is consistent, and even if we drop certain domain interpretation we will still have consistent theory of first order with model from sets universum, but also with models outside of it. Suppose that we get our objects from some big category. I believe that it does not changes nothing, does it? Then there are theories for which we have models which are not sets ( whilst to be consistent it still have to have models which are sets!).
Is there any obvious mistake in this hand-waving of mine?
From Wikipedia article about [First Order Logic](http://en.wikipedia.org/wiki/First-order_logic) I know that "*he definition above requires that the domain of discourse of any interpretation must be a nonempty set.*" but it is a remark pointing to "[empty domains](http://en.wikipedia.org/wiki/Empty_domain)" and "[free logic](http://en.wikipedia.org/wiki/Free_logic)" which obviously are not in case here. The only interesting link is this: [Interpretation\_(model theory)](http://en.wikipedia.org/wiki/Interpretation_%2528model_theory%2529) but it still is related to set or empty domain interpretations. But from category theory it seems to be possible to have theories which are interpreted in larger domains that sets in consistent way.
I would like to mention other questions in this or similar area here on mathoverflow, but I would like to say that I am not interested in category theory by other means than just an example of theory which is has other domain of interpretation than set.:
[Is there a relationship between model theory and category theory?](https://mathoverflow.net/questions/11974/is-there-a-relationship-between-model-theory-and-category-theory)
[Categorical foundations without set theory](https://mathoverflow.net/questions/8731/categorical-foundations-without-set-theory)
| https://mathoverflow.net/users/3811 | Theory interpreted in non-set domain of discourse may be consistent? | It appears that you yearn to study various first-order theories, but do not want to be constrained by any requirement that your models, or domains of discourse, be sets. There are several ways to take such a proposal.
On the one hand, many mathematicians have yearnings similar to yours, and this has led them to try to use category theory as a theoretical background for their mathematical investigations. Surely this is part of the attraction of category theory, and some promote category theory as a kind of alternative foundation of mathematics (that is, alternative to set theory) for precisely this kind of reason. But you say that you are not especially interested in adopting that view.
Another way to study mathematical structures that are not sets, while keeping a principally set-theoretic background, is to focus on the set-class distinction in set theory. If V is the universe of all sets, we can define certain classes in V, such as { x | φ(x) }, where φ is any property. Such a class is not always a set. For example, the Russell paradox is based on the observations that if the collection R = { x | x ∉ x } were a set, then R ∈ R iff R ∉ R, a contradiction. So R is not a set. But R is still a collection of sorts, and we call it a class. A *proper class* is a class that is not a set. In ZFC, one can treat classes and proper classes by manipulating their definitions. That is, the classes do not exist as objects within the set-theoretic universe, but rather as definable subcollections of the universe. The intuition is that proper classes are simply too big to be sets. Other proper classes would include the class V itself (consisting of all sets), the class of all ordinals, the class of all cardinals, the class of all groups, all rings, all monoids, etc. Each of these classes is too large to be a set, but each has a perfectly clear definition defining a family of objects.
There are other formalizations of set theory, such as Goedel-Bernays set theory GBC and Kelly-Morse set theory, that allow one to treat classes as objects. In these theories, there are two kinds of objects, sets and classes, and every set is a class, but there are classes that are not sets (such as those I listed above). It turns out that GBC is a conservative extension of ZFC, which means that the assertions purely about sets that are provable in GBC are exactly the same as the assertions about sets that are provable in ZFC. Kelly-Morse, in contrast, is not conservative over ZFC, and it implies, in particular, that their must be set models of ZFC.
Now, the point is that you could study magmas that are proper classes. These would not be sets, but would still exist and could be formally analyzed as mathematical structures in these various set theoretic backgrounds. For example, one magma is simply the set union operation: (a,b) maps to (a U b), defined on all pairs of sets a, b. This magma is not a set, simply because it is much too large. There are innumerably many other such examples.
| 6 | https://mathoverflow.net/users/1946 | 15813 | 10,591 |
https://mathoverflow.net/questions/15612 | 33 | Suppose a compact convex body $P \subset \Bbb R^3$ has only polygonal orthogonal projections onto a plane. Does this imply that $P$ is a convex polytope?
This question occurred to me when I was making exercises for [my book](http://www.math.ucla.edu/~pak/book.htm). I figured this is probably easy and well known, but the literature hasn't been any help. One remark: if the number of sides of all polygons is bounded by $n$, the problem might be easier. Furthermore, if $P$ is *assumed* to be a convex polytope, [this](http://www.cs.princeton.edu/~chazelle/pubs/ComplexityCuttingComplexes.pdf) elegant paper by Chazelle-Edelsbrunner-Guibas (1989) gives a (perhaps, unexpectedly large) sharp $\exp O(n \log n)$ upper bound on the number of vertices of $P$ (ht Csaba Toth who [generalized](http://math.ucalgary.ca/~cdtoth/cx-stabbing-rev.pdf) this to higher dimensions).
| https://mathoverflow.net/users/4040 | Do plane projections determine a convex polytope? | Theorem 4.1 of [this paper by Klee](http://www.ams.org/mathscinet-getitem?mr=105651) says yes. Moreover, the result generalizes to higher dimensions for projections of arbitrary dimension $\ge 2$.
| 21 | https://mathoverflow.net/users/1044 | 15815 | 10,593 |
https://mathoverflow.net/questions/15802 | 9 | A short while ago, Dvir proved the Kakeya conjecture over finite fields. Does this have any implications for restriction theorems over finite fields? I am aware only of implications going in the opposite direction (Mockenhaupt-Tao). Is there anything new known about restriction theorems over finite fields?
| https://mathoverflow.net/users/398 | Restriction theorems over finite fields | Not directly. The key connection between restriction and Kakeya in Euclidean settings is that thanks to Taylor expansion, a surface in Euclidean space looks locally flat, and so the Fourier transform of measures on that surface are a superposition of Fourier transforms of very flat measures, which by the uncertainty principle tend to propagate along tubes. The arrangement of these tubes is then governed by the Kakeya conjecture.
In finite fields, there is no notion of Taylor expansion, unless one forces it into existence by working on the tangent bundle of a surface, rather than a surface itself. So there is a connection between restriction and Kakeya on the former object, but not the latter. (This is discussed at the end of my paper with Mockenhaupt.)
| 13 | https://mathoverflow.net/users/766 | 15816 | 10,594 |
https://mathoverflow.net/questions/15805 | 4 | Given any CAT(0) space $X$, we can define a map $s:X\times X\times [0;1]\rightarrow X$, such that $s(x,y,-)$ is the constant speed geodesic from $x$ to $y$ . Any isometry $f$ of $X$ is compatible with that map in the sense, that $s(f(x),f(y),t)=f(s(x,y,t))$. Then one can ask, whether any self-homeomorphism of $X$, which is compatible with $s$ in the upper sense is already a isometry.
This is clearly wrong for $X=\mathbb{R}^n$, as all affine maps are compatible with $s$. So the question is, whether these are the only examples.
For example I think I can show, that the $n$-dimensional hyperbolic space ($n\ge 2$) is rigid in that sense.
EDIT: Due to the big amout of counterexamples one could better ask the following question:
Are the spaces $\mathbb{R}^n$ the only spaces, which have self homeomorphisms compatible with $s$ (in the upper sense), that are not self-similarities ?
| https://mathoverflow.net/users/3969 | Are isometries the only geodesic preserving maps in a CAT(0)-space? | The map which you call "geodesic preserving" is usually called "affine".
It seems that affine maps to the real line are well understood even for general length space.
For your later edit: you may always take two spaces which admit self-similar maps and consider map on the product which move each coordinate with different coefficients.
A. Lytchak says that the following is a well known open question:
>
> If such map exist then the space can be embedded into product of spaces and the map preserves product structure.
>
>
>
P.S. The "example" I gave before was not an example.
| 5 | https://mathoverflow.net/users/1441 | 15822 | 10,597 |
https://mathoverflow.net/questions/15800 | 20 | Which $6j$-symbols for quantised enveloping algebras are known explicitly?
The quantum $6j$-symbols for $sl(2)$ are well-known. The references are
[Masbaum and Vogel](http://projecteuclid.org/euclid.pjm/1102622100) and
[Frenkel and Khovanov](http://projecteuclid.org/euclid.dmj/1077242324).
What is known for other simple Lie algebras?
In case this seems somewhat vague here is a precise question. The data for a $6j$-symbol starts with a tetrahedral graph with edges labelled by highest weights. Then usually there is additional information needed at the vertices which I want to avoid. Take the example of $sl(n)$ and use partitions instead of highest weights. Label two opposite edges by a partition of the form $1^k$ (corresponding to an exterior power of the vector representation) and label the other four edges by partitions. Then associated to this data is a scalar.
Then I would expect this function to be determined by linear recurrence relations (i.e. D-finite or holonomic). Is this correct? and if so can you give recurrence relations?
Ideally we would also regard $n$ as an indeterminate.
**Background** In general $6j$-symbols arise for any semisimple abelian category which is also monoidal. They are the components of the natural transformation $(\otimes)(\otimes \times 1)\cong (\otimes)(1\times \otimes)$.
In more down to earth terms. If you know the Grothendieck ring of a semisimple abelian monoidal category and you attempt to construct this then the information you are missing is the $6j$-symbols. You can construct the abelian category and you can construct the tensor product functor but you don't have the associator.
For example it is well-known that the character table of a finite group does not determine the group. It does determine the category of representations as a semisimple abelian category. The $6j$-symbols are needed to make this a monoidal category.
For further discussion see:
[Character table does not determine group Vs Tannaka duality](https://mathoverflow.net/questions/11306/character-table-does-not-determine-group-vs-tannaka-duality/11346#11346)
| https://mathoverflow.net/users/3992 | Calculating 6j-symbols (aka Racah-Wigner coefficients) for quantum groups | We can also use partitions ($k$) (symmetric powers) instead of ($1^k$) on one or two of the edges. This still gives just scalars, but includes the full story for sl(2).
This problem seems to be equivalent to the problem of computing the exchange operator
in the tensor product of two (quantum) symmetric or exterior powers of the vector representation of the quantum sl(n), e.g. $S^kV\otimes S^mV$, in the standard basis of the symmetric (resp., exterior) powers (about exchange operators see my paper with Varchenko arXiv:math/9801135 and my ICM talk arXiv:math/0207008). This can be computed if you know the fusion operator for these representations, which can be computed efficiently using the ABRR (Arnaudon-Buffenoir-Ragoucy-Roche) equation, see e.g. the appendix to arXiv:math/9801135. I am not sure if the answer is completely worked out anywhere, but there are at least some answers. For instance, see the paper arXiv:q-alg/9704005 where something is done even in the elliptic setting (which relates to elliptic 6j symbols of Frenkel-Turaev). What they do is compute the matrix elements for $m=1$, but the general $m$ can be obtained using the fusion procedure. This should be a really nice computation with a nice answer of the type you are expecting. In particular, in a special case you'll get coefficients of Macdonald's difference operators attached to symmetric powers. In the exterior powers case (or a product of a symmetric and an exterior power) the answer will be simpler, since $k$ cannot get larger than $n$ (in this case you should perhaps get a pure product, and it should be completely covered in the above paper).
EDIT: Remark. Let $V,W$ be representations of the quantum group with 1-dimensional zero weight spaces. Then a natural basis in ${\rm Hom}(L\_\lambda, (V\otimes L\_\mu)\otimes W)$ is the compositions of intertwiners in one order, while a natural basis of ${\rm Hom}(L\_\lambda, V\otimes (L\_\mu\otimes W))$ is the composition of intertwiners in the other order. Thus, the 6j-symbol matrix (which is by definition the transition matrix between these two bases) is the exchange operator for intertwiners.
| 14 | https://mathoverflow.net/users/3696 | 15825 | 10,600 |
https://mathoverflow.net/questions/15830 | 11 | "Let Top be the category of topological spaces." If I see a definition like this, in which homeomorphic (isomorphic in the category) spaces are not identified together, then for each given topological space, how many times does it show up as an object in Top? Once? Countably many times? Uncountably many times? Is there a semantic crisis if we don't identify all homeomorphic topological spaces to the same object?
| https://mathoverflow.net/users/1874 | Confusion over a point in basic category theory | Well, such categories are actually proper classes. For instance, spaces that are homeomorphic to $\mathbb{R}$ will be given on ANY set of the same cardinality. So how many there are, depends on how many such sets there are...but there are quite a lot of sets. In fact, more than any individual set worth. (Though on the details there, I'm not sure about what happens for fixed cardinality)
We CAN identify them, and we get a [skeleton](http://ncatlab.org/nlab/show/skeleton). However, we usually don't do this, because there's no natural way to pick one object in each class, and one of the important catch phrases for category theory is "never make a choice." (Naturally, this is a vast oversimplification, but it's a good rule of thumb)
| 17 | https://mathoverflow.net/users/622 | 15831 | 10,605 |
https://mathoverflow.net/questions/15821 | 4 | I have a set *S* of real numbers, and I would like to create a new set *R* with exactly n real numbers (not necessarily from the set) that represent it best.
What I mean by best?
Well, I have query that asks for given point what is the closest point from S to that point, and when I ask same thing for set R I would like to have best fit to the real answers.
>
> To be specific, how do I minimize the [Hausdoff distance](http://en.wikipedia.org/wiki/Hausdorff_distance) between *S* and *R*?
>
>
>
I hope I've been clear enough in what i want.
I've heard of mathoverflow, so I said to myself why wouldn't I ask for help there.
Thank you in advance.
(**Edited** in light of the comments below.)
| https://mathoverflow.net/users/4102 | Approximating a set with fixed number of elements | This is the $k$-center problem (or in your notation, the $n$-center problem). you're given a set $S$ of points, and you want to find a set $R$ of $n$ points such that the set of balls of radius $r$ around each point in $R$ cover all of $S$, and $r$ is minimized.
Your metric space is the line, so this problem is relatively easy to solve. Here's a two-step approach: First, "guess" the optimal solution r (ie. pick some value of r). Now go from left to right, assigning centers greedily, which is to say, as far away from the previously placed center as possible, while covering all points. If you use up $n$ points before covering all of $S$, your guess was wrong, and you need to restart with a larger value of r. Else, you're done.
Now of course $r$ is a real number, but there are only discretely many "guesses", since the optimal r must be such that there are two points at distance exactly $r$ from a center (otherwise r is not optimal). so the total set of choices of r is merely the set constructed from measuring the pairwise distances and halving them.
All of this assumes you're in algorithms-land, which means that you have reasonable ways of representing points and comparing them.
p.s this algorithm is well known (not original).
| 5 | https://mathoverflow.net/users/972 | 15838 | 10,609 |
https://mathoverflow.net/questions/15767 | 6 | Let V be a complex affine variety given as the vanishing set of a set of polynomials with integral coefficients. I have 3 questions.
1)
Under what assumption will the dimension of V over C remain the same as the dimension of V computed over some finite field like Z\_P, assuming that the prime P does not divide any of the coefficients of the polynomials defining V?
2)
How can such a prime P be found if we have a set of polynomials whose vanishing set is V?
3)
Suppose that V has a singularity over Z\_P. Can we conclude that V is singular variety? In other words. Does having a singularity over a finite field immediately imply the variety is not smooth? If not, what guarantees that having a singularity modulo p implies a singularity over C?
| https://mathoverflow.net/users/4089 | Concerning the dimension of a complex variety modulo a prime | To make more precise the answer of Felip. You have a scheme $X=Spec(A)$ over $\mathbb Z$, where $A$ is a finitely generate $\mathbb Z$-algebra such that its generic fiber $X\_{\mathbb Q}$ (just consider your polynomials as polynomials with rational coefficients) gives $V$ by field extension $\mathbb{C}/\mathbb{Q}$. Of course, $X\_{\mathbb Q}$ has the same dimension as $V$, and $X\_{\mathbb Q}$ is smooth if and only $V$ is smooth.
**Questions 1-2**. You want to compare $\dim X\_p$ with $\dim X\_{\mathbb Q}$. In general $\dim X\_p\ge \dim X\_{\mathbb Q}$. The equality holds under some flatness condition at $p$. Namely, there is rather general result: if $f: Y\to Z$ is a flat morphism of finite type between noetherian schemes and suppose that $Z$ is integral and universally catenary (e.g.any scheme of finite type over a noetherian regular scheme, so any open subset of $Spec(\mathbb Z)$ is OK), then all [**EDIT**: non-empty] fibers $Y\_z$ of $f$ have the same dimension [**EDIT**: if the generic fiber of $f$ is equidimensional (i.e. all irreducibles components have the same dimension)].
Problem: your $X$ is not necessarily flat over $\mathbb Z$. But the flatness over $\mathbb Z$ (or any Dedekind domain) is easy to detect: it is equivalent to be torsion free. So consider the ideal $I$ equal to
$$ { a\in A \mid ka=0 \text{ for some non zero } k \in \mathbb Z \}$$
(don't know how to type "{" and "}"). Then $A/I$ is flat over $\mathbb Z$ and defines a closed scheme of $X$ which coincides with $X$ over an open subset of $\mathbb Z$. Actually, as $I$ is finitely generated, there exists a positive integer $N$ such that $NI=0$. Then $I=0$ over $Spec(\mathbb Z[1/N])$. So for any prime number $p$ prime to $N$, $X\_p$ will have dimension $\dim V$. Now you have to compute such a $N$... For hypersurface, $N$ is just the gcd of the coefficients (see comments in Felip's answer). I don't know whether efficient methodes exist in general. Of course if a prime $p$ does not divid any polynomial in the definning ideal of $X$, then this $p$ is OK. But you have to test this property for all polynomials and not just a set of generators (Example: the ideal generated by $T\_1+pT\_2, T\_1$, then $p$ is bad).
**Question 3**. Suppose $V$ is smooth (and connected for simplicity), and you are looking for the $p$ such that $X\_p$ is also smooth. You first proceed as in Question 1 to find an open subset $Spec(\mathbb Z[1/N]$ over which $X$ is flat. Let $d=\dim X\_{\mathbb Q}=\dim X\_p$ for all $p$ prime to $N$. Write
$$ X=Spec\big(\mathbb Z[T\_1, \ldots T\_n]/(F\_1,\ldots, F\_m)\big)$$
Then $X\_p$ is smooth is and only if the Jacobian matrix of the $F\_i$'s mod $p$ has rank $n-d$ at all points of $X\_p$. Equivalently, the ideal $J\subseteq \mathbb Z[T\_1,\ldots, T\_n]$ generated by $F\_1,...,F\_m$ and the rank $n-d$ minors of the Jacobian matrix is the unit ideal mod $p$. Therefore, the computation consists in determining the ideal $J$. As $X\_{\mathbb Q}$ is supposed to be smooth, $J$ is generated by a positive integer $M$. Now for all $p$ prime to $MN$, $X\_p$ is smooth of dimension $\dim V$.
[**EDIT**]: The assertion on the dimensions of the fibers needs some hypothesis on the generic fiber of $Y\to Z$. We must assume it is equidimensional. Otherwise $Y\_z$ has dimension equal to that of the generic fiber for $z\in Z$ belonging to the image of all irreducible components of $X$. For the initial question, it is better to assum the original variety $V$ is irreducible.
| 6 | https://mathoverflow.net/users/3485 | 15843 | 10,613 |
https://mathoverflow.net/questions/15443 | 3 | Why if one have an $\varepsilon$-expansive homeomorphism $T:X \rightarrow X$ ($X$ a compact metric space) and a given partition $D$ of $X$ which has diameter smaller than $\varepsilon$ the sequence of refined partitions $D\_n = \bigvee\_{i = -n}^n T^{-i} D$ has diameter converging to zero ?
Recall that a $\varepsilon$-expansive homeomorphism $T$ is such that given any two distinct points $x$ and $y$ there exist $n \in \mathbf{Z}$ such that $d(T^nx, T^ny) > \varepsilon$
I can see intituively why this is true, somehow the refined partitions have less an less points in its members precisely because they have diameter less than epsilon but $T$ keeps separating points (and i fact open sets of points) at distance greater than $\varepsilon$.
Thanks in advance!
| https://mathoverflow.net/users/4031 | Partitions and Expansiveness | Fix $\delta>0$ and let
$$
A\_n=\{(x,y): d(x,y)\ge\delta, d(T^ix,T^iy)\le\varepsilon,|i|\le n\}
$$
Sets $A\_n$ are closed, nested and, by expansiveness, $\cap A\_n=\varnothing$. Hence $A\_N=\varnothing$ for some $N=N(\delta)$. Clearly, $diam D\_N\le\delta$.
| 3 | https://mathoverflow.net/users/2029 | 15854 | 10,618 |
https://mathoverflow.net/questions/15841 | 76 | This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this case that τ is *coarser*
than σ, that σ is *finer* than τ or that
σ *refines* τ. (See [wikipedia on comparison of
topologies](http://en.wikipedia.org/wiki/Comparison_of_topologies).)
The least element in this order is the indiscrete topology and the largest topology is the discrete topology.
One can show that the collection of all topologies on a fixed set is a complete lattice. In the downward direction, for example, the intersection of any collection of
topologies on X remains a topology on X, and this intersection
is the largest topology contained in them all. Similarly,
the union of any number of topologies generates a smallest
topology containing all of them (by closing under finite
intersections and arbitrary unions). Thus, the collection of all topologies on X is a complete
lattice.
Note that the compact topologies are closed downward in
this lattice, since if a topology τ has fewer open sets than
σ and σ is compact, then τ is compact.
Similarly, the Hausdorff topologies are closed upward,
since if τ is Hausdorff and contained in σ, then
σ is Hausdorff. Thus, the compact topologies inhabit
the bottom of the lattice and the Hausdorff topologies the
top.
These two collections kiss each other in the compact
Hausdorff topologies. Furthermore, these kissing points,
the compact Hausdorff topologies, form an antichain in the
lattice: no two of them are comparable. To see this,
suppose that τ subset σ are both compact
Hausdorff. If U is open with respect to σ, then the
complement C = X - U is closed with respect to σ and
hence compact with respect to σ in the subspace
topology. Thus C is also compact with respect to τ in
the subspace topology. Since τ is Hausdorff, this
implies (an elementary exercise) that C is closed with respect to τ, and so U is
in τ. So τ = σ. Thus, no two distinct compact Hausdorff topologies are comparable, and so these topologies are spread out sideways, forming an antichain of the lattice.
My first question is, do the compact Hausdorff topologies
form a maximal antichain? Equivalently, is every topology
comparable with a compact Hausdorff topology? [Edit: François points out an easy counterexample in the comments below.]
A weaker version of the question asks merely whether every
compact topology is refined by a compact Hausdorff
topology, and similarly, whether every Hausdorff topology
refines a compact Hausdorff topology. Under what
circumstances is a compact topology refined by a unique
compact Hausdorff topology? Under what circumstances does a
Hausdorff topology refine a unique compact Hausdorff
topology?
What other topological features besides compactness and
Hausdorffness have illuminating interaction with this
lattice?
Finally, what kind of lattice properties does the lattice
of topologies exhibit? For example, the lattice has atoms,
since we can form the almost-indiscrete topology having
just one nontrivial open set (and any nontrivial subset
will do). It follows that every topology is the least upper
bound of the atoms below it. The [lattice of topologies is
complemented](http://www.smc.math.ca/cjm/v20/cjm1968v20.0805-0807.pdf).
But the lattice is not distributive (when X has at least
two points), since it embeds N5 by the
topologies involving {x}, {y} and the topology generated by
{{x},{x,y}}.
| https://mathoverflow.net/users/1946 | How do the compact Hausdorff topologies sit in the lattice of all topologies on a set? | **This is a community wiki of the answers in the comments.**
* The compact Hausdorff topologies do not generally form a maximal antichain. If X is infinite, split X into two infinite halves and put the discrete topology on one half and the indiscrete topology on the other half. (Comment by François G. Dorais) **Addendum:** Without sufficient Choice, the infinite set $X$ may be [amorphous](http://en.m.wikipedia.org/wiki/Amorphous_set). Amorphous sets are *precisely* the infinite sets for which this approach doesn't work. Very little Choice is needed to ensure that no such beast exists. (Edit by Cameron Buie)
* There is a maximal compact topology on a countable space which is not Hausdorff. See [Steen & Seebach 99](http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA118#v=onepage&q=&f=false). (Comment by Gerald Edgar)
* There is a minimal Hausdorff topology on a countable space which is not compact. See [Steen & Seebach 100](http://books.google.com/books?id=DkEuGkOtSrUC&lpg=PA118&ots=3hHBTJE-k7&dq=maximal%20compact%20topology&pg=PA119#v=onepage&q=&f=false). (Comment by François G. Dorais)
* Those examples can be lifted to any cardinality space, simply by using the disjoint sum with any given compact Hausdorff space. (Comment by Gerald Edgar)
* Every set admits a compact Hausdorff topology, by topologizing it as the one-point compactification of the discrete space structure on the complement of any point. (Answer below by Cameron Buie)
(*Feel free to edit and expand*)
| 39 | https://mathoverflow.net/users/2000 | 15857 | 10,620 |
https://mathoverflow.net/questions/15600 | 3 | There is an algorithm that give us cuboids in $\mathbb{R}^3$, say $Q\_1,Q\_2,\ldots$, such that $\cup\_{i=1}^{\infty} Q\_i$ is the simplex with vertices $(0,0,0), (1,0,0) , (0,1,0), (0,0,1)$, and the $Q\_i$'s are almost disjoint (i.e $\lambda(Q\_i\cap Q\_j)=0$ if $i\neq j$)?
In $\mathbb{R}^2$ there are many easy ways to fill a triangle with almost disjoint-rectangles but I had not find the way to generalize this to higher dimensions. Do you have any ideas?
This will be very helpful, for example, to approximate the cumulative distribution function of a sum of 3 or more random variables that are not necessarily independents.
| https://mathoverflow.net/users/4056 | How to fill a simplex with almost disjoint cuboids? | OK, since we finally have figured out what Andres is asking and since 600 characters is a bit too restrictive, I'll post this as an answer.
The following Asymptote code will draw the filling except I used the size 4 simplex instead of size 1 one here:
```
size(400);
import three;
import graph3;
pen[] q={red,green,magenta,blue,black};
q.cyclic(true);
int N=4;
triple O=(0,0,0),A=(4,0,0),B=(0,4,0),C=(0,0,4);
draw(O--A--B--C--A--C--O--B);
for(int n=0;N>n;++n)
{
real s=1/2^n;
for(real x=4-3*s;x>=0;x-=2s)
for(real y=4-x-3*s;y>=0;y-=2s)
{
draw(shift((x,y,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y+s,4-x-y-3*s))*scale3(s)*unitcube,q[n]);
draw(shift((x-s,y,4-x-y-2*s))*scale3(s)*unitcube,q[n]);
}
}
```
The unitcube is just $[0,1]^3$, the rest should be self-explanatory.
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