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https://mathoverflow.net/questions/34059 | 231 | Let $f$ be an *infinitely differentiable function* on $[0,1]$ and suppose that for each $x \in [0,1]$ there is an integer $n \in \mathbb{N}$ such that $f^{(n)}(x)=0$. Then does $f$ coincide on $[0,1]$ with some polynomial? If yes then how.
I thought of using Weierstrass approximation theorem, but couldn't succeed.
| https://mathoverflow.net/users/1483 | If $f$ is infinitely differentiable then $f$ coincides with a polynomial | The proof is by contradiction. Assume $f$ is not a polynomial.
Consider the following closed sets:
$$
S\_n = \{x: f^{(n)}(x) = 0\}
$$
and
$$
X = \{x: \forall (a,b)\ni x: f\restriction\_{(a,b)}\text{ is not a polynomial} \}.
$$
It is clear that $X$ is a non-empty closed set without isolated points. Applying Baire category theorem to the covering $\{X\cap S\_n\}$ of $X$ we get that there exists an interval $(a,b)$ such that $(a,b)\cap X$ is non-empty and
$$
(a,b)\cap X\subset S\_n
$$
for some $n$. Since every $x\in (a,b)\cap X$ is an accumulation point we also have that $x\in S\_m$ for all $m\ge n$ and $x\in (a,b)\cap X$.
Now consider any maximal interval $(c,e)\subset ((a,b)-X)$. Recall that $f$ is a polynomial of some degree $d$ on $(c,e)$. Therefore $f^{(d)}=\mathrm{const}\neq 0$ on $[c,e]$. Hence $d< n$. (Since either $c$ or $e$ is in $X$.)
So we get that $f^{(n)}=0$ on $(a,b)$ which is in contradiction with $(a,b)\cap X$ being non-empty.
| 181 | https://mathoverflow.net/users/2029 | 34067 | 22,055 |
https://mathoverflow.net/questions/34069 | 3 | Can anyone tell me why the endomorphism ring of a finite-length module is artinian? Bonus points if you can do it without using the radical, semisimplicity, Fitting's lemma or anything fancy. If you have to or it makes the proof easier, that's OK too, but I have reason to believe that there's a simple proof (namely Dennis and Farb give this as an exercise in Chapter 0 of their book Noncommutative Algebra).
| https://mathoverflow.net/users/3544 | endomorphism ring of a finite-length module | Never mind. I was more successful with Google this time. It turns out that the statement is simply false. In the comment above I gave an example showing that the endomorphism ring need not be left Artinian. The following paper contains a (much less trivial) example showing that the endomorphism ring also need not be right Artinian:
"Ring of Endomorphisms of a Finite Length Module"
R. N. Gupta and Surjeet Singh
Proceedings of the American Mathematical Society, Vol. 94, No. 2 (Jun., 1985), pp. 198-200.
Incidentally, the module in that example appears to be a nontrivial self-extension of a simple module.
| 4 | https://mathoverflow.net/users/4384 | 34074 | 22,060 |
https://mathoverflow.net/questions/34058 | 3 | I've been studying some category theory lately and in particular, I became acquainted with the notions of products and coproducts, which led me to ponder the following:
Consider the category of all complex Hilbert spaces (the morphisms being linear isometries). This category has coproducts, due to the direct sum construction: if $X\_{\alpha} , {\alpha\in\Lambda}$ is family of Hilbert spaces, define $X := \bigoplus\_{\alpha\in\Lambda}X\_{\alpha}$ as the set of all "$\Lambda$-tuples" $(x\_\alpha)\_{\alpha \in \Lambda}$ such that:
$x\_\alpha \in X\_\alpha \\ \forall \alpha$ and $\sum\_{\alpha \in \Lambda} \|x\_{\alpha}\|^2 < \infty$
Then one can define addition, scalar multiplication and an inner product on $X$ in an obvious way, and we have the canonical inclusion maps.
However, I don't see any way to make this construction into a product, though maybe there is another construction I don't know of.
I'm sorry if this question is elementary for category theorists, but to me it's not so obvious.
**EDIT**: Thanks for the replies. As it was pointed out, this category doesn't even admit finite products with morphisms being linear isometries. As I don't see any more natural choice for morphisms, I suppose there isn't any good answer to my question (other than "no" :)).
| https://mathoverflow.net/users/7392 | Does the category of Hilbert spaces possess a product? | The category you specified does not have products, because it doesn't have a product of zero objects. The product of zero objects is an final object, if it exists, but final objects need to have unique maps (in this case, isometries) from all other objects. Such a final Hilbert space would need to be at least as large as any other Hilbert space (hence nonzero), but it cannot have nonidentity automorphisms (such as minus one).
| 2 | https://mathoverflow.net/users/121 | 34075 | 22,061 |
https://mathoverflow.net/questions/34066 | 5 | Suppose $R$ is a Cohen-Macaulay ring. It is well known that if $I$ is an ideal of $R$ generated by $n$ elements, and $I$ has codimension $n$, then $R/I$ is also Cohen-Macaulay.
Now suppose that $I$ does not have codimension $n$, but (the scheme defined by) $R/I$ has several irreducible components, one of which has codimension $n$. Is (the coordinate ring of) that component necessarily Cohen-Macaulay?
Because being Cohen-Macaulay is a local condition, it is clear that the component is generically (in fact everywhere it does not intersect the other components) Cohen-Macaulay, but there is no reason obvious to me why this would extend to the whole component.
| https://mathoverflow.net/users/3077 | Irreducible components of quotients of Cohen-Macaulay rings of the "correct" dimension | Hi Alex, I think this fails for $n=2$. Start with a polynomial ring $S$ and height $2$ prime $P$ such that $S/P$ is not Cohen-Macaulay (for example let $P$ be the kernel of the map $S=k[a,b,c,d] \to k[x^4,x^3y,xy^3,y^4]$). Let $a,b$ be a regular sequence in $P$. Let $R=S[t]$ and $I=(ta,tb)$. Then $I$ has height $1$ and is $2$-generated but the components of $I$ is $(t)$ and the components of $(a,b)$, which include $P$. But $R/P$ is not Cohen-Macaulay because $S/P$ isn't.
| 6 | https://mathoverflow.net/users/2083 | 34085 | 22,067 |
https://mathoverflow.net/questions/34052 | 51 | How many functions are there which are differentiable on $(0,\infty)$ and that satisfy the relation $f^{-1}=f'$?
| https://mathoverflow.net/users/1483 | Function satisfying $f^{-1} =f'$ | Let $a=1+p>1$ be given. We shall construct a function $f$ of the required kind with $f(a)=a$ by means of an auxiliary function $h$, defined in the neighborhood of $t=0$ and coupled to $f$ via $x=h(t)$, $f(x)=h(a t)$, $f^{-1}(x)=h(t/a)$. The condition $f'=f^{-1}$ implies that $h$ satisfies the functional equation $$(\*)\quad h(t/a) h'(t)=a h'(at).$$ Writing $h(t)=a+\sum\_{k \ge 1} c\_k t^k$ we obtain from $(\*)$ a recursion formula for the $c\_k$, and one can show that $0< c\_r<1/p^{r-1}$ for all $r\ge 1$. This means that $h$ is in fact analytic for $|t|< p$, satisfies $(\*)$ and possesses an inverse $h^{-1}$ in the neighborhood of $t=0$. It follows that the function $f(x):=h(ah^{-1}(x))$ has the required properties.
| 38 | https://mathoverflow.net/users/8050 | 34095 | 22,069 |
https://mathoverflow.net/questions/34107 | 1 | Let $\kappa$ be a cardinal, and let $P$ be a poset. Let $\mathcal{P}\_\kappa(P)$ denote the poset of $\kappa$-small subposets of $P$ and let $\mathcal{P}\_\kappa^\downarrow(P)\subseteq\mathcal{P}\_\kappa(P)$ be the subposet consisting of those subposets that are downward-closed. Then according to a reliable source, when $\kappa$ is regular, we can show that $\mathcal{P}^\downarrow\_\kappa(P)$ is $\kappa$-filtered because given some $\kappa$-small family of $\kappa$-small subposets, $$A\_i:I\to \mathcal{P}\_\kappa(P)\quad |I|<\kappa$$
the downward closure of the union over this family, $\operatorname{Cl}^\downarrow(\bigcup\_{i\in I}A\_i)$, is $\kappa$-small (which gives a majorant for the family $A\_i$).
However, since I have no experience at all working with regular cardinals, I'm not really sure how to make heads or tails of this. Why does the regularity of $\kappa$ imply that the downward-closure of that union is $\kappa$-small?
| https://mathoverflow.net/users/1353 | The poset of k-small downward-closed subposets of a poset P is k-filtered when k is a regular cardinal? | A cardinal $\kappa$ is regular if (and only if) the union of fewer than $\kappa$ many sets of size less than $\kappa$ still always has size less than $\kappa$. That seems to be exactly what you have here. Also, the union of downward closed sets remains downward closed, so you don't need to take the downward closure of the union, as it is already downward closed.
Note, however, that the downward closure of a $\kappa$-small family might no longer be $\kappa$-small, if $P$ has large initial segments. For example, $P$ may have no $\kappa$-small downward closed subposets at all (this is true in the reverse ordinal $\kappa^\*$, which is $\kappa$ turned upside down).
| 2 | https://mathoverflow.net/users/1946 | 34108 | 22,074 |
https://mathoverflow.net/questions/34098 | 1 | Is the isomorphism class of a fixed cardinality a set(not a proper class)? Or a fixed ordinality for that matter?
By "isomorphism" I mean just bijection for cardinals and order preserving bijection for ordinals, in the category of sets.
| https://mathoverflow.net/users/5292 | Is the isomorphism class of a fixed cardinality a set? | The isomorphism classes of the set {} and the ordinal 0 are sets, each with a single element.
let $\kappa$ be an non-zero ordinal and $\alpha$ # $\beta$ be the [natural sum](http://en.wikipedia.org/wiki/Ordinal_arithmetic#Natural_operations) of the ordinals $\alpha$ and $\beta$. Since the natural sum is the restriction of the addition of [surreal numbers](http://en.wikipedia.org/wiki/Surreal_numbers) to ordinals, and the surreal numbers are an [ordered field](http://en.wikipedia.org/wiki/Ordered_field), the natural sum is cancellative and strictly increasing. Therefore, for any ordinal $\gamma$, the set {$\gamma$#0, $\gamma$#1, ... $\gamma$#$\kappa$} of ordinals that are the natural sum of $\gamma$ and an ordinal less than $\kappa$ is order-isomorphic to $\kappa$. These sets are different for different values $\gamma$, so this provides a bijection between the ordinals and a subclass of the isomorphism class of $\kappa$. If the isomorphism class of $\kappa$ was a set, this subclass would be a set and therefore the ordinals would also form a set. The ordinals cannot form a set, therefore the isomorphism class of $\kappa$ does not form a set.
Assuming the axiom of choice, every set is well-orderable. Given a well-ordering of a set there is a bijection between the order-isomorphism class of that set and a subclass of it's isomorphism(bijection) class. if the set is non-empty, then since it's order-isomorphism class is not a set it's isomorphism class is not a set either.
This can also be proved without the axiom of choice. Suppose that a set x is bijectible
with {y}$\sqcup$ z, where $\sqcup$ means disjoint union, for two sets y and z. Then for any set w, x is bijectible with {w}$\sqcup$z , and since these sets are different for different values of w, this gives a bijection of the class of all sets with a subclass of the isomorphism class of x. Since the class all sets cannot be a set, the isomorphism class of x cannot be a set. It remains to be shown that any non-empty set x is bijectible with {y}$\sqcup$z for two sets y and z. to see this, simply choose one element v from x (this does not require the axiom of choice, because we are only making one choice of one element from one set) and write x as the union of {v} and the set of other elements of x. these two sets are disjoint, so the obvious surjection from their disjoint union to their union is a bijection. The union is x, so v and the set of other elements of x are the y and z we were looking for, respectively.
| 1 | https://mathoverflow.net/users/7727 | 34109 | 22,075 |
https://mathoverflow.net/questions/34032 | 4 | $ B = (B\_t, \mathcal{F}\_t; t\ge 0 ) $ is a 1-d Brownian family on a
measurable space $(\Omega, \mathcal{F})$ with a family of probability
measures $\{\mathbb{P}^x\}$, i.e. $\mathbb{P}^x(B\_0
= x) = 1$, and $B$ is 1-d BM starting from $x$ under $\mathbb{P}^x$.
Let $\tau$ be a given stopping time w.r.t. underlying filtration, $f$ be a given continuous bounded real function. Consider $V(x) = \mathbb{E}^x [f(B\_\tau)]$, where
$\mathbb{E}^x$ is the expectation under $\mathbb{P}^x$.
[Question] Is $V(\cdot)$ continuous for any given stopping time
$\tau<\infty$? If not, is there any counter example? Or does continuity
hold with further conditions?
If $\tau$ is deterministic, then $V$ has no doubt to be continuous. I
am not sure, even if the problem is well formulated with the extension
to stopping time $\tau$. Thanks for any of your comments.
| https://mathoverflow.net/users/5656 | Continuity in intial state of Brownian Motion | Here is a simpler example that I hope convinces you
that $V$ need not be continuous, even in the one dimensional case.
Take one dimensional Brownian motion $(B\_t)$
and define the stopping time $\tau(\omega)=1\_{(B\_0(\omega)<0)}$.
Then, for any bounded measurable $f$, we have
$$V(x)=E\_x[f(B\_1)]1\_{(-\infty,0)}(x)+f(x) 1\_{[0,\infty)}(x).$$
The function $V$ can be made discontinuous at zero by choosing $f$ to
have a strict maximum at $x=0$, since then $E\_x[f(B\_1)] < f(0)$.
**Comment:** You really cannot expect the function $V$ to be continuous in general.
The values of a typical stopping time $\tau$ are intimately tied
up with the sample paths of the Brownian motion; in your
words $\tau$ is "strongly correlated'' with $\omega$.
It's in the definition of stopping time.
The only stopping times that are independent of the Brownian motion
are the deterministic ones.
| 5 | https://mathoverflow.net/users/nan | 34118 | 22,083 |
https://mathoverflow.net/questions/34120 | 5 | I was just going through the 3rd Proof of Sylow's theorem given in the "Topics In Algebra" Book by I.N. Herstein. It looked very interesting and i really liked its Philosophy. My question what is its significance, and how can it be applied to problems, or something else.
| https://mathoverflow.net/users/1483 | Sylow's theorem 3rd Proof Page 96 I.N.Herstein | This is the proof that uses the lemma that if a finite group $G$ has
a Sylow $p$-subgroup then so does each subgroup of $G$. To complete
the proof of existence of Sylow $p$-subgroups, it suffices to show
one can embed each group in a group with a Sylow $p$-subgroup.
By Cayley's theorem each finite $G$ embeds in $S\_n$ with $n=|G|$
and $S\_n$ embeds in $S\_{p^k}$ where $p^k\ge n$. One then writes
down a Sylow $p$-subgroup of $S\_{p^k}$ (essentially an iterated
wreath product of $C\_p$s).
But a slicker conclusion is to embed $S\_n$ in $GL\_n(p)$
(via permutation matrices),
as one sees with little effort that the upper triangular matrices
with $1$s on the diagonal form a Sylow $p$-subgroup of $GL\_n(p)$.
| 14 | https://mathoverflow.net/users/4213 | 34123 | 22,087 |
https://mathoverflow.net/questions/34088 | 42 | Let $M$ be a Riemannian manifold. There exists a unique *torsion-free* connection in the (co)tangent bundle of $M$ such that the metric of $M$ is covariantly constant. This connection is called the Levi-Civita connection and its existence and uniqueness are usually proven by a direct calculation in coordinates. See e.g. Milnor, Morse theory, chapter 2, \S 8. This is short and easy but not very illuminating.
According to C. Ehresmann, a connection in a fiber bundle $p:E\to B$ (where $E$ and $B$ are smooth manifolds and $p$ is a smooth fibration) is just a complementary subbundle of the vertical bundle $\ker dp$ in $T^\*E$. If $G$ is the structure group of the bundle and $P\to B$ is the corresponding $G$-principal bundle, then to give a connection whose holonomy takes values in $G$ is the same as to give a $G$-equivariant connection on $P$.
If $p:E\to B$ is a rank $r$ vector bundle with a metric, then one can assume that the structure group is $O(r)$; the corresponding principal bundle $P\to B$ will in fact be the bundle of all orthogonal $r$-frames in $E$. One can then construct an $O(r)$-equivariant connection by taking any metric on $P$, averaging so as to get an $O(r)$-equivariant metric and then taking the orthogonal complement of the vertical bundle.
Notice that in general one can have several $O(r)$-equivariant connections: take $P$ to be the total space constant $U(1)$-bundle on the circle; $P$ is a 2-torus and every rational foliation of $P$ that is non-constant in the "circle" direction gives a $U(1)$-equivariant connection. (All these connections are gauge equivalent but different.)
So I would like to ask: given a Riemannian manifold $M$, is there a way to interpret the Levi-Civita connection as a subbundle of the frame bundle of the tangent bundle of $M$ so that its existence and uniqueness become clear without any calculations in coordinates?
| https://mathoverflow.net/users/2349 | A geometric interpretation of the Levi-Civita connection? | To understand the existence and uniqueness of the LC connection, it is not possible to sidestep some algebra, namely the fact (with a 1-line proof) that a tensor $a\_{ijk}$ symmetric in $i,j$ and skew in $j,k$ is necessarily zero. The geometrical interpretation is this: once one has the $O(n)$ subbundle $P$ of the frame bundle $F$ defined by the metric, there exists (at each point) a unique subspace transverse to the fibre that is *tangent* both to $P$ and to a coordinate-induced section $\{\partial/\partial x\_1,\ldots,\partial/\partial x\_n\}$ of $F$.
| 23 | https://mathoverflow.net/users/3975 | 34129 | 22,090 |
https://mathoverflow.net/questions/34142 | 17 | This should be easy to prove but I have no idea how to do it:
If $X \subseteq \mathbb{R}^2$ is borel then $f(X)$ is borel where $f(x,y) = x$
Thanks
Tobias
| https://mathoverflow.net/users/8092 | Projection of Borel set from $R^2$ to $R^1$ | This is false; take a look at <https://en.wikipedia.org/wiki/Analytic_set> for a quick introduction. For details, look at Kechris's book on **Classical Descriptive Set Theory**. There you will find also some information on the history of this result, how it was originally thought to be true, and how the discovery of counterexamples led to the creation of descriptive set theory. Another good reference is **A second course on real functions**, by van Rooij and Schikhof.
Here are some additional details (that I added to [another thread](https://mathoverflow.net/a/48590/6085)). Recall that the *analytic* sets are the empty set, and the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$. This notion makes sense in arbitrary topological spaces, and has been particularly studied in [Polish spaces](https://en.wikipedia.org/wiki/Polish_space) (separable completely metrizable spaces).
* The analytic sets are closed under countable unions and intersections, and contain the Borel sets. In all *uncountable* Polish spaces, there are analytic sets whose complement is not analytic. Suslin proved that the Borel sets of uncountable Polish spaces are precisely those analytic sets whose complement is analytic.
* In Baire space $\mathcal N=\mathbb N^{\mathbb N}$ (the countable product of discrete $\mathbb N$), analytic sets are the projections of *closed* subsets of ${\mathcal N}^2$.
This is false with $\mathbb R$ in place of $\mathcal N$. Note that in that case, closed sets are countable union of compact sets, so their projections are $F\_\sigma$. Baire space plays a special role in the theory, since a nonempty set in a Polish space is analytic iff it is the continuous image of $\mathcal N$.
The actual results in ${\mathbb R}$ are as follows:
* A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.
* A set is analytic iff it is the projection of a $G\_\delta$ subset of $\mathbb R^2$.
* There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G\_\delta$ set $A$.
* A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)
| 22 | https://mathoverflow.net/users/6085 | 34143 | 22,099 |
https://mathoverflow.net/questions/34145 | 95 | The following question was a *research* exercise (i.e. an open problem) in R. Graham, D.E. Knuth, and O. Patashnik, "Concrete Mathematics", 1988, chapter 1.
It is easy to show that
$$\sum\_{1 \leq k } \left(\frac{1}{k} \times \frac{1}{k+1}\right) = \sum\_{1 \leq k } \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1.$$
The product $\frac{1}{k} \times \frac{1}{k+1}$ is equal to the area of a $\frac{1}{k}$by$\frac{1}{k+1}$ rectangle. The sum of the areas of these rectangles is equal to 1, which is the area of a unit square. **Can we use these rectangles to cover a unit square?**
>
> Is this problem still open?
>
>
> What are the best results we know about this problem (or its relaxations)?
>
>
>
| https://mathoverflow.net/users/7507 | Can we cover the unit square by these rectangles? | This problem actually goes back to Leo Moser.
The best result that I'm aware of is due to D. Jennings, who proved that all the rectangles of size $k^{-1} × (k + 1)^{-1}$, $k = 1, 2, 3 ...$, can be packed into a square of size $(133/132)^2$ ([link](https://doi.org/10.1016/0097-3165(94)90116-3)).
**Edit 1.** A web search via Google Scholar gave a reference to this [article](https://doi.org/10.1016/S0167-5060(08)70600-9) by V. Bálint, which claims that the rectangles can be packed into a square of size $(501/500)^2$.
**Edit 2.** The state of art of this and related packing problems due to Leo Moser is discussed in Chapter 3 of ["Research Problems in Discrete Geometry"](https://books.google.co.uk/books?id=cT7TB20y3A8C&printsec=frontcover&dq=Research+Problems+in+Discrete+Geometry&source=bl&ots=ampA5AGiJ3&sig=peuMukAVFcivXibr5W-8q1RWs2A&hl=en&ei=kN1VTIPQNsTKjAfVj4XDBA&sa=X&oi=book_result&ct=result&resnum=6&ved=0CDIQ6AEwBQ#v=onepage&q&f=false) by P.Brass, W. O. J. Moser and J. Pach. The problem was still unsettled as of 2005.
| 54 | https://mathoverflow.net/users/5371 | 34149 | 22,102 |
https://mathoverflow.net/questions/34087 | 4 | Suppose $(x\_\alpha)\_\alpha$ is an uncountable, linearly independent family of norm one vectors in a Banach space. Can one always select a basic sequence (or at least a minimal system) from this family? I suspect the answer is no but I cannot come up with an example.
Thank you!
| https://mathoverflow.net/users/7872 | Selecting basic sequences | Not a basic sequence. Consider $e\_0 \oplus e\_\gamma$ in $R\oplus H$ for $H$ a non separable Hilbert space, or, if you want a separable example, make $e\_\gamma$ a Hamel basis for a separable Hilbert space.
EDIT: Aug 2. Every separated sequence of unit vectors contains a minimal subsequence with bounded biorthogonal functionals. There remains the case where your linearly independent set has compact closure. I have an example of such a set where any minimal sequence in the set has only unbounded biorthogonal functionals, but I do not know whether such a set must contain a minimal sequence.
Too bad you did not ask this a day earlier when we could have discussed it face to face. I'll write something down when I get a chance.
EDIT: Aug 2. It was not so bad to write--I was able to do it on the plane.
If $x\_n$ is a separated subset of the unit sphere of $X$, then $x\_n$ has a minimal subsequence whose biorthogonal functionals are uniformly bounded. Indeed, if $x\_n$ does not have weakly compact closure, then it has a basic subsequence (see e.g. the book of Albiac-Kalton), so we can assume that $x\_n\to x$ weakly. If $x=0$, then $x\_n$ has a basic subsequence. If not, let $Q$ be the quotient map from $X$ onto $X/[x]$, where $[x]$ is the linear span of $x$. $Qx\_n\to 0 $ weakly and is bounded away from zero by the separation assumption, hence has a basic subsequence $Qx\_{n(k)}$, whence $x\_{n(k)}$ is minimal with uniformly bounded biorthogonal functionals.
I don't know what can happen when $A$ is a linearly independent subset of the unit sphere of $X$ that is totally bounded, but there exists such sets so that every minimal sequence in the set has biorthogonal functionals that are not uniformly bounded. Consider the Cantor set as the branches of the infinite binary tree and let the nodes index the unit vector basis of $c\_0$. Given a branch $t=(t(n))$ (where $t\_1<t\_2<\dots$ in the tree ordering) of the tree, let $x\_t=\sum 2^{-n} e\_{t(n)}$. By compactness (which is really just pigeonholing), any sequence $y\_k$ of $x\_t$-s has a subsequence that converges to some $x\_s$, which means that for any $n$, if $k\ge k(n)$ then $y\_k(j)=x\_s(j)$ for $1\le j \le n$. From this it is easy to see that $y\_k$ cannot be uniformly minimal.
In the above argument, try replacing the unit vector basis of $c\_0$ with an appropriate normalized countably linearly independent sequence that has no minimal subsequence. I think it is known that such sequences exist. Probably an example is in Kadec's book. Maybe this will give an example that has no infinite minimal subset.
| 7 | https://mathoverflow.net/users/2554 | 34153 | 22,104 |
https://mathoverflow.net/questions/34148 | 2 | Hi there,
Sorry if this has already been asked before. I tried googling for it, but perhaps I could not find the right words to search for. My question is: Which is the fastest way to compute A\*inv(B) [edit] where A and B are matrices?
| https://mathoverflow.net/users/4430 | Efficient computation of AB^-1 for matrices | What are the sizes of $\mathbf{A}$ and $\mathbf{B}$? This information is important.
Let me assume you mean you want the efficient *numerical* computation of the matrix $ \mathbf{A} \mathbf{B}^{-1}$.
The general strategy would be to do this: Let $\mathbf{J} = \mathbf{A} \mathbf{B}^{-1}$; therefore $\mathbf{J}\mathbf{B} = \mathbf{A}$. You must then rewrite this into $\mathbf{P}\mathbf{x} = \mathbf{Q}$ form (this depends on the dimensions of your matrices).
For instance, for $\mathbf{A} \in \mathbb{R}^{m \times n}$, $\mathbf{B} \in \mathbb{R}^{n \times n}$ and $\mathbf{J} \in \mathbb{R}^{m \times n}$, you can write a system of equations:
$J\_{i,\*} \cdot B\_{\*,j} = A\_{i,j}$
(Notation: for a matrix $\mathbf{X}$, we define $X\_{i,\*}$ as the $i$-th row vector and $X\_{\*,j}$ as the $j$-th column vector, and $X\_{i,j}$ as the element in the $i$-th row and $j$-th column).
From here, you can use a fast linear solver to solve the resulting linear equation system -- you will get a solution for the elements of $\mathbf{J}$.
By solving a linear system of equations and not taking the inverse directly, you're not only cutting down on the no. of operations required, you can exploit also properties like sparsity, inertia, etc. and have capabilities like pre-conditioning at your disposal.
| 4 | https://mathoverflow.net/users/7851 | 34156 | 22,107 |
https://mathoverflow.net/questions/34099 | 20 | Hi!
What text on both incompleteness theorems you would recommend for beginner?
Specifically, I'm looking for the text with the following properties:
1) The proofs should be finitistic, in Godel's tradition, i. e. formalizing "I'm unprovable" (not, for instance, via formalization of halting problem);
2) The text must be of reasonable length but with complete proofs, so that one can study them in a reasonable amount of time (e. g. only those forms of recursion theory theorems are proved which are precisely needed for incompleteness proofs);
3) The entire text should be motivated and discussing ideas (even those of philosophical character) before and between technical constructions.
I would be very thankful if you'll equip your suggestion with some short resume.
| https://mathoverflow.net/users/6307 | The best text to study both incompleteness theorems | First chapter of Jean-Yves Girard, "Proof Theory and Logical Complexity", Vol I, Bibliopolis, 1987
It satisfies all of your conditions, but it is not an elementary book. If I remember correctly, the authors (A.S. Troelstra and H. Schwichtenberg) of the book "Basic Proof Theory" which is published in 2001 wrote in their introduction that their intention was to fill the gap between this and all other (introductionary) books in proof theory. As far as I know, he never published the second volume.
| 7 | https://mathoverflow.net/users/7507 | 34163 | 22,112 |
https://mathoverflow.net/questions/33405 | 24 | ***This question is still wide open - all of the answers so far rely on magical calculations. I've only accepted an answer because, by bounty rules, otherwise one would be accepted automatically. I can't change the accepted answer, but it would be amazing to have more discussion on this question.***
I'd like a nice proof (or a convincing demonstration), for a surface in $\mathbb R^3$, that explains why the following notions are equivalent:
1) Curvature, as defined by the area of the sphere that Gauss map traces out on a region.
1.5) The integral of the product of principal curvatures.
2) The angle defect of parallel transport about a geodesic triangle.
(This equivalence may be considered as either a part of the Theorema Egregium or a part of Gauss-Bonnet. Proving that numbers 1 and 1.5 are the same is pretty easy).
**Motivation:** I'm teaching a five-day class for very bright high-school students. The idea is to give them an impression of what geometry is about. However, when I looked at Spivak's proof of this, it was much more of a messy calculation than I expected. I'd like, if at all possible, something more conceptual, ideally with a nice picture attached to it.
Since this doesn't have to be a perfectly complete class, I'll be perfectly happy with a good illustration of why this is true instead of a rigorous proof, if a conceptual *and* rigorous proof is completely out of the question.
One idea I had is to show the example of a sphere and the hyperbolic plane, and then explain that on very small scales the curvature is constant. However, then I would need a nice proof that the embeddings of the hyperbolic plane in $\mathbb R^3$ have curvature -1.
Thank you very much!
P.S. This question is related, but not quite the same (I hope), to this question:
[Equivalent definitions of Gaussian curvature](https://mathoverflow.net/questions/22410/equivalent-definitions-of-gaussian-curvature)
P.P.S. Thank you to whomever recommended to Berger's "Panoramic View of Riemannian Geometry". it was quite useful to me. I do not know why you deleted your answer.
That books claims there is no conceptual proof. However, I'd still be very happy with a nice illustration of why one should believe this, especially for negative curvature.
| https://mathoverflow.net/users/2467 | Why are two notions of Gaussian curvature are the same - what is the simplest & most didactic proof? | There is a short conceptual proof of Gauß-Bonnet due to Chern (see also ["A panoramic view of Riemannian geometry"](http://books.google.co.uk/books?id=d_SsagQckaQC&dq=panoramic+view+berger&printsec=frontcover&source=bn&hl=en&ei=1PJVTOb8I4Xu0wTXnuT6Ag&sa=X&oi=book_result&ct=result&resnum=5&ved=0CC8Q6AEwBA#v=onepage&q&f=false) by Berger). The argument assumes a basic familiarity with differential forms though.
Assume that the surface $S$ is oriented so that its canonical measure $dm$ is a 2-form. Let's consider the set of all unit vectors tangent to $S$, i.e. the unitary fiber $US$ of $S$. The canonical projection $p:US\to S$ associates with each unit vector a point on $S$ where it is tangent. Now, $US$ is a 3-manifold which posesses a canonical differential form $\zeta$. The exterior derivative $d\zeta$ is the form lifted by $p$ onto $US$ of the 2-form of the curvature $K(m)dm$. If the domain $D\subset S$ is simply connected, one can define a continuous field $\xi$ of unit vectors on $D$; therefore, $D$ can be lifted into $US$.
The Gauß-Bonnet formula follows directly from the Stokes formula applied to $\xi(D)$ since the canonical 1-form $\zeta$ is the geodesic curvature. This is actually more than you ask for because the boundary of $D$ does not have to consist of geodesics.
| 6 | https://mathoverflow.net/users/5371 | 34164 | 22,113 |
https://mathoverflow.net/questions/34160 | 18 | In an answer to a question [on MU](https://math.stackexchange.com/questions/540/what-is-the-riemann-zeta-function/832#832) about the Riemann zeta function, I sketched a proof that the probability distribution on $\mathbb{N}$ which assigns $n$ the probability
$$\frac{ \frac{1}{n^s} }{\zeta(s)}$$
(henceforth called the zeta distribution with parameter $s$) where $s > 1$ is the unique family of probability distributions on $\mathbb{N}$ satisfying the following three requirements:
* The exponent of $p$ and the exponent of $q$ in the prime factorization of $n$ are chosen independently for all pairs of primes $p \neq q$.
* The exponent of a particular prime $p$ is geometrically distributed.
* The probabilities are monotonically decreasing as a function of $n$.
The basic motivation for the first requirement is the Chinese Remainder Theorem. I can think of two motivations for the second requirement: first, that geometric distributions are the maximally entropic distributions on $\mathbb{N}\_{\ge 0}$ with a given mean, and second, that (if I'm not mistaken) one naturally gets a geometric distribution from Haar measure on the $p$-adic integers.
In fact, the distribution one gets from Haar measure on the $p$-adic integers is the one in which a $p$-adic integer is divisible by exactly $p^k$ with probability $(1 - p^{-1}) p^{-k}$. This is essentially the $s \to 1$ limit of the zeta distribution above, which I gave as a reason one might deduce that this limit is important from first principles.
It seems like it should be possible to combine the motivations for the first two requirements into a statement about Haar measure on the profinite integers $\prod\_p \mathbb{Z}\_p$, except that I don't know exactly what kind of statement I'd be looking for, so I thought I'd ask here.
**Question:** Complete the following statement. It is natural to study the $s \to 1$ limit of the zeta distribution because (some statement about Haar measure on the profinite integers, maybe with "Tate's thesis" thrown in somewhere).
| https://mathoverflow.net/users/290 | The Riemann zeta function and Haar measure on the profinite integers | ...because for sets with logarithmic density, the logarithmic density and the $s\to1$ zeta-measure agree. And for sets with natural density, the logarithmic and natural densities are the same.
| 7 | https://mathoverflow.net/users/935 | 34170 | 22,119 |
https://mathoverflow.net/questions/34106 | 0 | Let $(X,A)$ be a finite CW-pair $m=p^r$ for some prime $p$. Unspecified coefficient is in $\mathbb{Z}$.
From the universal coefficient theorem, We know that
$H^1(A;\mathbb{Z}\_m)=\textrm{Hom} (H\_1(A),\mathbb{Z}\_m)$ ---(1) and
$H^2(X,A;\mathbb{Z}\_m)=\textrm{Hom}(H\_2(X,A);\mathbb{Z}\_m)\bigoplus \textrm{Ext}(H\_1(X,A),\mathbb{Z}\_m)$. ---(2)
From the long exact sequence of pair, we have a coboundary map $\delta\colon H^1(A;\mathbb{Z}\_m)\to H^2(X,A;\mathbb{Z}\_m)$.
I know that $\pi\_1\circ \delta \colon H^1(A;\mathbb{Z}\_m)\to \textrm{Hom}(H\_2(X,A),\mathbb{Z}\_m)$ ($\pi\_1$ is a 1st factor projection from (2)) is same as the composition $H^1(A;\mathbb{Z}\_m)\cong \textrm{Hom}(H\_1(A);\mathbb{Z}\_m)\to \textrm{Hom}(H\_2(X,A);\mathbb{Z}\_m)$ (the first map is an isomorphism from (1) and the second map is obtained by taking $\textrm{Hom}(-,\mathbb{Z}\_m)$ to $\partial\colon H\_2(X,A)\to H\_1(A)$.)
Let's think $\pi\_2\circ \delta \colon H^2(A;\mathbb{Z}\_{m})\to Ext(H\_1(X,A);\mathbb{Z}\_m)$, where $m=p^r$ as before.
*Question* : Is it true that if $r$ is large, then $\pi\_2\circ\delta$ is trivial map or is it meaningless question because of unnaturality of splitting of (2)?
| https://mathoverflow.net/users/7776 | About universal coefficient theorem | There is a map (i.e. a commutative diagram) from the exact sequence $$0\to Ext(H\_{n-1}(A),G)\to H^n(A;G)\to Hom(H\_n(A),G)\to 0$$ to the exact sequence $$0\to Ext(H\_n(X,A),G)\to H^{n+1}(X,A;G)\to Hom(H\_{n+1}(X,A),G)\to 0$$All the groups are functors of both the pair $(X,A)$ and the abelian group $G$. All the maps are natural. The exact sequences split, but the splittings are not natural, so there is not a canonical map $Hom(H\_n(A),G)\to Ext(H\_n(X,A),G)$ to inquire about. However, by the commutative diagram there is a canonical map from a subgroup of the former to a quotient of the latter, namely from $$ker(Hom(H\_n(A),G)\to Hom(H\_{n+1}(X,A),G)=Hom(coker(H\_{n+1}(X,A)\to H\_n(A)),G)=Hom(P,G)$$ to $$coker(Ext(H\_{n-1}(A),G)\to Ext(H\_n(X,A),G))=Ext(ker(H\_n(X,A)\to H\_{n-1}(A)),G)=Ext(Q,G)$$ where $P=Im(H\_n(A)\to H\_n(X))$ and $Q=H\_n(X)/P$. This is in fact determined by the exact sequence $0\to P\to H\_n(X)\to Q\to 0$, part of the six-term exact sequence $$0\to Hom(Q,G)\to Hom(H\_n(X),G)\to Hom(P,G)\to Ext(Q,G)\to Ext(H\_n(X),G)\to Ext(P,G)\to 0$$
This map can easily be nonzero when $G=\mathbb Z/m$, for example if $H\_{n+1}(X,a)\to H\_n(A)\to H\_n(X)\to H\_n(X,A)\to H\_{n-1}(A)$ is the exact sequence $0\to \mathbb Z\to \mathbb Z\to \mathbb Z/p\to 0$ and $p$ divides $m$.
| 2 | https://mathoverflow.net/users/6666 | 34177 | 22,122 |
https://mathoverflow.net/questions/34180 | 9 | We know that in a normal extension of number fields $L/K$, for any prime $P$ of $K$ and any primes $Q\_1,Q\_2$ of $L$ lying over $P$, the ramification indices and inertial degrees are the same,
$$e(Q\_1|P)=e(Q\_2|P),\quad f(Q\_1|P)=f(Q\_2|P),$$
and that this is not necessarily the case if $L/K$ is not normal. Is it never the case when $L/K$ is not normal? That is, must there be at least one $P$ and $Q\_1,Q\_2$ for which $e(Q\_1|P)\neq e(Q\_2|P)$ or $f(Q\_1|P)\neq f(Q\_2|P)$?
| https://mathoverflow.net/users/1916 | Converse to basic result on prime decomposition | Forget the ramification indices: they're nearly always 1. Let's focus on the residue field degrees. Let $E$ be the Galois closure of $L$ over $K$. For a prime $P$ in $K$, write $f\_P(E/K)$ for the common residue field degree of all primes in $E$ over $P$. Since $E$ is determined (up to isom. as an extension of $K$) by $L$, one can imagine there might be a *formula* for $f\_P(E/K)$ in terms of the $f(Q\_i|P)$'s as $Q\_i$ runs over the primes in $K$ that lie over $P$. There is such a formula if $P$ is unramified in $E$: $f\_P(E/K) = \text{lcm } f(Q\_i|P)$. (This is basically because a composite of finite fields over a particular finite field has degree equal to the lcm of the degrees of the extensions.)
Now imagine that for all $P$ the numbers $f(Q\_i|P)$ are equal. Then their lcm is that common value, so when $P$ is unramified in $E$ we have $f\_P(E/K) = f(Q\_i|P)$ for all $Q\_i$ over $P$ in $L$. Then $f\_{Q\_i}(E/L) = 1$ for all $Q\_i$, so all but finitely many primes in $L$ are split completely in $E$. By Chebotarev's density theorem that implies $E = L$, so $L/K$ is Galois. (Strictly speaking we are using a result that is logically much weaker than Chebotarev: that the primes which split completely determine a Galois extension is due to Bauer from 1916 or so, before Chebotarev proved his general theorem.)
Since Chebotarev works with a density 0 set of primes taken out, instead of assuming for all $P$ that the numbers $f(Q\_i|P)$ (when $Q\_i|P$) are equal you only need to assume that for all but finitely many $P$ or even for all but a density 0 set of $P$.
| 15 | https://mathoverflow.net/users/3272 | 34185 | 22,128 |
https://mathoverflow.net/questions/34186 | 36 | This question is related to [this recent but currently
unanswered MO
question](https://mathoverflow.net/questions/33879/decidability-of-matrix-algebra)
of Ricky Demer, where it arose as a comment.
Consider the structure $R^n$ consisting of $n\times n$
matrices over the reals $\mathbb{R}$, $n$-dimensional row
vectors, column vectors and real scalars, with the ordered
field structure on the scalars. Thus, we can add and
multiply matrices; we can multiply anything by a scalar; we
can multiply matrices by vectors (on the suitable side);
and we can add and multiply vectors of the suitable shape.
The corresponding *matrix algebra* language has four
variable sorts - scalars, matrices, row vectors and column
vectors - together with the rules for forming terms so that
these expressions make sense in any $R^n$. In this
language, you can quantify over matrices, vectors and
scalars, form equations (and inequalities with the
scalars), but you cannot quantify over the dimension. The
idea is that an assertion in this language can be
interpreted in any dimension, one $R^n$ at a time. You have to make assertions that do not refer to the dimension; the
language is making assertions about matrices and vectors in
some fixed but unspecified dimension.
My question is whether truth in this real matrix algebra
obeys a 0-1 law as the dimension increases, that is:
>
> **Question.** Is every statement in the matrix algebra
> language either eventually true or eventually false in
> $R^n$ for all sufficiently large dimensions $n$?
>
>
>
To give some trivial examples:
* the statement asserting matrix commutativity
$\forall A,B\, AB=BA$ is true in dimension $1$ but false in all
higher dimensions.
* the statement that the dimension is at least 17, or at
most 25, or an odd number less than 1000, are all
expressible, since you can quantify over enough vectors
and the assertions that they are independent or that they
span are expressible. The truth values of these
statements all stabilize in sufficiently high dimension.
* the assertion that a particular real number is an
eigenvalue for a matrix is expressible.
But it isn't clear to me how one could express, for
example, that the dimension is even. (Edit: Gerry and Ryan below have explained how this is easily done.)
In the previous question, Ricky inquired
whether there is a decision procedure to determine which
assertions of matrix algebra are true for all $n$. For any
particular $n$, then Tarski's theorem on the decidability
of real-closed fields shows that the theory of the
structure $R^n$ is decidable: when $n$ is fixed, we may
translate any statement about matrices and vectors into
statements about real numbers by talking about the
components. (We may also add to the language the functions
that map a matrix or vector to the value of any particular
entry, as well as $det(A)$ etc.)
If my question here has a positive answer, and the
stabilizing bound is computable from the formula, then this
would provide an affirmative answer to Ricky's question,
since we could just determine truth in a large enough
$R^n$.
Lastly, I don't think it will fundamentally change the
problem to work in the complex field, since the
corresponding structure $C^n$ with complex matrices and
vectors is interpretable in $R^n$. For example, I think we
could freely refer to complex eigenvalues.
---
**Edit.** The real case was quickly dispatched by Gerry and Ryan, below. Let us therefore consider the complex case. So we have for each dimension $n$ the structure $C^n$ with $n\times n$ matrices, row vectors, column vectors and complex scalars. The question is: Does the truth of every statement of matrix algebra stabilize in $C^n$ for sufficiently large $n$?
Ricky proposed that we add Hermitian transpose (conjugation on scalars) to the language. This would would also allow us to refer to the real scalars. If we expand the language so that we are able to define the class of real matrices and vectors, however, then we can still express Gerry's and Ryan's solutions for a negative answer here.
---
**Edit 2.** As in the comments, let us say that the *truth set* of a formula $\phi$ in the language is the set of $n$ for which $\phi$ is true in dimension $n$. These truth sets form a Boolean algebra, closed under finite differences. Which sets of natural numbers are realizable as truth sets? (Note that there are only countably many truth sets.) And how does it depend on the field?
| https://mathoverflow.net/users/1946 | Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions? | One can get arithmetic progressions as *truth sets*, as in Joel's comment. Pick non-negative integers $a$ and $b$, pick a finite group $G$ which has at least one representation of degree $a$. Then there is a formula expression the statement "the vector space is a $G$-module which is a sum of irreducible representations of degree $a$ and exactly $b$ trivial summands".
**Later:** For example, the irreps of $G=(\mathbb Z\_3\times\mathbb Z\_3)\rtimes\mathbb Z\_3$ have degree 1 and 3. It is generated by two elements which have cube equal to the identity, and which commute with their commutator. For example, if we want dimensions to be divisible by $3$, we can say:
$(\exists A,B)(A^3=B^3=[A,[A,B]]=[B,[A,B]]=I \wedge \neg(\exists v,\lambda,\mu)(Av=\lambda v\wedge Bv=\mu v))$
(uppercase letters are matrices, lowercase letters are vectors, greek letters are scalars, and commutators are group commutators) A model for this is a $G$ which does not have one-dimensional submodules. This works for other prime values of $3$.
**Later:** A vector space $V$ has a structure of $M\_n(k)$-module iff $n\mid\dim V$. This can also be written in the language and it is much simpler that the first example!
| 9 | https://mathoverflow.net/users/1409 | 34199 | 22,136 |
https://mathoverflow.net/questions/34188 | 12 | What can be said about the limiting distribution of the sequence of fractional parts of $\{n^{a},n>0\}$ for $a\in(1,2)$. I ran a computer experiment for $n\sqrt{n}$ and it looks like uniformly distributed. Is there a simple proof?
| https://mathoverflow.net/users/3375 | Distribution of fractional parts of n^{3/2} | Exercise 2.23 in Kuipers and Niederreiter, Uniform Distribution Of Sequences: Use Theorem 2.7 to show that the sequence $(\alpha n^{\sigma})$, $n=1,2,\dots$, $\alpha\ne0$, $1\lt\sigma\lt2$, is u.d. mod 1.
They are using $(x)$ for the fractional part. Theorem 2.7 is Let $a$ and $b$ be integers with $a\lt b$, and let $f$ be twice-differentiable on $[a,b]$ with $f''(x)\ge\rho\gt0$ or $f''(x)\le-\rho\lt0$ for $x\in[a,b]$. Then $$\left|\sum\_{n=a}^be^{2\pi if(n)}\right|\le(|f'(b)-f'(a)|+2)\left({4\over\sqrt\rho}+3\right).$$ Theorem 2.7 is attributed to van der Corput, Zahlentheoretische Abschatzungen, Math. Ann. 84 (1921) 53-79.
| 8 | https://mathoverflow.net/users/3684 | 34202 | 22,139 |
https://mathoverflow.net/questions/34205 | 6 | There are short and sweet proofs of various forms of Stirling's approximation. But even the sweetest among them don't instill the same conviction in the reader as a direct bijective proof.
Computer scientists can often get away with a very weak form of Stirling's approximation:
$$(n/2)^{n/2} \leq n! \leq n^n$$
From this follows
$$(n/2)\ log(n) - (n/2)\ log(2) \leq log(n!) \leq n\ log(n)$$
and therefore $log(n!) = \Theta(n\ log(n))$. This is sufficient to establish the lower bound on comparison-based sorting algorithms and many other asymptotic bounds.
>
> Does anyone know a natural bijective proof of $(n/2)^{n/2} \leq n! \leq n^n$?
>
>
>
| https://mathoverflow.net/users/2036 | Bijective proof of weak form of Stirling's approximation | Think of all maps from the first $n/2$ elements of {$1,...,n$} to the last $n/2$. Say, let $a\_1< \ldots < a\_k \to z$. Make a cycle $a\_1 \to a\_2 \to \ldots \to a\_k \to z \to a\_1$. Do this for all $z$. The details are straightforward. This proves the lower bound.
| 6 | https://mathoverflow.net/users/4040 | 34227 | 22,153 |
https://mathoverflow.net/questions/34229 | 6 | In his book "Riemannian Geometry" do Carmo cites the Hopf-Rinow theorem in chapter 7. (theorem 2.8). One of the equivalences there deals with the cover of the manifold using nested sequence of compact subsets. This made me wonder whether the following lemma holds:
**Lemma:** Let $M$ be a compact Hausdorff space, and let $K\_i \subset M$ be a sequence of compact subsets such that: $K\_i\subset K\_{i+1}$ and $\cup\_{i=1}^{\infty} K\_i = M$. Then there exists an index $i\_0$ such that $K\_i = M$ for all $i \geq i\_0$.
Here is my proof to this:
**Proof**: Assume that $K\_i \neq M$ for all $i$, that is all $K\_i$'s are proper subsets of $M$. With out loss of generality we can then assume that $K\_i\subsetneq K\_{i+1}$. This implies that $\forall i$ there exists $x\_i$ such that $x\_i\notin K\_i$ but $x\_i\in K\_{i+1}$. Since $K\_i$ is compact subset of a Hausdorff space, there exist open $U\_i$ and $V\_i$, such that $K\_i \subset U\_i$, $x\_i\in V\_i$ and $U\_i \cap V\_i = \emptyset$.
Now, if $x\in \cup\_{i=1}^{\infty} U\_i$ then clearly $x\in M$ since $U\_i \subset M$. On the otehr hand, if $x\in M$, then $x\in K\_{i\_0}$ for some $i\_0$, and thus it is also in $U\_{i\_0}$. This yields that $M= \cup\_{i=0}^{\infty}U\_i$.
Let us now assume that $\cup\_{j=1}^n U\_{i\_j}=M$ is a finite cover. If $n\_0 = \max\_{j=1,\ldots,n}\{i\_j\}$ then $x\_{n\_0+1}\notin \cup\_{j=1}^nU\_{i\_j}$. This in turn means, that we cannot find a finite sub-cover of $M$ using the open cover $\cup\_{i=1}^{\infty}U\_i$. But this contradicts the compactness of $M$. This completes the proof. $\square$
Finally, here's my question. Is this lemma correct? Is my proof correct?
Thanks in advance and all the best!
Dror, Edit: As I verified with the author, he meant that the last equivalence is valid when the manifold is *not compact*. Thus, my false lemma, is irrelevant from the first place.
| https://mathoverflow.net/users/8047 | Compact cover of a Hausdorff compact space | Note: whilst typing this, Martin posted his answer. As I come to a completely different conclusion, I'd be very interested in knowing who's right!
---
False. Let $M = [0,1]$ and $K\_i = \{0\} \cup [\frac{1}{i},1]$.
The flaw in the proof is the assumption that the $U\_j$ are increasing; ie that $U\_j \subseteq U\_{j+1}$. Thus the sentence "If $n\_0=\max\_{j=1, \dots, n} i\_j$ then $x\_{n\_0+1} \notin \bigcup\_{j=1}^n U\_{i\_j}$." is not true.
| 12 | https://mathoverflow.net/users/45 | 34231 | 22,155 |
https://mathoverflow.net/questions/34232 | 7 | Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be an injection for $n>m$. Can $f$ be continuous? Why?
I got this question in mind when I was trying to find a continuous map from $\mathbb{R}^{2}$ to $\mathbb{R}$.
| https://mathoverflow.net/users/1483 | Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$ | If $f$ is injective and continuous from $\mathbb{R}^n$ to $\mathbb{R}^m$
where $n>m$ then $f$ restricts to a continuous bijection from $S^{n-1}$,
the unit sphere in $R^n$, to a compact subset $K$ of $\mathbb{R}^m$.
Thus you can embed $S^{n-1}$, and a foriori $S^m$ in $\mathbb{R}^m$.
But there are homological obstructions to embedding $S^m$ in $\mathbb{R}^m$.
Using the arguments of this excellent paper
Albrecht Dold,
A simple proof of the Jordan-Alexander complement theorem,
*Amer. Math. Monthly* **100** (1993), 856-85.
(essentially a cunning use of the Mayer-Vietoris theorem)
it would entail the homology of the space $\mathbb{R}^m-K$ being nonzero
in negative dimension, which is absurd.
**Added** As I replied in haste I forgot the sledgehammer
that cracks this little nut, namely [Alexander duality](http://en.wikipedia.org/wiki/Alexander_duality).
**Added later** In fact this result also follows from Brouwer's
Invariance of Domain.
This is Theorem 2B.3 on page 172 of [Hatcher's book](http://www.math.cornell.edu/~hatcher/AT/ATpage.html).
This implies that if one has an embedding from
$\mathbb{R}^n$ to itself, then its image is open. One gets such an
embedding by composing your putative embedding with the natural embedding of
$\mathbb{R}^m$ in $\mathbb{R}^n$. Adapting the proof, gives a swift proof
that the answer of your original question is no.
If you have a continuous injection from $\mathbb{R}^n$ to $\mathbb{R}^m$,
with $n>m$ then you have an embedding of $S^{n-1}$ into a nontrivial
hyperplane in $\mathbb{R}^n$. By the Jordan-Brouwer separation theorem
the image $K$ of this embedding separates $\mathbb{R}^n$ but it's easy to
see that since the image is in a hyperplane any two points of
the complement of $K$ can be connected by a path (exercise for reader :-)).
| 15 | https://mathoverflow.net/users/4213 | 34234 | 22,156 |
https://mathoverflow.net/questions/34228 | 7 | Let $R$ be a sheaf of rings on a topological space $X$. Assume $R \neq 0$. Does then $R$ have a maximal ideal? So this is a spacified analogon of the theorem, that every nontrivial ring has a maximal ideal. Currently I try to develope this sort of spacified commutative algebra and algebraic geometry. If anyone knows some literature about it, please let me know.
So let's try to imitate the known proof for rings and use Zorn's Lemma. For that, we need that for every linear ordered set $(J\_k)\_{k \in K}$ of proper ideals in $R$, their sum $\sum\_{k \in K} J\_k$ is also a proper ideal. Note that if we replace $R \neq 0$ by $R\_x \neq 0$ for all $x \in X$ and the notion proper by "stalkwise proper", then everything works out fine since stalks and sum commute. However, global sections do not commute with (infinite) sums. Anyway, let's try to continue:
Assume $\sum\_{k \in K} J\_k = R$, that is, $1$ is a global section of the sum. Then there is an open covering $X = \cup\_{i \in I} U\_i$, such that $1 \in \sum\_{k \in K} J\_k(U\_i) = \cup\_{k \in K} J\_k(U\_i)$. Thus we get a function $I \to K, i \mapsto k\_i$, such that $1 \in J\_{k\_i}(U\_i)$. If this function has an upper bound, say $k$, then we get a contradiction $J\_k=R$. Thus the function is unbounded. And now?
I think that this already indicates that there will be counterexamples, but I'm not sure. Also note that everything is fine when $X$ is quasi compact.
| https://mathoverflow.net/users/2841 | Does every nontrivial sheaf of rings have a maximal ideal? | Take $X=\mathbf Z$ with topology $(k,\infty)\cap\mathbf Z$ for $-\infty\leq k\leq\infty$, so that sheaves on $X$ may be identified with sequences $\dots\to F\_k\to F\_{k+1}\to\dots$, $k\in\mathbf Z$. Now take for $R$ the constant sheaf with value $\mathbf Q$. All ideals have the form $\dots\to0\_{k-1}\to0\_k\to\mathbf Q\_{k+1}\to\mathbf Q\_{k+2}\to\dots$ for some $-\infty\leq k\leq\infty$, so there are no maximal ideals.
| 8 | https://mathoverflow.net/users/2035 | 34235 | 22,157 |
https://mathoverflow.net/questions/34233 | -3 | We know that by Dirichlet's formula for the Divisor function $ \displaystyle \sum\limits\_{n \leq x} d(n) = x \log{x} + (2C-1)x + \mathcal{O}(\sqrt{x})$.
What is the best approximation available till date for the given formula. I know that finding the infimum of the $\mathcal{O}'s$ is an unsolved problem, but would like to see the closest approximation.
| https://mathoverflow.net/users/1483 | Dirichlet's Divisor Function | Use Wikipedia: <http://en.wikipedia.org/wiki/Dirichlet_divisor_problem> .
| 4 | https://mathoverflow.net/users/6153 | 34238 | 22,158 |
https://mathoverflow.net/questions/34241 | 5 | I'm trying to understand why the Laplacian operator is used in [blob detection](http://en.wikipedia.org/wiki/Blob_detection) in image analysis. I must admit that in trying to figure out why the Laplacian is useful in this application, I've really confused myself with the different uses of the word 'Laplace.' For instance, Wikipedia has many articles on this, and the ones I'm having trouble unifying conceptually are the [Laplace Transform](http://en.wikipedia.org/wiki/Laplace_transform) and the [Laplace Operator](http://en.wikipedia.org/wiki/Laplace_operator).
From co-workers and some [reading on the internet](http://homepages.inf.ed.ac.uk/rbf/HIPR2/log.htm), I have come to very shallowly think of my Laplacian convolutions on images as performing something similar to the second derivative, where the most quickly changing areas on the image are what become highlighted in the new, convolved image. From the page on the Laplace Operator this makes a lot of sense. This doesn't make sense to me from the page on the Laplace Transform. My question then, I think, is how are the Laplace Operator and the Laplace Transform related? If I can see, from the definition, that the Laplace Operator is basically doing the second derivative, I would think I should be able to see something similar from the Laplace Transform. But I don't. Am I mistaken in thinking that the Laplace Transform and the Laplace operator are the same thing? How are they related?
| https://mathoverflow.net/users/7445 | Laplacian operator and relation to the Laplace Transform | They are certainly not the same thing.
You might sometimes see them appear in the same context because transforms of Laplace-Fourier type are immensely useful for analyzing linear differential operators like the Laplacian. But the Fourier transform has better analytic properties, so that's the one you are more likely to see used.
Here's some intuition you might find helpful.
The discrete Laplacian computes the difference between a node's averaged neighbors and the node itself. It's often used in image processing and that gives an easy way to visualize it. The 1D case where the kernel is [1 -2 1] is especially simple:
In an area of constant color the Laplacian is zero. Indeed, even if you have linear variation it remains zero, e.g. in the neighborhood [1 2 3] the Laplacian's value at the center point is
$$1 \cdot 1 + (-2) \cdot 2 + 3 \cdot 1 = 0.$$
But quadratic and higher-order variation excites the Laplacian and results in non-zero values. Thus it's especially useful for detecting 'jumps'. That's why it's the weapon of choice in edge detection. It's often combined with a Gaussian to pre-filter out any small-scale features or noise that might cause spurious edges to be detected.
I should mention that the Laplacian in two dimensions and higher is significantly richer than the one-dimensional case might suggest. For one, not all two-dimensional images with a uniformly zero Laplacian are linear. But qualitatively a lot of the same intuition holds true as to how the Laplacian reacts to variation.
| 5 | https://mathoverflow.net/users/2036 | 34245 | 22,162 |
https://mathoverflow.net/questions/34240 | 7 | What I mean is this. By downward Lowenheim-Skolem theorem, first-order formula Q is a always true iff it is true in every countable structure. But is there some first-order formula Q which is true in every computable structure and false in some non-computable structure? My feeling is that of course the answer should be "yes", but I can't construct an example. I feel also that the questions of the sort has been widely studied. (For example, maybe some study of conditions under which first-order theory has computable model or hasn't). Do you know anything about that?
Thanks in advance.
| https://mathoverflow.net/users/6307 | Are computable models sufficient? | There is no computable, countable nonstandard model of Peano arithmetic. This result is known as Tennenbaum's theorem after Stanley Tennenbaum. There is an online paper at [1]. So if you take any sentence $R$ in the language of PA that is not true in the standard model but is consistent with PA, it will be true in some countable model of PA but not in the (unique, up to isomorphism) computable model of PA.
But I'm not sure if that addresses your question, because that answer is about models of a theory. In your question, it seems like you're talking about all countable models in the language of particular formula, rather than all models of a particular theory.
To make the answer fit that question, we want to replace PA with a finitely axiomatized subtheory of PA for which the only computable model is the standard model. There is a heuristic principle that we should be able to find a subtheory like this by examining the proof of Tennenbaum's theorem, and reference [1] confirms that this does work.
Let $T$ be the sentence that is the conjunction of the axioms of this subtheory. Let $R$ be the independent sentence from the first part, and look at the sentence $T \rightarrow \lnot R$. This will be true in every computable model (because the only computable model that satisfies $T$ is the standard model, which satisfies $\lnot R$). It will be false in any countable nonstandard model of PA in which $R$ holds, because $T$ is a subtheory of PA.
1: Richard Kaye, "Tennenbaum's Theorem for Models of Arithmetic", <http://web.mat.bham.ac.uk/R.W.Kaye/papers/tennenbaum/tennenbaum>
(Note: I corrected the last paragraph based on a comment by Sergei Tropanets.)
| 10 | https://mathoverflow.net/users/5442 | 34248 | 22,163 |
https://mathoverflow.net/questions/34237 | 28 | I've been skulking around MathOverflow for about a month, reading questions and answers and comments, and I guess it's about time I asked a question myself, so here is one has interested me for a long time.
Suppose $M$ is a compact even dimensional smooth manifold with two symplectic forms $\omega\_0$ and $\omega\_1$ When are they "isotopic", i.e., when does there exist a 1-parameter family of diffeos $\phi\_t$ of $M$, starting from the identity, such that $\phi\_1^\*(\omega\_0) = \omega\_1$? Of course a necessary condition is that $\omega\_0$ and $\omega\_1$ should define the same 2-dimensional cohomology classes. Is this also sufficient? One can ask the same question for volume forms. I asked Juergen Moser about this twenty-five years ago, and he came back with an elegant proof of sufficiency for the volume element case a few months later in a well-known paper in TAMS. He remarks in that paper as follows:
"The statement concerning 2-forms was also suggested by R. Palais. Unfortunately
it seems very difficult to decide when two 2-forms which are closed,
belong to the same cohomology class and are nondegenerate can be deformed
homotopically into each other within the class of these differential forms."
So my question is, what if any progress has been made on this question. Poking around here and in Google hasn't turned up anything. Does anyone know if there are any progress?
| https://mathoverflow.net/users/7311 | When are two symplectic forms "isotopic"? | It is known, for a long time now, that there exist examples of symplectic forms in the same cohomology class which are non-isotopic. I do not remember if there exists such example in the dimension $4$, but in dimension $6$ there are different examples. Here is an example constructed by Dusa McDuff:
Let $X$ be a product $S^2\times S^2\times T^2$ ($T^2$ is a torus $(\mathbb R/2\pi\mathbb Z)^2$ with angle coordinates $(\psi,\gamma)$) and $\omega$ is a sum $\omega\_1\oplus\omega\_2\oplus\omega\_3$ of area forms on factors. We suppose that total areas of the first and of the second factor coincides. Consider the map $\varphi \colon X \to X$, where $\varphi (x,y,\psi,\gamma) = (x, T\_{x,\psi}(y),\psi,\gamma)$, where $T\_{x,\psi}$ is the rotation around $x$ on the angle $\psi$. Then forms $\omega$ and $\varphi^\*(\omega)$ define the same cohomology class and non-isotopic.
Moreover, forms $\omega$ and $\varphi^\*(\omega)$ could be joined by a path in a space of symplectic structures.
There is a survey containing the statement of this result and helpful references:
<http://www.math.sunysb.edu/~dusa/princerev98.pdf>
| 24 | https://mathoverflow.net/users/2823 | 34249 | 22,164 |
https://mathoverflow.net/questions/34246 | 4 | I would like to ask whether there is a combinatorial proof of the following recurrence relation for Catalan numbers:
$$
C\_{n+1}=\frac{4n+2}{n+2} C\_n.
$$
Thanks!~
| https://mathoverflow.net/users/6594 | Combinatorial proof of a recurrence for the Catalan numbers | What you are asking is reported as fourth proof in the wiki article for the formula of the Catalan numbers:
<http://en.wikipedia.org/wiki/Catalan_number#Fourth_proof>
| 10 | https://mathoverflow.net/users/6101 | 34259 | 22,171 |
https://mathoverflow.net/questions/34252 | 8 | If I have a symmetric positive definite matrix A and a diagonal matrix B, and I know the eigenvalues of both A and B (by iterative numerical computation in A's case and trivially for B), is there any way I can rapidly find the eigenvalues of the matrix M=A+B?
(I would be surprised if it helps, but I actually have the stronger condition that A is Laplacian.
Unfortunately, the entries of the matrix B are large, and so B cannot be considered a small "perturbation" to A. Finally, I only really have the extremal eigenvalues of A, though I am only hoping to find the extremal eigenvalues of A+B.)
| https://mathoverflow.net/users/8105 | Eigenvalues of A+B where A is symmetric positive definite and B is diagonal | I doubt it. At least it shouldn't be easier than the case where you have the sum of two arbitrary positive definite matrices A',B' with known eigenvalues and eigenvectors. Then you could use an orthogonal basis of eigenvectors for B' and set $A=PA'P^{-1}$ and $B=PB'P^{-1}$. B would be diagonal and AB would have the same eigenvalues as A'B'. Couldn't one even make B=I by choosing an orthonormal basis?
| 4 | https://mathoverflow.net/users/8008 | 34263 | 22,174 |
https://mathoverflow.net/questions/34269 | 9 | Dear MO Community,
I am trying to understand Mazur's 1976 notes "Rational points on Modular Curves" (which can be found in Springer Lecture Notes in Mathematics 601).
Let N be a prime number, and let $X\_0(N)$ be the usual modular curve over $\mathbb{Q}$. Say that a point on it is 'CM' if the elliptic curve corresponding to the point has Complex Multiplication. I am interested in the following sorts of questions (which are suggested by what happens over $\mathbb{Q}$):
1. Let $K$ be a number field that is not $\mathbb{Q}$. For which N can I construct CM points on $X\_0(N)(K)$ that are not defined over $\mathbb{Q}$? Are there finitely many such N?
2. For such N, are there finitely many such CM points?
For example, let $K = \mathbb{Q}(\sqrt{-6})$, and let $R = O\_K = \mathbb{Z}[\sqrt{-6}]$. Set $E = \mathbb{C}/R$. Choose $N$ such that $R/NR$ has nontrivial radical, or equivalently such that $R/NR = \mathbb{F}\_N[\epsilon]$ for $\epsilon$ some nontrivial element in the radical. Then (E,ker $\epsilon$) determines a point $a\_E(N)$ on $X\_0(N)$ which "is defined over a subfield of index 2 in the ray class field of $R \otimes \mathbb{Q}$, with conductor equal to the conductor of $R$ (which in my example is 1). Is $a\_E(N)$ defined over $K$? What is this index-2 subfield to which Mazur refers?
I think my second question is equivalent to asking: For such N, are there finitely many imaginary quadratic orders $R$ such that the aforementioned index 2 subfield is K, and such that R/NR has nontrivial radical?
| https://mathoverflow.net/users/5744 | CM rational points on modular curves | Dear Barinder,
This is the subject of the arithmetic theory of complex multiplication. Given any CM
elliptic curve $E$, and any kind of level structure on $E$, this theory determines the precise
field of definition of $E$ with its given level structure. You can find a discussion of this
in Silverman II ("Advanced topics") and in Shimura's book.
In your example: the $j$-invariant of $\mathbb C/R$ lives in the totally real subfield of
the Hilbert class field of $K$ (which is $\mathbb Q(\sqrt{2})$, and is also the ray class field of conductor 1 that you mention). The $\Gamma\_0(N)$ level structure ker $\epsilon$
is defined over this same field. So this is the field of definition of $a\_E(N)$.
| 6 | https://mathoverflow.net/users/2874 | 34272 | 22,180 |
https://mathoverflow.net/questions/34264 | 11 | Consider a real valued function $g$ on an open interval $(a,b)$ which is the derivative of a function continuous on $[a,b]$ at each point of $(a,b)$. The function $g$ has the intermediate value property, so a monotone $g$ will have to be continuous, a general $g$ cannot have simple discontinuities, etc. With such constraints how badly can a derivative behave in terms of continuity, can it get much worse than the derivative of say, $x^2 \sin(1/x)
$?
| https://mathoverflow.net/users/6627 | Examples of badly behaved derivatives | Talking about how bad may be the derivative of an everywhere derivable function on the interval $[a,b]$, the natural example that occurs to my mind is: a Pompeiu derivative, that is, a derivative that vanishes in a dense set (these weird functions, however, constitue a closed linear space of $C^0[a,b]$, while it's not even obvious that they are closed wrto addition!). For an account on the subject you may refer e.g. to the above quoted book by Andrew M. Bruckner [*Differentiation of real functions*](https://doi.org/10.1007/BFb0069821): it's also in the CRM series now (here is a preview:
[Google Books](http://books.google.it/books?id=fXfEG-F2zJUC)).
For a quick reference, you may also have a look to the [wiki article](http://en.wikipedia.org/wiki/Pompeiu_derivative) (it's me who wrote it ;-) ).
| 11 | https://mathoverflow.net/users/6101 | 34277 | 22,184 |
https://mathoverflow.net/questions/34266 | 4 | The transformation formula for a Siegel modular form can be interpreted as the statement that the modular form is a holomorphic section of a line bundle over the period domain (the quotient of the Siegel upper half-plane by a subgroup of finite index in the integral symplectic group). I've seen brief mentions of this fact in several references, but haven't been able to find one in which this geometric point of view is developped in detail. Does anyone know such a reference?
In particular, I would like to know if holomorphic line bundles over period domains have been classified.
Thanks in advance
| https://mathoverflow.net/users/2183 | Siegel modular forms as sections of line bundles over the period domain | Modulo torsion, the Picard group of the quotient of the Siegel upper half plane by a finite-index subgroup of $\text{Sp}\_{2g}(\mathbb{Z})$ is just $\mathbb{Z}$. This result should probably be attributed to Borel.
For a calculation of the torsion and explicit line bundles corresponding to the various pieces (plus references), see my paper "The Picard group of the moduli space of curves with level structures", available [here](http://www.nd.edu/~andyp/papers/). The results you want are discussed starting on page 3.
| 6 | https://mathoverflow.net/users/317 | 34278 | 22,185 |
https://mathoverflow.net/questions/34281 | 4 | Is it possible to build set theory on first-order logic *without equality*?
For example, how could one show that if $x\_0=x\_1$ then $\left\{x\_0\}=\{x\_1\right\}$, where $x\_0$ and $x\_1$ are two sets? And the other way around? In my opinion, it is impossible using only the following facts:
1. axiom of extensionality: $x\_0=x\_1\leftrightarrow\forall x\_2\left(x\_2\in x\_0\leftrightarrow x\_2\in x\_1\right)$
2. reflexivity of equality: $x=x$
3. symmetry of equality: $x\_0=x\_1\rightarrow x\_1=x\_0$
4. transitivity of equality: $x\_0=x\_1\land x\_1=x\_2\rightarrow x\_0=x\_2$
5. class comprehension: $x\in\left\{y:\psi\left(y\right)\right\}\leftrightarrow\text{set}\left(x\right)\land\psi\left(x\right)$
6. definition of singleton: $\left\{x\right\}=\left\{y:y=x\right\}$
7. definition of sethood: $\text{set}\left(x\_0\right)\leftrightarrow\exists x\_1\left(x\_0\in x\_1\right)$
I hope I didn't forget any fact. Also, it would be better if I point out I'm using Morse-Kelley set theory.
If I use first-order logic *with equality*, I could prove the first part of the theorem because for any term $t\_0$ and $t\_1$, if $t\_0=t\_1$ then $f\left(t\_0\right)=f\left(t\_1\right)$ ($f$ is a unary logical function here). But how can I prove the other way around?
Thanks.
| https://mathoverflow.net/users/3554 | First-order logic without equality and set theory | The converse direction fails because, in Morse-Kelley set theory plus your facts, $\{x\}$ is empty whenever $x$ is a proper class. So $\{x\_0\}=\{x\_1\}$ for any two proper classes.
The forward direction, on the other hand, seems to follow immediately from the facts you listed. If $x\_0=x\_1$ then, since facts 2, 3, and 4 make equality an equivalence relation, any $y$ will be equal to $x\_0$ iff it is equal to $x\_1$. So by facts 5 and 6, $\{x\_0\}$ and $\{x\_1\}$ have the same members. Then by fact 1, $\{x\_0\}=\{x\_1\}$.
| 5 | https://mathoverflow.net/users/6794 | 34282 | 22,186 |
https://mathoverflow.net/questions/34280 | 2 | Why is $H^p(X, \mathcal{E}xt^q(F, I))=0$ for $p>0$ and $I$ an injective sheaf and $F$ an arbitrary sheaf? I'm trying to check the hypothesis in the grothendieck spectral sequence applied to the functor $\mathcal{E}xt(F, -)$ and global sections functor, and so I'm trying to see why the sheaf $\mathcal{E}xt(F, I)$ is acylic for the the global sections functor, which I think translates to my question in the first sentence.
| https://mathoverflow.net/users/8111 | Spectral sequence for Ext | $\mathcal{E}xt^q(F, I)$ is 0 for $q > 0$. On the other hand it follows easily by considering extensions by 0 that $\mathcal{H}om(F, I)$ is flabby, hence acyclic.
| 6 | https://mathoverflow.net/users/4790 | 34286 | 22,188 |
https://mathoverflow.net/questions/33711 | 17 | When $p$ is a prime $\equiv9\bmod16$, the class number, $h$, of $\mathbb Q(p^{1/4})$ is known to be even. In
[[Charles J. Parry,
A genus theory for quartic fields.
*Crelle's Journal* **314** (1980), 40--71]](http://dx.doi.org/10.1515/crll.1980.314.40)
it is shown that $h/2$ is odd when 2 is not a fourth power in $\mathbb Z/p\mathbb Z$. Does this still hold
when 2 is a fourth power?
Some years ago I gave an (unpublished) proof that this is true
provided the elliptic curve $y^2=x^3-px$ has positive rank, and in particular that it is
true on the B. Sw.-D. hypothesis. It's known that the above curve has positive rank for
primes that $\equiv5$ or $7\bmod16$, but to my knowledge $p\equiv9\bmod16$ remains untouched. But perhaps
there's an elliptic-curve free approach to my question?
| https://mathoverflow.net/users/6214 | If p is a prime congruent to 9 mod 16, can 4 divide the class number of Q(p^(1/4))? | Let $p \equiv 1 \bmod 8$ be a prime number, let $K = {\mathbb Q}(\sqrt[4]{p})$, and let $F$ be the quartic subfield of the field of $p$-th roots of unity. An easy exercise involving Abhyankar's Lemma shows that $FK/K$ is an unramified quadratic extension, hence the class number of $K$ is always even.
The field $KF$ has the quartic subfield $L = {\mathbb Q}(\sqrt{u})$, where $u$ is the fundamental unit of $k = {\mathbb Q}(\sqrt{p})$. An routine application of the ambiguous class number formula to $L/k$ shows that $L$ has odd class number (there are two ramified primes, one infinite and the other one above $2$; clearly $-1$ is not a norm residue at the infinite prime).
Now I claim that if $p \equiv 9 \bmod 16$, the class number of $KF$ is odd. By class field theory, this implies that the $2$-class number of $K$ must be $2$. An application of the ambiguous class number formula to $KF/L$ shows that the $2$-part of the ambiguous class group has order
$$ h = \frac{2}{(E:H)}, $$
where $E$ is the unit group of $L$ and $H$ its subgroup of units that are norms from all completions of $KF$: in fact, only the two prime ideals above $p$ are ramified in $KF/L$. Thus it is sufficient to show that $E \ne H$. I will show that $\sqrt{u}$ is a quadratic nonresidue modulo the primes $\mathfrak p$ above $p$. But if $u = T + U \sqrt{p}$ (replace $u$ by $u^3$ in order to guarantee that $T$ and $U$ are integers), then $(\sqrt{u}/{\mathfrak p})\_2 = (u/\mathfrak p)\_4 = (T/p)\_4 = (T^2/p)\_8 = (-1/p)\_8 = -1$
because $p \equiv 9 \bmod 16$; here we have used the congruence $T^2 \equiv -1 \bmod p$.
The reason why the case $(2/p)\_4 = -1$ is easier is because in this case, the ideal above
$2$ ramified in $K$ generates a class of order $2$ in the $2$-class group, whereas this prime generates a class with odd order if $(2/p)\_4 = +1$, which means that there is no strongly ambiguous ideal class in this case.
**Edit.** Paul Monsky has kindly written up this argument, filled in all the details, and made it available [here](http://arxiv.org/abs/1009.3990). Thanks!
| 16 | https://mathoverflow.net/users/3503 | 34292 | 22,190 |
https://mathoverflow.net/questions/34197 | 22 | Van der Corput [1] proved that there are infinitely many arithmetic progressions of primes of length 3 (PAP-3). (Green & Tao [2] famously extended this theorem to length $k$.)
But taking this in a different direction, are *all* odd primes in a PAP-3? That is, for every prime $p>2$, is there a $k$ such that $p+k$ and $p+2k$ are prime?
Unsurprisingly, the first 100,000 primes have this property; the largest value of $k$ needed is just 1584 (see [4] and also [5] where this is greatly extended). Heuristically, you'd expect a given prime to be in
$$\int\_2^\infty\frac{a\ dx}{\log(x\log x)}=+\infty$$
different PAP-3s, and there are no small prime obstructions, so the conclusion seems reasonable. On the other hand, it seems to involve Goldbach-like (or better, Sophie Germaine-like) additive patterns in the primes: in essence, we're looking for prime $q$, $2q-n$ for a fixed odd $n$, so I don't imagine this has been resolved.
Basically, I'm just looking for more information on this problem. Surely it's been posed before, but does it have a common name and/or citation? Have any partial results been proved? Perhaps this is a consequence of a well-known conjecture?
[1] A. G. van der Corput (1939). "Über Summen von Primzahlen und Primzahlquadraten", *Mathematische Annalen* **116**, pp. 1-50.
[2] Ben Green and Terence Tao (2008). "The primes contain arbitrarily long arithmetic progressions", *Annals of Mathematics* **167**, pp. 481–547. <http://arxiv.org/abs/math/0404188>
[3] Amarnath Murthy, <http://oeis.org/A084704>
[4] Giovanni Teofilatto, <http://oeis.org/A120627>
[5] Charles R Greathouse IV, <https://oeis.org/A190423>
| https://mathoverflow.net/users/6043 | Are all primes in a PAP-3? | This question is extremely close to this one
[Covering the primes by 3-term APs ?](https://mathoverflow.net/questions/2214/covering-the-primes-by-3-term-aps)
though not exactly the same.
For much the same reasons as described in the answer given there, the answer to your question is almost certainly yes, but a proof is beyond current technology, exactly as you suggest. I'm not aware that the problem has a specific name.
To show that 3 belongs to a 3PAP is of course trivial: it belongs to 3,5,7 or 3,7,11. Showing that there are infinitely many such 3PAPs is, as you point out, a problem of the same level as difficulty as the Sophie Germain primes conjecture or the twin primes conjecture.
For a general p, I find it extremely unlikely that you could show that there is a k > 0 such that p + k, p + 2k are both prime without showing that there are infinitely many. Proving this for even one value of p would be a huge advance.
I think you could show that almost all primes p do have this property using the Hardy-Littlewood circle method.
| 26 | https://mathoverflow.net/users/5575 | 34298 | 22,195 |
https://mathoverflow.net/questions/34291 | 3 | A group $G$ is residually finite if, for any two elements $g$ and $g^\prime$ in $G$, there is a finite group $G^\prime$ and a (group) homomorphism $f: G \rightarrow G^\prime$ such that $f(g)$ doesn't equal $f(g^\prime)$. The definition for a semigroup is analagous: just make $G$ and $G^\prime$ semigroups and make $f$ a semigroup homomorphism. I was wondering if there is a good reference which will answer questions like the following:
Is there a group $G$ which is not residually finite as a group but is residually finite as a semigroup (in other words there is a finite semigroup $S$ and a semigroup homomorphism from $G$ to $S$ which separates elements, but there is no finite group $G^\prime$ and a group homomorphism from $G$ to $G^\prime$ which separates elements)?
If $S$ is a residually finite semigroup and $G$ is a subgroup of $S$, then $G$ is residually finite as a semigroup. Is $G$ residually finite as a group?
Thanks!
| https://mathoverflow.net/users/3804 | Residual finiteness of groups versus residual finiteness of semigroups | I first posted this as a comment, but I guess that this is an answer.
If a group is residually finite as a semigroup, it is residually finite as a group. This is an immediate consequence of the following easy fact: if G is a group and φ:G→S is a semigroup homomorphism, then the image φ(G) is a group and φ is a group homomorphism from G to φ(G). I guess that the latter fact is in any textbook on semigroups, though I do not have one at hand.
| 7 | https://mathoverflow.net/users/7982 | 34299 | 22,196 |
https://mathoverflow.net/questions/34257 | 3 | We say that an object $X$ of a category $C$ is $\kappa$-compact (also $\kappa$-presentable and $\kappa$-accessible) for a cardinal $\kappa$ if $h^X(\cdot):=Hom(X,\cdot)$ commutes with all $\kappa$-filtered colimits.
In Makkai-Pare, a different but equivalent definition is given, that $X$ is $\kappa$-compact if for any $\kappa$-filtered functor $F:I\to C$ (i.e. the category $I$ is $\kappa$-filtered), any morphism $X\to \varinjlim F$ factors as $X\to F(i) \to \varinjlim F$, and any two factorizations $X\to F(i)\to \varinjlim F$ and $X\to F(i')\to \varinjlim F$ admit a majorant factorization $X\to F(i'')\to \varinjlim F$, that is, a factorization $X\to F(i'')\to \varinjlim F$ along with a commutative diagram
$$\begin{matrix}
X&\to&F(i)\\
\downarrow&&\downarrow\\
F(i')&\to&F(i'')\\
\end{matrix}$$
such that $$F(i)\to \varinjlim F=F(i)\to F(i'')\to \varinjlim F$$ and $$F(i')\to \varinjlim F=F(i')\to F(i'')\to \varinjlim F$$ are the natural maps.
Let's try to show that they are equivalent:
*Proof*.
'$\Rightarrow$' The existence of a factorization follows from the fact that the factorization exists in the category of sets, since giving a map $X\to \varinjlim F$ is the same as giving a map $$\*\to h^X\varinjlim\_i F(i)\cong\varinjlim\_i h^XF(i),$$ and since we can give the colimit as a quotient of the disjoint union, the point $\*\to \varinjlim\_i h^XF(i)$ maps through some $h^XF(i)$, which gives us the necessary commutative triangle upon inspection.
Question:
In the definition of Makkai-Pare, it seems (to me) that we should need the existence of a majorant factorization for any family of factorizations indexed by a $\kappa$-small set. Why is this not the case?
Also, how do we prove the existence of the majorant factorization? I would guess that it follows from some property of filtered colimits in the category of sets, but I'm not familiar with what that is.
Edit: I would look in Makkai-Pare, but the only version I can find is on google books, and it cuts off literally right after the statement of the second definition. If anyone has it and is feeling generous, I'd very much appreciate any "assistance" you can give in finding it ;) (I'm currently not at school, so the library is not an option).
| https://mathoverflow.net/users/1353 | Equivalence of the two definitions of k-compactness/k-presentability | The definitions are equivalent as stands; no extra conditions (eg majorants for infinite sets of factorisations) are needed. This is essentially because colimits in $\mathbf{Sets}$ are computed finitarily.
One way to present the colimit $\underset{i}{\varinjlim} Hom(X,F(i))$ (see eg Mac Lane *CWM*) is as
1. the coproduct of all the sets $Hom(X,F(i))$, quotiented by
2. the relation $\sim$ defined by: for $f \colon X \to F(i)$ and $g \colon X \to F(i')$, take $f \sim g$ if there is some $i''$ and a commutative square as in your original post.\*
Looking at it this way, the two conditions in Makkai and Paré’s definition say that the canonical map $\underset{i}{\varinjlim} Hom(X,F(i)) \to Hom(X,\underset{i}{\varinjlim} F(i))$ is
1. surjective;
2. injective;
so together they say exactly that it’s an iso, which is what the usual definition says.
\* For a general colimit, we’d need to use “the equivalence relation generated by $\sim$” (which is still something finitary), but if the colimit is filtered, so *a fortiori* if it’s $\kappa$-filtered, then $\sim$ is already an equivalence relation.
| 3 | https://mathoverflow.net/users/2273 | 34305 | 22,200 |
https://mathoverflow.net/questions/34318 | 0 | Are there any examples other than using dimension for vector spaces where the easiest way to show that two objects are isomorphic is by using a classification theorem and showing that they must both be in the same class? (homeomorphisms count too)
| https://mathoverflow.net/users/nan | Isomorphism by classification | Genus for surfaces would be a simple example.
Connectedness for compact $1$-dimensional manifolds would be another!
| 3 | https://mathoverflow.net/users/1465 | 34320 | 22,211 |
https://mathoverflow.net/questions/34332 | 32 | As defined in many modern algebra books, a homomorphism of unital rings must preserve the unit elements: $f(1\_R)=1\_S$. But there has been a minority who do not require this, one prominent example being Herstein in *Topics in Algebra*.
What are some of the most striking consequences of not requiring ring homomorphisms to be unital? For example, what aspects of algebraic geometry would need to be reworked if we no longer required it? What interesting theorems or techniques arise in the not-necessarily-unital theory which do not apply (or are degenerate) for unital homomorphisms?
| https://mathoverflow.net/users/1916 | Consequences of not requiring ring homomorphisms to be unital? | Let's suppose that our rings are commutative (which is the case that is immediately relevant to algebraic geometry).
If $\phi:A \to B$ is a (possibly non-unital) homomorphism, then $e := \phi(1\_A)$ is an
idempotent in $B$, and so we get a decomposition $B = eB \times (1-e)B,$
and the map $\phi$ factors as $A \to eB \to B,$ where the first map is unital, and the second map is simply the inclusion, which is the inclusion of a direct factor.
On Specs, we thus get the composite of the map Spec $eB \to $ Spec $A$,
composed with the "map" (this is not necessarily an honest map of schemes, because it corresponds to the possibly non-unital map $e B \to B$)
Spec $B \to$ Spec $eB$, which just vaporizes the open and closed subset Spec $(1-e)B$ of Spec $B$,
and is the identity on the open and closed subset Spec $eB$.
So the upshot is that nothing much new happens in algebraic geometry, except that we allow maps which are only defined on some open and closed subset of a given scheme. Of course,
this is a big *except*, because these are not honest maps at all (they are simply not defined on some part of their "domain"). There doesn't seem to be any reason to add them into the mix, which is surely one reason why this generalized notion of homomorphism is not used much in practice.
P.S. One could argue another way, beginning with geometry, and passing to algebra by remembering that rings are rings of functions. If we have a map $\phi:X \to Y$ of spaces
(of some type, e.g. affine schemes, or anything else), then surely the constant function 1 on $Y$ will pull-back to the constant function 1 on $X$. Thus the induced homomorphism on rings of functions will have to be unital, and so one simply has no cause to consider non-unital homomorphisms in the geometric setting.
P.P.S. The argument in the first paragraph shows that allowing non-unital homomorphisms in the category of commutative rings is the same as adding, in addition to unital homomorphisms, homomorphisms of the form $B\_1 \to B\_1\times B\_2,$ given by $b\_1\mapsto (b\_1,0),$ for any pair of commutative rings $B\_1$ and $B\_2$. So it's not really a very exciting change from the purely algebraic point of view either.
| 36 | https://mathoverflow.net/users/2874 | 34335 | 22,217 |
https://mathoverflow.net/questions/34342 | 1 | I have the system of multi-variable polynomial (quadratic) equations with real coefficients. The number of equations is given scales as $K$ and the number of unknowns goes as $K^2$. So for for large $K$, this is an underdetermined system.
Can I conclude that I can always find $K$ large enough so that this system has at least one real solution or where should I look for counterexamples?
| https://mathoverflow.net/users/8119 | Real solutions to underdetermined system of polynomial equations | However large $n$ may be, the equation
$x\_1^2 + \ldots + x\_n^2 = -1$ has no real solution.
(Similarly, the homogeneous equation $x\_1^2 + \ldots + x\_n^2 = 0$ has no nontrivial solution.)
If you want to go there, the theorem which gives you a necessary and sufficient condition for a system of real polynomial equations to have a real solution is the **Real Nullstellensatz**. See for instance Section 3.9` of
[http://alpha.math.uga.edu/~pete/modeltheory2010Chapter3.pdf](http://alpha.math.uga.edu/%7Epete/modeltheory2010Chapter3.pdf)
(*Very* roughly, this theorem says that the above is essentially the only way that a system of $n$ real polynomial equations in more than $n$ variables can fail to have a real solution.)
| 5 | https://mathoverflow.net/users/1149 | 34344 | 22,222 |
https://mathoverflow.net/questions/34339 | 6 | Arinkin has a theorem which says that an abelian variety can be reconstructed from its derived category of coherent D-modules.
D.Orlov conjectured that this theorem is true for any variety.
My question is:
Is this conjecture proved or disproved? I wonder know the related work, examples and any other related observations, comments.
Thanks
| https://mathoverflow.net/users/1851 | Reconstruction from category of D-modules on variety | As far as I understand from your statement of the conjecture, the conjecture is false, although there are similar statements that are true. If I understand correctly, a weaker question (more likely to have the answer yes) would be "can one recover a variety from its category of D-modules."
For a non-example of the weaker question, if $X = Spec(\mathbb{C}[x])$ and $Y = Spec(\mathbb{C}[x^2,x^3])$, then D(X) and D(Y) are Morita equivalent. If X is a smooth curve and Y is another curve, then D(X) is Morita equivalent to D(Y) iff X and Y are homeomorphic (in the example above, the normalization map gives a homeomorphism $X \to Y$). If $X = Spec(\mathbb{C}[x])$, then the natural numbers parameterize isomorphism classes of curves Y with D(X) Morita equivalent to D(Y).
A similar-sounding statement which is true is "If X and Y are smooth curves, they are isomorphic iff D(X) and D(Y) are isomorphic (as algebras)." A paper with these and many more facts can be found here <http://arxiv.org/abs/math/0304320>
| 6 | https://mathoverflow.net/users/2669 | 34352 | 22,228 |
https://mathoverflow.net/questions/34350 | -3 | Let $R$ be a commutative ring. If $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$ is an exact sequence of modules, we have that $\operatorname{Supp}M=\operatorname{Supp}M'\cup \operatorname{Supp}M''$. Furthermore can we obtain that $\dim M=\dim M'+\dim M''$? In fact, I want to know that given a module $M$ with $\dim M=n$, is it possible to construct an exact sequence as above with $\dim M'=0$ and $\dim M''=n-1$?
| https://mathoverflow.net/users/5775 | An elementary question about the Krull dimension of modules | The dimension of $M$ is the maximum of the dimensions of $M'$ and $M''$, so what you are asking for cannot happen (at least not in the noetherian case). To Dylan: the dimension of a module is the dimension of its support.
| 1 | https://mathoverflow.net/users/4790 | 34357 | 22,232 |
https://mathoverflow.net/questions/34316 | 18 | OK, the title is opinionated and contentious, but I have a definite
question. I know that the title refers to the Bourbaki volume
*Groupes et Algèbres de Lie* (Chapters 4-6), published in 1968, but
>
> who said that it is the only great book that Bourbaki ever wrote?
>
>
>
The only reference I can find is the 2009 Prize Booklet for the AMS-MAA
Joint Meetings, where no source is given, but I'm sure I've seen the
claim somewhere else.
**Edit.** I have rolled back the title of this question to almost its original
form, because putting the title in quotes misled some people into thinking
I sought a source for the exact phrase "the only great book that Bourbaki
ever wrote." Rather, I wanted a source (not necessarily unique) for the *idea*
that Chapters 4-6 of *Groupes et Algèbres de Lie* is Bourbaki's one great
book. Gerald's answer and Jim's comment together are exactly what I wanted.
| https://mathoverflow.net/users/1587 | The only great book that Bourbaki ever wrote? | Google found this:
Notices of the AMS, September 1998, p. 979:
Bill Casselman's review of POLYHEDRA by Cromwell,
we find the phrase "the one great book by Bourbaki"
| 24 | https://mathoverflow.net/users/454 | 34362 | 22,236 |
https://mathoverflow.net/questions/34374 | 5 | Just like the title. I want a simple proof of the statment in the title.
$\mathbb{Q}\_p$ is the p-adic field.
I wonder which module (or vector space) will be chosen as the space for the representation. Is this statement true for arbitarily module/vector space?
Thanks!
| https://mathoverflow.net/users/8122 | Any finite dimensional admissible(smooth) irreducible representation of GL(2,Q_p) is 1-dim | What does "admissible" mean for you? Does it imply smoothness (stabilisers are open)? If not then I think the statement might be false (choose some hopelessly discontinuous injection from $\mathbf{Q}\_p$ into $\mathbf{C}$ and then consider the induced "natural" representation of $GL(2,\mathbf{Q}\_p)$ on $\mathbf{C}^2$; that's definitely irreducible, and the space of vectors fixed by some compact open will definitely be finite-dimensional).
But assuming smoothness too, it is true that any finite-dimensional smooth irreducible representation of $GL(2,\mathbf{Q}\_p)$ is 1-dimensional. The proof is: if $V$ is such
a thing, then choose a basis for $V$ and for each basis vector choose a compact
open subgroup stabilising it. The intersection of these guys is still compact and
open, and fixes everything. So the kernel of the representation contains a compact
open subgroup. But this is a bit worrying because the kernel is normal. Now use
the fact that the normal subgroup generated by matrices $(1,e;0,1)$ and $(1,0;e,1)$
for $e$ small is still the whole of $SL(2,\mathbf{Q}\_p)$ to deduce that $SL(2,\mathbf{Q}\_p)$
is in the kernel, and now the action has to factor through $GL(2,\mathbf{Q}\_p)/SL(2,\mathbf{Q}\_p)=\mathbf{Q}\_p^\times$. But now Schur's Lemma, which works for smooth irreducible
representations, says $V$ is 1-dimensional.
| 14 | https://mathoverflow.net/users/1384 | 34381 | 22,247 |
https://mathoverflow.net/questions/34384 | 1 | It is well-known theorem that every locally compact, homogeneous, metric space is complete.
Does anybody know example of complete, homogeneous, metric space which is not locally compact?
| https://mathoverflow.net/users/7421 | Сomplete homogeneous space which is not locally compact | A Banach space is homogeneous since the metric is arising from a norm. An infinite dimensional Banach space has the property that its unit ball is not compact; therefore the space is not locally compact.
For a concrete example, take the space of continuous real functions on an interval with the supremum norm.
| 2 | https://mathoverflow.net/users/2938 | 34386 | 22,251 |
https://mathoverflow.net/questions/34390 | 12 | It happens occasionally that one can prove that a given set is not empty by proving that it is actually large. The word "large" here may refer to different properties.
For example, one can prove that a certain set is not empty by proving that its cardinality is big, as in the proof that there exist transcendental numbers : The set of algebraic numbers is countable, but the set of real numbers is uncountable, so there is uncountably many transcendental numbers.
One could also prove that a certain set is not empty by proving, for example, that it has positive measure, that it is dense, etc.
What are some good examples of such proofs?
| https://mathoverflow.net/users/1162 | On proving that a certain set is not empty by proving that it is actually large | Many existence proofs which exploit the idea of Baire category.
For instance, existence of a metrically transitive automorphism of the closed unit square was first obtained by the category method (see ["Measure-preserving homeomorphisms and metrical transitivity"](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=oxtoby&s5=metrical%2520transitivity&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) by Oxtoby and Ulam)
. Another classical example is due to Banach who proved that every function from a residual subset of $C[0,1]$ is nowhere differentiable.
A nice and elementary [book](http://books.google.co.uk/books?id=wUDjoT5xIFAC&pg=PR7&lpg=PR7&dq=measure+and+category&source=bl&ots=4EMy1X1RUN&sig=cd5KeoXCfve3gNwvk2wyrOmTLmU&hl=en&ei=VS5YTIfsH9GT4gaesIj_Bg&sa=X&oi=book_result&ct=result&resnum=5&ved=0CDIQ6AEwBA#v=onepage&q&f=false) by Oxtoby discusses these and many other applications of the category method.
| 16 | https://mathoverflow.net/users/5371 | 34396 | 22,257 |
https://mathoverflow.net/questions/34394 | 7 | In his book *"The geometry of geodesics"* H. Busemann defines the notion of a G-space to be a space which satisfies the following axioms:
1. The space is metric
2. The space is finitely compact, i.e., a bounded infinite set has at least one accumulation point
3. [*metric convexity*] For every $x\neq z$ there exists a third point $y$ different from $x$ and $z$ such that $d(x,y)+d(y,z)=d(x,z)$
4. [*local prolongation*] To every point $p$ there corresponds $\rho\_p>0$ such that for every two point $x,y\in S(p,\rho\_p)$ there exists a point $z$ such that $d(x,y)+d(y,z)=d(x,z)$
5. [*uniqueness of prolongation*] If $d(x,y)+d(y,z\_1)=d(x,z\_1)$ and $d(x,y)+d(y,z\_2)=d(x,z\_2)$ and $d(y,z\_1)=d(y,z\_2)$ then $z\_1=z\_2$.
Busemann conjectured that every $G$-space is a topological manifold. My question is does every topological/smooth/Riemannian manifold is also a $G$-space?
As for connected complete Riemannian manifold, I figured out that **1** holds since by the metric. **3** holds since every two points can be joined by a minimal geodesic, and then we can pick $y$ to be a point on it. **4** holds since it is a manifold and locally it is homeomorphic to some Euclidean space. Unfortunately, even in this case, I couldn't figure out **5** and **2**.
| https://mathoverflow.net/users/8047 | G-spaces and manifolds | On a \**complete*\*smooth Riemannian manifold,
1. Any bounded (with respect to the distance function induced by the Riemannian metric) closed set in a manifold is compact.
2. This is telling you that there is a minimal geodesic joining $x$ to $y$ that, when extended, is also a minimal geodesic joining $x$ to $z\_1$. And there is another minimal geodesic joining $x$ to $y$ that when extended is a minimal geodesic joining $x$ to $z\_2$. But if there are two distinct geodesics joining $x$ to $y$, neither is minimal beyond $y$. So the two geodesics have to be the same and therefore $z\_1 = z\_2$.
CORRECTION: "complete" added to assumption above.
For a smooth manifold, you need to construct a distance function to get a G-space. One way to do this is to construct a complete Riemannian metric. I'm not certain that this can be done, but offhand if you take a locally finite covering by open sets diffeomorphic to the Euclidean ball, use the standard Euclidean metric on each ball (where each ball has radius $1$), and use a partition of unity subordinate to this cover to glue together these metrics, it seems to me that the resulting metric is complete.
For a topological manifold, I don't know.
| 11 | https://mathoverflow.net/users/613 | 34398 | 22,259 |
https://mathoverflow.net/questions/34402 | 7 | Let X be the Cantor set, which we view as the space $2^\mathbb{N}$ (the set of all infinite binary sequences), equipped with the product topology. We can construct a Borel probability measure $\mu$ on this space by defining $\mu(C\_{a\_i})=1/2$, where the $C\_{a\_i}=\{x\in X | x\_i=a\_i\}$ are the open subbase cylinders of the product topology, and extending to a $\sigma$-algebra in the standard fashion.
Now, consider the Hilbert space $L^2(X,\mu)$. We can obtain orthonormal bases for it using the measure-space isomorphism between $(X,\mu)$ and $[0,1]$ (with Lesbesgue measure) via the binary decimal representations of real numbers. However, the ordinary bases (e.g., the trigonometric basis) on $L^2([0,1])$ are quite ugly when viewed on the Cantor set.
>
> Is there an orthonormal basis for $L^2(X,\mu)$ with nice properties (continuity? simply expressible functions?) relative
> to the structure of the Cantor set?
>
>
>
| https://mathoverflow.net/users/3831 | Nice orthonormal basis for L^2(Cantor set) | Since the Cantor set with your measure is also the compact group $(\mathbb{Z}/2)^\mathbb{N}$ with Haar measure, a natural orthonormal basis is the (continuous) characters $\alpha:X\to S^1$, namely the finite products of coordinates $c\_n(x)$, $n\in\mathbb{N}$ if you view $\mathbb{Z}/2$ as `{-1,1}`. These form the discrete group $(\mathbb{Z}/2)^{(\mathbb{N})}$.
If you view $X$ as the Cantor middle third, $c\_n(x)$ corresponds to $a\_n(x)-1$, the $n$-th base $3$ digit of $x$ minus 1 (all digits are 0 or 2 by definition). These correspond to Walsh functions mentioned by Willie Wong when you use the measure isomorphism $X\to I$, which maps $x$ to ${1\over2} \sum\_n a\_n(x) 2^{-n}$.
Another possible model is $\mathbb{Z}\_2$, the compact group of [2-adic integers](http://en.wikipedia.org/wiki/2-adic_integers), and the characters are then identified to power-of-two roots of unity, forming a group isomorphic to $\mathbb{Z}[1/2]/\mathbb{Z}$. ~~This seems to lead to the same basis, although indexed differently.~~ EDIT: as remarked by Greg Kuperberg in a comment, this can't be true.
| 16 | https://mathoverflow.net/users/6451 | 34409 | 22,266 |
https://mathoverflow.net/questions/34358 | 4 | My knowlege in group theory is very limited and this question is out of pure curiousity:
Given a random finite group, how many conjugacy classes will it probably have?
I can try to make this question more precise:
Define $a\_n$ to be the average number of conjugacy classes in finite groups of order less or equal $n$.
What is the asymptotic behaviour of the sequence $a\_n$?
| https://mathoverflow.net/users/2837 | Number of conjugacy classes in generic finite group? | In elementary terms, you have to analyze the following class equation
$ n = 1 + h\_2 + ... + h\_r $
where
* n is the order of the group G
* $h\_k$ denotes the number of elements in the k-th conjugacy class, and $ n = c\_k.h\_k$.
Dividing by n, you get
$1 = \frac{1}{n} + \frac{1}{c\_2} + ... + \frac{1}{c\_r} $ which has a finite number of solutions.
Christine Ayoub in her paper *On the number of conjugate classes in a group* (Proc. Internat. Conf Theory of Groups Canberra 1967) has worked out this analysis for p-groups and there are probably more recent papers on this aspect, which Scott and others allude to in the comments. See for example
1. *MR2557143 Keller, Thomas Michael .
Lower bounds for the number of
conjugacy classes of finite groups.
Math. Proc. Cambridge Philos. Soc.
147 (2009), no. 3, 567--577.*
Another way of looking at your question is to see that the number of conjugacy classes is the same as the number of irreducible representations. The [character table](http://en.wikipedia.org/wiki/Character_table) is always square. Therefore, one could ask "*what are the number of irreducible characters Irr(G) in a finite group of order n?*". The number of linear characters are [G:G'] where G'=[commutator subgroup](http://en.wikipedia.org/wiki/Commutator_subgroup) but the nonlinear ones are tougher and there are papers establishing various bounds for these.
1. *MR2526321 (2010d:20010) Aziziheris,
Kamal ; Lewis, Mark L. Counting
the number of nonlinear irreducible
characters of a finite group. Comm.
Algebra 37 (2009), no. 5,
1572--1578.*
2. *MR0689258 (84d:20014) Wada,
Tomoyuki . On the number of
irreducible characters in a finite
group. Hokkaido Math. J. 12
(1983), no. 1, 74--82.*
3. *MR0798751 (87a:20006) Wada,
Tomoyuki . On the number of
irreducible characters in a finite
group. II. Hokkaido Math. J. 14
(1985), no. 2, 149--154.*
| 4 | https://mathoverflow.net/users/5372 | 34413 | 22,269 |
https://mathoverflow.net/questions/34364 | 8 | I have a question about graph width measures of undirected simple graphs. It is well-known that cographs (graphs which can be built by the operations of disjoint union and complementation, starting from isolated vertices) have cliquewidth at most 2. (Courcelle et al, Upper bounds to the clique width of graphs). Now consider some fixed non-negative integer k, and consider the class of graphs $\mathcal{G} \_k$ of graphs such that for every $G = (V,E) \in \mathcal{G} \_k$ there is a set $S$ of at most k vertices such that $G[V - S]$ is a cograph. Since the graph class $\mathcal{G} \_k$ can also be seen as the class of graphs that can be built out of cographs by adding at most $k$ vertices, this class has also been called cographs + $kv$.
My question is: what is a tight bound on the cliquewidth of graphs in $\mathcal{G}\_k$?
It is known that if a graph $G$ is obtained from $H$ by deleting $k$ vertices then $cw(H) \leq 2^k (cw(G) + 1)$. This shows that if a cograph $G$ can be obtained from a graph $H$ by deleting $k$ vertices, then $cw(H) \leq 2^k (3 + 1)$, and hence the cliquewidth of a graph in $\mathcal{G}\_k$ is at most $2^k 4$. I am unsure whether this exponential dependency on $k$ is necessary. In this context I would also be interested in the maximum decrease in the cliquewidth by deleting a vertex; i.e. if we delete a single vertex from a graph, how much can the cliquewidth decrease?
| https://mathoverflow.net/users/5200 | Cliquewidth of Cographs + kv | I think the exponential bound is necessary. Here is why.
Consider the disconnected graph $G$ on $n$ vertices, labelled $1$ to $n$. Denote the set of vertices that new vertex $i$ is connected to by $f(i)$, a subset of $\lbrace 1,2,\ldots,n \rbrace$.
The question then becomes: is it possible to find such a function $f$ so that the set $\lbrace f(i) \mid 1 \le i \le k \rbrace$ of subsets of $V(G)$ generates a set of subsets of size $2^k$?
With "generates" I mean the closure under the operation of taking pairwise subsets.
The idea here is that to distinguish between any subsets one needs to label them differently. However, due to the way the cliquewidth operations are restricted, this can only happen if their intersection is constructed first, and then the remaining vertices.
As far as I can tell, this scenario is possible, but clearly the above is quite far from a rigorous construction of an example.
| 4 | https://mathoverflow.net/users/7252 | 34425 | 22,276 |
https://mathoverflow.net/questions/34424 | 28 | [This Wikipedia article](http://en.wikipedia.org/wiki/List_of_finite_simple_groups) states that the isomorphism type of a finite simple group is determined by its order, except that:
* L4(2) and L3(4) both have order 20160
* O2n+1(q) and S2n(q) have the same order for q odd, n > 2
I think this means that for each integer g, there are 0, 1 or 2 simple groups of order g.
Do we need the full strength of the [Classification of Finite Simple Groups](http://en.wikipedia.org/wiki/Classification_of_finite_simple_groups) to prove this, or is there a simpler way of proving it?
*(Originally asked at [math.stackexchange.com](https://math.stackexchange.com/questions/1423/number-of-finite-simple-groups-of-given-order-is-at-most-2-is-a-classification)).*
| https://mathoverflow.net/users/4947 | Number of finite simple groups of given order is at most 2 - is a classification-free proof possible? | It is usually extraordinarily difficult to prove uniqueness of a simple group given its order, or even given its order and complete character table. In particular one of the last and hardest steps in the classification of finite simple groups was proving uniqueness of the Ree groups of type $^2G\_2$ of order $q^3(q^3+1)(q-1)$, (for $q$ of the form $3^{2n+1}$) which was finally solved in a series of notoriously difficult papers by Thompson and Bombieri. Although they were trying to prove the group was unique, proving that there were at most 2 would have been no easier.
Another example is given in the paper by Higman in the book "finite simple groups" where he tries to characterize Janko's first group given not just its order 175560, but its entire character table. Even this takes several pages of complicated arguments.
In other words, there is no easy way to bound the number of simple groups of given order, unless a lot of very smart people have overlooked something easy.
| 53 | https://mathoverflow.net/users/51 | 34432 | 22,281 |
https://mathoverflow.net/questions/34441 | 2 | I don't know much about the theory of Hilbert spaces but a research project has me working with them a little bit. In particular requiring an operator to be Hilbert-Schmidt is a recurring condition.
According to wikipedia one nice thing about H-S operators is that on a separable Hilbert space $H$ the set of H-S endomorphisms forms a Hilbert space that is naturally isomorphic to $H\otimes H^\*$.
So I'm wondering what else is nice about H-S operators. And what else works with H-S operators that wouldn't work for another class of operators. Fredholm operators also come up a lot. It $T$ is H-S and $S$ is Fredholm what can be said about the composition?
| https://mathoverflow.net/users/7 | Hilbert Schmidt operators | I think the perfect reference for you is Lars Hörmander's *The Analysis of Linear Partial Differential Operators*, vol III (the chapter on elliptic operators). There you'll find in perfect Hörmander style all you need about Fredholm, Hilbert-Schmidt, trace class operators. (As to the composition, H-S is a two-sided ideal; if S is any bounded operator TS and ST are H-S).
| 3 | https://mathoverflow.net/users/6101 | 34444 | 22,287 |
https://mathoverflow.net/questions/34363 | 6 | The complete elliptic integral of the first kind
$K(m)=\int\_0^{\pi/2}\frac{\mathrm{d}t}{\sqrt{1-m\sin^2t}}$
is easily computed via the arithmetic-geometric mean iteration; to wit,
$K(m)=\frac{\pi}{2M(1,\sqrt{1-m})}$
where $M(a,b)$ is the arithmetic-geometric mean of $a$ and $b$. With a little more trickery, the iteration can be hijacked to compute the complete elliptic integral of the second kind $E(m)$ as well.
In a number of applications, it happens that one needs both the values of $K(m)$ and its complement $K(1-m)$ (and sometimes similarly for $E(m)$ and $E(1-m)$).
My question is, apart from having to do an AGM iteration for each of $K(m)$ and $K(1-m)$, is there an algorithm (maybe a modification of the basic AGM iteration) that simultaneously generates both $K(m)$ and its complement? I would also be interested in seeing also an extension of this algorithm, if one exists, for computing $E(m)$ as well (after which $E(1-m)$ is easily computed via Legendre's relation).
| https://mathoverflow.net/users/7934 | Simultaneously computing a complete elliptic integral and its complement | There are two possible ways to attack this problem
1. Both K and K' can be expressed in
terms of the Theta function as
described here
<http://mathworld.wolfram.com/EllipticModulus.html>.
If you compute $\Theta\_3$, you can get
both at the same time.
2. The other way is to observe that
both K and K' are expressible in terms of the hypergeometric function $\_2F\_1(\frac{1}{2}, \frac{1}{2} ; 1; m)$. They are solutions of the
same self-adjoint Gauss
hypergeometric differential
equation (since the equation is invariant under the transformation (m $\rightarrow$ 1-m))
$(k^3 - k)\frac{d^2y}{dk^2} + (3k^2
-1)\frac{dy}{dk} + ky = 0 $
By virtue of this fact, both K and K' are connected. You will find the following series expansion for K'(k)
derived in [Borwein's book Pi and
AGM](http://rads.stackoverflow.com/amzn/click/047131515X) Section 1.3
$ K'(k) = \frac{2}{\pi} log \frac
{4}{k} K(k) - 2 [(\frac{1}{2})^2(\frac{1}{1.2}k^2 + (\frac{1.3}{2.4})^2(\frac{1}{1.2} + \frac{1}{3.4})k^4 + (\frac{1.3.5}{2.4.6})^2(\frac{1}{1.2} + \frac{1}{3.4} + \frac{1}{5.6})k^6 $.....(infinite series) + ]
You may also find Chapter 5 of [Armitage and Eberlein's book on Elliptic Functions](http://rads.stackoverflow.com/amzn/click/0521780780) useful.
EDIT1: I put in the complete series expansion for K'(k).
---
Regarding the computation of E(k), E(k) and K(k) are connected by the differential equation $ \frac{dK}{dk} = \frac{E - (1-k^2)K}{k(1-k^2)} $ which is how the [Legendre relation](http://mathworld.wolfram.com/LegendreRelation.html) you mention above comes about.
Again Borwein has the solution for this problem(buy the book!). Exercise 3 in Sec 1.4 has the formula based on the quartic AGM iteration
$ E(k) = K(k)[1 - \sum\_{n=0}^{\infty} 4^n [\alpha\_n^4 - (\frac{\alpha\_n^2+\beta\_n^2}{2})^2 ] $ where
* $\alpha\_n = (a\_{2n})^{\frac{1}{2}}
and \beta\_n = (b\_{2n})^{\frac{1}{2}}$
and $ a\_n, b\_n$ and $c\_n $ satisfy the AGM relation
| 4 | https://mathoverflow.net/users/5372 | 34457 | 22,296 |
https://mathoverflow.net/questions/34446 | 2 | Given an undecidable collection of first-order sentences, is there necessarily a complete undecidable theory containing it? A direct attempt to prove it seems to require some control over the completions which need not exist, but I don't see a counterexample.
On a more concrete side, Macyntire proved that the theory of all pairs of real closed fields is undecidable. I am interested to know if there is a particular pair of real closed fields with undecidable theory.
| https://mathoverflow.net/users/25726 | Undecidable completion of undecidable theory, and pairs of RCF | My recollection is that Macintyre proved there are $2^{\aleph\_0}$ complete theories of pairs
$(K,L)$ where $L\subset K$ are real closed fields. This is in his thesis but I don't think
he published it anywhere else. There are later papers of Francoise Delon and Walter Bauer
that develop this further. On the other hand there is an earlier theorem of Robinson's that
if $L\subset K$ are real closed and $L$ is dense in $K$ then the theory of $(K,L)$ is decidable.
| 5 | https://mathoverflow.net/users/5849 | 34461 | 22,299 |
https://mathoverflow.net/questions/34469 | 8 | It is believed that $BPP$ has no complete problems. Even for $BPP^O$ for a suitable oracle $O$ it is believed not to have complete problems, unless P=BPP. I wonder if the class MA (the randomized version of NP) has complete problems. For example, IP has complete problems (given that it is equal to PSPACE). There is a similar [post](https://mathoverflow.net/questions/27572/are-there-complexity-classes-with-provably-no-complete-problems) asking about complexity classes with no complete problems. Here, I'm interested specifically on complete problems for MA. If the answer is positive, can you give some examples? I've tried google and the complexity zoo, but with no success.
| https://mathoverflow.net/users/7692 | Complete problems for randomized complexity classes | In general, for randomized classes complete problems tend to be either promise problems or approximation problems (which means they don't technically satisfy the conditions for being complete problems).
If you allow approximation problems, you can get complete problems in BPP. For example, for BPP you can ask: given a Turing machine $M$, approximate the probability that it accepts within $t$ steps on input $x$, where $t$ is polynomial in the size of the input. It's clear that you can approximate this probability with a BPP machine (using simulation), and it's also obvious that any BPP language can be reduced to this problem. Technically, this problem isn't in BPP since it's not a language (i.e., its output is not $\{0,1\}$), so it's not BPP-complete.
You can turn this into a promise problem by imposing the "promise" that the acceptance probability either be greater than $\frac{2}{3}$ or less than $\frac{1}{3}$.
These complete promise problems or approximation problems play the same role that complete problems play for non-randomized languages, and they really deserve more respect from computer scientists. For quantum computing, the natural complete problems also tend to be approximation (or promise) problems, and they have been quite useful in the theory of quantum computing.
There is a natural promise problem from quantum computing [(stoquastic Hamiltonian)](http://arxiv.org/abs/quant-ph/0611021) which is MA-complete (and not trivially so). However, I don't believe there are any languages (non-promise problems with $\{0,1\}$ answers) known to be MA-complete.
| 20 | https://mathoverflow.net/users/2294 | 34472 | 22,305 |
https://mathoverflow.net/questions/34415 | 6 | I am reading the paper *Hausdorff dimension for Horseshoes,* by McCluskey and Manning. In the following theorem
**Theorem:**
Let $\Lambda$ be a basic set for a $C^1$ axiom A diffeomorphism $f:M^2\to M^2$ with $(1,1)$ splitting
$$
T\_{\Lambda}M=E^s\oplus E^u.
$$
Define $\phi:W^u(\Lambda)\to\mathbb R$ by
$$
\phi(x)=-\log\|Df\_x|\_{E^u\_x}\|.
$$
Then the Hausdorff dimension of $W^u(x)\cap\Lambda$ is given by the unique $\delta$ for which
$$
P\_{f|\_{\Lambda}}(\delta\phi)=0. \qquad \qquad (1)
$$
The authors compute the Hausdorff dimension of $\Lambda\cap W^u\_x$ by using the Bowen's equation (1). I read the proof but I was not able to figure out the intuition behind the Bowen's equation in this theorem. Could you give me explanation (do not need to be rigorous) about that or point out a reference ?
| https://mathoverflow.net/users/2386 | Relation between Hausdorff dimension and Bowen's equation | I think there are several different ways to make intuitive sense of this, so I'll have a bit of a go at each of them, and hope you find it helpful.
**First explanation:** *At the global level for similarity maps.*
Let $M$ be a manifold, and consider a conformal map $f\colon M\to M$. (By conformal we mean that $Df\_x$ is a scalar multiple of an isometry.) The simplest case to consider is the one where $\|Df\_x\|$ is constant everywhere, say $\|Df\_x\| = e^\lambda$ for some $\lambda > 0$; for example, let's consider the case where $M$ is the $n$-dimensional torus $\mathbb{R}^n/\mathbb{Z}^n$ and $f(x)=e^\lambda x$.
In this case, $\lambda$ is the Lyapunov exponent of every point in $M$, and you can easily check that if $B(x,n,\delta)$ is the Bowen ball of radius $\delta$ and order $n$, then $B(x,n,\delta) = B(x,\delta e^{-\lambda n})$. The metric balls $B(x,\delta e^{-\lambda n})$ are used in the definition of Hausdorff dimension, while the Bowen balls $B(x,n,\delta)$ are used in the definition of entropy. If instead of the classical definition of entropy one uses Bowen's definition as a Caratheodory dimension characteristic (see Pesin's book, for example), then the definition of entropy is a verbatim transcription of the definition of Hausdorff dimension, with Bowen balls replacing metric balls. In particular, it is pretty straightforward to see that
$$
\text{Hausdorff dimension} = \frac{\text{topological entropy}}{\text{Lyapunov exponent}}. \qquad \qquad \text{(1)}
$$
At some level, everything else is just a fancy generalisation of (1). Indeed, in the case mentioned, the potential $\log \|Df\_x\|$ is equal to $\lambda$ everywhere, and so the pressure function is given by
$$
P\_\Lambda (-t\log \|Df\_x) = h\_\mathrm{top}(\Lambda, f) - t \lambda
$$
using basic properties of pressure. The unique root of this equation is $t=h\_\mathrm{top}(\Lambda, f) / \lambda$, which by (1) is exactly the Hausdorff dimension.
**Second explanation:** *At the local level for arbitrary maps*.
Now consider an arbitrary conformal map, so $\|Df\_x\|$ may vary from point to point. One way of computing both Hausdorff dimension and topological entropy/pressure is by using (invariant) measures, so let $\mu$ be an invariant measure. Then if you know something about the pointwise dimensions and local entropies of $\mu$, you can use that knowledge to calculate (or at least estimate) global dimensional quantities like Hausdorff dimension, entropy, and pressure. But at the local level, subject to some bounded distortion estimates, it's not too difficult to show that
$$
d\_\mu(x) = \frac{h\_\mu(x)}{\lambda(x)};
$$
that is, that
$$
\text{pointwise dimension} = \frac{\text{local entropy}}{\text{Lyapunov exponent}},
$$
a nice counterpart to (1) in this more general setting.
**Third explanation:** *As a generalisation of Moran's equation.*
If you build a Cantor set on the interval (or indeed, in $\mathbb{R}^n$) by using $k$ basic sets at every stage of the construction, each of which is scaled down by a factor of $\lambda\_k$ from the basic set at the previous stage that contains it, then Moran's theorem says that the Hausdorff dimension of the resulting Cantor set is the unique value of $t$ such that
$$
\lambda\_1^t + \cdots + \lambda\_k^t = 1. \qquad \qquad \text{(2)}
$$
If you take the log of both sides, you get a special case of Bowen's equation. Probably the easiest case to see this in is when $k=2$ and $\lambda=1/3$; then the unique solution of (2) is $t=\log 2/\log 3$, as expected. So one can think of Bowen's equation as the natural generalisation of Moran's equation to the setting where the ratio coefficients can vary at each stage of the construction of a Cantor set.
**References:** We included a discussion and proof of Moran's Theorem in chapter 2 of *Lectures on Fractal Geometry and Dynamical Systems*, by Yakov Pesin and Vaughn Climenhaga, and the relationship (1) and its siblings appear in various places in the first four chapters of that book. And if it's not too gauche to continue to refer to my own work, I'll point out that "Bowen's equation in the non-uniform setting", which is [here](http://arxiv.org/abs/0908.4126) on the arXiv, contains some more precise and detailed formulations of a bunch of the things I said above.
| 10 | https://mathoverflow.net/users/5701 | 34475 | 22,307 |
https://mathoverflow.net/questions/34442 | 4 | I would be glad to see a reference to the following easy lemma in additive combinatorics: if $A\_1$ and $A\_2$ are two sets of remainders modulo $n^2$, each has cardinality $n > 1$ and all elements of $A\_i$ are different modulo $n$ (for $i=1,2$), then $A\_1+A\_2$ is not equal to the set of all remainders modulo $n^2$.
Maybe, it is a partial case of more general and deep:) result.
| https://mathoverflow.net/users/4312 | Sum of sets modulo a square | There must be an easier proof but here is a nice approach which can indeed lead to deeper results (feel free to edit for math display, I tried):
Techniques with characteristic polynomials and roots of unity can be very powerful. I like the way that the appropriate lemmas are explained in my paper with Ethan Coven "Tiling the Integers with Translates of One Finite Set" <http://arxiv.org/abs/math/9802122> or Journal of Algebra v 212 (1988) p 161-174. One does not need their full generality for this problem but perhaps for deeper results.
I'll sketch this result which implies what was asked for: Suppose that A and B are sets of size #A and #B so that A+B is a complete set of residues mod N=#A#B. Let p be a prime dividing N. Then exactly one of the sets has its members equally distributed mod p.
digression: Lemma 3.2 from the paper above (not needed here) shows that at least one of the following is true:
1) No member of A-A is relatively prime to #B
2) No member of B-B is relatively prime to #A
end of digression
Consider the corresponding polynomials $A(x)=\sum\_{a \in A}x^a$ and $B(x)=\sum\_{b \in B}x^b$. Then
i) A(1)=#A and B(1)=#B
ii) A(x)B(x) is a sum of N distinct powers of x, one from each residue class.
iii) $A(x)B(x)=(x^N-1)Q(x)+\frac{x^N-1}{x-1}$ for some polynomial Q(x).
iv) Every irreducible polynomial dividing $\frac{x^N-1}{x-1}$ divides at least one of $A(x)$ and $B(x)$
As an example consider A={0,9,13,16,29,32} B={0,10,12,22,24,34} with A+B a complete set of residues mod N=36.
$$\frac{x^{36}-1}{x-1}=(x+1)(x^2+x+1)(x^2+1)(x^2-x+1)(x^4+x^2+1)(x^4-x^2+1)(x^{18}-x^9+1)$$
evaluated at $x=1$ this becomes 36=2 \* 3 \* 2 \* 1 \* 3 \* 1
In general the irreducible polynomial divisors of $\frac{x^N-1}{x-1}$ are the cyclotomic polynomials corresponding to the divisors of N. Evaluated at x=1 each is either 1 (composite divisor) or a prime p (prime power divisor) and the primes have product N. Since A(1)B(1)=N and A(x)B(x) is divisible by all the prime power cyclotomic divisors of $\frac{x^N-1}{x-1}$ and these evaluated at 1 also have product N, each divides just one of A(x) or B(x) and all other polynomial divisors evaluate to 1 at 1. In particular: for each prime divisor of N, only one of A(X), B(x) divides by $\frac{x^p-1}{x-1}$ and only that one has corresponding set equidistributed mod p.
In our example A is a complete set of residues mod 6 so A(x) divides by (1+x) and by (1+x+x^2).
Since A(1)=6 , A(x) can't have either of (1+x^2) and (1+x^2+x^4) as factors. But they do divide A(x)B(x) and hence they divide B(x). This means that neither (1+x) nor (1+x+x^2) can divide B(x), again since B(1)=6. Hence, B is **not** equidistributed mod 2 (or mod 3) and certainly not mod 6.
By the way, $B(x)=(x^{10}+1)(x^{24}+x^{12}+1)$ and $A(x)=(x^{13}+1)(x^{32}+x^{16}+1)$ (mod $x^{36}-1$)
| 2 | https://mathoverflow.net/users/8008 | 34481 | 22,310 |
https://mathoverflow.net/questions/34484 | 2 | Let $X=Spec(R)$ be an affine scheme. Let $Y$ be a closed subset of $X$ and denote by $U$ its complement. Assume $U$ is quasicompact. Then $U= \cup\_{i=1}^{n} D(f\_i)$, where $f\_i \in R$.
Denote the inclusions by $j: U \to X$ and $i:Y \to X$. Let $F \in D\_{qcoh}(X)$, where $D\_{qcoh}(X)$ stands for the derived category of complex with quasicoherent cohomology (If you don't like $D\_{qcoh}(X)$ replace it with $D^{b}\_{qcoh}(X)$). Assume that $j^{\ast} F=0$.
Is the cohomology of $F$ $(f\_1,\ldots,f\_n)$-torsion? And how to prove it?
A qick way to get convince (not a proof). Assume that $U=D(f)$. Since $j^{\ast} G=0$ where $G$ is a quasicoherent sheaf, $j^{\ast} G=0$ is equivalent to $G(X)\_f=0$. which implies that $G(X)$ is $(f)$-torsion.
Thanks
| https://mathoverflow.net/users/8144 | Sheaf cohomology and torsion | The answer is yes, assuming by $(f\_1,\ldots, f\_n)$-torsion you mean that each element of each cohomology group of $F$ is killed by some power of this ideal.
There are several ways to see this. The most straightforward is probably to note that the inclusion $j\_i\colon D(f\_i) \to X$ factors via $j\colon U \to X$ so $j\_i^\*F \cong F\_{f\_i} \cong 0$ for each $i=1,\ldots,n$. Your example, the fact that localization commutes with taking cohomology, and that there are finitely many $f\_i$ shows that this is enough.
| 3 | https://mathoverflow.net/users/310 | 34486 | 22,313 |
https://mathoverflow.net/questions/22041 | 0 | Hi Everyone
I have some difficulties deriving the Stochastic Differential Equation for the following problem, any help or reference would be appreciated.
We are given a Brownian Motion $B\_t$ and we note $M\_t=\sup\_{s\le t}B\_s$.
Moreover we have a smooth real valued function $F(t,x,y)$ (for example a $C^{1,2,1}$) over $\mathbb{R}^+\times \mathbb{R} \times \mathbb{R}^+$ and we are looking for the SDE followed by $F(t,M\_t-B\_t,M\_t)$
I have a hard time trying to express $dF$.
Best regards
| https://mathoverflow.net/users/2642 | Looking for a version of Itô's Lemma | Let $F(t,m-b,m)=G(t,m,b)$ (this way it's easier to write). As long as $M$ has bounded variation, we can happily write (I skip arguments, which are $t,M\_t,B\_t$)
$$
dG(t,M\_t,B\_t) = G'\_t dt + G'\_m dM\_t + G'\_b dB\_t + \frac12 G''\_{bb} dt.
$$
| 2 | https://mathoverflow.net/users/8146 | 34492 | 22,318 |
https://mathoverflow.net/questions/34452 | 34 | Suppose that $X$ is a smooth projective variety (eg $P^n$) and $E$ is a vector bundle (eg the tangent bundle). If the characteristic is zero, then taking symmetric powers "commutes" with taking duals:
$
Sym\_m(E)^\*
$
and
$Sym\_m(E^\*)$ are canonically isomorphic.
This is not true in characteristic $p>0$ (one has a canonical isomorphism with a divided power, instead.) But even in characteristic $p$, if the bundle is trivial, then there are non-canonical isomorphisms, and it is not hard to show that the Chern classes are the same.
Question: Is $Sym\_m(E)^\*$ isomorphic to $Sym\_m(E^\*)$ (non-canonically) in general?
| https://mathoverflow.net/users/5771 | Symmetric powers and duals of vector bundles in char p | Here's another look at the case $m=2$. Maybe it can be used to shed more light on Torsten's nice example. $S^2E$ is part of an exact sequence $0\to \Lambda^2E\to E\otimes E\to S^2E\to 0$. $\Gamma^2E$ is part of an exact sequence $0\to \Gamma^2E\to E\otimes E\to \Lambda^2E\to 0$. The composition of $E\otimes E\to \Lambda^2E \to E\otimes E$ is $1-T$ where $T$ is the involution $x\otimes y\mapsto y\otimes x$. Its kernel $\Gamma^2E$ may be considered as the symmetric bilinear forms on $E^\*$, while its cokernel $S^2E$ is the quadratic forms. If $2=0$ then the equation $(1-T)(1+T)=0$ says that the image of $1-T$ is contained in the kernel of $1-T$; we have $\Lambda^2E$ injecting into $\Gamma^2E$. We have in fact an exact sequence $0\to \Lambda^2E\to\Gamma^2E\to E'\to 0$, and another one $0\to E'\to S^2E\to \Lambda^2E\to 0$, where I am writing $E'$ for "$E$ twisted by Frobenius".
If you're not in characteristic $2$ then there's no reason for $S^2E$ (or $\Gamma^2E$) to have a proper nontrivial subbundle.
Added: In the case when $E$ is the tangent bundle of $P^2$, or alternatively the rank $2$ quotient bundle of a trivial rank $3$ bundle which, as Torsten mentions, is the tangent bundle twisted by a line bundle, I believe it is not hard to work out by hand that the only global maps $E\otimes E\to E\otimes E$ are the linear combinations of the identity and the involution $v\otimes w\mapsto w\otimes v$. Of these, the only ones that kill the image of $\Lambda^2 E$ and so give a map $S^2E=coker(\Lambda^2E\to E\otimes E)\to E\otimes E$ are the multiples of $v\otimes w\mapsto v\otimes w+w\otimes v$, so that the only maps $S^2E\to \Gamma^2E=ker(E\otimes E\to\Lambda E)$ are the multiples of the usual one. This argument works over any ground ring, and shows that the two bundles are isomorphic only if $2$ is invertible.
| 14 | https://mathoverflow.net/users/6666 | 34512 | 22,326 |
https://mathoverflow.net/questions/34510 | 1 | How to see that the cup products vanish on suspensions?
| https://mathoverflow.net/users/8152 | when cup product is a zero homomorphism | 13.66 in Switzer's Algebraic Topology: Homotopy and Homology. The idea is to use the fact that $\Sigma X$ decomposes into two copies of $CX$, say $A$ and $B$, glued along the common boundary of $X$. For any two cohomology classes $x$ and $y$ in $\tilde{E}^\* \Sigma X$, you can uniquely pull $x$ back to a class $x'$ on the relative pair $(\Sigma X, A)$ and $y$ back to a class $y'$ on $(\Sigma X, B)$. Cupping is natural w.r.t the two relative inclusions $i\_A: (\Sigma X, \{x\_0\}) \to (\Sigma X, A)$ and $i\_B: (\Sigma X, \{x\_0\}) \to (\Sigma X, B)$, and so you get the calculation $x \smile y = i\_A^\*(x') \smile i\_B^\*(y') = i^\*(x' \smile y')$, where $i: (\Sigma X, \{x\_0\}) \to (\Sigma X, \Sigma X)$ is another relative inclusion and $x' \smile y'$ a class on the pair $(\Sigma X, \Sigma X)$ --- but that guy has trivial reduced cohomology.
| 12 | https://mathoverflow.net/users/1094 | 34514 | 22,328 |
https://mathoverflow.net/questions/34478 | 12 | (Just to be precise, in this question, the word "knot" means "ambient isotopy class of a (EDIT: smooth) knot in $S^3$.") A knot in $S^3$ is called prime if it is not the [connected sum](http://en.wikipedia.org/wiki/Connected_sum#Connected_sum_of_knots) of two other non-trivial knots in $S^3$. Clearly any knot is a sum of prime knots, and it is a theorem that this decomposition is unique. One downside of this is that there are infinitely many prime knots (for example, all non-trivial torus knots are prime). Here's a vague version of the question - Is there a way to trade off the uniqueness result (and add finitely many operations) in exchange for starting with a finite list of knots?
To try to make my question a little more precise, I'll define an "operation" as a function that takes a list of knots as input and outputs a finite list of knots. I'm not sure how to say this well, but I'd like to avoid very dumb operations like "fix an ordered list of all knots $K\_1,K\_2,\ldots$ and if $K\_i$ is input, output $K\_{i+1}$." However, I am interested in dumb answers, just not as dumb as that :-)
>
> Is there finite list of knots $L$ and a finite list of operations $O$ on knots with the property that if $S$ is the smallest set of knots which contains $L$ and is closed under the operations $O$, then $S$ contains all knots?
>
>
>
For example, the first paragraph, phrased in this language, is
L = all prime knots
O = input two knots and output their connected sum
S = all knots
| https://mathoverflow.net/users/2669 | Is there a procedure for obtaining all knots in S^3? | If you allow yourself to consider knotted trivalent graphs, instead of just knots, then you can start with just the tetrahedron and Mobius bands, and use the operations unzip, bubbling, and connect sum to get all knotted trivalent graphs. This result is due to Dylan Thurston and Dror Bar-Natan and is written up by Dylan in a [GT monograph](http://arxiv.org/abs/math/0311458). Their motivation was the lack of a good candidate for an answer to your question which stayed entirely in the language of knots. However for many situations KTGs are just as good as knots (e.g. click on a random link at [Dror's wiki](http://katlas.math.toronto.edu/drorbn/index.php?title=Main_Page)).
| 9 | https://mathoverflow.net/users/22 | 34523 | 22,334 |
https://mathoverflow.net/questions/34524 | -3 | I'm looking for some knowledge on probability, I've scoured the net but I can't really grasp the answer.
I was having a discussion with a co-worker about roulette probability. He says that at any given spin the probability that the outcome being red or black is equal (not taking into account the 0, which is neither).
My understanding of probability is that you should take into account the whole set of past outcomes. So if the outcome is red three times in a row, the probability that the next outcome is black will get bigger.
So to get a definitive answer I've created a roulette simulator and an artificial player. The player only bets when the outcome was the same three times in a row, then he bets the opposite. So if the outcome was red three times, he bets black.
To my surprise, the win/loss ratio was practically equal given a large enough simulation.
To finalize, my question is: how come that past outcomes have exactly zero influence on the probability of any given outcome?
I get the feeling (seeing some other (related) questions) that this may not be the place to ask, but would you then be so kind to at least get me in the right direction or point me to some resources explaining this? Thanks!
| https://mathoverflow.net/users/8154 | Roulette probability | (It's true that this question will probably be closed soon.)
Ask yourself this question: Does the roulette ball or table have a memory? If not, then past events cannot possibly affect the next probability. "No memory", or more technically independent outcomes, is a standard hypothesis in probability problems like the one you are interested in.
The deeper question (but still not at the research level) is how statistics always seem to even out in the long term, even if there is no memory forcing them to do so. This is the content of the Central Limit Theorem of probability theory, and is closely related to the Second Law of Thermodynamics, whose rigorous treatment comes from statistical mechanics. The short answer is this: Statistics always even out in practice because in the long term there are many, many, many combinations with nearly even statistics, compared to just a handful of combinations that are greatly skewed. To be more concrete: (black, red, black, black, red, black, red, red, red) has precisely the same probability as (black, black, black, black, black, black, black, black, black), but no one ever asks about that precise first sequence; instead it gets lumped together with the 125 other sequences that have the same overall statistics, whereas all-black has no statistical compadres to share the burden of occurring more than one time in 512.
| 2 | https://mathoverflow.net/users/7936 | 34531 | 22,337 |
https://mathoverflow.net/questions/34476 | 4 | Let $X$ be a scheme defined over $\mathbb{C}$ with an involution $\sigma$. How to get
a $X\_{\mathbb{R}}$ scheme defined over $\mathbb{R}$ such that $X\_{\mathbb{R}}\times\_\mathbb{R} \mathbb{C} = X$ ? How are the real algebraic bundle on $X\_{\mathbb{R}}$ and complex bundle on
$X$ related?
| https://mathoverflow.net/users/8141 | Real algebraic variety and real algebraic bundle | The paper by
Atiyah
on [K-theory and reality](http://qjmath.oxfordjournals.org/cgi/reprint/17/1/367)
Quart. J. Math. Oxford Ser. (2) 17 1966 367--386. MR0206940 discusses the topological analogue of this question, showing how to relate complex vector bundles over a space with vector bundles over the fixed points of an involution. The idea is to define a sort of intermediate K-group KR of vector bundles with involution, and compare it to the two other K-groups.
| 3 | https://mathoverflow.net/users/51 | 34534 | 22,339 |
https://mathoverflow.net/questions/34527 | 3 | This question is concerned with a possible lemma which would be very useful in one of my current research projects, but which I am currently unable to prove. The project as a whole relates to the asymptotic growth rates of certain matrix products, but once all the dynamical parts are finished with, I need to obtain some quantitative information about the behaviour of the subderivatives of a certain convex function. This would be greatly simplified if the following lemma were true:
>
> **Lemma A**. Let $f \colon [0,1] \to \mathbb{R}$ be a continuous convex function, and let $\varepsilon>0$. Suppose that for every $t \in (0,1)$, for every subderivative $f'(t)$ of $f$ at $t$ the inequality
> \[ f'(t)\leq \frac{f(t)-f(0)}{t} + \varepsilon\]
> holds. Then there exist subderivatives $f'(0)$ and $f'(1)$ of $f$ (at 0 and 1 respectively) such that $|f'(0)-f'(1)| \leq C\varepsilon$, where $C>0$ is a constant which does not depend on $f$ or on $\varepsilon$.
>
>
>
In visual terms, this means that for any $t \in (0,1)$ the slope of the straight line connecting $(0,f(0))$ and $(t,f(t))$ is constrained to be close to the "gradient" of $f$ at $t$, and I would like to deduce from this that the entire graph of $f$ does not admit very many different "slopes".
This problem can be re-stated in the following form, which I personally find somewhat easier to think about. Let us define $F \colon (0,1] \to \mathbb{R}$ to be the function which describes the left derivatives of the above function $f$. This $F$ is a well-defined monotone increasing function which is continuous on the left.
>
> **Lemma B.** Fix $\varepsilon>0$. Let $F \colon (0,1] \to [0,\infty)$ be a monotone increasing function, continuous with respect to limits from the left, such that for every $t>0$ we have $F(t) \leq \frac{1}{t}\int\_0^tF(s)ds + \varepsilon$. Is it necessarily the case that $\sup F - \inf F \leq C\varepsilon$ for some constant $C>0$ which does not depend on $F$ or on $\varepsilon$?
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>
I've spent a couple of weeks thinking about this on-and-off without making very much headway, and I am beginning to get a little frustrated at not being able to solve a problem in one-dimensional real analysis! Of course, this does not preclude the possibility that the problem is actually very simple for someone equipped with the correct tools. So my question is: does anyone know whether either of the above two lemmas is true, or have any pointers which might be useful in establishing a proof?
| https://mathoverflow.net/users/1840 | A differential inclusion relating to the slope of a convex function | Unfortunately the lemma is false. Given a candidate $C$, let $\varepsilon = 1$ and $F(t) = \ln(te^C+1)$. Then the hypotheses of Lemma B hold but the conclusion fails.
| 4 | https://mathoverflow.net/users/7936 | 34537 | 22,341 |
https://mathoverflow.net/questions/34520 | -1 | let $g(s)$ be real-valued function defined on $[0,T]$ such that $g(T)=0$ and suppose that $g$ is a "nice function"
Assume that $0<\gamma<1$, $v$ is a positive number, and
$$\frac{dg}{ds}+(v\gamma) g +(1-\gamma)(e^{\rho s}g)^{\frac{1}{\gamma-1}}g=0$$
Find a closed form for $g$?
| https://mathoverflow.net/users/5672 | A differential equation | This seems to be a Bernoulli differential equation. Please cf. <http://en.wikipedia.org/wiki/Bernoulli_differential_equation> for the solution (in your case $n= \frac{\gamma}{\gamma-1}$).
| 0 | https://mathoverflow.net/users/6415 | 34538 | 22,342 |
https://mathoverflow.net/questions/34521 | 18 | I have been using the following result:
Given a polynomial $f(x,t)$ of degree $n$ in $\mathbb{Q}[x,t]$, if a rational specialization of $t$ results in a separable polynomial $g(x)$ of the same degree, then the Galois group of $g$ over $\mathbb{Q}$ is a subgroup of that of $f$ over $\mathbb{Q}(t)$.
However, I have been unable to prove this for myself, and cannot seem to find a proof of it anywhere. Is there an elementary proof? And if not, can anyone direct me to a source containing one, or at least explain the general principle?
My need to understand the result arose from considering the following:
If I specialize $t$ such that $g$ factorizes as $x^k.h(x)$, where $h$ is an irreducible polynomial of degree $n-k$, is it legitimate to surmise that the Galois group of $h$ over $\mathbb{Q}$ is a subgroup of the original?
If nothing else, I'd be very grateful for an answer to this!
| https://mathoverflow.net/users/4078 | Proof of the result that the Galois group of a specialization is a subgroup of the original group? | Here is a broader setup for your question. Let $A$ be a Dedekind domain with fraction field $F$, $E/F$ be a finite Galois extension, and $B$ be the integral closure of $A$ in $E$.
Pick a prime $\mathfrak p$ in $A$ and a prime $\mathfrak P$ in $B$ lying over $\mathfrak p$.
The decomposition group $D(\mathfrak P|\mathfrak p)$ naturally maps by reduction mod $\mathfrak P$ to the automorphism group $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$ and Frobenius showed this is surjective. The kernel is the inertia group, so if $\mathfrak p$ is unramified in $B$ then we get an isomorphism from $D(\mathfrak P|\mathfrak p)$ to $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$, whose inverse is an embedding
of the automorphism group of the residue field extension into $\text{Gal}(E/F)$.
If we take $A = {\mathbf Z}$ then we're in the number field situation and this is where Frobenius elements in Galois groups come from.
In your case you want to take $A = {\mathbf Q}[t]$, so $F = {\mathbf Q}(t)$.
You did not give any assumptions about $f(x,t)$ as a polynomial in ${\mathbf Q}[x,t]$. (Stylistic quibble: I think it is better to write the polynomial as $f(t,x)$, specializing the first variable, but I'll use your notation.) Let's assume $f(x,t)$ is absolutely irreducible, so the ring $A' = {\mathbf Q}[x,t]/(f)$ is integrally closed. [EDIT: I should have included the assumption that $f$ is smooth, as otherwise $A'$ will *not* be integrally closed, but this "global" int. closed business is actually not so important. See comments below.] Write $F'$ for the fraction field of $A'$. After a linear change of variables we can assume $f(x,t)$ has a constant coefficient for the highest power of $x$, so $A'$ is the integral closure of $A$ in $F'$.
Saying for some rational $t\_0$ that the specialization $g(x) = f(x,t\_0)$ is separable in ${\mathbf Q}[x] = (A/(t-t\_0))[x]$ implies the prime $(t-t\_0)$ is unramified in $A'$. Let $E$ be the Galois closure of $F'/F$ and $B$ be the integral closure of $A$ in $E$. A prime ideal that is unramified in a finite extension is unramified in the Galois closure, so $(t-t\_0)$ is unramified in $B$. For any prime $\mathfrak P$ in $B$ that lies over $(t-t\_0)$, the
residue field $B/\mathfrak P$ is a finite extension of $A/(t-t\_0) = \mathbf Q$ and since
$E/F$ is Galois the field $B/\mathfrak P$ is normal over $A/(t-t\_0)$. These residue fields have characteristic 0, so they're separable: $B/\mathfrak P$ is a finite Galois extension of $\mathbf Q$. I leave it to you to check that $B/\mathfrak P$ is the Galois closure of $g(x) = f(t\_0,x)$ over $\mathbf Q$. Then the isomorphism of $D(\mathfrak P|(t-t\_0))$ with $\text{Aut}((B/\mathfrak P)/\mathbf Q) = \text{Gal}((B/\mathfrak P)/\mathbf Q)$ provides (by looking at the inverse map) an embedding of the Galois group of $g$ over $\mathbf Q$ into the Galois group of $f(x,t)$ over $F = {\mathbf Q}(t)$.
I agree with Damiano that there are problems when the specialization is not separable. In that case what happens is that the Galois group of the residue field extension is identified not with the decomposition group (a subgroup of the Galois group of $E/F$) but with the quotient group $D/I$ where $I = I(\mathfrak P|\mathfrak p)$ is the inertia group, and you don't generally expect a proper quotient group of a subgroup to naturally embed into the original group.
| 19 | https://mathoverflow.net/users/3272 | 34558 | 22,355 |
https://mathoverflow.net/questions/31786 | 27 | For a lattice in $\mathbb{R}^2$, if we include each edge independently with probability $p$ (i.e. *bond percolation*), it is well known that there is a critical probability $0 < p\_c < 1$ depending on the lattice, such that if $ p > p\_c$ then there is almost surely an infinite component (i.e. with probability $1$), and if $p < p\_c$ then almost surely all components are finite.
The square lattice is special because it is self-dual. This allows one to guess that in this particular case $p\_c = 1/2$, although the proof of this fact, originally due to Harry Kesten, is fairly subtle.
One wonders if there is an analogous fact for "two-dimensional" percolation in $\mathbb{R}^4$. In particular, consider the hypercubic lattice as a $1$-dimensional cubical complex, and then attach two-dimensional square "plaquettes" independently with probability $p$. Since the hypercubic lattice is self-dual, we might expect some kind of phase transition when $p=1/2$. My hope is that topology would give us the right language to talk about this.
One possibility, and perhaps the nicest answer I can imagine, was suggested to me by Russ Lyons: If $p > 1/2$ then there are almost surely embedded planes, and if $p< 1/2$ then there are almost surely not. Here the embedded plane should be the union of closed $2$-cells. It turns out that once $p > 1/2$ in $\mathbb{R}^2$, there are not only infinite components, which implies embedded half-lines, but embedded lines as well.
Another possibility is that once $p > 1/2$ there are almost surely bounded $1$-dimensional cycles in homology $H\_1 ( - , \mathbb{Z} / 2\mathbb{Z})$, which are boundaries of unbounded two-dimensional complexes, and that when $p < 1/2$ there are almost surely not. (I believe that this is similar to the "plaquette percolation" studied by Jennifer Chayes in her Ph.D. thesis, but the work I know of is in $\mathbb{R}^3$.)
I am not an expert in percolation theory, and would really like to know if anyone knows about any previous work in this area, or any standing conjectures. (Or is there any obvious reason that either of the possibilities I suggested could be ruled out?)
| https://mathoverflow.net/users/4558 | What is the right notion of self-dual (two-dimensional) percolation in R^4? | The paper "PLAQUETTES, SPHERES, AND ENTANGLEMENT" by GEOFFREY R. GRIMMETT AND ALEXANDER E. HOLROYD does not deal with the self-dual problem, but nevertheless could be of interest for you. If shows that "The high-density plaquette percolation model in d dimensions contains a surface that is homeomorphic to the (d − 1)sphere and encloses the origin." Here's the link: <http://www.statslab.cam.ac.uk/~grg/papers/sphere8.pdf>
| 5 | https://mathoverflow.net/users/6415 | 34564 | 22,359 |
https://mathoverflow.net/questions/34561 | 8 | Let $M$ be a matroid admitting a coordinatization over a complex vector space. If we know that the complex coordinatization space for $M$ is connected, then may we conclude that the matroid admits a coordinatization over the real numbers?
The only examples that I am able to construct which do not have real coordinatizations have disconnected coordinatization spaces.
Note: Some texts refere to 'coordinatization over a vector space' as 'realizability over a vector space.'
| https://mathoverflow.net/users/8161 | Realization space of matroids | The answer should be no. Here is the reason: It is perfectly possible to have a connected algebraic variety, defined over $\mathbb{R}$, which has no $\mathbb{R}$-points. For example, $\{ (x,y) : x^2+y^2=-1 \}$. Using [Mnev's universality theorem](http://en.wikipedia.org/wiki/Mnev%2527s_universality_theorem), you should be able to build a matroid whose realization space is [stably equivalent](http://en.wikipedia.org/wiki/Mnev%2527s_universality_theorem#Stable_equivalence_of_semialgebraic_sets) to this variety, and thus has the same property. I have not checked the details here.
| 6 | https://mathoverflow.net/users/297 | 34567 | 22,362 |
https://mathoverflow.net/questions/34433 | 12 | A functor $C \to D$ between categories induces a morphism of presheaf categories $Pre(D) \to Pre(C)$. This functor has a left adjoint given by left Kan extension and I am interested in knowing when this left adjoint preserves pull-back squares.
I'm interested in any conditions that make this happen, but I am particularly interested in a special case. Let me say a little more about the context I am working in, and why I am interested. In the situation that this came up $C$ is a "lluf" subcategory of $D$, that is $C$ has the *same objects* as $D$ and the functor $C \to D$ is *faithful*. In that case it is good to call the functor $U:Pre(D) \to Pre(C)$ the forgetful functor. It automatically preserves limits and colimits. Let L be its left-adjoint.
Since C has the same objects as D, this forgetful functor is also *conservative*, meaning that it reflects isomorphisms. So by general non-sense (specifically [Beck's monadicity theorem](http://ncatlab.org/nlab/show/monadicity+theorem)) this is a monadic adjunction. This means that $Pre(D) = T-alg$ is the category of T-algebras in $Pre(C)$ where T is the monad $T= UL$.
I'm trying to understand conditions under which this monad is *cartesian* in the sense [described at the n-lab](http://ncatlab.org/nlab/show/cartesian+monad). This means, among other things, that the monad T is supposed to send fiber products to fiber products. This is equivalent to having L send fiber products to fiber products. I want to understand when this happens. Does it always happen? Are there reasonable conditions on C or D that ensure that this happens?
Notice that I am not asking for L to be "left-exact", i.e. to commute with all finite limits. This property is generally much too strong. In particular L will not usually preserve terminal objects. This means it doesn't preserve products, but should instead send products to fiber products over $L(1)$.
Here is an example. Let $C = pt$ be the singleton category and $D = G$ be the one object category with morphisms a group G. There is a unique inclusion $C \to D$ which is obviously faithful. The forgetful functor
$$U:Pre(D) \to Pre(C)$$
sends a G-set to its underlying set. The left adjoint L sends a set S to the free G-set $S \times G$. This doesn't preserve terminal objects, but it does commute with fiber products. What is more, the monad $T=UL$ is a classic example of a cartesian monad in the n-lab sense.
I've played around with this, but can't seem to get it to work. I feel like this is going to be a well known result or there is going to be a counter example which sheds light on the situation.
>
> **Question**: In the context I described above (where $C \to D$ is lluf), does the left adjoint $$L: Pre(C) \to Pre(D)$$ always commute with fiber products? If not what is a counter example, and are there conditions one can place on C and D to ensure that L does commute with fiber products?
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>
| https://mathoverflow.net/users/184 | An elementary question about adjunctions between presheaf categories preserving pullbacks. | $\newcommand{\C}{\mathbf{C}} \newcommand{\D}{\mathbf{D}} \newcommand{\Lan}{\mathrm{Lan}} \newcommand{\yon}{\mathbf{y}} \newcommand{\CC}{[\C^\mathrm{op},\mathbf{Set}]} \newcommand{\DD}{[\D^\mathrm{op},\mathbf{Set}]}$
Expanding on my comment above:
Define: a category is *semi-filtered* iff every pair of arrows $x \leftarrow z \rightarrow y$ can be completed to a commutative square, and every parallel pair of arrows $f,g \colon x \to y$ have some $h : y \to w$ with $hf = hg$; equivalently, if every finite connected diagram has some co-cone under it. (Afaik, this isn't standard terminology; I don't recall seeing this property discussed, but it almost certainly has been.)
It's *filtered* if moreover it's non-empty and every pair of objects $x,y$ is connected by some $x \to w \rightarrow y$; equivalently, if every finite diagram has some co-cone under it.
**Answer**: for $f \colon \C \to \D$, the left Kan extension $f\_\* \colon \CC \to \DD$ will preserve pullbacks (equivalently, all connected finite limits) exactly if the opposite of its every comma category $(f \downarrow d)$ is semi-filtered.
This is a close variant of the standard lemma (see e.g. Mac Lane and Moerdijk *Sheaves in Geometry and Logic*) that $f\_\*$ preserves pullbacks and the terminal object (equivalently, all finite limits) iff the opposite of every $(f \downarrow d)$ is filtered, i.e. if $f$ is *flat*.
**Proof**: the values of $f\_\*$ can be computed as colimits over the opposites of comma categories (see [this answer](https://mathoverflow.net/questions/32791/how-is-the-right-adjoint-f-to-the-inverse-image-functor-f-described-for-f/32808#32808)). But in $\mathbf{Set}$, finite limits commute with filtered colimits; and similarly, pullbacks commute with semi-filtered colimits.
(The first of these facts is standard. The second follows because in a semi-filtered category, each connected component is filtered; so a semi-filtered colimit is a coproduct of filtered colimits; and pullbacks commute with both coproducts and filtered colimits.)
| 7 | https://mathoverflow.net/users/2273 | 34577 | 22,369 |
https://mathoverflow.net/questions/23815 | 13 | The spectral theorem for a real $n \times n$ symmetric matrix $A$ says that $A$ is diagonalizable with all eigenvalues real. If $A$ happens to have non-negative integer entries, it can be interpreted as the adjacency matrix of an undirected graph $G$, and the spectral theorem gives us information about how the sequences $A\_{ij}^n$ behave, which count the number of walks of length $n$ from vertex $i$ to vertex $j$. In particular, it says that $A\_{ij}^n$ has the form $\sum\_{k=1}^{n} a\_{ijk} \lambda\_k^n$ for some real $\lambda\_k$ and some $a\_{ijk}$.
If $A$ is not symmetric, on the other hand, two things can happen that don't happen in the above case:
* The $\lambda\_k$ may fail to be real. In other words, there can be "periodicity" of period greater than $2$ in the sequences $A\_{ij}^n$. This happens, for example, if $A$ is a directed cycle graph.
* The coefficients $a\_{ijk}$ may be polynomials in $n$. This happens, for example, if $A$ is a directed path graph with loops based at each vertex.
Is it possible to prove "combinatorially" that neither of these things can happen when $A$ is symmetric? (What I mean is that, given only that you know what $A\_{ij}^n$ looks like in terms of eigenvalues, what can you prove just by looking at $G$?) For example, it is straightforward to prove that the coefficient of the largest positive eigenvalue is constant by a path-counting argument which I described [here](http://qchu.wordpress.com/2010/04/27/the-mckay-correspondence-i/). I think a path-counting argument can also in principle prove that the $\lambda\_k$ are real via some kind of mixing argument which could show that the only constraint on the length of a long walk from a vertex to itself is its length mod 2 (due to the presence of even cycles). I think such an argument can at least show that the eigenvalues of maximum absolute value must be real, but I don't know how easy it is to deduce information about the other eigenvalues.
If you have a solution with "adjacency matrix" replaced by "Laplacian," I'd also be interested in that.
| https://mathoverflow.net/users/290 | Combinatorial proof of (a special case of) the spectral theorem? | I believe the answer is yes and that this might be found somewhere in the literature on the graph reconstruction problem.
Let me denote the characteristic polynomial of the graph $G$ as $\phi(G,x)$ and let its cofactors be $\phi\_{ij}(G,x)$. In the paper ["Walk Generating Functions, Christoffel-Darboux Identities and the Adjacency Matrix of a Graph"](http://journals.cambridge.org/action/displayFulltext?type=1&fid=2197932&jid=CPC&volumeId=1&issueId=01&aid=1771436), C.D. Godsil proves the following lemma (5.2) which is a graph theoretic analogue to the Christoffel-Darboux formula for orthogonal polynomials, the identity can be proven by purely combinatorial means:
Let $V(G)=\{1,2,\dots,n\}$ be the vertices of $G$, then for any pair $i,j$ one has
$$\sum\_{k\in V(G)}\phi\_{ik}(G,x)\phi\_{jk}(G,y)=\frac{\phi\_{ij}(G,x)\phi(G,y)-\phi\_{ij}(G,y)\phi(G,x)}{y-x}.$$
Taking $i=j$ one obtains
$$\sum\_{k\in V(G)}\phi\_{ik}(G,x)\phi\_{ik}(G,y)=\frac{\phi\_{ii}(G,x)\phi(G,y)-\phi\_{ii}(G,y)\phi(G,x)}{y-x}.$$
If there was a graph with a complex eigenvalue then pick such a graph with the smallest number of vertices. Let the eigenvalue be $\eta$, and substitute in our identity $x=\eta, y=\bar{\eta}$. We see that the RHS vanishes and so must $\sum\_{k\in V(G)}|\phi\_{ik}(G,\eta)|^2$, so in particular $\phi\_{ii}(G,\eta)=0$. But $\phi\_{ii}(G,x)=\phi(G/i,x)$ and so $G/i$ would then be a graph with one less vertex having $\eta$ as an eigenvalue, contradicting our minimality assumption.
| 6 | https://mathoverflow.net/users/2384 | 34580 | 22,370 |
https://mathoverflow.net/questions/32020 | 9 | The sequence
[A059710](http://www.research.att.com/~njas/sequences/A059710)
starts 1,0,1,1,4,10,35,...
This satisfies the polynomial recurrence relation
$$ (n+5)(n+6)a(n)=2(n-1)(2n+5)a(n-1)+(n-1)(19n+18)a(n-2)+14(n-1)(n-2)a(n-3) $$
I have a $q$-analogue of this sequence. The first few terms are:
$$1$$
$$0$$
$$1$$
$$q^{3}$$
$$q^{6} + q^{4} + q^{2} + 1$$
$$q^{9} + q^{8} + 2 q^{7} + 2 q^{6} + 2 q^{5} + q^{4} + q^{3}$$
$$q^{14} + q^{13} + 4 q^{12} + 2 q^{11} + 5 q^{10} + 4 q^{9} + 5 q^{8} +
2q^{7} + 5 q^{6} + q^{5} + 2 q^{4} + q^{3} + q^{2} + 1$$
$$q^{21} + q^{19} + 2 q^{18} + 4 q^{17} + 5 q^{16} + 9 q^{15} + 10 q^{14}
+ 13 q^{13} + 13 q^{12} + 14 q^{11} + 12 q^{10} + 12 q^{9} + 8 q^{8} + 7
q^{7} + 4 q^{6} + 3 q^{5} + q^{4} + q^{3}$$
These are $q$-analogues since if you put $q=1$ you get the original sequence.
Would anyone like to suggest a $q$-analogue of the polynomial recurrence relation?
I have asked a closely related question in
[17610](https://mathoverflow.net/questions/17610/)
I can calculate a few more terms than I have posted here.
Since you asked, the polynomial is constructed as follows: take $V$ to be the seven dimensional representation of $G\_2$; take the invariant tensors in $\otimes^nV$;
take the Frobenius character of this representation of $S(n)$; take the fake degree polynomial of this symmetric function (almost the principal specialisation).
**Further information** In response to Will's comment:
Evaluating at $q=-1$ gives
$$ 1,0,1,-1,4,-2,13,-10,55,-40,241,-190,\ldots $$
Reducing modulo $1+q+q^2$ gives
$$1,0,1,1,1,1,5,3,5,19,15,19,\ldots$$
Reducing modulo $1+q^2$ gives
$$1,0,1,-q,0,0,q,q-1,3,0,2q+3,-q-1,\ldots$$
Reducing modulo $(1-q^5)/(1-q)$ gives
$$1,0,1,q^3,-q^3,0,0,0,0,-q^3-q-1,3,0,\ldots$$
Reducing modulo $1-q+q^2$ gives
$$1,0,1,-1,1,1,1,-1,1,-1,1,-1,\ldots$$
As requested by Jacques, I have put the first fifteen polynomials in a file which you
should be able to access here
[G2 polynomials](http://www.warwick.ac.uk/~masdbn/cspG2.txt)
I have put the first forty polynomials of a second example in a file which you
should be able to access here
[A1 polynomials](http://www.warwick.ac.uk/~masdbn/cspA1.txt)
These are $q$-analogues of the
[Riordan numbers](http://www.research.att.com/~njas/sequences/A005043)
The linear recurrence relation is given there as
$$ (n+1)\*a[n] = (n - 1)\*(2\*a[n - 1] + 3\*a[n - 2]) $$
| https://mathoverflow.net/users/3992 | Finding recurrence relation for a sequence of polynomials | Using [FriCAS](http://fricas.sourceforge.net/), one can indeed guess a q-recurrence, given the first 50 terms or so. It is not nice, though. The command issued is
`guessHolo(q)(cons(1, [qRiordan n for n in 1..60]), debug==true, safety==10)`
for the q-differential equation (a linear combination with polynomial coefficients of $f(x), f(qx),\dots,f(q^5 x)$, degree in $x$ is 6), or
`guessPRec(q)(cons(1, [qRiordan n for n in 1..48]), debug==true, safety==2)`
for the q-recurrence.
| 5 | https://mathoverflow.net/users/3032 | 34589 | 22,376 |
https://mathoverflow.net/questions/34591 | 3 | The statement is simple:
What is the probability that a set of n-1 transpositions generates the symmetric group, $S\_n$?
The motivation is that I remembered reading that this was an open problem somewhere on the internet, and then I solved it. I'm curious to see other people's solutions, because I think it's a nice problem, and don't quite believe that it is hard enough to be open.
| https://mathoverflow.net/users/5312 | Probability of generating the symmetric group | A solution (assuming that all transpositions are distinct and are choosen uniformly among all
${n\choose 2}$ possible transpositions) can be given as follows:
A set of $n-1$ transpositions $(a\_1,b\_1),\dots,(a\_{n-1},b\_{n-1})$ on the set $\lbrace 1,\dots,n\rbrace$ generates the whole symmetric group
of $\{1,\dots,n\}$ if and only if the graph with vertices $\lbrace 1,\dots,n\rbrace$ and edges $\lbrace a\_i,b\_i\rbrace$ is a tree.
The probability to generate $S\_n$ is thus the same as the probability to get a tree with $n$ vertices $V$ when choosing randomly
$n-1$ edges with endpoints in $V$.
By Cayley's theorem, there are $n^{n-2}$ different trees with vertices $\lbrace 1,\dots,n\rbrace$.
Since there are ${{n\choose 2}\choose n-1}$ different graphs with $n-1$ edges and vertices $\{1,\dots,n\}$, the
probability is given by
$n^{n-2}/{{{n\choose 2}\choose n-1}}$.
If repetitions are allowed, one gets $n^{n-2}/{{n\choose 2}+n-2\choose n-1}$ (assuming uniform probability for all distinct multisets).
| 11 | https://mathoverflow.net/users/4556 | 34594 | 22,379 |
https://mathoverflow.net/questions/34592 | 29 | It is well known that if $K$ is a finite index subgroup of a group $H$, then there is a finite index subgroup $N$ of $K$ which is normal in $H$. Indeed, one can observe that there are only finitely many distinct conjugates $hKh^{-1}$ of $K$ with $h \in H$, and their intersection $N := \bigcap\_{h \in H} h K h^{-1}$ will be a finite index normal subgroup of $H$. Alternatively, one can look at the action of $H$ on the finite quotient space $H/K$, and observe that the stabiliser of this action is a finite index normal subgroup of $H$.
But now suppose that $K$ is a finite index subgroup of *two* groups $H\_1$, $H\_2$ (which are in turn contained in some ambient group $G$, thus $K \leq H\_1 \leq G$ and $K \leq H\_2 \leq G$ with $[H\_1:K], [H\_2:K] < \infty$). Is it possible to find a finite index subgroup $N$ of $K$ which is simultaneously normal in both $H\_1$ and in $H\_2$ (or equivalently, is normal in the group $\langle H\_1 H\_2 \rangle$ generated by $H\_1$ and $H\_2$)?
The observation in the first paragraph means that we can find a finite index subgroup $N$ which is normal in $H\_1$, or normal in $H\_2$, but it does not seem possible to ensure normality in both $H\_1$ and $H\_2$ simultaneously. However, I was not able to find a counterexample (though it has been suggested to me that the automorphism groups of trees might eventually provide one).
By abstract nonsense one can assume that the ambient group $G$ is the amalgamated free product of $H\_1$ and $H\_2$ over $K$, but this does not seem to be of too much help.
I'm ultimately interested in the situation in which one has finitely many groups $H\_1,\ldots,H\_m$ rather than just two, but presumably the case of two groups is already typical.
| https://mathoverflow.net/users/766 | Existence of simultaneously normal finite index subgroups | I think the answer to your question is no. Take $G=PSL\_d(\mathbb{Q}\_p)$. It is a simple group. Take $H\_1=PSL\_d(\mathbb{Z}\_p)$ and take $H\_2=H\_1^g$ for some $g \in G$ so that $H\_1 \ne H\_2$. Now, if I am not mistaken $H\_1$ and $H\_2$ are maximal in $G$ so together they generate $G$. Also, $G$ commensurates $H\_1$ since $H\_1$ is open in $G$ and profinite. So $K=H\_1 \cap H\_2$ is open and of finite index in both $H\_1$ and $H\_2$. But as $G$ is simple, $K$ contains no non-trivial normal subgroup of $G$.
I am sure you can do something similar with Lie groups and lattices.
| 33 | https://mathoverflow.net/users/5034 | 34603 | 22,385 |
https://mathoverflow.net/questions/33460 | 13 | Let $G$ be a torsion-free group and $C$ the ring of complex numbers. The zero divisor (idempotent, resp.) conjecture is that there is no nontrivial zerodivisor (idempotent, resp.) in $CG$.
The wiki page:
<http://en.wikipedia.org/wiki/Group_ring>
says "This conjecture (zero divisor conjecture) is equivalent to K[G] having no non-trivial idempotents under the same hypotheses for K and G.
"
Is this obvious true? Are there some reference for this claim?
| https://mathoverflow.net/users/1546 | Zero divisor conjecture and idempotent conjecture | Passman showed that whenever there are zero-divisors in a group ring one also has (non-zero) nilpotent elements. He shows that for any field $k$ and any torsionfree group $G$, the ring $kG$ is a prime ring, i.e. the zero-ideal is a prime ideal.
Now, if $a,b \in kG$ are non-zero and $ab=0$, then there exists $c \in kG$ such that $bca\neq 0$. (This is just another way of saying that the product of the two-sided ideal $(b)$ with the two-sided ideal $(a)$ cannot give the zero ideal, since $(0)$ is a prime ideal.)
Now, we see that $(bca)^2 = bcabca =0$. Hence, $kG$ contains nilpotent elements. The other direction is obvious.
| 11 | https://mathoverflow.net/users/8176 | 34616 | 22,395 |
https://mathoverflow.net/questions/34628 | 5 | Given a linear representation $\rho$ of $SL\_n(\mathbb C)$ of finite dimension $m$,
the image $\rho(U)$ of a maximal unipotent Jordan block $U\in SL\_n$ decomposes
into generally several Jordan blocks of size $m\_1,\dots,m\_k$.
Is it possible to describe the partition $m=m\_1+m\_2+\dots+m\_k$, say in terms
of the highest weight vector associated to an irreducible representation $\rho$?
| https://mathoverflow.net/users/4556 | Dimensions of Jordan blocks associated to representations | The answer is yes. The Jordan block decomposition of the generic nilpotent in $SL\_m$ on a representation is the same as the decomposition of any representation under the principal $SL\_2$ (which is a map of $SL\_2$ to $SL\_n$ which sends the generic nilpotent in $SL\_2$ to a generic one in $SL\_n$; people usually have a particular one in mind, but they are all the same up to conjugation by Jacobson-Morozov).
This can be extracted from the formula for the character of the principal $SL\_2$ usually called the "quantum Weyl dimension formula"
$$\chi(V\_\lambda)=\frac{\prod\_{\alpha\in \Delta^+}q^{\langle\rho+\lambda,\alpha\rangle}-q^{-\langle\rho+\lambda,\alpha\rangle}}{\prod\_{\alpha\in \Delta^+}q^{\langle\rho,\alpha\rangle}-q^{-\langle\rho,\alpha\rangle}}$$
One "only" needs to expand this out in terms of the characters of the irreducible $SL\_2$ reps $\frac{q^{n+1}-q^{-n-1}}{q-q^{-1}}$.
| 5 | https://mathoverflow.net/users/66 | 34633 | 22,402 |
https://mathoverflow.net/questions/1940 | 9 | Suppose M is a von Neumann algebra.
Denote by L its maximal noncommutative localization,
i.e., the Ore localization with respect to the set of all left and right regular elements,
i.e., elements whose left and right support equals 1.
Denote by A the set of all closed unbounded operators with dense domain
affiliated with the standard representation of M on a Hilbert space, i.e., L^2(M),
also known as the standard form of M.
Von Neumann proved that if M is finite, then L and A are canonically isomorphic.
What can we say about the relationship of L and A when M has type III?
I am also interested in the properly infinite semifinite case.
| https://mathoverflow.net/users/402 | Maximal localizations of von Neumann algebras | I think the question is not well-posed or has a negative answer.
One first has to deal with the question whether the left-right-regular elements satisfy the Ore condition, or equivalently, we have to ask: Can we find common denominators? If one is not in the finite case, this is not possible.
For $B(H)$, let us take injective bounded operators $T$ and $S$ such that the images are dense but intersect only in the zero vector. In order to find a common denominator, we need to find an operator $R$ (bounded, injective, dense image) and bounded operators $X$ and $Y$ such that $R = TX$ and $R= SY$. This cannot possibly work since $T$ and $S$ have disjoint image.
Since $B(H)$ sits inside any type $III$-factor, no Ore localization in the above sense exists.
| 4 | https://mathoverflow.net/users/8176 | 34635 | 22,404 |
https://mathoverflow.net/questions/34615 | 0 | Hi
I have a time series of probabilites, vector X
I need to convert the probabilites to uniform numbers.
As I understand it if I put the series into the cdf the output is thus uniform.
The problem is I do not know what the cdf is for my series so how is this done ?
Every question/example I see seems to say ...'data follows norm dist' or some such but when you don't know what the distribution is how is this possible?
Any help appreciated as this seems v confusing to me.
Thks vm.
| https://mathoverflow.net/users/8178 | Transforming to uniform numbers | Replace each data point by its percentile. E.g., if $x\_{27}$ is the 45th largest of 7289 data points, let $u\_{27} = 45/7289$.
| 0 | https://mathoverflow.net/users/7651 | 34637 | 22,406 |
https://mathoverflow.net/questions/29636 | 10 | In the Markl, Schneider and Stasheff text, topological operads are an indexed collection of spaces $O(n)$ for $n \in \{1,2,3,\cdots\}$ satisfying some axioms. In May's text, the index set is allowed to include zero.
1) Is there a standard terminology for operads with and without $O(0)$?
2) Is there standard terminology for topological operads where $O(0)$ is a point, vs. $O(0)$ not being a point?
Although it's less important I'd be curious if people have examples where these distinctions are interesting.
Since any operad acts on its $O(0)$ part perhaps the $O(0)$ part should be called something like its "base"? But then "baseless operad" would sound kind of pejorative.
| https://mathoverflow.net/users/1465 | Operad terminology - Operads with and without O(0). | I can second Jeffrey's comment, reduced is used to say that O(1) is just the monoidal unit (it allows us to use the Boardman Vogt resolution in homotopy theory). It's my opinion that this terminology will probably stick.
I would also say that a $\mu$ in O(n) had arity n.
That the O(0) part of an operad is referred to as the 'constants' of the operad makes a lot of sense, every algebra for O must contain O(0) and the composition of those must behave in a certain way.
Calling O(0) the point also makes sense, because in the category of algebras O(0) will be the initial object.
*Here my comment has become too long, just as I've got to the point of my comment:*
The comments to the question tend to prefer terminology that relates to the behaviour of the operad (eg "reduction", because a unit lowers the arity). My personal preference (and I think the literature follows it), is that terminology should have more of a relation to the category of algebras than to the operad itself.
So my vote is that you call O(0) the initial of O. And you call an operad without O(0) initial-less or uninitiated.
| 5 | https://mathoverflow.net/users/109 | 34643 | 22,411 |
https://mathoverflow.net/questions/19528 | 4 | I'm looking for adaptive controllers (adaptive in both step size and order) for stiff integrators. I have asymptotically correct error estimates for the current method and all candidate methods of order 1 higher and lower than the current method. My naive controllers have occasional problems with either oscillating between different methods despite smooth long-term behavior, or getting stuck (e.g. with a high order method and unreasonably short time steps).
For the curious, these are IRKS general linear methods, see Butcher, Jackiewicz, and Wright 2007.
| https://mathoverflow.net/users/4954 | Adaptive controllers for stiff ODE and DAE integrators | Have a look at [chapter 8 of Jackiewicz's book](http://books.google.fr/books?id=SjNfr7gfUZEC&lpg=PP1&dq=General%2520Linear%2520Methods%2520for%2520Ordinary%2520Differential%2520Equations.&pg=PA417#v=onepage&q&f=false), especially section 8.10 for a general background. There's some [matlab code](http://www.math.auckland.ac.nz/~hpod/atlas/irks.m) by Podhaisky too, used to [do this](http://www.math.auckland.ac.nz/~hpod/atlas/), but no control here.
And then, the [theses of Butcher's recent student are here](http://www.math.auckland.ac.nz/~butcher/theses/), which discuss implementation details, in particular Huang's chapter 3 should be very useful to you.
Older fortran code by Hairer do implement both order and stepsize control: see [RADAU and DR\_RADAU here](http://www.unige.ch/~hairer/software.html), it's not for IRKS but gives a well-tested framework that could be suitably modified.
| 2 | https://mathoverflow.net/users/469 | 34644 | 22,412 |
https://mathoverflow.net/questions/34525 | 2 | Let X be a tensored and cotensored V-category, where V is a fixed complete, cocomplete, closed symmetric monoidal category.
Define $C:=Span(X)$ to be the category of spans in X (this is the functor category $X^{Sp}$ where $Sp$ is the walking span). We notice that $C$ is automatically "tensored" over $V$ (by computing the tensor product pointwise). Then C has a natural V-enriched structure given as follows: $Map\_C(a,b)$ is the object of $V$ representing the functor $M\_{ab}(\gamma):= Hom\_C(\gamma \otimes a, b)$ (such an object exists by the adjoint functor theorem and since the tensor product is cocontinuous).
We can give another description of the mapping space as: $$Map\_C(a,b)=Map\_X(A,B)\underset{Map\_X(A,B'\times B'')}{\times} (Map\_X(A',B')\times Map\_X(A'',B''))$$
Where $a=A'\leftarrow A \to A''$ and $b=B'\leftarrow B \to B''$.
To prove that these two descriptions are equivalent, I applied Yoneda's lemma to the second definition of $Map\_C(a,b)$, which gives us $$Hom\_V(Q,Map\_C(a,b))=Hom\_X(Q\otimes A, B)\underset{Hom\_X(Q\otimes A,B'\times B'')}{\times}(Hom\_X(Q\otimes A',B')\times Hom\_X(Q\otimes A'',B''))$$
Which by the ordinary fiber product in the category of sets is precisely the set of triplets of arrows $(Q\otimes A\to B,(Q\otimes A'\to B',Q\otimes A''\to B''))$ giving the commutativity of the natural transformation diagram in $X$. This construction is obviously functorial in $Q$ for fixed $a$ and $b$.
Surely there must be a better way to do this, presumably without relying so heavily on the definition of the fiber product in the category of sets. What does such a proof look like? I assume there must be a simpler proof, because this fact was asserted as though it were trivial in a book I'm reading.
Question: What's a slicker way to prove that the two definitions are equivalent?
| https://mathoverflow.net/users/1353 | A better way to compute the mapping spaces of the category of spans in an enriched tensored category? | Because of the way you've chosen to write your second description, I don't think you're going to be able to avoid using something about the fiber product in Set. But there is a general fact here: any V-functor V-category [A,X], where A and X are V-categories, inherits any V-enriched (weighted) limits that X has, constructed pointwise. Tensors are a particular kind of V-weighted limit, and your category C is the V-functor V-category [V[Sp],X], where V[-] denotes the free V-category on an ordinary category, whose hom-objects are coproducts of copies of the unit object of V. The property that V-valued homs represent the functor $M\_{a b}$ is a reformulation of the definition of tensors as a V-enriched limit, so the question then simply becomes, why is your second description an equivalent description of the canonical V-enrichment structure on C?
Now the V-valued homs of such a functor category are "always" given by writing down the Set-valued homs as a limit of homsets in Set and reinterpreting it as a limit of hom-objects in V. The universally applicable way to do this is with an end, as Finn says, but any other way that is equivalent in Set will be equivalent in V as well, for the same Yoneda reason as in your proof. So the question becomes simply, is the set of natural transformations between two spans (considered as functors out of Sp) described by the analogous pullback of homsets in Set? And there you have to know something about pullbacks in Set.
| 2 | https://mathoverflow.net/users/49 | 34648 | 22,414 |
https://mathoverflow.net/questions/34595 | 18 | It is well known that a modular form of weight k and level \Gamma is a global section of k-power of a Hodge line bundle over some modular curve. e.g. H^0(X,E^k).
My question is
***How to characterize Eisenstein series among such sections using geometric datas?***
For example, we know cusp forms are just sections of H^0(X,E^k(-cusps)).But how about Eisenstein series?
Actually in his Introduction to "Abelian Varieties" 1970, Mumford writes:
"It is interesting to ask whether further ties between the analytic and algebraic theories exist: e.g. an algebraic defintion of the Eisenstein series as a section of a line bundle on the moduli space. ..."
Could somebody explain the analytic-algebraic-representation aspects of Eisenstein series in some detail?
Thank you!
| https://mathoverflow.net/users/4245 | Eisenstein series as sections of line bundles on moduli spaces | Here is one construction:
We have the exact sequence
$$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k})
\to H^0(\text{cusps}, \omega^{\otimes k}\_{| \text{cusps}}).$$
(Here I am using $\omega$ for what you called $E$; this is the traditional notation
for modular forms people.) It is easy to define a Hecke action on the third $H^0$ so that this
exact sequence is Hecke equivariant.
The right hand map is surjective if $k > 2$, and its image has codimension one when
$k = 2$. In any event, write $\mathcal I$ to denote the image, so that
$$0 \to H^0(\omega^{\otimes k}(-\text{cusps})) \to H^0(\omega^{\otimes k})
\to \mathcal I \to 0$$
is short exact. One then shows that this short exact sequence has a unique Hecke
equivariant splitting; i.e. there is a uniquely determined Hecke equivariant subspace
$\mathcal E \subset H^0(\omega^{\otimes k})$ such that $\mathcal E$ projects isomorphically
onto $\mathcal I$. This space $\mathcal E$ is the space of weight $k$ Eisenstein series (for whatever level we are working at).
| 13 | https://mathoverflow.net/users/2874 | 34653 | 22,418 |
https://mathoverflow.net/questions/34620 | 24 | The standard examples of schemes that are not quasi-compact are either non-noetherian or have an infinite number of irreducible components. It is also easy to find non-separated irreducible examples. But are there other examples?
**Question**: Let $X$ be a locally noetherian scheme and assume that $X$ is irreducible (or has a finite number of irreducible components) and separated. Is $X$ quasi-compact (i.e., noetherian)?
If the answer is no in general, what conditions on $X$ are sufficient? Locally of finite type over a noetherian base scheme $S$? Fraction field finitely generated over a base? What if $X$ is regular? In general, the question is easily reduced to the case where $X$ is normal and integral.
It certainly feels like the answer is yes when $X$ is locally of finite type over $S$. Idea of proof: Choose an open dense affine $U\subseteq X$, choose a compactification $\overline{U}$ and modify $X$ and $\overline{U}$ such that the gluing $Y=X\cup\_U \overline{U}$ is separated. Then, $Y=\overline{U}$ (by density and separatedness) is proper and hence quasi-compact.
**Remark 1**: If $X\to S$ is a proper morphism, then the irreducible components of the Hilbert scheme Hilb(X/S) are proper. The subtle point (in the non-projective case) is the quasi-compactness of the components (which can be proven by a similar trick as outlined above).
**Remark 2**: If $X\to S$ is universally closed, then $X\to S$ is quasi-compact. This is question [23337](https://mathoverflow.net/questions/23337/is-a-universally-closed-morphism-of-schemes-quasi-compact).
| https://mathoverflow.net/users/40 | When is an irreducible scheme quasi-compact? | There are smooth counterexamples. Let $S\_0$ be a smooth separated irreducible scheme over a field $k$ with dimension $d > 1$, and $s\_0 \in S\_0(k)$. Blow up $s\_0$ to get another such scheme $S\_1$ with a $\mathbf{P}^{d-1}\_k$ over $s\_0$. Blow up a $k$-point $s\_1$ over $s\_0$ to get $S\_2$, and keep going. Get pairs $(S\_n, s\_n)$ so that the open complement $U\_n$ of $s\_n$ in $S\_n$ is open in $U\_ {n+1}$ and is strictly contained in it. Glue them together in the evident manner, to get a smooth irreducible $k$-scheme. It is locally of finite type, but is not quasi-compact (since the $U\_n$ are an open cover with no finite subcover). This is separated (either by direct consideration of affine open overlaps, or by using the valuative criterion).
| 21 | https://mathoverflow.net/users/3927 | 34656 | 22,421 |
https://mathoverflow.net/questions/34658 | 53 | For smooth $n$-manifolds, we know that they can always be embedded in $\mathbb R^{2n}$ via a differentiable map. However, is there any corresponding theorem for the topological category? (i.e. Can every topological manifold embed continuously into some $\mathbb R^N$, and do we get the same bound for $N$?)
| https://mathoverflow.net/users/8188 | Is there a Whitney Embedding Theorem for non-smooth manifolds? | I'm not sure about $\mathbb{R}^{2n}$, but you can embed them in $\mathbb{R}^{2n+1}$ using dimension theory. The theorem is that every compact metric space whose [covering dimension](http://en.wikipedia.org/wiki/Lebesgue_covering_dimension) is $n$ can be embedded in $\mathbb{R}^{2n+1}$. The example of non-planar graphs (which are $1$-dimensional) shows that this is the best you can do in general.
The classic source for this is Hurewitz-Wallman's beautiful book "Dimension Theory", which I recall being pretty readable to me when I was an undergraduate, though I haven't looked at it in a while. There is also a nice discussion of this in Munkres's book on point-set topology -- when I last taught a point-set topology class, I used this as one of the capstone theorems in the course.
| 57 | https://mathoverflow.net/users/317 | 34659 | 22,422 |
https://mathoverflow.net/questions/34646 | 8 | Let $A$ be a $C^\*$-algebra or some norm-closed algebra of operators on a Hilbert space.
In the old paper
Hille, E. *On Roots and Logarithms of Elements of a Complex Banach Algebra*, Math. Annalen, Bd. 136, S. 46--.57 (1958)
the question is studied, which elements $x \in A$ have the property that the exponential function is open at $x$ (i.e. every neighborhood of $x$ maps to a neighborhood of $\exp(x)$). Under some conditions on the spectrum of $x$, Hille shows that the answer is positive.
>
> **Question:** Is there an example of a unital $C^\*$-algebra (or operator algebra) with the property that the exponential function is not open?
>
>
>
More specifically: Is the exponential function open for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$?
Is there an example of a Banach algebra where the exponential function is not open?
EDIT: Jonas Meyer has clarified that $B(H)$ is a counterexample. It remains unclear what happens in general, or whether there is any description of the class of Banach algebras, where the exponential map is open. In particular, it remains unclear for the algebra of upper triangular operators on $\ell^2 {\mathbb N}$.
| https://mathoverflow.net/users/8176 | Is the set of exponentials open? | The exponential map is not open on the von Neumann algebra of all bounded linear operators on an infinite dimensional separable complex Hilbert space. A [1987 article by Conway and Morrel](http://www.ams.org/mathscinet-getitem?mr=870760) shows that the spectrum of an element of the interior of the image of the exponential map does not separate 0 from infinity. On the other hand, a bilateral shift *U* has spectrum equal to the unit circle centered at the origin, and every Borel logarithm on the circle applied to the unitary operator $U$ yields a preimage point for *U* under the exponential map. Hence *U* is in the image but not in the interior of the image.
I learned about Conway and Morrel's article from [this answer](https://mathoverflow.net/questions/154/can-you-describe-the-image-of-the-exponential-map-bh-bh/342#342) by David Speyer.
| 5 | https://mathoverflow.net/users/1119 | 34670 | 22,428 |
https://mathoverflow.net/questions/34674 | 11 | I've often heard that we can give examples of CW complexes that aren't homeomorphic to the realization of any simplicial set (although I haven't heard that there exist Kan complexes that aren't isomorphic to the total singular complex of a CGWH space. Are there?) Would someone mind providing an example of one (and an example for the opposite statement as well, if it is true)?
| https://mathoverflow.net/users/1353 | Example of a CW complex not homeomorphic to the realization of a simplicial set? | The mapping cylinder of a really messy continuous map $I\to I$
The nerve of the category in which there are two objects and each Hom set is a singleton.
| 13 | https://mathoverflow.net/users/6666 | 34677 | 22,432 |
https://mathoverflow.net/questions/34663 | 1 | Some times ago I posted [this](https://mathoverflow.net/questions/32138/reference-for-some-elementary-facts-about-principal-bundles) question here. There I carelessly assumed that if you have a set of sections of a vector bundle which span every fiber pointwise, they also generate the module of smooth sections over the smooth functions on the base manifold, I think that is true if your set of sections is finite. But in my situation for the horizontal bundle this is not the case. So I want to ask the following question:
Let $(P,\pi,B,G)$ be a principal bundle with total space $P$, base $B$, projection $\pi$ and structure group $G$. Let $\gamma \colon TP \to \mathrm{Lie}(G)$ a connection one-form, which induces the horizontal space $HP$. For a vector field $X \in \Gamma^\infty(TB)$ I write $X^h \in \Gamma^\infty(HP)$ for the horizontal lift of $X$ with respect to $\gamma$.
Is it true that $\Gamma^\infty(HP) = C^\infty(P)-\mathrm{Span}(X^h \mid X \in \Gamma^\infty(TB))$, i.e. that the horizontal lifts generate the $C^\infty(P)$-module of horizontal vector fields.
If so, can you give me a proof?
(For a vector bundle $E \to M$ I denote the smooth sections on $E$ by $\Gamma^\infty(E)$. $C^\infty(M)$ are the smooth functions on the manifold $M$)
Note that every manifold is assumed to be real, finite dimensional, $T\_2$, paracompact and smooth.
| https://mathoverflow.net/users/7538 | module of sections of the horizontal bundle | There is a result (for which I don't recall a reference, but you might prove it maybe with the smooth Serre-Swan theorem. See for example Nestruev, Smooth manifolds and observables), which says that for the sections of the pullback of a vector bundle $E\to M$ along a map $\phi:N\to M$ one has
$$\Gamma(\phi^\*(E))=C^\infty(N)\otimes\_{C^\infty(M)}\Gamma (E)$$
This is a canonical isomorphism obtained from the natural map $\Gamma(E)\to\Gamma(\phi^\*(E))$. Combining this with the fact that the horizontal bundle in your case is canonically isomorphic with the pullback of $TB$ to $P$ would give you a proof.
| 2 | https://mathoverflow.net/users/745 | 34678 | 22,433 |
https://mathoverflow.net/questions/33237 | 33 | Recall that a certain type of object admits an ADE classification if there is a notion of equivalence relative to which equivalence classes of objects of the given type can be placed in one-to-one correspondence with the collection of simply laced Dynkin diagrams. The simply laced Dynkin diagrams have themselves been classified into two infinite families (denoted $\mathrm{A}\_n$ and $\mathrm{D}\_n$) and three exceptional examples (denoted $\mathrm{E}\_6, \mathrm{E}\_7$, and $\mathrm{E}\_8$). See the [Wikipedia page](http://en.wikipedia.org/wiki/ADE_classification) for more details.
There is a veritable laundry list of objects that admit an ADE classification. Examples include (modulo a number of qualifiers that I don't want to get into):
* Semisimple Lie algebras
* Conformal field theories
* Tame quivers
* Platonic solids
* Positive definite quadratic forms on graphs
The list goes on. Accoding to the aforementioned wikipedia page, Vladmir Arnold asked in 1976 if there is a connection between these different kinds of objects which really explains why they all admit a common classification. The page also makes an offhand comment about how such a connection might be suggested by string theory.
I am hoping that somebody can explain some of the progress that has been made (if any) on Arnold's question. A good answer to this question is not one which explains the proofs that various different types of objects admit ADE classifications, nor one which aimlessly extends the list above. Rather, I would like to see someone take a collection of objects which on the surface are unrelated but which all have ADE classifications and then outline a deeper connection between them which at least suggests that they might all have a common classification. Bonus points if anyone can justify wikipedia's invocation of string theory in this context.
| https://mathoverflow.net/users/4362 | Is there a common genesis for ADE classifications? | I will first address the string theory part of the question.
String theory provides examples of physical systems admitting several descriptions that provide natural bridges between Kleinian singularities (and therefore Platonic solids), ALE spaces, quiver diagrams, ADE diagrams and two dimensional Conformal Field Theories.
The scene is given by **compactifications of string theory on Kleinian orbifolds** $M\_\Gamma=\mathbb{C}^2/\Gamma$ where $\Gamma$ is a discrete subgroup of $SU(2)$. The space $M\_\Gamma$ admits a **Kleinian singularity** at the origin. After studying this physical system, one is less surprised to see that Kleinian singularities, quiver diagrams, ALE spaces, ADE diagrams and 2 dimensional Conformal Field Theories all admit the same ADE classifications since they provide different descriptions of the same underlying physical system.
Michael Douglas and Gregory Moore have studied the compactification of string theory on Kleinian orbifold $M\_\Gamma$ using D-branes as probes of the geometry.
D-branes are extended objects on which strings can end.
D-branes provide a physical description of the geometry in terms of **supersymmetric gauge theories**. Such supersymmetric gauge theories are efficiently summarized by a **quiver diagram** with a very natural physical interpretation: the nodes correspond to D-branes with specific gauge groups on them and the links between the nodes are open strings ending on the branes.
The minimal energy configurations (the vacua) of these supersymmetric gauge theories are obtained finding the extrema of a potential whose construction is equivalent to the **hyperkhäler quotient** construction of **Asymptotic Locally Euclidian Spaces** (ALE spaces) first obtained by Kronheimer. ALE spaces are HyperKähler four dimensional real manifolds whose anti-self-dual metrics are asymptotic to a Kleinian orbifold $M\_\Gamma=\mathbb{C}^4/ \Gamma$. Physically ALE spaces described **gravitational instantons**.
ALE spaces provide small resolutions of the Kleinian singularities where the singular point is replaced by a system of spheres whose intersection matrix is equivalent to the Cartan matrix of an **ADE Dynkin diagram**. One can also consider Yang-Mills instantons on such spaces. The gauge group associated with the Yang-Mills instantons is given by the type of ADE diagram obtained by the resolution of the singularity. This was analyzed in the math literature by Kronheimer and Nakajima. Physically the ALE instantons moduli space is equivalent to the vacua of the gauge theory description of D-branes located at the singularities.
The link between D-branes on ALE spaces (or equivalently Kleinian singularities) and the ADE classification of **two dimensional Conformal Field Theories** (CFT) was studied by Lershe, Lutken and Schweigert. Although the geometry is singular, the CFT description is smooth. The 2 dimensional CFT is coming directly from the string description: as a string evolves it described a 2 dimensional surface called the string worldsheet. D-branes enter the CFT as boundary states. In the description of the CFT, one recovers **Arnold's ADE list of simple isolated singularities**.
**Updates**
I would like to comment on the non-stringy part of the question. This is motivated by the comments of Victor Protsak.
If one removes all the string theory interpretation in the discussion above. What is left is
Kronheimer's description of ALE spaces. Kronheimer's construction provides a beautiful realization of McKay's correspondence between Kleinian singularities, their crepant resolutions and ADE diagrams. This is reviewed in chapter 7 of Dominic Joyce's book "Compact Manifolds with Special Holonomy". From that perspective, the string theory description provides a physical interpretation of Kronheimer's construction and adds a natural link with quiver diagrams and 2 dimensional Conformal Field Theories.
| 8 | https://mathoverflow.net/users/4046 | 34680 | 22,435 |
https://mathoverflow.net/questions/34673 | 48 | What is a good reference for the following fact (the hypotheses may not be quite right):
>
> Let $X$ and $Y$ be projective varieties over a field $k$. Let $\mathcal{F}$ and $\mathcal{G}$ be coherent sheaves on $X$ and $Y$, respectively. Let $\mathcal{F} \boxtimes \mathcal{G}$ denote $p\_1^\*(\mathcal{F}) \otimes\_{\mathcal{O}\_{X \times Y}} p\_2^\* \mathcal{G}$. Then
> $$H^m(X \times Y, \mathcal{F} \boxtimes \mathcal{G}) \cong \bigoplus\_{p+q=m} H^p(X,\mathcal{F}) \otimes\_k H^q(Y, \mathcal{G}).$$
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Note: Wikipedia leads me to believe that this may be related to Theorem 6.7.3 in EGA III2, but I find this theorem quite intimidating. Although I would be willing to study this if there is no more basic reference, I would at least like some confirmation that I am studying the right thing.
| https://mathoverflow.net/users/5094 | Kunneth formula for sheaf cohomology of varieties | The treatment in EGA is indeed intimidating, but in fact over a field the formula is not hard to prove. You only need $X$ and $Y$ to be separated schemes over $k$, and $\mathcal{F}$ and $\mathcal{G}$ to be quasi-coherent. Then cover $X$ and $Y$ by affine open subsets $\{U\_i\}$, and $\{V\_j\}$, and write down the Čech complex for $\mathcal{F}$ and $\mathcal{G}$ with respect to these two coverings, and the Čech complex of $\mathcal{F} \boxtimes \mathcal{G}$ with respect to the covering $U\_i \times V\_j$. It is not hard to see that the last is the tensor product of the first two; then the thesis follows from Eilenberg-Zilber (or however you want to call the fact that the cohomology of the tensor product of two complexes over a field is the tensor product of the cohomlogies).
| 32 | https://mathoverflow.net/users/4790 | 34683 | 22,437 |
https://mathoverflow.net/questions/34640 | 12 | Can someone please explain the difference between local rigidity and infinitesimal rigidity? Does either version of rigidity imply the other?
In particular, I'm thinking about Weil's rigidity theorem for hyperbolic metrics on manifolds of dimension $\geq 3$. I've seen it referred to as both local and infinitesimal, which further adds to my confusion about the distinction.
| https://mathoverflow.net/users/8183 | Local vs. infinitesimal rigidity | Infinitesimal rigidity implies local rigidity, but not conversely. Local rigidity means a representation has no deformations, whereas infinitesimal rigidity means the natural tangent space to the character variety is 0-dimensional (this tangent space is a certain cohomology group with twisted coefficients). Weil proved both in the context you mention. See e.d. David Fisher's [survey paper](http://library.msri.org/books/Book54/files/02fisher.pdf), where Weil's theorem is Theorem 3.2.
See also Section 5 of [this paper](https://projecteuclid.org/journals/experimental-mathematics/volume-15/issue-3/Computing-Varieties-of-Representations-of-Hyperbolic-3-Manifolds-into-SLfR/em/1175789760.full).
| 14 | https://mathoverflow.net/users/8197 | 34684 | 22,438 |
https://mathoverflow.net/questions/34702 | 3 | $A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. Then is it true that $\mbox{depth }M\le\mbox{depth }A$ just like $\mbox{dim }M\le\mbox{dim }A$? I don't see any relation between an $M$-sequence and an $A$-sequence. At least I know it is true when $\mbox{inj.dim }M<\infty$, from the relation $\mbox{depth }M\leq\mbox{dim }M\leq\mbox{inj. dim }M=\mbox{depth }A\leq\mbox{dim }A$. But what happens when $\mbox{inj.dim }M=\infty$? Another inequality I'm not quite sure about when $\mbox{inj.dim }M=\infty\ $: is it true that $\mbox{dim }M\leq\mbox{depth }A$?
| https://mathoverflow.net/users/5292 | Depth and dimension | $A=k[[x,y]]/(x^2,xy)$ then depth$(A)=0$. Let $M=R/(x)=k[[y]]$ then $y$ is a nonzerodivisor on $M$.
| 11 | https://mathoverflow.net/users/7763 | 34708 | 22,451 |
https://mathoverflow.net/questions/34704 | 9 | $A$ a Noetherian local ring, $M\neq 0$ a finite $A$-module. I'm not quite sure about the relation between finiteness of projective and injective dimensions of $M$. Does the finiteness (or infiniteness) of one necessarily imply the finiteness (or infiniteness) of another?
| https://mathoverflow.net/users/5292 | Projective & injective dimensions | To complement Mariano's answer: If finite projective dimension implies finite injective dimension for any module $M$, then $R$ better have finite injective dimension (the converse is also quite easy).
The local rings $R$ which have finite inj. dim. over themselves are also known as [Gorenstein rings](http://arxiv.org/abs/math/0209199). In fact, a theorem by Foxby says that $R$ possesses a module of both finite proj. and inj. dim. if and only if $R$ is Gorenstein.
| 12 | https://mathoverflow.net/users/2083 | 34709 | 22,452 |
https://mathoverflow.net/questions/33831 | 9 | The following definition is from:
* Dmitry Roytenberg, "AKSZ-BV formalism and Courant algebroid-induced topological field theories", *Letters in Mathematical Physics*, 2007 vol. 79 (2) pp. 143-159, MR2301393.
>
> A *graded manifold* $M$ over base $M\_0$ is a sheaf of $\mathbb Z$-graded commutative algebras ${\rm C}(M)$ over a smooth manifold $M\_0$ locally isomorphic to an algebra of the form ${\rm C}^\infty(U) \otimes {\rm S}(V)$ where $U \subseteq M$ is an open set, $V$ is a graded vector space whose degree-zero component $V\_0$ vanishes, and ${\rm S}(V)$ is the free graded-commutative algebra on $V$.
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Presumably there is an equivalent definition that builds in the axioms for $M\_0$ to be a smooth classical manifold — we could start with the topological space with a sheaf of Frechet algebras locally isomorphic to something.
Recall, from, for example the MO question [Algebraic description of compact smooth manifolds?](https://mathoverflow.net/questions/5344/algebraic-description-of-compact-smooth-manifolds), that all classical manifolds are *affine* in the sense that, although presented as sheaves, a complete invariant is the algebra of global functions.
>
> **Question:** Is it true that the algebra of global sections of the sheaf ${\rm C}(M)$ is a complete invariant? I.e. can I recover a graded manifold from its algebra of global smooth functions?
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The follow-up question would be to describe, a la the above-linked MO question, which algebras are algebras of smooth functions on a graded manifold. The orthogonal follow-up question is the same one for "dg manifolds", which is a graded manifold with a square-zero degree $-1$ vector field, thought of as a derivation of the sheaf algebras of functions: does knowing the derivation on global sections determine it on all local sections?
---
Since no one has yet posted an answer, let me generalize the question to include:
>
> **Question:** Is the above definition the right one?
>
>
>
Now let me motivate this generalization. First, the answer to my original question above is trivially "yes" provided that the underlying manifold $M\_0$ can be read from the algebra, as then I would have access to the right partitions of unity to really get my hands on the whole sheaf. This certainly happens when the "fiber" (whose linear functions are) $V$ has only positive gradings: then the manifold is recovered from the degree-zero subalgebra. But if $V$ has both positive and negative degrees, then the degree-zero subalgebra is too big. Instead, the base $M\_0$ should be recovered as the maximal degree-zero quotient algebra, and I have less intuition for whether such a thing should exist.
But then it's clear that the sheaf in the above definition is not, generally, a sheaf of Frechet algebras. For example, consider the vector space $\mathbb R^{1t^2 + 1t^{-2}}$ with one dimension in degree $2$ and one in degree $-2$, and call the coordinate functions $x$ and $y$. Then there is a quadratic map $\mathbb R^{1t^2 + 1t^{-2}} \to \mathbb R^1$ corresponding to the subalgebra of polynomials in $xy$, and this degree-zero subalgebra is not Frechet-complete. Rather, to exhibit the map as a map of *smooth* rather than algebraic spaces requires that ${\rm C^\infty}(\mathbb R^{1t^2 + 1t^{-2}})$ include the Frechet algebra ${\rm C^\infty}(\mathbb R^1)$ as a subalgebra.
So what's going on? If I add the words "Frechet completion" in the correct spot in the above definition, is it the answer to the first question "yes"?
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There is another, older meaning of "affine", namely "properly embeds into (finite-dimensional) affine space". Before adding this extra content, the title of this question included a parenthetical "and in what sense". So at the risk of making this question long enough that the correct answer is a good reference:
>
> **Question:** For the correct definition of "graded manifold", is it true that every graded manifold embeds smoothly into some finite-dimensional graded vector space"?
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>
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| https://mathoverflow.net/users/78 | Is every graded manifold affine, and is this definition of graded manifold the right one? | First of all, you're completely right in that I didn't need to stipulate that $M\_0$ be a manifold in the definition: it follows from a graded manifold being a locally ringed space with a particular local model (it was a physically motivated survey article, so I wasn't much concerned with foundational issues). It also follows that a map of graded manifolds must be *polynomial* in coordinates of nonzero degree, so things like $e^{xy}$ on $\mathbb{R}[-2]\times\mathbb{R}[2]$ are not allowed. Furthermore, the body of $\mathbb{R}[-2]\times\mathbb{R}[2]$ is a single point, so there are no non-constant maps from $\mathbb{R}[-2]\times\mathbb{R}[2]$ to $\mathbb{R}$; in particular, your example is not a map of graded manifolds. In general, the body $M\_0$ of a graded manifold $M$ can be recovered as the spec of $C^0(M)/(I\cap C^0(M))$, where $I$ is the ideal generated by coordinates of non-zero degree -- much like it is done for supermanifolds. Of course, in the non-negatively graded case, $M\_0$ can also be recovered as $0\cdot M$.
Now, manifolds as well as supermanifolds and graded (super)manifolds are "affine" in the sense you asked. More precisely, taking global sections of the structure sheaf defines a fully faithful (contravariant) functor to $\mathbb{R}$-algebras (notice that you don't need a compactness assumption on the body, nor any topology on the algebra, which is a blessing for those who, like me, are put off by functional analysis). For manifolds this is classical (see eg. Cor 35.10, p. 301 of Kolar-Michor-Slovak
<http://www.emis.de/monographs/KSM/kmsbookh.pdf> ), the super/graded case follows easily. The differentials are okay as well, since they're nothing but vector fields.
Finally, whether my definition is the "right" one depends on the application one has in mind, I guess. For me, coordinates of positive degree generate the symmetries, while those of negative degrees serve to cut out the zero locus of the homological vector field (the solution set of Maurer-Cartan or Euler-Lagrange equations) and resolve its singularities. The relevant completions are then pro/ind completions, not something as awful as Frechet. Of course, that doesn't rule out applications for which the latter would be relevant.
ADDENDUM. Oh yes, graded manifolds are affine in the other sense as well, at least the non-negatively graded ones are (I haven't thought through the general case). It follows from the fact that they are isomorphic to total spaces of graded vector bundles over manifolds, by a result analogous to Batchelor's theorem for supermanifolds.
| 6 | https://mathoverflow.net/users/8203 | 34721 | 22,458 |
https://mathoverflow.net/questions/34719 | 11 | A quantitative form of the twin prime conjecture asserts that the the number of twin primes less than $n$ is asymptotically equal to $2\, C\, n/ \ln^2(n)$ where $C$ is the so-called twin prime constant. A variety of sieve methods (originating with Brun) can be used show that the number of twin primes less than $n$ is at most $A\, n/ \ln^2 (n) $ for some constant $A>2C$. My question is: What is the smallest known value of $A$? I'd also be interested in learning what the best known constants are for the prime k-tuple conjecture?
| https://mathoverflow.net/users/630 | What is the best known upper bound for the number of twin primes? | J Wu, Chen's double sieve, Goldbach's conjecture, and the twin prime problem, Acta Arith 114 (2004) 215-273, MR 2005e:11128, bounds the number of twin primes above by $2aCx/\log^2x$, with $C=\prod p(p-2)/(p-1)^2$, and $a=3.3996$; I don't know whether there have been any improvements.
| 19 | https://mathoverflow.net/users/3684 | 34723 | 22,460 |
https://mathoverflow.net/questions/34697 | 7 | Let $\mathcal{C}$ be a strict 2-category. A corollary of the bicategorical Yoneda lemma says that any pseudofunctor $\mathcal{C} \to \operatorname{Cat}$ is pseudonaturally equivalent to a strict 2-functor. I would like to know if the "next level" of strictification is true; namely, is it true that any pseudonatural transformation of strict 2-functors $\mathcal{C} \to \operatorname{Cat}$ is isomorphic (via an invertible modification) to a 2-natural transformation?
The specific case of this that I am interested is when $\mathcal{C}$ is the delooping of a monoidal category: the first statement says that any left $\mathcal{C}$-module can be made into a "strict" left $\mathcal{C}$-module, while the second would imply that any morphism of left $\mathcal{C}$-modules can be strictified.
| https://mathoverflow.net/users/396 | Is a pseudonatural transformation of strict 2-functors to Cat isomorphic to a 2-natural transformation? | No, it is not true. For a counterexample, let C be the delooping of the group Z/2 (regarded as a discrete monoidal category). A strict C-module is then a category equipped with an involution, a strict C-module morphism is a functor preserving the involution strictly, and a pseudo C-module morphism preserves the involution up to coherent isomorphism. In particular, a strict C-module morphism must map an object fixed by the involution to another fixed object, but a pseudo C-module morphism can map a fixed object to one which is only fixed up to isomorphism. Thus, if A is a C-module with fixed objects and B is a C-module with no fixed objects, there can be pseudo C-module morphisms from A to B, but there cannot be any strict C-module morphisms from A to B.
It is, however, true that any strict (or even pseudo) 2-functor $F:C\to Cat$ is pseudonaturally equivalent to a strict 2-functor $F':C\to Cat$ which has the property that any pseudonatural transformation $F'\to G$ is isomorphic to a strict 2-natural transformation. An $F'$ with this property is called *flexible* (if you're a 2-category theorist) or *cofibrant* (if you're a homotopy theorist). The possibility of flexible replacement follows from generalities about 2-monads: there is a strict 2-monad T on the 2-category $Cat^{ob(C)}$ for which strict T-algebras are strict 2-functors $C\to Cat$, strict T-morphisms are strict 2-natural transformations, and pseudo T-morphisms are pseudonatural transformations. The general coherence theorems of Power and Lack (see the papers "A general coherence result" and "Codescent objects and coherence") apply to this 2-monad and specialize to the statement I quoted above. The homotopy theorist can instead construct a model structure on the 2-category of strict 2-functors and strict 2-natural transformations in which the cofibrant objects are cofibrant/flexible; see Lack's paper "Homotopy-theoretic aspects of 2-monads."
| 13 | https://mathoverflow.net/users/49 | 34731 | 22,464 |
https://mathoverflow.net/questions/34754 | 17 | If we use homology and cohomology over a field $k$, if a space has homology and cohomology groups of finite type in each degree, then $H\_\ast(X;k)$ is dual to $H^\ast(X;k)$ using the universal coefficient theorem for cohomology.
Now, suppose I have a fibration $F \to E \to B$ such that $F$ and $B$ have homology and cohomology over $k$ of finite type in each degree and $\pi\_1(B) = 0$ for simplicity. Certainly, the $E\_2$-page of the cohomology Serre spectral sequence will be dual to the $E^2$-page homology Serre spectral sequence. My first question is: Is it also true that the differentials for the cohomology spectral sequence are dual as a linear map to the differentials of the homology spectral sequence, and vice versa?
Secondly, the cohomology Serre spectral sequence is a multiplicative one. Is the homology one comultiplicative? If so, is the product for cohomology dual to the coproduct for homology?
Finally, if all of this holds, to which extend can it generalized?
| https://mathoverflow.net/users/798 | Are the homology and cohomology Serre spectral sequences dual to each other? | Yes, this is the case. This is easiest to see using the exact couple formalism. Suppose you have an exact couple, meaning a long exact sequence consisting of maps $i: D \to D$, $j: D \to E$, $k: E \to D$, where all the terms are (possibly graded) vector spaces over a field. Because dualization is exact (as is taking levelwise duals of graded objects), you can verify that the sequence of maps $i^\*: D^\* \to D^\*$, $k^\*: D^\* \to E^\*$, $j^\*:E^\* \to D^\*$ also forms an exact couple and the associated spectral sequence is dual to the original spectral sequence.
The "short" construction of the Serre spectral sequence is obtained in the case where B is a CW-complex (one can reduce to this case by replacing B with a weakly equivalent object), and filtering the total space by the preimages of the skeleta. The long exact sequences in homology and cohomology associated to this filtration are dual to each other, and so when one creates the graded vector spaces forming the exact couple by summing up one gets that the homology and cohomology exact couples are dual to each other.
| 19 | https://mathoverflow.net/users/360 | 34756 | 22,479 |
https://mathoverflow.net/questions/34771 | -1 | Do Gorenstein rings necessarily have finite projective dimensions?
| https://mathoverflow.net/users/5292 | Do Gorenstein rings necessarily have a finite projective dimension (as a module over itself)? | Take $A=k[x]/(x^2)$ for a field $k$. This is a self-injective $k$-algebra (that is, it is an injective module over itself), so it is Gorenstein. Yet the residue field is of infinite projective dimension.
| 2 | https://mathoverflow.net/users/1409 | 34780 | 22,495 |
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