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https://mathoverflow.net/questions/22369 | 7 | I would just like a clarification related to closed subschemes.
If $(X,{\cal O}\_X)$ is a locally ringed space and $A\subset X$ is any subset with the subspace topology then $i^{-1}{\cal O}\_X$ will be a sheaf of rings on $A$ where $i:A\rightarrow X$ is the inclusion map. (Recall that the inverse image $i^{-1}{\cal O}\_X$ is the sheafification of the presheaf $U \mapsto \lim\_{V\supset i(U)} {\cal O}\_X(V)$ for $U\subseteq A$ open, where the inductive limit is over all open subsets $V$ of $X$ containing $U$.)
Is the reason why we don't do this (and instead start talking about closed subschemes, etc. etc.) just that $(A,i^{-1}{\cal O}\_X)$ need not be a scheme even when $X$ is?
Put differently: given any closed subset of a scheme there will be many ways to make it a closed subscheme. What is the relation between the locally ringed spaces on a closed subset making it a closed subscheme and the locally ringed space I have described above which we obtain by pulling back the structure sheaf via the inclusion map.
| https://mathoverflow.net/users/1148 | Closed subschemes and pulling back the structure sheaf via the inclusion map | It might help to consider the extreme case when $x$ is a closed point of $X$,
and $i$ is the inclusion $\{x\} \hookrightarrow X$. The pullback $i^{-1}\mathcal O\_X$
is then the stalk of $\mathcal O\_X$ at $x$, i.e. the local ring $A\_{\mathfrak m}$,
if Spec $A$ is an affine n.h. of $x$ in $X$, and $\mathfrak m$ is the maximal ideal in $A$
corresponding to the closed point $x$.
Now a single point, with a local ring $A\_{\mathfrak m}$ as structure sheaf, is not a scheme
(unless $A\_{\mathfrak m}$ happens to be zero-dimensional).
Moreover, the restriction map from sections of $\mathcal O\_X$ over $X$ to
section of $i^{-1}\mathcal O\_X$ over $x$ is *not* evaluation of functions at $x$
(which corresponds to reducing elements of $A$ modulo $\mathfrak m$), but is rather
just passage to the germs of functions at $x$.
The idea in scheme theory is that sections of $\mathcal O\_X$ should be functions,
and restriction to a closed subscheme should be restriction of functions. In particular,
restriction to a closed point should be *evaluation* of the function (if you like, the constant term of the Taylor series of the function), not passage to the germ (which is like
remembering the whole Taylor series).
If you bear this intuition in mind, and think about the case of a closed point, you will
soon convince yourself that the general notion of closed subscheme is the correct one:
If we restrict functions to the locus cut out by an ideal sheaf $\mathcal I$, or
(in the affine setting) by an ideal $I$ in $A$, then two sections will give the
same function on this locus if they coincide mod $\mathcal I$ (or mod $I$ in the
affine setting), and so it is natural to define the structure sheaf to then be
$\mathcal O\_X/\mathcal I$ (or to take its global sections to be $A/I$ in the affine
settin), rather than $i^{-1}\mathcal O\_X$.
| 25 | https://mathoverflow.net/users/2874 | 22383 | 14,767 |
https://mathoverflow.net/questions/22392 | 1 | Consider the following polynomial series:
$S(x) = \sum\_{i=1}^{\infty}(-1)^{i+1}x^{i^{2}}$
Between 0 and 1, this looks like a well-behaved function - is there any way to write this function in this interval without using a series?
Given $0 < S(x) < 1$, I need to solve the equation for $x$ (in the 0 to 1 interval), but an analytic solution would be much nicer than a numerical one...
| https://mathoverflow.net/users/5579 | Polynomial series | I don't know what you mean by "polynomial series" since your function doesn't seem to have much to do with polynomials (perhaps you could elaborate?).
$S(x) = -\frac12 \theta(\frac12, \frac{\log x}{\pi i }) - 1$, where $\theta$ is [Jacobi's theta function](http://en.wikipedia.org/wiki/Theta_function). I don't know of any nice algebraic methods to take inverses of modular forms (even if you restrict to real $x$, i.e., $\tau$ purely imaginary), so I'm not sure if this is the sort of answer you want.
| 4 | https://mathoverflow.net/users/121 | 22394 | 14,775 |
https://mathoverflow.net/questions/22398 | 11 | In the literature and in some posts here, there has been variation in the undefined use of the term "block" for a category of modules over a ring, or more abstractly an abelian category (all of which are categories of modules by Freyd-Mitchell). This raises the natural question:
>
> What is meant by a "block" in an abelian category?
>
>
>
The concept originates to some extent in the modular representation theory of finite groups or their group algebras pioneered by Richard Brauer.
Here a block is just an indecomposable two-sided ideal of the group algebra,
corresponding to a primitive central idempotent. But in later developments the
language of homological algebra plays a greater role than the group algebra or its center: the category of modules decomposes into a direct sum of subcategories,
which are as small as possible relative to permitting no nontrivial extensions among their simple objects (irreducible representations). This approach generalizes well to other situations, where a center or central characters may be elusive and where the decomposition may be infinite, etc. By now "blocks" occur in many areas of representation theory influenced by classical Lie theory: algebraic groups, restricted enveloping algebras, quantum analogues,
finite $W$-algebras, Cherednik algebras, Kac-Moody algebras and groups, Lie superalgebras.
There is some inconsistency in the literature about allowing "blocks" which might be further decomposed into direct sums. In classical or Kac-Moody Lie theory
this usually reflects the special influence of infinitesimal/central characters. But full centers in Lie theory and related quantum groups may be unknown or unneeded, e.g., the approach Kac took to his analogue of the Weyl character formula for integrable modules of an affine Lie algebra relied just on a single Casimir-type operator (which works equally well in the classical finite-dimensional case). In Jantzen's book *Representations of Algebraic Groups* (AMS, 2003), the discussion of blocks for algebraic group schemes in II.7.1 is careful but not completely general.
In practice looser definitions of "block" than the homological one work well enough in many settings, but it creates some confusion when the word is used with no definition at all. Is there a single convention which reduces in familiar cases to older usage? In a recent comment to another post I paraphrased Humpty Dumpty ("my remote ancestor"), who actually said: "When *I* use a word it means just what I choose it to mean --- neither more nor less."
But communication is better when the short and convenient word "block" starts out with a common meaning.
| https://mathoverflow.net/users/4231 | What is a "block" in an abelian category? | Here's a definition of blocks taken from [Comes-Ostrik](http://arxiv.org/abs/0910.5695) (which just happened to be the first paper that came to mind that I knew talks about blocks, it's not a standard reference for this):
>
> Let A denote an arbitrary F-linear category. Consider the weakest equivalence relation on the set of isomorphism classes of indecomposable objects in A where two indecomposable objects are equivalent whenever there exists a nonzero morphism between them. We call the equivalence classes in this relation blocks. We will also use the term block to refer to a full subcategory of A generated by the indecomposable objects in a single block.
>
>
>
| 12 | https://mathoverflow.net/users/22 | 22400 | 14,777 |
https://mathoverflow.net/questions/22390 | 2 | Is there a general theory of embeddings of the (total variety of) the tangent bundle on a (nonsingular) projective variety into projective space? I suppose what I really mean is (and to be more precise about the projective space), if $\mathbb{P}$ is the *projective completion* of $T\_X\rightarrow X$, then what can be said about the relative dimension of "infinity": $D=\mathbb{P}\backslash T\_X$?
I apologize if this question is vague. Any thoughts or references will be greatly appreciated.
Thanks!
| https://mathoverflow.net/users/5395 | How far is the tangent bundle from projective space? | The simplest way to get a "projective completion" is to consider the projectivization on $X$ of $T\_X \oplus L$ for some line bundle $L$ on $X$. In this case the complement will be the projectivization of $T\_X$ and will have codimension 1. Sometimes you can contract this completion to get smaller complement, e.g. if $X$ is a curve (choose $L$ in such a way that $\omega\_X\otimes L$ is very ample, then $P(T\_X \oplus L) = P(T\_X\otimes L^{-1} \oplus O)$ which is a blowup of the projective cone over $X$ in the embedding given by $\omega\_X\otimes L$).
| 7 | https://mathoverflow.net/users/4428 | 22407 | 14,783 |
https://mathoverflow.net/questions/22410 | 3 | I'm trying to find out more about geometry of surfaces and, in particular, Gaussian curvature. I understand that it can be defined in terms of the principal curvatures (extrinsically) and also intrinsically, and that the result that it can be equivalently defined in these two ways was a significant result. Does anyone know where I can find a nice, clear proof of the equivalence of these definitions and perhaps some historical/background information on their importance? Thanks in advance.
| https://mathoverflow.net/users/5431 | Equivalent definitions of Gaussian curvature | I like the presentation of the Theorema Egregium in *A Comprehensive Introduction to Differential Geometry* (Volume 2) by Michael Spivak. A translation of the original paper by Gauss and the historical background can be found in *General investigations of curved surfaces of 1827 and 1825* by James Caddall Morehead and Adam Miller Hiltebeitel (the 2005 Dover edition).
| 5 | https://mathoverflow.net/users/5371 | 22414 | 14,789 |
https://mathoverflow.net/questions/22339 | 7 | Baer's criterion can be generalized as follows: Let $A$ be an abelian category satisfying (AB3-5) with a generator $R$ and let $T : A^{o} \to Set$ be a continuous functor such that $T(R) \to T(I)$ is surjective for all subobjects $I \subseteq R$. Then for every monomorphism $M \to N$ the map $T(N) \to T(M)$ is surjective. If $T$ is representable and $A=R-Mod$, this becomes the usual Baer's criterion. The proof is [simple](http://maddin.110mb.com/baer.pdf).
* Does anyone has come across this theorem?
* Are there applications to non-representable functors?
* Can we somehow put the proof into a general pattern of the form: A statement about monomorphisms can be proven on a "generating system"?
* There are striking similarities with the proof that the cohomology of flabby sheaves vanishes. Is there a common generalization?
| https://mathoverflow.net/users/2841 | Baer's criterion for functors | Regarding the original question (with $A=R$-$\mathbf{Mod}$), I think that
by SAFT *any* continuous functor $A^{\mathrm{op}}\to \mathbf{Set}$ is representable,
and hence the assertion in the original question does not generalize
Baer's theorem.
In detail (with $A=R$-$\mathbf{Mod}$):
(\*) $R$ is a generator in $A$, and hence a cogenerator in
$A^{\mathrm{op}}$.
(\*) $A$ is co-well-powered, because there is a bijection between the
quotient objects of $M\in A$ and the set of submodules of $M$, and the
latter set is small (since by assumption $M$ is small). It follows
that $A^{\mathrm{op}}$ is well-powered.
(\*) $A$ is small cocomplete (as is any $\tau$-algebra, for
$\tau=$(operations, identities)), and hence $A^{\mathrm{op}}$ is small
complete.
(\*) Both $A^{\mathrm{op}}$ and $\mathbf{Set}$ have small hom-sets.
So, all the conditions of SAFT hold for a functor
$A^{\mathrm{op}}\to\mathbf{Set}$,
and hence any such continuous functor has a left adjoint. Now, if a
functor $G\colon A^{\mathrm{op}}\to\mathbf{Set}$ has a left adjoint
then it is surely representable: Saying that a functor $G$ is
representable is like saying that there is a universal arrow from a
one-object set $1$ to $G$ (Prop. 3.2.2, p. 60 in Mac Lane), and for
this we can take the unit $\eta\_1\colon 1\to G(F1)$ (with $F$ the left
adjoint of $G$).
(See also the discussion on Watt's theorem on p. 131 of Mac Lane).
I am not sure about the general case of the edited question
(where $A$ is an arbitrary abelian category with a generator +
AB3--AB5). Cocompleteness holds by AB3 (as I have seen in
[Wikipedia](http://en.wikipedia.org/wiki/Abelian_category)
), but I do
not know enough to say anything about the question of being co-well-powered.
| 2 | https://mathoverflow.net/users/2734 | 22421 | 14,793 |
https://mathoverflow.net/questions/22423 | 2 | Let's say a morphism $f:X\to Y$ is **compactifiable** if it admits a factorization $f = pj$ with $j:X\to P$ an open immersion and $p:P\to Y$ proper.
In SGA 4 Exp. XVII, Deligne says that Nagata proved that any morphism of separated integral northerian schemes is compactifiable but that he didn't understand the proof.
My questions:
* Where can I find a proof of Nagata's theorem?
* What about the complex analytic setting?
| https://mathoverflow.net/users/1985 | Compactifiable morphisms | Brian Conrad has written up a proof of Nagata's theorem, starting from notes of Deligne: <http://math.stanford.edu/~conrad/papers/nagatafinal.pdf>. About the analytic case, I have no idea.
| 5 | https://mathoverflow.net/users/4790 | 22424 | 14,795 |
https://mathoverflow.net/questions/22075 | 12 | This is not my field, a friend needs the answer for the following question. Suppose we have a decreasing probability function, $p: N \rightarrow [0,1]$ such that $sum\_n p(n) = \infty$. Take the graph where we connect two integers at distance d with probability $p(d)$. Will this graph be connected with probability one?
I see that if the sum is convergent, then we almost surely have an isolated vertex (unless $p(1)=1$), so this would be "sharp". One possible approach would be to take the path that starts from the origin and if it is at n after some steps, then next goes to the smallest number that is bigger than $n$ and is connected to $n$ and to show that this path has a positive density with probability one. Is this second statement true?
I am sure that these are easy questions for anyone who knows about this.
| https://mathoverflow.net/users/955 | Connectedness of random distance graph on integers | All right. Here goes, as promised. We shall work with a big circle containing a huge number $N$ of points and a sequence of probabilities $p\_1,\dots,p\_L$ such that $\sum\_j p\_j=P$ is large (so we never connect points at the distance greater than $L$ but connect points at the distance $d\le L$ with probability $p\_d$). If $N\gg L$ and $p\_j<1$ for all $j$, the probability of a connected path going around the entire circle is extremely small, so the problem is essentially equivalent to the one on the line. I chose the circle just to make averaging tricks technically simple (otherwise one would have to justify some exchanges of limits, etc.). Fix $\delta>0$.
Our aim will be to show that with probability at least $1-2\delta$, we have $\sum\_{j\in E\_0} p\_{|j|}\ge P$ where $E\_0$ is the connected component of $0$ and integers are understood modulo $N$, provided that $P>P(\delta)$. This, clearly, implies the problem (just consider the connected component of $0$ in the subgraph with even vertices only; whatever it is, the edges going from odd vertices to even vertices are independent of it, so we get $0$ joined to $1$ with probability $1$ in the limiting line case with infinite sum of probabilities).
We shall call a point $x$ good if $\sum\_{y\in E\_x}p\_{|y-x|}\ge P$. We will call a connected component $E$ with $m$ points good if at least $(1-\delta)m$ its points are good.
Fix $m$. Let's estimate the average number of points lying in the bad components. To this end, we need to sum over all bad $m$-point subsets $E$ the probabilities of the events that the subgraph with the set of vertices $E$ is connected and there are no edges going from $E$ elsewhere and then multiply this sum by $m$. For each fixed $E$ these two events are independent and, since $E$ is bad, there are at least $\delta m$ vertices in $E$ for which the probability to not be connected with a vertex outside $E$ is at most $e^{-P}$ (the total sum of probabilities of edges emanating from a vertex is $2P$ and only the sum $P$ can be killed by $E$). Thus, the second event has the probability at most $e^{-\delta P m}$ for every bad $E$ and it remains to estimate the sum of probabilities to be connected.
We shall expand this sum to all $m$-point subsets $E$. Now, the probability that subgraph with $m$ vertices is connected does not exceed the sum over all trees with the set of vertices $E$ of the probabilities of such trees to be present in the graph. Thus, we can sum the probabilities of all $m$-vertex trees instead.
We need an efficient way to parametrize all $m$-trees. To this end, recall that each tree admits a route that goes over each edge exactly twice. Moreover, when constructing a tree, in this route one needs to specify only new edges, the returns are defined uniquely as the last edge traversed only once by the moment. Thus, each $m$ tree can be encoded as a starting vertex and a sequence of $m-1$ integer numbers (steps to the new vertex) interlaced with $m-1$ return commands. For instance, (7;3,2,return,-4,return,return) encodes the tree with vertices 7,10,12,6 and the edges 7--10, 10--12, 10--6. Well I feel a bit stupid explaining this all to a combinatorist like you...
Now when we sum over all such encodings, we effectively get $N$ (possibilities for the starting vertex) times the sum the products of probabilities over all sequences of $m-1$ integers multiplied by the number of possible encoding schemes telling us the positions of the return commands. (actually a bit less because not all sequences of integers result in a tree). Since there are fewer than $4^{m-1}$ encoding schemes, we get $4^{m-1}(2P)^{m-1}$ as a result. Thus the expected number of bad $m$-components is at most $N\cdot 4^{m-1}(2P)^{m-1}e^{-\delta Pm}$. Even if we multiply by $m$ (which is not really necessary because each tree is counted at least $m$ times according to the choice of the root) and add up over all $m\ge 1$, we still get less than $\delta N$ if $P$ is large.
Now we see that the expected number of bad points is at most $2\delta N$ (on average at most $\delta N$ points lie in the bad components and the good components cannot contain more than $\delta N$ points by their definition). Due to rotational symmetry, we conclude that the probability of each particular point to be bad is at most $2\delta$.
The end.
| 9 | https://mathoverflow.net/users/1131 | 22425 | 14,796 |
https://mathoverflow.net/questions/22393 | 9 | Is the following statement true?
>
>
> >
> > Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps $R\to S\_{\mathfrak{p}}$ are formally smooth for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally smooth.
> >
> >
> >
>
>
>
I've looked all over, and I have not been able to find a proof or counterexample for this claim. It might even be an open problem. This statement is true for formally unramified and formally étale morphisms, but the proof for formally étale morphisms uses the fact that the module of Kähler differentials is $0$, so it doesn't seem like the same approach will work for the formally smooth case.
| https://mathoverflow.net/users/1353 | Is formal smoothness a local property? | I think this works. Suppose we have a ring $R$ and an $R$-module $M$ all of
whose localisations are projective and consider $S=S^\ast\_RM$, the symmetric algebra
on $M$. Then $R \rightarrow S$ is formally smooth precisely when $M$ is
projective. Let $\mathfrak p$ be a prime ideal of $S$ and $\mathfrak q$ the
prime ideal of $R$ lying below $\mathfrak p$. Then $S\_{\mathfrak p}$ is a
localisation of $S\_{\mathfrak q}=S^\ast\_{R\_{\mathfrak q}}M\_{\mathfrak q}$ which is a
formally smooth $R\_{\mathfrak q}$-algebra and hence $S\_{\mathfrak p}$ is a
formally smooth $R$-algebra (localisations being formally étale). To show that
the statement is false it thus is enough to find a non-projective $R$-module all
of localisation are projective.
If $R$ is a boolean ring (i.e., $r^2=r$ for all $r\in R$) then all its
localisations are isomorphic to $\mathbb Z/2$ all of whose modules are
projective. Hence it suffices to find a boolean ring $R$ and a non-projective
$R$-module. This is easy: Let $S$ be a totally disconnected compact topological
space and $s\in S$ a non-isolated point and put $R$ equal to the boolean ring of
continuous maps $S \rightarrow \mathbb Z/2$. Evaluation at $s$ gives a ring
homomorphism $R \rightarrow \mathbb Z/2$ making $\mathbb Z/2$ an $R$-module. If
it were projective there would be a module section $\mathbb Z/2 \rightarrow
R$. Let $e$ be the image of $1$. Then for any $f\in R$ we have $fe=f(s)e$. This
means that $e$ must be the characteristic function of $s$. Indeed, as $e \mapsto
1$ under evaluation at $s$ we have $e(s)=1$. On the other hand, if $t\neq s$
there is an $f\in R$ with $f(t)=1$ and $f(s)=0$ giving $e(t)=f(t)e(t)=f(s)e(t)=0$.
Hence $\{s\}=\{t\in S:e(t)=1\}$ is open so that $s$ is isolated.
Note that we may take for $S$ any infinite compact totally disconnected set as
some point of it must be non-isolated. Nice examples are the Cantor set (all of whose points are non-isolated) and the
spectrum of $\prod\_T\mathbb Z/2$ for any infinite set $T$ (which is the
ultrafilter compactification of $T$, any non-principal ultrafilter is
non-isolated I think).
| 16 | https://mathoverflow.net/users/4008 | 22432 | 14,803 |
https://mathoverflow.net/questions/22453 | 3 | I'm trying to understand the proof of the Oppenheim conjecture using Ratner's theorem, and I don't immediately see why $SO(2,1)$ is generated by unipotents. Why is $SO(2,1)$ generated by unipotents? More generally, are there equivalent conditions to being generated by unipotents that are easier to check?
| https://mathoverflow.net/users/5598 | How Can I Tell when A Subgroup of a Lie Group is Generated by Unipotents? | Groups like SO$(2,1)$ have been studied in several frameworks: geometry and generation of classical groups over various ground fields, where unipotent elements tend to appear as transvections; real Lie groups, where the structure theory of groups like this is well developed (as in Helgason's old book, for example, republished by AMS); real points of (almost) simple algebraic groups, where papers by Borel and Tits have worked out the abstract group structure and generation in considerable generality. From the last point of view, I guess the crucial point about your question is that the group is *isotropic* over $\mathbb{R}$ and thus noncompact as a Lie group unlike SO(3). In particular, it has nontrivial unipotent elements (here of infinite order as group elements); the subgroup they generate in the algebraic group setting is closed and normal as
well as defined over $\mathbb{R}$.
Your example is old and references can be quoted, but you need to keep in mind the different approaches possible. What is simplest here depends a lot on your viewpoint.
P.S. The blog entry by Terry Tao suggested by Jack Schmidt gives a nice way to visualize the 3-dimensional situation, but the general perspectives I've sketched are useful in higher dimensions.
| 6 | https://mathoverflow.net/users/4231 | 22458 | 14,818 |
https://mathoverflow.net/questions/22459 | 17 | Let $p$ be an odd prime and $G := \langle x,y : x^p = y^p = (xy)^p = 1 \rangle$. I want to show that $G$ is infinite and wonder if there is a good way to prove this. I'm familiar with the basics of geometric group theory, but don't know if they can be used here.
Obviously $G^{ab} = \mathbb{Z}/p \times \mathbb{Z}/p$. In particular, $x,y,xy$ have order $p$. The same is true for $yx, x^{p-1} y^{p-1}$ and $y^{p-1} x^{p-1}$. My approach uses the same idea which is used in Serre's book *Trees* for elements in amalgamated sums. Define $M$ to be the set of formal words of the form $...x^i y^j ...$ (alternating powers of $x$ and $y$), where the exponents are in $[0,p-1]$ and $(xy)^p, (yx)^p, (x^{p-1} y^{p-1})^p, (y^{p-1} x^{p-1})^p$ are no subwords. There is a obvious action from the free group $\langle x,y \rangle$ on $M$. Now it should be obvious that $x^p, y^p$ and $(xy)^p$ act as the identity, but in fact, a proof requires many many cases and would somehow include a solution for the word problem for $G$. Perhaps this is as tedious as making $M$ ad hoc to a group. When this is done, the action extends to $G$. The obvious surjective map $M \to G$ is then injective because the action of $G$ on the empty word yields a inverse map. Now $M$ is infinite, for example it contains all the powers of $x y^2$.
I hope there is a better proof. Perhaps there is a nice action of $G$ on a topological space which makes you see that $x y^2$ has infinite order? By the way, I don't want to use heavy theorems from group theory (Burnside problem etc.)!
Feel free to add other interesting properties of $G$.
| https://mathoverflow.net/users/2841 | 〈x,y : x^p = y^p = (xy)^p = 1〉 | Many techniques discussed here: [group-pub](http://lists.maths.bath.ac.uk/sympa/arc/group-pub-forum/2010-01/msg00005.html)
EDIT: Some of the ideas, in the above thread, I like the most:
If $q$ is a prime congruent to $1$ mod $p$, then consider the Frobenius group $H\rtimes K$ with $H$ cyclic of order $q$ and $K$ cyclic of order $p$. Then if $a$ generates $H$ and $b$ generates $K$, you can show $b$ and $ab$ are such that $b$, $ab$ and $ab^2$ have order $p$, so this group is a quotient of your group. Choosing $q$ arbitrarily large suffices to show your group is infinite.
You can also show that the two infinite permutations on $\mathbb{Z}$ $a=...(-p+1,-p+2,...,-1,0)(1,2,...,p)(p+1,...,2p)... $ $b=...(-p+2,-p+3,...,0,1)(2,...,p+1)(p+2,...,2p+1)...$
are such that $a$, $b$, and $ab$ have order $p$, and yet generate an infinite group (it acts transitively on $\mathbb{Z}$).
Fox calculus can be used to show $\lt\lt xy^{-1}\gt\gt$ has infinite abelianization.
| 8 | https://mathoverflow.net/users/1446 | 22460 | 14,819 |
https://mathoverflow.net/questions/22465 | 13 | Does it make sense to talk about, say, the free division ring on 2 generators? If so, does the free division ring on countably many generators embed into the free division ring on two generators?
| https://mathoverflow.net/users/5036 | Free division rings? | If you had a "free division ring" $F$ on a set $X$, and any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".
| 18 | https://mathoverflow.net/users/3959 | 22469 | 14,826 |
https://mathoverflow.net/questions/22472 | 4 | As the title says, is there a mathematical object referred to as "ivy" or "ivy type" or similar?
I have a type of graph where this name fits perfectly, but I don't want it to clash with something already defined.
(I could in this paper call it a "reduced graph" or "contracted graph" but the above definition would make more sense.)
| https://mathoverflow.net/users/1056 | Is there a mathematical object called "ivy"? | Ivy does not appear to be a common term in mathscinet. Integrable vector Young functions are called IVY-functions in [MR2055989](http://www.ams.org/mathscinet-getitem?mr=2055989) and related papers. Otherwise all occurrences are the plant or a person.
| 14 | https://mathoverflow.net/users/3710 | 22474 | 14,829 |
https://mathoverflow.net/questions/22477 | 12 | I'm looking for an example of a set S, and a sigma algebra on it, which has no atoms.
Motivation: It seems to me that a lot of definitions in probability and stochastic processes - conditional probability, filtrations, adapted processes - become a lot simpler if phrased in terms of a sample space partitioned into atoms.
In the book I'm reading, this is done for finite sample spaces.
But the only problem I find with extending the definitions to general spaces seems to be that, in general, you might have a sigma algebra without any atoms.
(Note: All this definitions can be made without introducing a measure, so objections on grounds of uncountability don't apply.)
Does anyone have an example ?
2) Related question: What if you replace sigma algebra by algebra - closed under finite, rather than countable unions ?
I suspect that the algebra of subsets of R generated by open intervals has no atoms, but can't prove it rigorously.
| https://mathoverflow.net/users/4279 | Sigma algebra without atoms ? | In your second question, you are asking merely for an [atomless Boolean algebra](http://en.wikipedia.org/wiki/Boolean_algebras_canonically_defined), of which there are numerous examples. One easy example related to the one given on the Wikipedia page is the collection of periodic subsets of the natural numbers N. That is, the subsets $A\subset N$ such that there is a finite set $a\subset k$ for some $k$ and $kn+m\in A$ if and only if $m\in a$ for $m<k$. This is closed under finite intersections, unions and complements, but has no atoms, since any nonempty periodic set can be made smaller by reducing it to a set with a larger period.
For your main question, a similar idea works, when generalized to the transfinite, to produce the desired atomless $\sigma$-algebra. Namely, consider the collection of periodic subsets of $\omega\_1$, the first uncountable ordinal. This ordinal is bijective with the set of reals, if the Continuum Hypothesis holds, but in any case (under AC), it is bijective with a subset of the reals, so one can take the underlying set of points here to be a set of real numbers. By periodic here I mean the collection of sets $A\subset \omega\_1$, such that there is a set $a\subset\alpha$ for some countable ordinal $\alpha$, such that $\alpha\beta+\xi\in A$ if and only if $\xi\in a$, where $\xi<\alpha$. Note that if $\alpha$ is fixed, every ordinal has a unique representation as $\alpha\beta+\xi$ for $\xi<\alpha$, since this is just dividing the ordinals into blocks of length $\alpha$. This means that $A$ consists of the pattern in $a$ repeated $\omega\_1$ many times. This collection of sets is easily seen to be a Boolean algebra and atomless, but it is also $\sigma$-closed, since for any countably many such $A$, I can find a common countable period since the collection of multiples of any fixed $\alpha$ form a club subset of $\omega\_1$. Thus, the intersection (or union) is again periodic, as desired. There are no atoms, since any nonempty periodic set can be made smaller by reducing it to have a larger period.
There are numerous atomless Boolean algebras arising in the forcing technique of set theory, used by Cohen to prove the independence of the Continuum Hypothesis and many others subsequently.
| 18 | https://mathoverflow.net/users/1946 | 22482 | 14,832 |
https://mathoverflow.net/questions/22484 | 4 | The title says is all.
To motivate the problem, here is a theorem for finite sets.
Theorem: If S is a finite set, then it can be proved that the atoms of any sigma algebra on S form a partition of S.
I am trying to extend this theorem to a countable set.
* It is easy to show that the atoms must be disjoint - this does not need finiteness.
-The part that does use finiteness is to show that every point is S belongs to some atom.
The idea is: If x is any element of S, one can create a nested sequence of proper subsets containing x. Since S is finite, there must be a smallest set in the sequence and that's an atom.
-This part breaks down for countably infinite sets
Thinking further, I have found that you can still extend the theorem if the sigma algebra F, has the following property:
Every member of F contains an atom of F
Proof: Since atoms are disjoint, there are only countably many of them.
Hence, if you consider the complement of all the atoms, you are still left with a set in F.
If this set is nonempty, it must contain an atom. Contradiction !
*So, the only way the theorem can fail to extend is if you have a member of F (necessarily infinite) which has no atoms.*
But I'm not sure if that would be consistent with the requirements of a sigma algebra.
Finding such a set would give you a countably infinite set with a sigma algebra on it without any atoms. Hence my question.
| https://mathoverflow.net/users/4279 | Is there a sigma algebra without atoms on a countably infinite set ? | There is an old result of Tarski which says that any algebra A of sets which is
1. κ-complete (i.e. A is closed under unions and intersections of fewer than κ sets from A), and
2. satisfies the κ-chain condition (i.e. there is no family of κ many pairwise disjoint nonempty sets in A),
then A is necessarily atomic.
The proof of atomicity is along the lines of what Jonas suggested, but with a wellordered descending chain. Given a0 ∈ A and x ∈ a0. Starting from a0, recursively construct a strictly descending wellordered chain (aζ: ζ < δ) of sets in A, each containing x, for as long as possible. Since the differences aζ \ aζ+1 are pairwise disjoint and nonempty, this construction must terminate with δ < κ. Therefore, the intersection a = ∩ζ<δ aζ is an element of A by κ-completeness and x ∈ a. This intersection must in fact be an atom in A, otherwise we could extend the chain further. It follows that a0 is the union of all atoms below a0 and hence that A is atomic.
In your case, you have an ω1-complete algebra of sets. If the underlying set of the algebra is countable, then there certainly cannot be a family of ω1 many pairwise disjoint nonempty sets in A. Note also that the case κ = ω shows that finite algebras of sets are atomic.
| 5 | https://mathoverflow.net/users/2000 | 22489 | 14,835 |
https://mathoverflow.net/questions/22452 | 8 | Encouraged by
[Does anyone have an electronic copy of Waldspurger's "Sur les coefficients de Fourier des formes modulaires de poids demi-entier"?](https://mathoverflow.net/questions/16354/)
I realized I could ask for this rare item here.
Again, it is Yury G. Teterin's 1984 (Russian) preprint "Representation of numbers by spinor genera."
For whatever reason (possibly length?) it never made it into the usual
Zapiski Nauchnykh Seminarov Leningradskogo Otdeleniya Matematicheskogo Instituta im. V. A. Steklova AN SSSR.
Let me give some detail. I do not know the number of pages. The preprint probably never appeared elsewhere, at least not under the same title. It was probably never translated into English. No American mathematics library admits to having it. The easiest place to find mention of the preprint is in Math Reviews MR0732548 (86d:11042) which is a review, by Oleg M. Fomenko, of Schulze-Pillot's 1984 "Thetareihen positiv definiter quadratischer Formen." (I do not know how to make the MR reference a link).
The preprint is mentioned in a later related item by the same author that has been translated, see
<http://www.springerlink.com/content/t701481j73531761/>
as well as one by Elena P. Golubeva
<http://www.springerlink.com/content/w740g82624753417/>
Finally, I have emailed Teterin with no result so far. If nothing happens for a long time I could write to O. M. Fomenko and see what happens. The three people in St. Petersburg that I have mentioned are
<http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=25046>
<http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=22736>
<http://www.mathnet.ru/php/person.phtml?option_lang=eng&personid=33524>
Well, thanks for any assistance. Individual replies are always welcome, one of my email addresses can be found through <http://www.ams.org/cml>
EDIT, May 2021: the pdf is on one of my websites
[KAP](http://zakuski.math.utsa.edu/%7Ekap/)
direct link
[TETERIN](http://zakuski.math.utsa.edu/%7Ekap/Teterin_1984.pdf)
Let me point out that there are the strongly related Golubeva\_Fomenko\_1984.pdf in English translation, as well as Schulze\_Pillot\_1984\_Darstellungsmasse.pdf and
Schulze\_Pillot\_1984\_Thetareihen.pdf in these cases still in German. Also Math Reviews of most of the papers for which I lack a translation into English.
| https://mathoverflow.net/users/3324 | Does anyone have access to a copy of Yury G. Teterin's 1984 (Russian) preprint "Representation of numbers by spinor genera" | Yura Teterin left mathematics and POMI but may be he is reading his POMI e-mail <http://www.pdmi.ras.ru/~yuri/>
also one can try to ask for a scan at POMI editorial dept (Vera Simonova [email protected]) or at the library [email protected]
| 9 | https://mathoverflow.net/users/2702 | 22491 | 14,836 |
https://mathoverflow.net/questions/22454 | 9 | Let $m,n$ be positive integers with $m \leqslant n$, and denote by $\mu\_M$ the minimal polynomial of a matrix.
Do we know for which $m$ the set $E\_m$ of $M \in \mathfrak{M}\_n(\mathbb{R})$ such that $\deg(\mu\_M) = m$ is connected?
| https://mathoverflow.net/users/3958 | Connected subset of matrices ? | The answer is always yes. Indeed the set is path-connected.
Let $C(f)$ denote the companion matrix associated to the monic
polynomial $f$. Every matrix $A$ is similar to a matrix in rational
canonical form:
$$B=C(f\_1)\oplus C(f\_1 f\_2)\oplus\cdots\oplus C(f\_1 f\_2,\cdots f\_k)$$
where here $\oplus$ denotes diagonal sum. Then $m$ is the degree of
$f\_1 f\_2\cdots f\_k$. Starting with $B$ deform each $f\_i$ into a power
of $x$. We get a path from $B$ to
$$B'=C(x^{a\_1})\oplus C(x^{a\_2})\oplus\cdots\oplus C(x^{a\_k})$$
inside $E\_m$. There's a path from $B'$ in $E\_m$ given by
$$(1-t)C(x^{a\_1})\oplus (1-t)C(x^{a\_2})\oplus\cdots\oplus C(x^{a\_k})$$
ending at
$$B\_m=O\oplus C(x^m).$$
Thus there is a path in $E\_m$ from $A$ to $UB\_mU^{-1}$
where $U$ is a nonsingular matrix. If $\det(U)\ne0$ then there is
a path in $GL\_n(\mathbf{R})$ from $U$ to $I$ and so a path in $E\_m$
from $A$ to $B\_m$. If $m< n$ then there is a matrix $V$ of negative determinant
with $VB\_m V^{-1}=B\_m$ so that we may take $U$ to have positive determinant.
The only case that remains is when $m=n$. In this case $E\_m$ contains
diagonal matrices with distinct entries, and each of these commutes
with a matrix of negative determinant.
| 5 | https://mathoverflow.net/users/4213 | 22496 | 14,841 |
https://mathoverflow.net/questions/22399 | 2 | For any integer $n\ge 3$, let $P(x)=\sum\limits\_{i(=2k)\ge 0}^{n}\binom{n}{2k}(1-x)^k$, $Q(x)=\sum\limits\_{i(=2k+1)\ge 0}^{n}\binom{n}{2k+1}(1-x)^{k+1}$. Define $\frac{P(x)}{Q(x)}\equiv\sum\limits\_{i=0}^{\infty}c\_ix^i $. I like to know $c\_0> c\_1> c\_2>\cdots$, but I don't know how to show it.
| https://mathoverflow.net/users/3818 | Decreasing coefficients? | The polynomials $P\_n$ and $Q\_n$ can be written as
$$
P\_n(x)=(1+\sqrt{1-x})^n+(1-\sqrt{1-x})^n,
\qquad
Q\_n(x)=\sqrt{1-x}\bigl((1+\sqrt{1-x})^n-(1-\sqrt{1-x})^n\bigr).
$$
In particular, they are both solutions to the linear difference equation
$P\_{n+1}(x)=2P\_n(x)-xP\_{n-1}(x)$ implying that their quotient
$f\_n(x)=P\_n(x)/Q\_n(x)$ satisfies the nonlinear recursion
$$
f\_{n+1}(x)=\frac{1+f\_n(x)}{1+(1-x)f\_n(x)}
$$
(already observed experimentally by Martin Rubey).
The property saying that the coefficients of the power series
$f(x)=c\_0+c\_1x+c\_2x^2+\dots$ satisfy $c\_0\ge c\_1\ge c\_2\ge\dots$
can be rephrased as $c\_0-(1-x)f(x)\ge0$, where the record
$a\_0+a\_1x+a\_2x^2+\dots\ge0$ means that all $a\_i\ge0$. This makes
the original problem equivalent to $1-(1-x)f\_n(x)\ge0$ for all $n$.
This follows by induction on $n$. The base of induction is clear
verification (already done by the author for a few first $n$).
Assuming that it is true for a given $n$ and using that the
non-negative expansion $1-(1-x)f\_n(x)$ has the vanishing constant term,
we conclude that the (formal) power series
$$
\frac1{1-\frac12\bigl(1-(1-x)f\_n(x)\bigr)}
=1+\sum\_{k=1}^\infty\frac1{2^k}\bigl(1-(1-x)f\_n(x)\bigr)^k
$$
is non-negative. It remains to apply the above recursion for $f\_n(x)$:
$$
1-(1-x)f\_{n+1}(x)
=\frac x{1+(1-x)f\_n(x)}
=\frac x2\cdot\frac1{1-\frac12\bigl(1-(1-x)f\_n(x)\bigr)},
$$
and the desired property follows.
| 10 | https://mathoverflow.net/users/4953 | 22498 | 14,843 |
https://mathoverflow.net/questions/22499 | 0 | I want to show, that $GL\_n(\mathbb{Z}/b\mathbb{Z})$ operates transitively on
$X = \{ (v\_1, \ldots, v\_n) \in (\mathbb{Z}/b\mathbb{Z})^n \ | \ v\_1\mathbb{Z}/b\mathbb{Z} + \ldots + v\_n\mathbb{Z}/b\mathbb{Z} = \mathbb{Z}/b\mathbb{Z}\} $
($b$ is an integer.)
IOW
1. $A \in GL\_n(\mathbb{Z}/b\mathbb{Z}), x \in X \Rightarrow Ax \in X$
2. For all $x,y \in X$ exists $A \in GL\_n(\mathbb{Z}/b\mathbb{Z})$ such that $x = Ay$.
I have no idea, where to start.
Thanks
-elfinit
| https://mathoverflow.net/users/5607 | Operation of GL_n(Z/bZ) | Any transformation
$$(v\_1,\ldots,v\_n)\mapsto (v\_1,\ldots,v\_{j-1},v\_j+av\_k,v\_{j+1},\ldots,v\_n)$$
for $j\ne k$ is achievable by means of some such matrix. It suffices to reduce
an admissible vector to $(1,0,\ldots,0)$ by means of a sequence of such reductions.
I would do it in three stages
1. Make $v\_n$ into a unit in $\mathbb{Z}/b\mathbb{Z}$;
2. Make $v\_1=1$;
3. Make all $v\_j=0$ for $j>1$.
| 1 | https://mathoverflow.net/users/4213 | 22500 | 14,844 |
https://mathoverflow.net/questions/22478 | 73 | Take, for example, the Klein bottle K. Its de Rham cohomology with coefficients in $\mathbb{R}$ is $\mathbb{R}$ in dimension 1, while its singular cohomology with coefficients in $\mathbb{Z}$ is $\mathbb{Z} \times \mathbb{Z}\_2$ in dimension 1. It is in general true that de Rham cohomology ignores the torsion part of singular cohomology. This is not a big surprise since de Rham cohomology really just gives the dimensions of the spaces of solutions to certain PDE's, but I'm wondering if there is some other way to directly use the differentiable structure of a manifold to recover torsion. I feel like I should know this, but what can I say...
Thanks!
| https://mathoverflow.net/users/4362 | Can analysis detect torsion in cohomology? | You can compute the integer (co)homology groups of a compact manifold from a Morse function $f$ together with a generic Riemannian metric $g$; the metric enters through the (downward) gradient flow equation
$$ \frac{d}{dt}x(t)+ \mathrm{grad}\_g(f) (x(t)) = 0 $$
for paths $x(t)$ in the manifold.
After choosing further Morse functions and metrics, in a generic way, you can recover the ring structure, Massey products, cohomology operations, Reidemeister torsion, functoriality.
The best-known way to compute the cohomology from a Morse function is to form the Morse cochain complex, generated by the critical points (see e.g. Hutchings's [Lecture notes on Morse homology](http://math.berkeley.edu/~hutching/)). Poincaré duality is manifest.
Another way, due to [Harvey and Lawson](http://arxiv.org/abs/math.DG/0101268), is to observe that the de Rham complex $\Omega^{\ast}(M)$ sits inside the complex of currents $D^\ast(M)$, i.e., distribution-valued forms. The closure $\bar{S}\_c$ of the the stable manifold $S\_c$ of a critical point $c$ of $f$ defines a Dirac-delta current $[\bar{S}\_c]$. As $c$ varies, these span a $\mathbb{Z}$-subcomplex $S\_f^\ast$ of $D^\*(M)$ whose cohomology is naturally the singular cohomology of $M$.
The second approach could be seen as a "de Rham theorem over the integers", because over the reals, the inclusions of $S\_f\otimes\_{\mathbb{Z}} \mathbb{R}$ and $\Omega^{\ast}\_M$ into $D^\ast(M)$ are quasi-isomorphisms, and the resulting isomorphism of $H^{\ast}\_{dR}(M)$ with $H^\ast(S\_f\otimes\_{\mathbb{Z}}\mathbb{R})=H^\ast\_{sing}(X;\mathbb{R})$ is the de Rham isomorphism.
| 43 | https://mathoverflow.net/users/2356 | 22506 | 14,850 |
https://mathoverflow.net/questions/22473 | 11 | Hi, everyone:
For the sake of context, I am a graduate student, and I have taken classes in
algebraic topology and differential geometry. Still, the 2 proofs I have found
are a little too terse for me; they are both around 10 lines long, and each line
seems to pack around 10 pages of results. Of course, I am considering cases for
"reasonable" spaces, being the beginner I am at this point.
It would also be great if someone knew of similar results for H\_1 (equiv. H\_3).
Thanks in Advance.
| https://mathoverflow.net/users/5566 | Request: intermediate-level proof: every 2-homology class of a 4-manifold is generated by a surface. | Torsten's answer was good, but there are also more elementary answers. Here's one, which is essentially a big transversality argument, followed by a mild de-singularization.
Let me consider $M$ to be a triangulated 4-manifold, and represent a class in $H\_2(M)$ as a sum of 2-faces of the triangulation. For each 2-face appearing with multiplicity $n$ in the sum, take $n$ copies of the face, pushed off of each other slightly and meeting only at the edges.
Now, along each edge (1-face) of the triangulation, since we started with a 2-cycle the total number of triangles meeting there is $0$. (This is a signed count if we are working in $H\_2(M; \mathbb{Z})$, and means that there are an even number if we are in $H\_2(M; \mathbb{Z}/2)$.) Along each edge, pair up the incident triangles in an arbitrary way that's compatible with the orientations, and resolve the intersections along the interior of the edge according to that pairing. (A neighborhood of the edge looks like $D^3 \times I$; the incident triangles are coming in from fixed directions, i.e., at fixed points in $S^2 \times I$; so given any pairing of the points on $S^2$, we can just join them up and avoid the edge altogether. It's easier to think about what happens in a 3-manifold, where it's very similar, you just have to be more picky about how you pair the incident triangles.)
After the last step, we have a surface $S$ with codimension-2 singularities at points in $M$. For each such singularity, consider a small ball $B$ in $M$ and consider $S \cap \partial B$. This is a link $L$ in a 3-manifold (oriented or not, depending on which homology we look at). Replace $S \cap B$ with a Seifert surface for $L$, and we're done. (You could also use any surface with boundary $L$ inside the 4-ball $B$ instead of the Seifert surface, of course. Frequently you can get lower genus that way.)
| 14 | https://mathoverflow.net/users/5010 | 22509 | 14,851 |
https://mathoverflow.net/questions/22523 | 1 | If we have function $y=L(x\_1,x\_2,x\_3,...,x\_n)$, and function $z=R(x\_1,x\_2,x\_3,...,x\_n)$. How to compute the derivative $\frac{dy}{dz}$?
Shall I do $\frac{dy}{dz} = \sup\_{g\in \Re^n}\frac{\bigtriangledown\_x L \cdot g}{\bigtriangledown\_x R \cdot g}$?
Is there any mathematical term associated with this kind of derivatives?
| https://mathoverflow.net/users/2013 | Implicit derivative? | There is no reason to expect that such a thing as $dy/dz$ exists. If you rephrase everything in terms of differentials, you have $$dy=\sum\_{i=1}^n\frac{\partial L}{\partial x\_i}dx\_i,\quad dz=\sum\_{i=1}^n\frac{\partial R}{\partial x\_i}dx\_i$$ where $dx\_1,\ldots,dx\_n$ are linearly independent, and so you find $dy=a\, dz$ if and only if $\nabla L=a\nabla R$. But normally, the two gradient won't be parallel, so you cannot define $dy/dz$. You *can* do so in any given direction, though: Your fraction $(\nabla L\cdot g)/(\nabla R\cdot g)$ is a perfectly good expression for $dy/dz$ *as measured in the direction given by $g$*.
| 8 | https://mathoverflow.net/users/802 | 22526 | 14,861 |
https://mathoverflow.net/questions/22522 | 1 | In the paper on MinWise independent permutations ([MinWise independent permutations](http://www.cs.princeton.edu/courses/archive/spr04/cos598B/bib/BroderCFM-minwise.pdf)), the authors say that it is often convenient to consider permutations rather than hash functions (Pg-3).
While I understand that for a set X to be minwise independent, all elements in it must have the same probability of becoming the minimum element of its image under a randomly chosen hash function ∏, the part that I don't understand is where the authors say that ∏ could be conveniently taken as a permutation instead of a hash function.
Is there any example of it that anybody could provide that could help me understand this part better? How is the mapping of an element from the set X using a permutation supposed to function? I understand the purpose. But since I cant figure out the permutation analogy, I'm unable to find out how this works in practice.
What I understand of this paper is something like this - To reduce the dimensionality of a set, we can generate (a fixed number of) minwise independent permutations, and for all these different permutations we can find out a minimum hash value. Since each element of the permutation is equally likely to be the minimum hash value, this minimum hash value can be then taken to (in a way) represent the set, hence representing the document on a scale of our choice. So no matter what number of words documents contain, if we generate 100 minwise independent permutations, we can represent the document in 100 words (or items from the universe set).
Thanks.
| https://mathoverflow.net/users/5611 | How is a permutation taken as an equivalent of a hash function in MinWise independent permutations? | For the purpose of min-wise hashing, the only goal of the permutation (or hash function) is to map the initial bag of numbers into another bag, so that we can then pick off the min as the 'hash value'. So for this purpose, you should think of the permutation as merely a function that takes [1..n] and maps it to [1..n] (sending every element to its image under the permutation).
Are you concerned about the relative sizes of the domain and range ?
| 1 | https://mathoverflow.net/users/972 | 22527 | 14,862 |
https://mathoverflow.net/questions/22462 | 81 | This is a question I've asked myself a couple of times before, but its appearance on MO is somewhat motivated by [this thread](https://mathoverflow.net/questions/22359/why-havent-certain-well-researched-classes-of-mathematical-object-been-framed-by), and sigfpe's comment to Pete Clark's answer.
I've often heard it claimed that combinatorial species are wonderful and prove that category theory is also useful for combinatorics. I'd like to be talked out of my skepticism!
I haven't read Joyal's original 82-page paper on the subject, but browsing a couple of books hasn't helped me see what I'm missing. The Wikipedia page, which is surely an unfair gauge of the theory's depth and uses, reinforces my skepticism more than anything.
As a first step in my increasing appreciation of categorical ideas in fields familiar to me (logic may be next), I'd like to hear about some uses of combinatorial species to prove things in combinatorics.
I'm looking for examples where there is a clear advantage to their use. To someone whose mother tongue is not category theory, it is not helpful to just say that "combinatorial structures are functors, because permuting the elements of a set A gives a permutation of the partial orders on A". This is like expecting baseball analogies to increase a brazilian guy's understanding of soccer. In fact, if randomly asked on the street, I would sooner use combinatorial reasoning to understand finite categories than use categories of finite sets to understand combinatorics.
**Added for clarification:** In my (limited) reading of combinatorial species, there is quite a lot going on there that is combinatorial. The point of my question is to understand how the *categorical* part is helping.
| https://mathoverflow.net/users/4367 | What are some examples of interesting uses of the theory of combinatorial species? | Composition of species is closely related to the composition of symmetric collections of vector spaces ("S-modules"), which is a remarkable example of a monoidal category everyone who had ever encountered operads necessarily used. Applying ideas coming from this monoidal category interpretation has various consequences for combinatorics as well. For example, look at papers of Bruno Vallette on partition posets ([here](https://arxiv.org/abs/math/0405312) and [here](https://arxiv.org/abs/math/0410051)): I believe that already the description of the $S\_n$ action on the top homology of the usual partition lattice was hard to explain from the combinatorics point of view - and for many other lattices would be impossible without the Koszul duality viewpoint.
| 30 | https://mathoverflow.net/users/1306 | 22537 | 14,867 |
https://mathoverflow.net/questions/22530 | 6 | This is probably really easy, but I just need someone to help me get mentally unstuck. As part of a description of the [McKay correspondence](http://www.valdostamuseum.org/hamsmith/McKay.html), I want to show that if $G$ is a finite subgroup of $SU(2)$ and $V$ the corresponding 2-dimensional representation, then $\dim \text{Hom}(W, W \otimes V) = 0$ for any irreducible representation $W$ of $G$. I suspect the result is true in slightly greater generality, but it clearly can't always be true. Since $\dim \text{Hom}(W, W \otimes V) = \dim \text{Hom}(W \otimes W^{\ast}, V)$, the result is false if, for example, $V \simeq W \otimes W^{\ast}$ or is a direct summand thereof for some $W$.
So I am wondering when, for a given $G$ and $V$, it is always true that $\dim \text{Hom}(W, W \otimes V) = 0$ for all irreducible representations $W$. One can easily reduce to the case that $V$ is irreducible.
| https://mathoverflow.net/users/290 | For a representation V of a finite group G, when is Hom(W, W⊗V) trivial for all irreps W? | One necessary condition is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)
I don't know if this is also sufficient. This boils down to looking at groups with trivial center.
In any case, you originally asked about finite subgroups of SU(2) and its fundamental representation. These are all classified, and none of them have trivial center, so all satisfy this property. Proof of the last bit: any subgroup $G$ of SU(2) gives a subgroup $\overline{G}$ of SO(3) by projection. If $\overline{G}$ has even order, it has an element of order 2, which necessarily lifts to an element of order 4 whose square is $-1$, so $-1 \in G$. The only subgroups of SO(3) of odd order are the odd cyclic groups, which lift to Abelian groups. Some references are is [these notes by Dolgachev](http://www.math.lsa.umich.edu/~idolga/McKaybook.pdf) or an [earlier MO question](https://mathoverflow.net/questions/16026/the-finite-subgroups-of-sl2-c).
| 5 | https://mathoverflow.net/users/5010 | 22540 | 14,869 |
https://mathoverflow.net/questions/22552 | 6 | The golden ratio $\phi=\frac{1+\sqrt5}2$ is [sometimes said](http://en.wikipedia.org/wiki/Continued_fraction#A_property_of_the_golden_ratio_.CF.86) to be one of the most difficult numbers to approximate with rational numbers, because its continued fraction development $$\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}}$$has all entries equal to $1$, so that its convergents are as far from $\phi$ as possible. This can be quantified and made precise (I guess: I'd love to know the precisely how!) It shares this property with all the numbers $\frac{a+b\phi}{c+d\phi}$ with $\left(\begin{smallmatrix}a&b\\\c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb Z)$ (and no others?)
Are there other numbers which are characterized by such extremal properties? Is there a *second worst* aproximable number?
| https://mathoverflow.net/users/1409 | Numbers characterized by extremal properties | Certainly there is one type of study of this question, coming under the heading of the Markoff Spectrum, going back to the German tranliteration of Markov. There is a very nice book called "The Markoff and Lagrange Spectra" by Thomas W. Cusick and Mary E. Flahive. It is this topic that lead to the Markoff Numbers, which are any of the triple $x,y,z$ of a positive integer solution to $$ x^2 + y^2 + z^2 = 3 x y z , $$ see
<http://en.wikipedia.org/wiki/Markov_number>
If you are curious, there is my paper with Kaplansky in the Illinois Journal, pdf at:
<http://zakuski.math.utsa.edu/~jagy/bib.html>
I am especially proud of section 8.3, called "Other Families," as I was able to construct a mirror image of our "Markov Ratios"
$$ 9 - \frac{4}{m^2} $$
in forms that gave
$$ 9 + \frac{4}{m^2} $$
To clarify, any indefinite binary quadratic form with integer coefficients and non-square discriminant has a nonzero "minimum" which is the smallest nonzero absolute value of any value obtained with integral values for the arguments. Our "Markov Ratio" was simply the discriminant divided by the square of this minimum.
Well, your Golden Ratio has the smallest possible Markov Ratio, with 5, the form being equivalent to $x^2 + x y - y^2.$ The second smallest is 8, form $2 x^2 + 4 x y - 2 y^2$ which is visibly imprimitive.
By the way, the "reduced" forms with the rather inflated middle coefficient (see chapter 3, Duncan A. Buell, Binary Quadratic Forms) correspond with purely periodic continued fractions, which I gradually came to realize are the best way to compute things: the cycle of a reduced form eventually arrives back at its precise starting point, so there is no need to compare with a long list of previously computed forms. If you simply compute the continued fraction for a number such as $\sqrt{5}$ you get some non-repeated coefficients or "quotients" $a\_0 ; a\_1 $ at the start.
If you look at the Cusick and Flahive book, you will see how the Markoff and Lagrange spectra coincide with "Markov Ratio" below 9 but are rather different above, where the description using binary forms is no longer adequate.
For the hasty:
<http://en.wikipedia.org/wiki/Markov_spectrum>
| 13 | https://mathoverflow.net/users/3324 | 22555 | 14,877 |
https://mathoverflow.net/questions/22549 | 27 | Suppose $x\in \mathbb{R}$ is irrational, with irrationality measure $\mu=\mu(x)$; this means that the inequality $|x-\frac{p}{q}|< q^{-\lambda}$ has infinitely many solutions in integers $p,q$ if and only if $\lambda < \mu$. A beautiful theorem of Roth asserts that algebraic numbers have irrationality measure $2$. For $\lambda<\mu$, let $\mathcal{Q}(x,\lambda) \subset \mathbb{N}$ be the (infinite) set of all $q$ occuring in solutions to the aforementioned inequality.
Question: For which pairs $(x,\lambda)$ does $\mathcal{Q}(x,\lambda)$ have *positive relative density* in the positive integers? For which pairs $(x,\lambda)$ does the cardinality of $\mathcal{Q}(x,\lambda) \cap [1,N]$ grow like a positive power of $N$?
| https://mathoverflow.net/users/1464 | How often are irrational numbers well-approximated by rationals? | $\mathcal{Q}(x,\lambda)$ has positive relative density if and only if $\lambda\le 1$.
This follows from Weyl's Theorem on Uniform Distribution. (There is a nice concise proof in Cassels' "Diophantine Approximation".)
Weyl's Theorem: Let $I\subset \mathbb{R}$ be an interval of length $\epsilon \le 1$. Let $S\_N(I)$ be the set of all integers $q$ in the interval $[1,N]$ such that for some integer $p$, it holds that $xq-p\in I$. Then
$$\frac{Card(S\_N(I))}{N} \to \epsilon
\text{ as } N\to\infty.$$
Here's a proof-sketch, using Weyl's Theorem, that if $\lambda > 1$ then $\mathcal{Q}(x,\lambda)$ has relative density zero:
Fix $\epsilon > 0$, and take $I$ (in Weyl's Theorem) to be the interval $(-\epsilon,\epsilon)$. Suppose $\lambda>1$. Let $q\in \mathcal{Q}(x,\lambda)$; so for some $p\in \mathbb{Z}$, $$|xq-p| < q^{1-\lambda}.$$
There is an integer $M$, depending only on $\epsilon$, such that $|xq-p| < \epsilon$ whenever $p$ and $q$ satisfy the above inequality and $q\ge M$. Therefore $$\mathcal{Q}(x,\lambda)\cap [M,N]\subset S\_N(I).$$ It follows from Weyl's Theorem that the relative density of $\mathcal{Q}(x,\lambda)$ does not exceed $2\epsilon$. Since $\epsilon$ is arbitrary, the relative density of $\mathcal{Q}(x,\lambda)$ must be zero.
This can be proved in a more elementary but laborious way using the "Ostrowski Number System", which is explained in the Rockett and Szusz book on continued fractions.
| 23 | https://mathoverflow.net/users/5229 | 22566 | 14,885 |
https://mathoverflow.net/questions/22553 | 15 | A Theorem of Cartier (e.g. [Mumford](http://books.google.com/books?id=-5weWX_YD6sC&printsec=frontcover&dq=curves+on+an+algebraic+surface&source=bl&ots=1r4O86OtJy&sig=PIH5b5EipbZYJWXrWy77NvltSP4&hl=en&ei=4PjUS9uMKMOC8gbbi_mCDA&sa=X&oi=book_result&ct=result&resnum=3&ved=0CBIQ6AEwAg#v=onepage&q&f=false), Lecture 25) states that every separated, finite type group scheme $G/k$ over a field $k$ of characteristic $0$ is reduced. Does this result remain valid if we drop the assumption that $G/k$ is separated and of finite type?
Frans Oort (MR0206005) observed that one can use limit formalism to argue that every affine group scheme over $k$ is reduced.
**Edit:** BCnrd pointed out that group schemes over a field are automatically separated. Furthermore, the proof of Cartier's Theorem in Mumford's book remains valid for a locally Noetherian $k$-group scheme.
| https://mathoverflow.net/users/5337 | Are group schemes in Char 0 reduced? (YES) | The answer is yes - *every* group scheme over a field of characterstic zero is reduced: see [Schémas en groupes quasi-compacts sur un corps et groupes henséliens](http://www.math.u-psud.fr/~biblio/numerisation/docs/P_PERRIN-109/pdf/P_PERRIN-109.pdf) (especially Thm. 2.4 in part II and Thm. 1.1 and Cor. 3.9 in part V of the 1st part), and for a summary of the relevant results see 4.2 (in particular 4.2.8) of [Approximation des schémas en groupes, quasi compacts sur un corps](http://archive.numdam.org/article/BSMF_1976__104__323_0.djvu).
| 12 | https://mathoverflow.net/users/781 | 22573 | 14,888 |
https://mathoverflow.net/questions/22579 | 47 | I think a major reason is because Lie algebras don't have an identity, but I'm not really sure.
| https://mathoverflow.net/users/4692 | What are the reasons for considering rings without identity? | The reason is simple: There are many non-unital rings which appear quite naturally.
If $X$ is a locally compact space (in the following every space is assumed to be Hausdorff), then $C\_0(X)$, the ring of continuous complex-valued functions on $X$ vanishing at infinity, is a $C^\ast$-algebra which is unital if and only if $X$ is compact. If $X = \mathbb{N}$, this is just the ring of sequences converging to $0$. Gelfand duality yields an anti-equivalence between unital commutative $C^\ast$-algebras and compact spaces, and also between (possibly non-unital) commutative $C^\*$-algebras (with "proper" homomorphisms) and locally compact spaces (with proper maps). In a very similar spirit ($\mathbb{C}$ is replaced by $\mathbb{F}\_2$), there is an anti-equivalence between unital boolean rings and compact totally disconnected spaces, and also between Boolean rings and locally compact totally disconnected spaces. One-point-Compactification on the topological side corresponds here to the unitalization on the algebraic side. Perhaps we have the following conclusion: As locally compact spaces appear very naturally in mathematics (e.g. manifolds), the same is true for non-unital rings.
If $A$ is a ring (possibly non-unital), its unitalization is defined to be the universal arrow from $A$ to the forgetful functor from unital rings to rings. An explicit construction is given by $\tilde{A} = A \oplus \mathbb{Z}$ as abelian group with the obvious multiplication so that $A \subseteq \tilde{A}$ is an ideal and $1 \in \mathbb{Z}$ is the identity. Because of the universal property, the module categories of $A$ and $\tilde{A}$ are isomorphic. Thus many results for unital rings take over to non-unital rings.
Every ideal of a ring can be considered as a ring. Important examples also come from functional analysis, such as the ideal of compact operators on a Hilbert space.
| 71 | https://mathoverflow.net/users/2841 | 22584 | 14,895 |
https://mathoverflow.net/questions/22582 | 4 | 1) I was wondering if there exists a criterion (of (a) combinatorial or (b) geometric nature) for a sum of simple vectors $V\in\wedge^k(\mathbb R^n)$,
$V=e\_{a\_{11}}\wedge\cdots\wedge e\_{a\_{1k}} + \cdots + e\_{a\_{m1}}\wedge\cdots\wedge e\_{a\_{mk}}$, where $e\_1,\cdots, e\_n$ is a basis of $\mathbb R^n$
to be a simple vector: in other words, I want to know what should one verify in order to be sure that $V$ can be written as $V=v\_1\wedge\cdots\wedge v\_k$ for some vectors $v\_i\in\mathbb R^n$.
2) What happens if we allow noninteger coefficients? In other words, how does the answer change if we consider question 1) for
$V=c\_1e\_{a\_{11}}\wedge\cdots\wedge e\_{a\_{1k}} + \cdots + c\_me\_{a\_{m1}}\wedge\cdots\wedge e\_{a\_{mk}}$, where now $c\_i\in\mathbb R$.
3) any reference or hint about some (name of) branch of mathematics dealing with this is also welcome!
| https://mathoverflow.net/users/5628 | Criterion for being a simple vector | You are looking for the Plucker relations, a collection of quadratic polynomials in the $c\_m$'s which vanish if and only if $V$ can be written as $v\_1 \wedge v\_2 \wedge \cdots \wedge v\_k$. You should be able to read about them in most books on algebraic geometry. For example, Griffiths-Harris or Miller-Sturmfels will definitely discuss this.
If you are comfortable with representation theory of $GL\_n$, they can be described very briefly: Let $\omega\_k$ be the weight of the highest weight vectors in $\bigwedge^k(\mathbb{R}^n)$. There is (up to scalar multiple) a unique $GL\_n$ equivariant map from $\bigwedge^k(\mathbb{R}^n)^{\otimes 2}$ to the representation of $GL\_n$ with highest weight $2 \omega\_k$. Let the kernel of this map be $K$. The plucker relations say that $V \otimes V$ is in $K$.
It is not difficult to write them down explicitly, but there are a lot of signs to get right, so I'll just refer you to the sources cited above.
There is no significant simplification in assuming that the $c$'s are $0$ or $1$.
Since you tagged this combinatorics and hyperplane arrangements, you may want to know the following: If $V$ is simple, then the set of $k$-tuples $(a\_{m1}, \ldots, a\_{mk})$ for which $c\_m$ is nonzero form the bases of a matroid. There is no useful converse to this statement. Essentially by definition, a matroid $M$ is realizable if and only if you can find vectors $v\_1$, $v\_2$, ..., $v\_k$ such that the bases of $M$ correspond to the nonzero terms in $v\_1 \wedge v\_2 \wedge \cdots \wedge v\_k$. But there is no good criterion for a matroid to be realizable.
| 13 | https://mathoverflow.net/users/297 | 22588 | 14,896 |
https://mathoverflow.net/questions/22514 | 5 | What can be said and done about the "SIGN-Gordon equation"?
$$\varphi\_{tt}- \varphi\_{xx} + \text{sgn}(\varphi) = 0.$$
It came up [here](https://mathoverflow.net/questions/22281/can-i-relate-the-l1-norm-of-a-function-to-its-fourier-expansion/).
| https://mathoverflow.net/users/1837 | Sign-Gordon Equation | Turns out it appears in literature as the "signum-Gordon equation". For example the paper
[Signum-Gordon wave equation and its self-similar solutions](http://th-www.if.uj.edu.pl/acta/vol38/pdf/v38p3099.pdf)
| 10 | https://mathoverflow.net/users/2384 | 22590 | 14,898 |
https://mathoverflow.net/questions/22211 | 35 | Some time ago I mentioned a certain open question in an MO answer, and Pete Clark suggesting posting the question on its own. OK, so here it is:
First, the setup. Let $X$ be a projective scheme over a field $k$. By Grothendieck, there is a locally finite type $k$-scheme $A = {\rm{Aut}}\_ {X/k}$ representing the functor assigning to any $k$-scheme $T$ the group of $T$-automorphisms of $X\_T$. (Artin proved a related result with projectivity relaxed to properness, even allowing $X$ to be an algebraic space.) The construction uses Hilbert schemes, so at most countably many geometric connected components can occur.
In some cases the automorphism scheme is connected (such as for projective space, when the automorphism scheme is ${\rm{PGL}}\_n$), and in other cases the geometric component group $\pi\_0(A) = (A/A^0)(\overline{k})$ can be infinite. For the latter, a nice example is $X = E \times E$ for an elliptic curve $E$ without complex multiplication over $\overline{k}$; in this case $A$ is an extension of ${\rm{GL}}\_ 2(\mathbf{Z})$ by $E \times E$, so $\pi\_0(A) = {\rm{GL}}\_ 2(\mathbf{Z})$. This latter group is finitely presented.
Question: is the geometric component group $\pi\_0(A)$ of the automorphism scheme $A$ of a projective $k$-scheme $X$ always finitely generated? Finitely presented? And with projectivity relaxed to properness, and "scheme" relaxed to "algebraic space"?
Feel free to assume $X$ is smooth and $k = \mathbf{C}$, since I believe that even this case is completely wide open.
Remark: Let me mention one reason one might care (apart from the innate appeal, say out of analogy with finite generation of Neron-Severi groups in the general proper case). If trying to study finiteness questions for $k$-forms of $X$ (say for fppf topology, which amounts to projective $k$-schemes $X'$ so that $X'\_K = X\_K$ for a finite extension $K/k$), then the language of ${\rm{H}}^1(k, {\rm{Aut}}\_{X/k})$ is convenient. To get hands on that, the Galois cohomology of the geometric component group intervenes. So it is useful to know if that group is finitely generated, or even finitely presented.
| https://mathoverflow.net/users/3927 | Finiteness property of automorphism scheme | I wanted to add some things to the comments I had already made but the list of
comments have become very large and the comments I have already made are
becoming more and more difficult to follow so I'll put everything (including the
things I have already said) here instead even though it is not an answer to the
question.
Let us first consider the case of a minimal surface $X$ (by minimal I mean $K\_X$
nef). Dolgachev (Dolgachev: Reflection groups in algebraic geometry is a good
reference even though the proof is only referenced there not given) gave a kind of
structure theorem for the image $A\_X$ of $\mathrm{Aut}(X)$ in
$\mathrm{Aut}(S\_X)$, where $S\_X$ is the orthogonal complement of $K\_X$ in
$\mathrm{NS}(X)$ modulo torsion. His result says that there is a normal subgroup
$W\_X$ of $\mathrm{Aut}(S\_X)$ generated by reflections in $-2$-curves and the
group $P\_X$ generated by $A\_X$ and $W\_X$ is a semi-direct product and of finite
index in $\mathrm{Aut}(S\_X)$. Note that it is possible to have $W\_X=\{e\}$ and
then $A\_X$ itself is of finite index and hence an arithmetic (*and* thus
finitely presented). It is also possible to have $W\_X$ of finite index and then
$A\_X$ is finite (*and* thus finitely presented). However, there are
intermediate cases where both $A\_X$ and $W\_X$ are infinite. Still $A\_X$ is a
quotient of $P\_X$ and hence is finitely generated. I do not know if it is always
finitely presented. Borcherds (Coxeter groups, Lorentzian lattices, and $K3$
surfaces. Internat. Math. Res. Notices 1998) gives examples where it is (and
where it is even nicer) but also examples where it is finitely generated but not
arithmetic.
[[Added]]
I now realise that finite presentation is always true: For that we
only need to show that $W\_X$ is normally generated in $\mathrm{Aut}(X)$ by a
finite number of elements and for that it is enough to show the same thing for
$\mathrm{Aut}(S\_X)$. We know that $W\_X$ is generated by reflections in $-2$-elements.
There are however only a finite number of conjugacy classes $-2$-elements. For
this it is, by standard lattice theoretic arguments, enough to prove that there
are only a finite number of isomorphism classes of orthogonal
complements. However, the discriminant of such a complement is bounded in terms
of the rank and discriminant of $S\_X$ and there are only a finite number of
forms of bounded rank and discriminant.
[[/Added]]
A further step would be to blow up points of $X$ (still assumed minimal). As
$X$ is the unique minimal model any automorphism of the blowing up is given by
an automorphism of $X$ that permutes the blown up points (and the subgroup
fixing the points is commensurable with the full automorphism group). In
the case of abelian or hyperelliptic surfaces blowing up just one point is
pointless as it just serves to kill off the connected component of
$\mathrm{Aut}(X)$ so in that case the first interesting case is blowing up two
points.
Consider the case of blowing up two points when $X$ is abelian. So we have two
points on $X$ one of which we can assume is $0$ and the other we'll call $x$. An
automorphism of $X$ that fixes both of these points will be en automorphism of
$X$ as abelian variety that fixes pointwise the closed subgroup $A$ generated by
$x$. The group fixing $x$ will then have finite index in the the group fixing
$A$ pointwise. For any abelian subvariety $A$ of $X$, the subgroup of
$\mathrm{Aut}(X)$ fixing all the points of $A$ is an arithmetic subgroup (in a
not necessarily semi-simple group) and in particular is finitely presented.
The same argument works for abelian varieties of any dimension. There one of
course also has the option of blowing up positive dimensional varieties, assume
$S$ is a smooth closed subvariety. This
time the automorphism group is the subgroup of automorphisms $X$ that fixes
$S$. We thus get an induced action on $S$ and the kernel of that action has the
same structure as before. Unless I am mistaken, the automorphisms of $S$ that
extend to $X$ are of finite index in $\mathrm{Aut}(S)$ (look at $\mathrm{Alb}(S)
\rightarrow X$ and split it up to isogeny). Hence the finite generation etc for
the blowing up is reduced to finite generation for $S$ (and conversely for $X$
replaced by $\mathrm{Alb}(S)$).
Consider now the case of $X$ still minimal but non-abelian or hyper-elliptic and
look at blowing up of one point $x$. For a general point of $X$ (in the sense of
being outside a countable number of proper subvarieties) the automorphism group
is trivial and hence finitely generated. The situation for arbitrary $x$ seems
unclear but one thread of the discussion started concerning itself with whether
for a general $X$ there is a characterisation (up to commensurability) of
$\mathrm{Aut}(X)$ similar to the minimal case: Look at all automorphisms of the
integral cohomology of $X$ that preserves multiplicative structure, Hodge
structure, Chern classes (of the tangent bundle) and effective cones (spanned by
effective cycles). Is the image of the automorphism group of $X$ of finite index
in this group? I think the answer is no (and I hope that what I present here is
a proof). For that we need to recall some facts on Seshadri constants
(Lazarsfeld: Positivity in Algebraic Geometry, I is my reference). Given a point
$x$ the Seshadri constants $\epsilon(L;x)$ for $L$ nef (but also for $L$
restricted to be ample) determine (and are determined by) the nef cone of the
blowing up at $x$; $L-rE$ is nef precisely when $0\leq r\leq \epsilon(L;x)$.
Switching tack, there is a subset $U$ of $X$ which is the intersection of a
countable number of open non-empty subsets of $X$ such that $\epsilon(L;x)$ is
constant on $U$ for all ample $L$. Indeed, $\epsilon(L;x)$ can be expressed
(loc. cit.: 5.1.17) in terms of whether or not $kL$ separates $s$-jets at $x$
and for fixed $k$ and $s$ the separation is true on an open subset.
The conclusion is that there is a $U$ which is the intersection of a countable
number of non-empty open subsets for which the nef cone of the blowing up of
$X$ at $x$ is independent of $x$ when one expresses it in the decomposition
$\mathrm{NS}(X)\bigoplus\mathrm Z E$. If we assume now that $K\_X$ is numerically
trivial we have that the first Chern class of the tangent bundle of the blowup
of $X$ at some $x$ equals $E$ (up to torsion) and hence the group above will
preserve the decomposition $\mathrm{NS}(X)\bigoplus\mathrm Z E$ and fix $E$ so
come from an automorphism of $\mathrm{NS}(X)$. The only further condition we put
on it is that it preserve the nef cone but for $x\in U$ this cone is independent
of $x$. As we can further arrange it so that $x\in U \implies \varphi(x)\in U$
(as $\mathrm{Aut}(X)$ is countable) we get that all elements of
$\mathrm{Aut}(X)$ give structure preserving automorphisms of the cohomology of
the blowup of $X$ at $x$. However, as observed before, at the price of shrinking
$U$ we can assume that that the automorphism group of the blowing up is
trivial. Hence, if we let $X$ be for instance a K3-surface with infinite
automorphism group we get an example.
| 16 | https://mathoverflow.net/users/4008 | 22596 | 14,903 |
https://mathoverflow.net/questions/22587 | 2 | This is a modification of the somewhat naive question that I asked below.
Suppose $X$ is a real Banach space of cotype-2, and $u\_1, u\_2, ... u\_n$ are unit vectors in this space. For $\gamma = ((\gamma\_1, \gamma\_2, ..., \gamma\_n))$ an arbitrary element of $\mathbb{R}^n$, suppose that $\|\gamma\_1 u\_1 + \gamma\_2 u\_2 + \cdots \gamma\_n u\_n\| \geq \|\gamma\|\_{\ell^\infty}$. Then is there a bound independent of $n$ (but depending on the cotype constant) of
$$\inf\_\epsilon \sup\_{\beta \geq 0} \frac{\sum \beta\_i\cdot \frac{1}{\sqrt{n}}}{\|\sum \beta\_i \epsilon\_i u\_i \|}$$
where $\epsilon = ((\epsilon\_1,\epsilon\_2, ..., \epsilon\_n))$ varies over all sequences of $\pm 1$, and $\beta \geq 0$ means that $\beta\_i \geq 0$ for all $i$.
My suspicion is that this is actually false, but I haven't constructed an explicit counterexample.
*The original question I asked was resolved by two answers below. (I had left out a $\sqrt{n}$ term that I meant to include, but even with this, the answer is no.)*
Suppose $X$ is a real Banach space of cotype-2, and $u\_1, u\_2, ... u\_n$ are unit vectors in this space. For $\gamma = ((\gamma\_1, \gamma\_2, ..., \gamma\_n))$ an arbitrary element of $\mathbb{R}^n$, suppose that $\|\gamma\_1 u\_1 + \gamma\_2 u\_2 + \cdots \gamma\_n u\_n\| \geq \|\gamma\|\_{\ell^\infty}$. (It is important that this inequality be strict, not just an 'up to a constant' sort of bound. What is going on becomes more clear if a picture is drawn.) Is there a bound independent of $n$ on
$\sup\_\beta \frac{\sum \beta\_i}{\|\beta\_1 u\_1 + \cdots + \beta\_n u\_n\|}?$
If not, what if we impose the additional condition that the $\beta\_i$ all be non-negative? One might guess the bound is linear with the cotype-2 constant.
| https://mathoverflow.net/users/5621 | A bound on linear functionals over cotype 2 spaces | Did you mean to have a factor of ${\sqrt n}$ on the right hand side? Consider $u\_i$ orthonormal in a Hilbert space.
In fact, the best constant is of order $n$, but if you restrict the $u\_i$ to be C-unconditional, then obviously the sup is bounded by C times the cotype 2 constant times ${\sqrt n}$.
Further comments:
(1) Your inequality says that the $u\_i$ are an Auerbach basis for the space they span, which means that there are functionals $u\_i^\*$ biorthogonal to $u\_i$ and having norm one. If you have the inequality but only up to the constant $1/C$, the biorthogonal functionals have norm at most $C$. You can renorm the space with $(1/C\|x\|) \vee \max\_i|u\_i^\*(x)|$ to make the $u\_i$ an Auerbach basis--at worst this multiplies the cotype 2 constant by $C$.
(2) You can replace the top of your left hand side with $\sum |\beta\_i|$ because your hypotheses on the $u\_i$ does not change if you multiply some of them by $-1$. To see that you cannot do better than order $n$ as an upper estimate, consider $u\_i :=(1/2)(e\_i-e\_{i+1})$ in $\ell\_1$.
Added May 2, 2010:
In your modified question, you get a bound independent of $n$ under the stronger hypothesis that $X^\*$ has type 2. Indeed, let $u\_k^\*$ be the (norm one) biorthogonal functionals to $u\_k$ and choose signs $\epsilon\_k$ so that $y^\*:= n^{-1/2}\sum\_{k=1}^n \epsilon\_k u\_k^\*$ has norm at most $T\_2(X^\*)$. Then for any $\beta\_k\ge 0$, $\|\sum\_{k=1}^n \epsilon\_k\beta\_k u\_k\|$ is at least as large as $T\_2(X^\*)^{-1} n^{-1/2} \sum\_{k=1}^n \beta\_k$.
Since for an $n$ dimensional space $E$, $T\_2(E^\*)\le C \log n C\_2(E)$, you do get at worst a $\log n$ growth rate of your ratio.
In your revised question, do you know it is important that the $u\_k$ form an Auerbach basis for their span?
| 3 | https://mathoverflow.net/users/2554 | 22605 | 14,910 |
https://mathoverflow.net/questions/22607 | 0 |
>
> **Possible Duplicate:**
>
> [solving f(f(x))=g(x)](https://mathoverflow.net/questions/17614/solving-ffxgx)
>
>
>
Here is a nice interview question for computer science people:
Write a unary function `f` such that
`f(f(x)) = -x`
Constraints:
1. The function should be pure (i.e. it should have no state and every time its called it should output the same value for the "same" input.).
2. Complex number arithmetic is not allowed. So `f(x) = ix` is not allowed.
3. You can use plain mathematics or use a program. Choice is yours.
Although I am a student of computer science but I am unable to figure out it's mathematical aspect.
Any help ?
| https://mathoverflow.net/users/5633 | Interview Question | Divide the domain into separate sets of four, each set having the form {x,y,-x,-y}. Now, let f simply rotate within these sets one step. That is, map x to y, and y to -x, and -x to -y, and -y to x. Thus, doing it twice maps every x to -x, as desired. (Also let f(0)=0.)
| 10 | https://mathoverflow.net/users/1946 | 22611 | 14,914 |
https://mathoverflow.net/questions/22616 | 1 | If X is a set and A is a subset of X containing at least two elements, then certainly for any element $a \in A$, the principal ultrafilter of $a$ contains the principal filter of A (which is NOT an ultrafilter). Are all ultrafilters containing the principal filter of A of this form, or are there non-principal examples? (I'm assuming axiom of choice etc.)
| https://mathoverflow.net/users/4528 | Ultrafilters containing a principal filter | If a filter $X$ contains any set $A$, then it contains the principal filter of $A$. Thus you are really asking: for which subsets $A$ of a set $X$ can a free ultrafilter contain $A$?
It is a standard exercise to see that such free ultrafilters exist iff $A$ is infinite.
Let me briefly sketch the proof:
1. If an ultrafilter contains a finite union of sets $A\_1,\ldots,A\_n$, then it contains $A\_i$ for at least one $i$. Thus an ultrafilter which contains any finite set is principal.
2. If $A$ is infinite, consider the family $F$ of subsets which either contain $A$ or have finite complement. Then $F$ satisfies the finite intersection condition, so is a subbase for a filter $\mathcal{F}$ (see e.g. Exercise 5.2.5 of [http://alpha.math.uga.edu/~pete/convergence.pdf](http://alpha.math.uga.edu/%7Epete/convergence.pdf)). Then it can be extended to an ultrafilter which, since it contains the Frechet filter, is free.
| 7 | https://mathoverflow.net/users/1149 | 22617 | 14,916 |
https://mathoverflow.net/questions/22624 | 62 | I am working on my [zero knowledge proofs](https://en.wikipedia.org/wiki/Zero-knowledge_proof) and I am looking for a good example of a real world proof of this type. An even better answer would be a Zero Knowledge Proof that shows the statement isn't true.
| https://mathoverflow.net/users/5601 | Example of a good Zero Knowledge Proof | The classic example, given in all complexity classes I've ever taken, is the following: Imagine your friend is color-blind. You have two billiard balls; one is red, one is green, but they are otherwise identical. To your friend they seem completely identical, and he is skeptical that they are actually distinguishable. You want to prove to him (I say "him" as most color-blind people are male) that they are in fact differently-colored. On the other hand, you do not want him to learn which is red and which is green.
Here is the proof system. You give the two balls to your friend so that he is holding one in each hand. You can see the balls at this point, but you don't tell him which is which. Your friend then puts both hands behind his back. Next, he either switches the balls between his hands, or leaves them be, with probability 1/2 each. Finally, he brings them out from behind his back. You now have to "guess" whether or not he switched the balls.
By looking at their colors, you can of course say with certainty whether or not he switched them. On the other hand, if they were the same color and hence indistinguishable, there is no way you could guess correctly with probability higher than 1/2.
If you and your friend repeat this "proof" $t$ times (for large $t$), your friend should become convinced that the balls are indeed differently colored; otherwise, the probability that you would have succeeded at identifying all the switch/non-switches is at most $2^{-t}$. Furthermore, the proof is "zero-knowledge" because your friend never learns which ball is green and which is red; indeed, he gains no knowledge about how to distinguish the balls.
| 164 | https://mathoverflow.net/users/658 | 22628 | 14,923 |
https://mathoverflow.net/questions/22619 | 12 | I keep running across papers that refer to a set of lecture notes by Robert MacPherson at MIT during the fall of 1993 on Perverse Sheaves. There might also be a set of notes from lectures in Utrecht in 1994 taken by Goresky. For references to these elusive, unpublished notes see the work of Maxim Vybornov and A. Polishchuk.
Does anyone have a copy or know how I could obtain a copy?
Thanks!
| https://mathoverflow.net/users/1622 | Perverse Sheaves - MacPherson Lecture Notes | The notes on Friedman's page are great -- they were very helpful when I was learning about perverse sheaves as a graduate student, since they explain how to think about middle perversity perverse sheaves on complex analytic spaces with complex stratifications (the case of most interest for most people) without having to first define the derived category. Unfortunately the main result there, although it is a theorem, is not fully proved in the notes. One of Bob's students, Mikhail Grinberg, had a project to fill in the details and flesh out the notes into a book, but he left mathematics for a job at Renaissance Technologies. He taught a course at MIT around 2000 where he went through the steps needed to fill in the proof. Perhaps someone has usable notes from that class?
The 1993 course was quite different. Again the goal was to see perverse sheaves and their abelian category structure without going through the derived category, but he was working with arbitrary perversities on a regular cell complex. By putting weird "phantom dimensions" on the cells and putting arrows between cells of adjacent phantom dimensions you get a quiver whose representations are perverse sheaves constructible for the cell complex structure. But it's hard to use this to compute beyond simple examples, because the cell complexes would have to be very large, and it's hard to define the condition that gives constructibility for a coarser stratification like a complex analytic one.
| 16 | https://mathoverflow.net/users/3889 | 22649 | 14,932 |
https://mathoverflow.net/questions/22635 | 36 | **Disclaimer**
Of course not, I'm aware of Gödel's second incompleteness theorem. Still there is something which does not persuade me, maybe it's just that I've taken my logic class too long ago. On the other hand, it may turn out I'm just confused. :-)
**Background**
I will be talking about models of set theory; these are sets on their own, so a confusion can arise, since the symbol $\in$, viewed as "set belonging" in the usual sense, may have a different meaning from the symbol $\in$ of the theory. So, to avoid confusion, I will speak about levels.
On the first level is the set theory mathematicians use all day. This has axioms, but is not a theory in the usual sense of logic. Indeed, to speak about logic we already need sets (to define alphabets and so on). In this *naif* set theory we develop logic, in particular the notions of theory and model. We call this theory **Set1**.
On the second level is the *formalized* set theory; this is a theory in the sense of logic. We just copy the axioms of the *naif* set theory and take the (formal) theory which has these strings of symbols as axioms. We call this theory **Set2**.
Now Gödel's result tells us that if **Set2** is consistent, it cannot prove its own consistence. Well, here we need to be a bit more precise. The claim as stated is obvious, since **Set2** cannot prove anything about the sets in the first level. It does not even know that they exist.
So we repeat the process that carried from **Set1** to **Set2**: we define in **Set2** the usual notions of logic (alphabets, theories, models...) and use these to define another theory **Set3**.
A correct statement of Gödel's result is, **I think**, that
>
> if **Set2** is consistent, then it cannot prove the consistence of **Set3**.
>
>
>
**The problem**
Ok, so we have a clear statement which seems to be completely provable in **Set1**, and indeed it is. This doesn't tell us, however that
>
> if **Set1** is consistent, then it cannot prove the consistence of **Set2**.
>
>
>
So I'm left with the doubt that what one can do "from the outside" may be a bit more than what one can do in the formalized theory. Compare this with Gödel's first incompleteness theorem, where one has a statement which is true for natural numbers (and we can prove it from the outside) but which is not provable in **PA**.
So the question is:
>
> is there any reason to believe that **Set1** cannot prove the consistence of **Set2**? Or I'm just confused and what I said does not make sense?
>
>
>
Of course one could just argue that **Set1**, not being formalized, is not amenable to mathematical investigation; the best model we have is **Set2**, so we should trust that we can always "shift our theorems one level". But this argument does not convince me: indeed Gödel's first incompleteness theorem shows that we have situations where the theorem in the formalized theory are strictly less then what we can see from the outside.
**Final comment**
In a certain sense, it is far from intuitive that set theory should have a model. Because models are required to be sets, and sets are so small...
Of course I know about universes, and how one can use them to "embed" the theory of classes inside set theory, so sets may be bigger than I think. But then again, existence of universes is not provable from the usual axioms of set theory.
| https://mathoverflow.net/users/828 | Can we prove set theory is consistent? | Would you accept it if Set1 just proved the existence of a model for Set2 (in the same way that Set1 proves the consistency of formalized Peano arithmetic by providing a model of it)?
If so, and if you accept in Set1 that there is an inaccessible cardinal κ, then the set Vκ is a model of ZFC, provably in Set1. Most set theorists today seem to believe that there are inaccessible cardinals and much bigger "large cardinals" in the universe of sets. So they count this as a proof of the consistency of ZFC, just as they count the existence of the standard natural numbers as a proof of Peano arithmetic.
You do not even need to assume Set1 has an inaccessible cardinal. For example, you could assume Set1 contains all of Morse-Kelley set theory and Set2 consists of ZFC, and then Set1 would prove the consistency of Set2.
What you cannot do is prove the consistency of Set2 within Set1 using only techniques that can be formalized within Set2. This is no different than with Peano arithmetic: we can formally prove that Peano arithmetic is consistent, but not using methods that can themselves be formalized in Peano arithmetic. The fact that you are interested in set theory only makes the problem seem more complicated; the underlying phenomenon is not much different.
| 21 | https://mathoverflow.net/users/5442 | 22656 | 14,937 |
https://mathoverflow.net/questions/22658 | 15 | Why are inverse images of functions more central to mathematics than the image?
I have a sequence of related questions:
1. Why the fixation on continuous maps as opposed to open maps? (Is there an epsilon-delta definition of open maps in metric spaces?)
2. Is there an inverse-image characterization of homomorphisms in algebraic categories? (What kind of map do you get if you look at a map from a group to another group, where inverse images of subgroups are subgroups?)
3. Inverse images have better set-theoretic properties than the image (for instance, commuting with unions, intersections, etc..) This clearly is a direct consequence of definition of a function. There is an asymmetry in the definition of a function (the domain and codomain behave differently with respect to the function). I think this also has consequences for differences between existence and uniqueness of left and right inverses for one-to-one or onto functions. Why this asymmetry? What are the historical reasons for the asymmetry? Whats sort of mathematics do we have if the definition of a function was purely symmetric? (For instance, f(a) may give multiple values, just like f^-1(a) may have multiple values).
4. Is it accurate to say the definitions for monomorphisms and epimorphisms in category theory correct for the asymmetry? (And hence, the notion of epimorphisms and onto-morphisms in concrete categories don't coincide)
| https://mathoverflow.net/users/5651 | Why are inverse images more important than images in mathematics? | Open sets can be identified with maps from a space to the Sierpinski space, and maps out of a space pull back under morphisms. (In other words, if you believe that the essence of what it means to be a topological space has to do with functions out of the space, you are privileging inverse images over images. A related question was discussed [here](https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets/).) I think essentially this kind of reasoning underlies the basic appearances of inverse images in mathematics. For example, in the category of sets, subsets can be identified with maps from a set to the two-point set, and again these maps pull back under morphisms. This should be responsible for the nice properties of inverse image with respect to Boolean operations.
Your third question was asked, closed, and deleted once; I started a blog discussion about it [here](http://qchu.wordpress.com/2010/03/25/the-definition-of-a-function/).
| 12 | https://mathoverflow.net/users/290 | 22666 | 14,943 |
https://mathoverflow.net/questions/22642 | 21 | ### Background on why I want this:
I'd like to check that suspension in a simplicial model category is the same thing as suspension in the quasicategory obtained by composing Rezk's assignment of a complete Segal space to a simplicial model category with Joyal and Tierney's "first row functor" from complete Segal spaces to quasicategories. Rezk's functor first builds a bisimplicial set (I will say simplicial space) with a very nice description in terms of the model category, then takes a Reedy fibrant replacement in the category of simplicial spaces. Taking this fibrant replacement is the only part of the process which doesn't feel extremely concrete to me.
To take a Reedy fibrant replacement $X\_\*'$ for a simplicial space $X\_\*$ you can factor a sequence of maps of simplicial sets as weak equivalences followed by fibrations. The first such map is $X\_0 \rightarrow \*$. If I factor that as $X\_0 \rightarrow X\_0' \rightarrow \*$ the next map to factor is $X\_1 \rightarrow X\_0' \times X\_0'$ and factors as $X\_1 \rightarrow X\_1' \rightarrow X\_0' \times X\_0'$. (The maps are all from $X\_n$ into the $n$th matching space of the replacement you're building, but these first two matching spaces are so easy to describe I wanted to write them down explicitly.)
---
### Background on the question:
The small object argument gives us a way to factor any map $X \rightarrow Y$ in a cofibrantly generated model category as a weak equivalence $X \rightarrow Z$ followed by a fibration $Z \rightarrow Y$. (In fact, as an acyclic cofibration followed by a fibration.) However, the object $Z$ this produces is in general hard to understand.
If the model category we're interested in is the usual one on simplicial sets (weak equivalences are weak homotopy equivalences on the geometric realizations, fibrations are Kan fibrations, and cofibrations are inclusions) we have some special ways of factoring maps $X \rightarrow \*$ as a weak equivalence followed by a fibration. Probably the most familiar is using the singular chains on the geometric realization of $X$ to make $X \rightarrow S(\mid X\mid ) \rightarrow \*$. This is again big and rather hard to understand.
Another method is Kan's Ex$^\infty$ functor, which I learned about in Goerss and Jardine's book. Ex is the right adjoint to the subdivision functor, which is defined first for simplices in terms of partially ordered sets, so the subdivision functor and Ex are both very combinatorial. There is a natural map $X \rightarrow$ Ex $X$, and Ex$^\infty X$ is the colimit of the sequence $X \rightarrow$ Ex $X \rightarrow$ Ex$^2 X \rightarrow$... . It turns out that $X \rightarrow $Ex$^\infty X \rightarrow \*$ is a weak equivalence followed by a fibration.
I would like a way of factoring a map $X \rightarrow Y$ of simplicial sets as a weak equivalence followed by a fibration that is similar in flavor to the use of Kan's Ex$^\infty$ functor to find a fibrant replacement for a simplicial set $X$.
---
### Question:
>
> In the standard model category structure on simplicial sets, is there a combinatorial way of factoring a map as a weak equivalence followed by a fibration?
>
>
>
| https://mathoverflow.net/users/3413 | Is there a combinatorial way to factor a map of simplicial sets as a weak equivalence followed by a fibration? | I don't know if this does what you want, but it's one thing I know how to do.
If $X\to Y$ is a map between *Kan complexes*, then you can build a factorization using the path space construction. Thus, $P=Y^I\times\_Y X$, and $P\to Y$ is given by evaluation, while $X\to P$ is produced using the constant path. Here $I=\Delta[1]$, and $Y^I$ is the mapping object in simplicial sets. Since $Y$ is a Kan complex, $P\to Y$ is a Kan fibration (using the fact that $Y^I\to Y^{\partial I}$ is a Kan fibration).
For a general map $f: X\to Y$, consider $Ex^\infty(f): Ex^\infty X\to Ex^\infty Y$. This is a map between Kan complexes, and therefore the path space construction on $Ex^\infty(f)$ gives a Kan fibration $P\to Ex^\infty Y$. Now pull back along $Y\to Ex^\infty Y$ to get a Kan fibration $Q\to Y$. Because pullbacks of fibrations in simplicial sets always are homotopy pullbacks, it should be possible to see that $X\to Q$ is a weak equivalence.
(I think this construction goes back to Quillen's book on model categories.)
| 16 | https://mathoverflow.net/users/437 | 22667 | 14,944 |
https://mathoverflow.net/questions/22131 | 4 | I want to know if the following fact has a standard name and/or reference
>
> Let $X$ be a subset of $\mathbb R^2$ and $B$ be a disc of the same area as $X$.
> Set $X\_\epsilon$ to be the $\epsilon$-neigborhood of $X$. Then $$area\\,X\_\epsilon\ge area\\,B\_\epsilon.$$
>
>
>
| https://mathoverflow.net/users/5524 | Name for an inequality of isoperimetric type | This inequality is essentially equivalent to the [Classical isoperimetric inequality](http://en.wikipedia.org/wiki/Isoperimetric_inequality). If you have a measurable body $X$ in $\mathbb{R}^n$ and a ball $B\subset \mathbb R^n$ of same volume then you have the following:
$$Area(X)=\lim\_{\epsilon \to 0} \frac{\operatorname{Vol}(X\_{\epsilon})-\operatorname{Vol}(X)}{\epsilon}$$
$$Area(B)=\lim\_{\epsilon \to 0} \frac{\operatorname{Vol}(B\_{\epsilon})-\operatorname{Vol}(B)}{\epsilon}$$
Proving that $Area(X)\geq Area(B)$ follows from $\operatorname{Vol}(X\_{\epsilon})\geq \operatorname{Vol}(B\_{\epsilon})$, which is your inequality. ($n=2$)
Now this follows from the [Brunn Minkowski inequality](http://en.wikipedia.org/wiki/Brunn%E2%80%93Minkowski_theorem) because
$$\operatorname{Vol}(X\_{\epsilon})=\left(\operatorname{Vol}(X+\epsilon B)^{1/n}\right)^n \geq \left(\operatorname{Vol}(X)^{1/n}+\epsilon \operatorname{Vol}(B)^{1/n}\right)^n=\operatorname{Vol}(B\_{\epsilon})$$
| 3 | https://mathoverflow.net/users/2384 | 22671 | 14,948 |
https://mathoverflow.net/questions/22676 | 5 | Both in abstract algebra and measure theory is there term ring/algebra, but their definition are different and we can not deduce one from the other, the only requirement in definition they share is that they be closed under two operations: addition and multiplication in abstract algebra while difference and union in measure theory. Why we use the same term for these two different notions in two different branches of mathematics? Is there any connection between them? Thanks!
To be more specific, my intention is to compare "ring in measure theory" with "ring in abstract algebra", not "ring" with "algebra". I use notation "ring/algebra" in the post because the term algebra has the same situation, that is, compare "algebra in measure theory" with "algebra in abstract algebra".
| https://mathoverflow.net/users/5072 | terminology about ring/algebra in abstract algebra and measure theory | A [ring of sets](http://en.wikipedia.org/wiki/Ring_of_sets) is a ring(usual definition) with the operations intersection(multiplication) and symmetric difference(addition). A [sigma ring](http://en.wikipedia.org/wiki/Sigma_ring) is a special kind of a ring of sets. Now a [sigma algebra](http://en.wikipedia.org/wiki/%CE%A3-algebra) (which would possibly more appropriately be called sigma field) is a sigma ring where every element has a complement (multiplicative inverse) is a sigma ring with unity (equivalent to every element has a complement),thus it is a Boolean algebra with respect to intersection and union, and a Boolean ring with respect to intersection and symmetric difference. Every Boolean ring is an algebra over $\mathbb F\_2$(thank you Mark Meckes for the correction). I hope this connection is enough to justify the use of these terms.
| 9 | https://mathoverflow.net/users/2384 | 22679 | 14,952 |
https://mathoverflow.net/questions/22638 | 7 | There exists information on the Picard (and Brauer) group of a reductive algebraic group over a number field k. For example, Sansuc shows (in his big Crelle paper of 1980) that if G is connected and semisimple over a number field k, then Pic G is the group of k-rational points of the character group of the fundamental group of G. In particular, if G is semisimple and simply connected, then Pic G=0. My question is: do there exist results of this type over more general bases? For example, a natural generalization of the equality "Pic G=0 if G semisimple and simply connected over a number field" would be "Pic G=Pic U if G is a semisimple and simply connected group scheme over a Dedekind scheme U". Is the latter true?
| https://mathoverflow.net/users/5641 | Picard groups of reductive group schemes | Assuming you know that Pic of the generic fibre is trivial, this seems to follow immediately from the localisation sequence: since G is a group there is a section, so the map Pic U to Pic G is an injection. On the other hand (since G is smooth so Pic = Cl) there is an exact sequence:
$\ \ \ \oplus\_x \mathbb{Z}\_x \to Pic \ G \to Pic \ G\_K \to 0 $
where X runs over the closed points of U and $G\_K$ denotes the generic fibre. Since $Pic\ G\_K$ is trivial it follows from this that the map Pic U to Pic G is surjective.
(Note that the only property of the closed fibres that is used is that they are reduced and irreducible.)
| 3 | https://mathoverflow.net/users/519 | 22684 | 14,956 |
https://mathoverflow.net/questions/22659 | 25 | (Alternate title: *Find the Adjoint: Hopf Algebra edition*)
I was chatting with Jonah about his question [Hopf algebra structure on $\prod\_n A^{\otimes n}$ for an algebra $A$](https://mathoverflow.net/questions/22536/hopf-algebra-structure-on-prod-n-a-otimes-n-for-an-algebra-a). It's very closely related to the following question:
>
> For a $k$-algebra $A$, is there a Hopf Algebra $H(A)$ such that for any Hopf algebra $B$, we have
>
> $\mathop{Hom}\_{k\text{-alg}}(B,A)\cong \mathop{Hom}\_{k\text{-Hopf-alg}}(B,H(A))$?
>
>
>
In other words, does the forgetful functor from $k$-Hopf-algebras to $k$-algebras have a right adjoint? There are a few related questions, any of which I'd be interested in knowing the answer to:
* Does the forgetful functor from Hopf algebras to *augmented* algebras (sending the counit to the augmentation) have a right adjoint?
* Does the forgetful functor from Hopf algebras to algebras with distinguished anti-automorphism (sending the antipode to the anti-automorphism) have a right adjoint?
* Does the forgetful functor from Hopf algebras to algebras with augmentation *and* distinguished anti-automorphism have a right adjoint?
Unfortunately, I don't feel like I can motivate this question very well. My motivation is that the better I know which forgetful functors have adjoints, the better I sleep at night.
| https://mathoverflow.net/users/1 | Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint? | The article [here](http://arxiv4.library.cornell.edu/pdf/0905.2613) proves that the forgetful functor from $k$-Hopf algebras to $k$-algebras has a right adjoint. The main tool they use is the Special Adjoint Functor Theorem.
| 18 | https://mathoverflow.net/users/4517 | 22685 | 14,957 |
https://mathoverflow.net/questions/22694 | 4 | In coordinate-free language, my question is as follows. Let $M$ be an $n$-dimensional manifold with volume form, and let $\mathcal D$ be a smooth (integrable, if necessary) distribution with constant rank $k\leq m$ (by which I mean that $\mathcal D$ is a subbundle of the tangent bundle, spanned by $k$ many every-linearly-independent smooth vector fields). Given a point $p\in M$, is there necessarily a neighborhood $U\ni p$ and vector fields $w\_1,\dots,w\_k$ on $U$ so that each $w\_a$ is divergence-free (with respect to the volume form) and so that $\mathcal D|\_U = \text{span}\{w\_1,\dots,w\_k\}$?
Since my question is local, I will ask it again on $\mathbb R^n$ in coordinates. I will let $x^i$, $i=1,\dots,n$ be the standard coordinates on $\mathbb R^n$. Suppose you are given smooth vector fields $v\_a(x) = \sum\_{i=1}^n v\_a^i(x) \frac{\partial}{\partial x^i}$ for $a=1,\dots,k$, where $k\leq n$, and suppose moreover that for each $x$, the set of vectors $\{v\_1(x),\dots,v\_k(x)\}$ is linear independent. (Put another way, $v$ is an everywhere-full-rank $(k\times n)$-matrix-valued function.) I'm looking for a smooth ${\rm GL}(k)$-valued-function $m(x) = \{m^a\_b(x)\}\_{a,b=1}^k$ in a neighborhood of $0\in \mathbb R^n$ so that for each $b=1,\dots,k$,
$$ \sum\_{i=1}^n \sum\_{a=1}^k \frac{\partial}{\partial x^i}\bigl[ m^a\_b(x)\,v^i\_a(x) \bigr] = 0 $$
If such an $m$ (or, if you prefer, $\{w\_1,\dots,w\_k\}$) exists, the natural questions are:
* Locally, how unique is it? Certainly it can by multiplied by any constant invertible $k\times k$ matrix. Is this it? I.e., if we fix the matrix (spanning set) at $0$ ($p$), does it fix the value in an open neighborhood?
* Can "local" be replaced by "global"? (I'm not even sure whether a rank-$k$ smooth distribution, which for me is a subbundle that's locally spanned by $k$ everywhere-independent vector fields, can be globally spanned by $k$ everywhere-independent vector fields.)
Finally, I should remark that for my application, my distribution is integrable — does this matter for whether the answer is yes? Also, for my application, not only do I only need the result locally, but I actually am working in the formal neighborhood of a point, so for my purposes it's fine if all smooth functions are replaced by formal power series. Occasionally, but rarely, questions of existence of solutions to differential equations are easy in formal land but hard even in smooth land.
If you're feeling extra smart, the last generalization I'd be interested in knowing the answer to is what happens if the rank of the distribution can drop. I.e. fix $k\leq n$, and suppose that we have vector fields $v\_1,\dots,v\_k$, which may be linearly dependent at $0$. When is there an (invertible) $(k\times k)$-matrix-valued function $m$ so that the above holds? The answer is a resounding "no" when $k=n=1$, but I have no intuition for higher values.
My motivation is that integrable distributions are the most general thing (that I know of, anyway), that can act by smooth "symmetries" of a manifold. I would like to know if every smooth symmetry can be taken to be volume-preserving. For a different interpretation of this general question, see e.g. Moser, On the volume elements on a manifold, Trans. Amer. Math. Soc., 1965, vol. 120, pp. 286--294, MR0182927, in which it is proven that any globally-volume-preserving diffeomorphism is smoothly isotopic through globally-volume-preserving diffeomorphisms to a locally-volume-preserving diffeomorphism.
| https://mathoverflow.net/users/78 | Does a (smooth, constant-rank, integrable) distribution have a (local) basis of divergence-free vector fields? | For any non-zero vector field $v$ and a volume form $\omega$ there exists (locally) a positive function $f$ such that $L\_{fv}\omega=0$ (Indeed, one can take coordinates such that $v=\frac{\partial}{\partial x\_1}$, $\omega=A(x\_1,...,x\_n)dx\_1\wedge ...\wedge dx\_n$. In these coordinates the condition $L\_{fv}\omega=0$ is equivalent to an ordinary differential equation $\frac{\partial f}{\partial x\_1}+\frac{1}{A}\frac{\partial A}{\partial x\_1}f=0$, I am using the homotopy formula $L\_u=i\_ud+di\_u$ here with $u=fv$). Hence, for any distribution (of constant rank) there exists a local basis of divergent-free vector fields - take any basis of vector fields and multiply each vector field from it on a suitable $f$.
| 8 | https://mathoverflow.net/users/2823 | 22696 | 14,962 |
https://mathoverflow.net/questions/22673 | 22 | In my mind, [algebraic topology](http://en.wikipedia.org/wiki/Algebraic_topology) is comprised of two components:
1. [Chain complex](http://en.wikipedia.org/wiki/Chain_complex) information, which is **intrinsic** information concerning how your object may be built up out of simple "lego blocks".
2. [Characteristic classes](http://en.wikipedia.org/wiki/Characteristic_class) (bundle information) which give information on how your object might stably embed in some sufficiently big standard object.
Chain complexes make sense over any [abelian category](http://en.wikipedia.org/wiki/Abelian_categories).
I have no corresponding intuitive understanding of what the "natural setting" for characteristic classes should be. The classical theory looks to me like a concession to the sad fact that, at its very basis, manifolds are locally modeled on Euclidean space and are not intrinsically defined objects. This is reflected in the central role played by specific concrete spaces such as the [Thom spaces](http://en.wikipedia.org/wiki/Thom_space) MSO(n), the [real and complex Grassmanians](http://en.wikipedia.org/wiki/Grassmannian) used to define [Wu classes](http://en.wikipedia.org/wiki/Stiefel%E2%80%93Whitney_class#Wu_classes), and the classifying spaces BU and BO concerning which we have [Bott periodicity](http://en.wikipedia.org/wiki/Bott_periodicity_theorem).
I realize that I have no understanding of any of this. Part of this feeling is because I really don't understand what forces us to consider these specific concrete spaces, to the exclusion of all others. If constants appearing in physics ought to be conceptually explained, I'd like to understand these "constants" in mathematics. Can one work with characteristic classes in a more general setting, to parallel abelian categories? What about over number fields, over arbitrary rings, or in finite characteristic? Can I replace Lie groups such as SO(n), U(n), and O(n) by [groups of Lie type](http://en.wikipedia.org/wiki/Group_of_Lie_type) for instance, and still have a "useful" theory?
My question is then:
> What is the most general categorical setting for a "useful" theory of characteristic classes? In particular, are all of those special concrete spaces really necessary, and if so, why?
| https://mathoverflow.net/users/2051 | Natural setting for characteristic classes? | Here is a perspective that might help to put characteristic classes into a more general framework. I like to think that there are two levels of the theory. One is geometric and the other is about extracting information about the geometry through algebraic invariants.
Bear with me if this sounds to elementary and obvious at first.
1. **The geometric side**: We have some class of bundle type objects which admit a theory of classifying spaces. This allows us to swap bundles over $X$ for maps of $X$ into some fixed space, which I will call $B$ for the moment. Equivalent bundles over $X$ give equivalent maps to $B$.
2. **The algebraic side**: We study maps from $X$ to $B$ by looking at their effect on some type of cohomology theory. The point is that we push the problem of studying maps $X \to B$ forward into an algebraic category where we have a better hope of extracting information.
The passage from geometric to algebraic certainly throws some information away; this is the price for moving to a more computable setting. But in the right circumstances the information you want might still be available.
Now, a general framework for this might be the following. Bundles in the abstract are objects that are local over the base and can be glued together. This is precisely what stacks are meant to describe. So think of bundles simply as objects that are classified by maps of $X$ to some stack. This can make sense in any category where you have a notion of coverings (a Grothendieck topology), so we don't have to stick with just ordinary topological spaces here. If you know how to talk about coverings of chain complexes then you can probably make a chain level version. But more concretely, we could also be talking about principal $G$-bundles for just about any sort of a group $G$. Or we could talk about fibre bundles with fibre of some particular type (in my own work, surface bundles come up quite a lot).
As an aside, if you happen to be working with spaces and you want to get back to the usual setting of classifying spaces like grassmannians and $BO$ or $BU$ then there is a way to get there from a classifying stack. Take its homotopy type; i.e, if $B$ is a stack, then choose a space $U$ and a covering $U \to B$, then form the iterated pullbacks $U\times\_B \cdots \times\_B U$ which give a simplicial space - the realization of this simplicial space will be the homotopy-theoretic classifying space).
Now, we have some class of bundle objects classified by a stack $B$. To have a "useful" theory of characteristic classes we need a cohomology theory in this category for which
1. We can compute enough of the cohomology of $B$ and the map induced by $X \to B$.
2. Enough information is retained at the level of cohomology to tell us things we want to know about morphisms $X \to B$.
It is very much an art to make a choice of cohomology theory that helps with the problem at hand.
I just want to point out that if you are working with vector bundles, then you needn't think of characteristic classes only as living in singular cohomology classes. A vector bundle represents a K-theory class, and you can think of that class as *the* K-theory characteristic class of the bundle.
**Addendum:** Just to say something about why we work with things like $BO$ instead of $BO(n)$, let me point out that it is a matter of putting things into the same place so we can compare them. Real rank n vector bundles have classifying maps $BO(n)$, and if you want to compare a map to $BO(n)$ with a map to $BO(m)$ then a natural thing to do is map them both to $BO(n+m)$. And then, why not go all the way to $BO(\infty)=BO$? It's just a matter of not having to compare apples and oranges.
| 7 | https://mathoverflow.net/users/4910 | 22704 | 14,967 |
https://mathoverflow.net/questions/22702 | 4 | ***Question*** : are the continuous characters of the form
* $\eta : \mathbb{Z}\_p^\* \to \mathbb{Z}\_p^\*$, or
* $\eta : (1+p\mathbb{Z}\_p)^{\times} \to \mathbb{Z}\_p^\*$ (i.e., on the principal units in $\mathbb{Z}\_p^\*$)
well understood? Can such characters be classified in either case ?
I'm hoping to find an analytic classification ; i.e. to describe such characters as functions, or more precisely, how the functions $z\mapsto z^s$ for $s\in\mathcal{O}\_{\mathbb{C}\_p}$ 'sit' inside the set of characters $\eta : (1 + p\mathbb{Z}\_p)^\times \to \mathbb{Z}\_p^\*$ (i.e., how 'far' is a generic character from some character of this type ?).
| https://mathoverflow.net/users/4399 | Classifying continuous characters $X \to \mathbb{Z}_p^*$, $X=\mathbb{Z}_p^*$ or $(1+p\mathbb{Z}_p)^{\times}$ ? | Yes, these are very well-understood! Here's what they are. If $p$ is odd then $\mathbf{Z}\_p^\times$ is a direct product of $\mu$, the subgroup of $p-1$th roots of unity, and $1+p\mathbf{Z}\_p$, the principal units. A continuous character of the product is a product of continuous characters, so that reduces the first part to the second part. As for the second part, the principal units are topologically generated by $1+p$ so it suffices to say where $1+p$ should go. Note however that $1+p$ can't go to an arbitrary element of $\mathbf{Z}\_p^\times$ because you need that if $(1+p)^{n\_i}$ tends to 1 in $\mathbf{Z}\_p$ then $s^{n\_i}$ tends to 1 in $\mathbf{Z}\_p$, where $s$ is the image of $1+p$. You can check that, for example, $s=-1$ does not have this property (because the $n\_i$ can be even or odd and still tend to zero $p$-adically). But it's also not hard to check that $s$ has this property iff $s$ is a principal unit. I do this in Lemma 1 of my paper "On p-adic families of automorphic forms" [here](http://www2.imperial.ac.uk/~buzzard/maths/research/papers/GL1.pdf) but this is most certainly standard and not due to me.
So in summary, for $p>2$, characters of the principal units biject with $1+p\mathbf{Z}\_p$ non-canonically, the dictionary being "image of $1+p$", and characters of the full unit group biject with the product of this and the cyclic group of order $p-1$, that being the characters of $\mu\_{p-1}$.
For $p=2$ the two questions are the same, and the same trick, appropriately modified, works. The group $1+4\mathbf{Z}\_2$ is procyclic, generated by 5, and its characters biject with the principal units, the dictionary being "image of 5". For the full unit group the characters biject with the principal units product +-1, because $\mathbf{Z}\_2^\times$ is just a product $\pm1\times (1+4\mathbf{Z}\_2)$.
| 8 | https://mathoverflow.net/users/1384 | 22705 | 14,968 |
https://mathoverflow.net/questions/22707 | 5 | Suppose $M$ and $M'$ are two $R$-moules (I am most interested in the case of $R$ a DVR). If $M\otimes M'$ is a semi-simple module (i.e., every submodule is a direct summand) then is it true that the tensor factors $M$ and $M'$ are semi-simple ?
I.e., if this is not true then it would be nice to see a counter-example.
| https://mathoverflow.net/users/4399 | If a tensor product of modules is semi-simple, are the tensor factors semi-simple ? | No; if $R$ is a DVR with uniformizing paramenter $t$, the $R$-module $(R/tR) \otimes (R/t^2R) = R/tR$ is semisimple, but $R/t^2R$ is not.
| 19 | https://mathoverflow.net/users/4790 | 22709 | 14,970 |
https://mathoverflow.net/questions/22706 | 2 | In my Analysis class, we started to prove a theorem that said:
>
> Let a > 1. So there is a unique increasing function $f:(0,\infty)\to\mathbb{R}$ so that:
>
>
> 1. $f(a) = 1$
> 2. $f(xy) = f(x) + f(y)\quad\forall x, y > 0$
>
>
> First we supposed its existence. Then through 2 it follows that f is a group homomorphism between the multiplicative group $(0,\infty)$ and the additive group $\mathbb{R}$.
>
>
> $f(x) = f(x\cdot1) = f(x)+f(1)$, so $f(1)=0=f(x\cdot x^{-1})=f(x)+f(x^{-1})$
>
>
> Then $f(x^{-1}) = -f(x)$.
>
>
> We affirm that $f(x^n) = n f(x)\quad\forall x>0, n \in \mathbb{N}$ (and then we proved by induction).
>
>
> Let x > 0 and $n\in \mathbb{N}^\*$. So exists $m \in \mathbb{Z}$ so that $a^m \le x^n \le a^{m+1}$
>
>
> So $f(a^m) \le f(x^n) \le f(a^{m+1})$, i.e. $m\le nf(x) \le m+1$
>
>
> And finally $m/n \le f(x) \le (m+1)/n$
>
>
> Let $A\_x = \{ \frac{m}{n} : m \in \mathbb{Z},\: n \in \mathbb{N},\: a^m \le x^n\}$
>
>
> So $f(x) = \sup A\_x$
>
>
>
He said that this means f is unique, but I can see no reason why. I've omitted some lemmas and parts of the proof, but kept all the results. It's quite clear f is $log\_ax$, but why it is unique?
Edit: I can't get LaTeX to work. It all seems fine while editing, but wrong in the question page.
| https://mathoverflow.net/users/5660 | Uniqueness of the logarithm function | Suppose that at some point $x$ the function $f(x)$ assumes a different value $f(x)\ne\sup A\_x$.
If $f(x)>\sup A\_x$, take a rational number $m/n$ in the interval $(\sup A\_x,f(x))$. It does not belong to $A\_x$ (otherwise it's at most $\sup A\_x$), so $a^m>x^n$ implying $m>nf(x)$, hence $f(x)$ is less than $m/n$, a contradiction.
If $f(x)<\sup A\_x$, then take any $m/n\in A\_x$ which is greater than $f(x)$ (if all such $m/n\in A\_x$ are less or equal than $f(x)$, then $\sup A\_x$ is less or equal than $f(x)$ as well by the definition of the supremum). In this case $m/n\in A\_x$ implies $a^m\le x^n$, so that $m\le nf(x)$ or $f(x)\ge m/n$, again a contradiction.
Therefore, the only choice for the function $f(x)$ to satisfy (1), (2) and the increasing property is to assume the value $\sup A\_x$ at the point $x$.
| 5 | https://mathoverflow.net/users/4953 | 22710 | 14,971 |
https://mathoverflow.net/questions/22722 | 20 | I am looking for a proof of the following fact:
*If $R$ is a principal artin local ring and $M$ a finitely generated $R$-module, then $M$ is a direct sum of cyclic $R$-modules.*
(Apparently such rings $R$ are called, e.g. in Zariski-Samuel, special principal ideal rings.)
I almost didn't ask this question for fear that I might just be missing something obvious, but I've been unable to come up with a proof myself and I can't find one anywhere (if I am just being stupid I certainly don't mind being told). Zariski-Samuel proves a structure theorem for principal ideal rings (they are products of PID's and special PIR's), as well as the fact that a submodule of a principal ideal ring generated by n elements is generated by $\leq n$ elements. I thought perhaps the statement above could be deduced from this last fact, in a similar manner to the way one proves the corresponding result for finite modules over PID's, but the obstruction to this (as I see it) is that submodules of free $R$-modules will not in general be free (for instance, the maximal ideal of $R$ is not free, assuming $R$ is not a field, i.e., has length $\geq 2$). Essentially the naive induction on the number of generators doesn't seem to work because I can't be sure that the relevant exact sequence splits...again, maybe I'm just being foolish. I think a proof might be found in chapter VIII of Bourbaki's Algebra text, but this chapter isn't in my copy (I think maybe it's only available in French).
Incidentally, it's straightforward to show that, if such a decomposition exists, the number of times a factor of the form $R/(\pi^i)$, where $(\pi)$ is the maximal ideal of $R$ and $1\leq i\leq k$ ($k$ being the length of $R$, i.e., the index of nilpotency of $(\pi)$) is uniquely determined as the length of a quotient of $M$, for instance.
The reason I'm interested in this is because I want to know that the isomorphism type of $M$ is completely determined by the function sending $i$, $1\leq i\leq k$, to the length of $M[\pi^i]$ (the kernel of multiplication by $\pi^i$), which, assuming such a decomposition exists, is definitely the case.
Edit: I realized that my module M can (being artinian) be written as a finite direct sum of indecomposable submodules, so I guess this reduces my question to: must an indecomposable submodule of $M$ be a cyclic?
| https://mathoverflow.net/users/4351 | Why are finitely generated modules over principal artin local rings direct sums of cyclic modules? | Let $I$ be the annihilator of $M$, by assumption $I=(\pi^i)$ for some $i$. One can view $M$ as an $R/I$ module and furthermore, embed $0 \to R/I \to M$. But $R/I$ is also principal artin local, so it is a quotient of a DVR by an element (by Hungerford's [paper](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-25/issue-3/On-the-structure-of-principal-ideal-rings/pjm/1102986148.full), in particular it is 0-dim [Gorenstein](https://en.wikipedia.org/wiki/Gorenstein_ring). So $R/I$ is an injective module over itself, and the embedding splits. You are done.
| 15 | https://mathoverflow.net/users/2083 | 22729 | 14,979 |
https://mathoverflow.net/questions/22629 | 48 | Are there primes of every Hamming weight? That is, for every integer $n \in \mathbb{Z}\_{>0}$ does there exist a prime which is the sum of $n$ distinct powers of $2$?
In this case, the [Hamming weight](https://en.wikipedia.org/wiki/Hamming_weight) of a number is the number of $1$s in its binary expansion.
Many problems of this sort have been considered, but perhaps not in such language. For instance, the question "Are there infinitely many [Fermat primes](https://en.wikipedia.org/wiki/Fermat_number)?" corresponds to asking, "Are there infinitely many distinct primes with Hamming weight exactly $2$?" Also related is the question of whether there are infinitely many [Mersenne primes](https://en.wikipedia.org/wiki/Mersenne_prime).
These examples suggest a class of such problems, "Do there exist infinitely many primes which are the sum of exactly $n$ distinct powers of two?"
Since this question is open even for the $n=2$ case, I pose a much weaker question here.
What is known is that for every $n \leq 1024$ there is such a prime.
The smallest such prime is listed in the Online Encyclopedia of Integer Sequences [A061712](https://oeis.org/A061712).
The number of zeros in the smallest such primes are listed in [A110700](https://oeis.org/A110700). The number of zeros in a number with a given Hamming weight is a reasonable measure of how large that number is. The conjecture at OEIS is quite a bit stronger than the question I pose.
Is there a theorem ensuring such primes for every $n \in \mathbb{Z}\_{>0}$?
| https://mathoverflow.net/users/5597 | Are there primes of every Hamming weight? | Fedja is absolutely right: this has been proven, for sufficiently large $n$, by Drmota, Mauduit and Rivat.
Although it looks at first sight as though this question is as hopeless as any other famous open problem on primes, it is easy to explain why this is not the case. Of the numbers between $1$ and $N := 2^{2n}$, the proportion whose digit sum is precisely $n$ is a constant over $\sqrt{\log N}$. These numbers are therefore quite "dense", and there is a technique in prime number theory called the method of bilinear sums (or the method of Type I/II sums) which in principle allow one to seriously think about finding primes in such a set. This is what Drmota, Mauduit and Rivat do.
I do not believe that their method has currently been pushed as far as (for example) showing that there are infinitely many primes with no 0 when written in base 1000000.
Let me also remark that they depend in a really weird way on some specific properties of these digit representation functions, in particular concerning the sum of the absolute values of their Fourier coefficients, which is surprisingly small. That is, it is not the case that they treat these Hamming sets as though they were "typical" sets of density $1/\sqrt{\log N}$.
I think one might also mention a celebrated paper of Friedlander and Iwaniec, <https://arxiv.org/abs/math/9811185>. In this work they prove that there are infinitely many primes of the form $x^2 + y^4$. This sequence has density just $c/N^{1/4}$ in the numbers up to $N$, so the analysis necessary to make the bilinear forms method work is really tough. Slightly later, Heath-Brown adapted their ideas to handle $x^3 + 2y^3$. Maybe that's in some sense the sparsest explicit sequence in which infinitely many primes are known (except of course for silly sequences like $s\_n$ equals the first prime bigger than $2^{2^n}$).
Finally, let me add the following: proving that, for some fixed $n$, there are infinitely many primes which are the sum of $n$ powers of two - this is almost certainly an open problem of the same kind of difficulty as Mersenne primes and so on.
| 56 | https://mathoverflow.net/users/5575 | 22738 | 14,984 |
https://mathoverflow.net/questions/22735 | 4 | Let K denote either the field of real numbers or the complex fields. An analytic function over $K^n$ is a function that can be represented locally by a convergent power series in n variables with coefficients in K.
My question is that can we take K to be other fields? It seems that such a field K should satisfy some criteria:
1. It is a metric space or at least a topological space.
2. There should be complete, in the sense that Cauchy nets converge.
3. The above 2 points probably force that K should have cardinality at least the size of the continuum.
Can there be other K where a reasonable theory of analytic functions can be developed? Say for cardinalities larger than the continuum? Probably this invokes some model theory.
| https://mathoverflow.net/users/nan | Analytic Functions over Fields other than Real or Complex Numbers | There are several rich theories of analysis on non-archimedian theories. Neal Koblitz' book on $p$-adic analysis is a good introduction. Non-archimedian analysis by Bosch, Güntzer and Remmert is more encyclopedic. Berkovich's Spectral Theory and Analysis over Non-archimedian Fields introduces his beautiful theory of analytic spaces allowing for a reasonable algebraic topological theory. In Goss's book Basic Structures of Function Field Arithmetic there is a good introduction to analysis in positive characteristic.
Your suggestion that this subject might have something to do with model theory is apt. As the above references show, the theory may be developed without model theory, but it has been studied intensively via model theory giving interesting results about quantifier elimination, uniformity across the $p$-adics, and establishing a basis for motivic integration. You might want to look at the paper by van den Dries and Denef, $p$-adic and real subanalytic sets. Ann. of Math. (2) 128 (1988), no. 1, 79--183.
| 13 | https://mathoverflow.net/users/5147 | 22746 | 14,991 |
https://mathoverflow.net/questions/22634 | 3 | I have a problem wherein I have defined a function $I\_r(t) = \int e^{(2r-1)at} \int e^{(2r-3)at} \cdots \int e^{at} dt\cdots dt$, and $I\_r(0) = 0$, for $r = 1,2,3,\ldots$.
I find that $e^{-ar^2t} I\_r(t) = \left(1-e^{-at}\right)^r q(t)$, where $q(exp(-at))$ is a polynomial of $e^{-at}$. Is there a general technique for evaluating repeated integrals of this type that allows me to write $q$ in a nice clean way?
If I took $I^\*\_r(t) = \int e^{at} \int e^{at}\cdots \int e^{at} dt\cdots dt$ with $I^\*\_r(0) = 0$, and multiplied by $e^{-art}$, I would get $e^{-art}I^\*\_r(t) = (1-e^{-at})^r$. I am looking for a nice closed-form solution where I have a quadratic in $r$.
This is related to the derivation of a discrete probability distribution where the transition rate function is quadratic w/r.t. the number of events per cell.
| https://mathoverflow.net/users/5640 | "Nice" Solution to repeated integral | I don't think you'll get a closed form but you get get an expression that involves products and sums that isn't too horrible and is easy to evaluate for particular cases. It'll take me forever to type it up here with nice formatting but I'll describe the approach in a way that's easy to reproduce.
We're starting with the number 1 and repeatedly performing these two operations: integration and multiplication by $e^{nat}$ for various $n$. Switch to working with Laplace transforms. In Laplace transform space, integration (starting from zero) is simply multiplication by $1/s$ and multiplication by the exponential takes $f(s)$ to $f(s-na)$.
So the Laplace transform of $I\_1$ is $1/(s(s-a))$, the Laplace transform of $I\_2$ is $1/(s(s-3a)(s-4a))$ and so on. Notice how when we keep going, the denominator is made up of factors of the form $s-(r^2-i^2)a$. So it's easy to write the Laplace transform of $I\_r$ using product notation. You now simply need to invert the Laplace transform. To do this, we need to write our rational function in $s$ as partial fractions. The factors in the denominator are all distinct so this is the easy case. After the inverse transform you should end up with something like
$e^{r^2t}\sum\_{i=0}^r\exp(-i^2t)/\prod\_{j\ne i}(i^2-j^2)$
I expect that I've made a slip somewhere as I only have a few moments spare. That's why I said 'something like'. But the method is sound and if you put in more work than me you'll get a definite result.
Update: Did some testing in Mathematica. I think the expression above is correct.
| 2 | https://mathoverflow.net/users/1233 | 22757 | 14,998 |
https://mathoverflow.net/questions/22745 | 3 | A special case says it all ... Let $ w\_1 < w\_2 < \ldots < w\_{12} $ be an increasing sequence of $12$ integers ("weights") such that the total weight $W=\sum\_{k=1}^{12}w\_k$ is even.
Say that $I \subseteq \lbrace 1,2, \ldots ,12 \rbrace$ is an exact subset iff
the sum $\sum\_{k \in I}w\_k$ equals $\frac{W}{2}$. My question is : is there
a sequence for which $ \lbrace 1,2,5,7,10,12 \rbrace $ is exact and is
the only exact subset (up to complementation) ?
| https://mathoverflow.net/users/2389 | Unique way to partition into two parts of equal weight | The answer is **yes**. Consider the sequence
$100, 200, 201, 202, 500, 601, 700, 701, 801, 1000, 1194, 1200.$
It is easy to see that $X=\{1,2,5,7,10,12\}$ and $Y=\{3,4,6,8,9,11\}$ are exact. Moreover, we claim that they are the only exact subsets. To see this, note that for every subset $X'$ of $X$, the sum of the corresponding sequence values is divisible by $100$. However, for every proper subset $Y'$ of $Y$, the sum of the corresponding sequence values is not divisible by $100$. Thus, $X$ and $Y$ are the only exact subsets.
| 10 | https://mathoverflow.net/users/2233 | 22764 | 15,002 |
https://mathoverflow.net/questions/22753 | 6 | Either the following is a really stupid question or it is a really really stupid question, but here goes:
Does there exist a classification of $\ell$-adic 2-dimensional representations of $\mathrm{Gal}(\bar{\mathbb{Q}}\_p/\mathbb{Q}\_p)$, where $\ell\neq p$?
I did a quick search of the internet that came up rather empty.
What about the subtler case of $\ell=p$?
References?
| https://mathoverflow.net/users/2147 | Classification of l-adic representations | When $\ell \neq p,$ these are rather straightforward to classify (except when $p = 2$);
see Tate's article in the second volume of Corvallis, for example.
The idea is that if $\rho$ is irred., then (unless $p = 2$), it must be induced from a character of a quadratic extension; thus the classification is given by local class field theory for quadratic extensions of $\mathbb Q\_p$. (When $p = 2$, there are some exceptional
irreps. that are not induced.)
If $\rho$ is reducible, it is an extension of characters. The characters of $\mathbb Q\_p^{\times}$ are classified by local class field theory of $\mathbb Q\_p$. There are lots
of ways to compute the possible extensions; Tate local duality/local Euler char. formula gives
one way.
When $\ell = p$, these are classified in terms of etale $(\phi,\Gamma)$-modules. To learn about this, you can e.g. read one of many expository articles on Laurent Berger's website.
(In fact there are many recent papers by Berger, Breuil, and Colmez involving $(\phi,\Gamma)$-modules, all online, and most of them include an introductory page or two recalling the basics of the theory.)
Pete is correct that this $\ell = p$ case is also the starting point of $p$-adic Langlands, just as the case $\ell \neq p$ is related to classical local Langlands.
However, as the above discussion shows, you don't need any Langlands theory to classify these reps.
Added: As JT points out in another answer, the (potentially) semi-stable representations also admit a nice classification, in terms of weakly admissible filtered $(\phi,N)$-modules.
Note that $(\phi,\Gamma)$-modules are themselves pretty nice objects. What is perhaps the most complicated part of the story is how, in the case of a potentially semi-stable representation, one compares its $(\phi,\Gamma)$-module description to its weakly admissible filtered $(\phi,N)$-module description. In the case of crystalline reps., this comparison is made via the theory of Wach modules. In general, it plays an important role in $p$-adic local Langlands, as well as in local Iwasawa theory. Laurent Berger has a number of papers discssing it (beginning with his thesis), and in the case of two-dimensional pst representations it is the subject of the most technical part (Chapter VI) of Colmez's recent [long text](http://www.math.jussieu.fr/~colmez/kirilov.pdf) on $p$-adic local Langlands.
| 11 | https://mathoverflow.net/users/2874 | 22765 | 15,003 |
https://mathoverflow.net/questions/22767 | 11 | Consider a (non-stellated) polygon in the plane. Imagine that the edges are rigid, but that the vertices consist of flexible joints. That is, one is allowed to move the polygon around in such a way that the vertices stay a fixed distance from their adjacent neighbors. Such a system is called a **polygonal linkage**.
As the linkage varies in its embedding in the plane, the area of the interior varies. The question is, **When is the area maximized**?
I have a specific answer I suspect is correct, but I am having trouble showing. I believe it is true that every polygonal linkage has as embedding where all the vertices lie on one circle (this isn't hard to show in the case when the linkage starts non-stellated). My claim is that **the area is maximized exactly when all the vertices lie on a circle**.
I can show this for a 4-sided polygon, but with techniques that do not generalize.
Also, my requirement that the polygon be non-stellated was only so that it was clear that there *was* a way to flex it to have all vertices on a circle. This question extends to the stellated case, but the question there is whether every stellated linkage can be flexed to one which is non-stellated.
| https://mathoverflow.net/users/750 | Is the area of a polygonal linkage maximized by having all vertices on a circle? | This is a theorem of Cramer. See "[On the impossibility of one rule-and-compass construction](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.30.980)" by Vladimir Jankovic.
For the quadrilateral case the quickest proof is using [Brahmagupta's formula](http://en.wikipedia.org/wiki/Brahmagupta%27s_formula#Extension_to_non-cyclic_quadrilaterals)
$$Area=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2 \theta}$$ where $a,b,c,d$ are the sides, $s$ is the half perimeter and $\theta$ is half the sum of opposite angles.
Edit: I wonder if this argument works:
Pick four consecutive vertices and move the linkage made of these four vertices till it's cyclic. There will be a subsequence of the polygons we get after such operations which converges, by the Weierstrass theorem. In the limit the polygon will be cyclic otherwise you can find four consecutive vertices not on a circle and increase the area again.
| 5 | https://mathoverflow.net/users/2384 | 22775 | 15,010 |
https://mathoverflow.net/questions/22678 | 0 | In a ring **R** (nonempty class of sets closed under difference and finite union), any sequence (here means a function on natural numbers $\mathbb N$) {$E\_i$} in **R** can be disjointlized to a disjoint sequence {$F\_i$} such that $\bigcup E\_i=\bigcup F\_i$ by traditional induction using the equation $F\_i=E\_i-\bigcup \limits\_{j < i}E\_j$. But for arbitrary uncountable sequence {$E\_\alpha$} in **R**, I either do not know if it is still possible to turn {$E\_\alpha$} into a disjoint sequence with the same union or have no idea how to use transfinite induction to prove it if disjointlization is possible, can you help me with this problem? Thanks!
| https://mathoverflow.net/users/5072 | disjointlize an arbitrary sequence in a ring? | Let's take the usual ring of finite unions $E$ of half-open rectangles $[a,b)\times [c,d)$ on the plane. The closed half-plane $x+y\ge 0$ is a union of continuum of such rectangles (all possible rectangles contained in that closed half-plane) but, since each $E$ contained in this half-plane can intersect the boundary line by only finitely many points, you cannot get the half-plane as a union of countably many ring elements. On the other hand, any disjoint family of ring elements is at most countable.
| 6 | https://mathoverflow.net/users/1131 | 22778 | 15,012 |
https://mathoverflow.net/questions/22768 | 4 | I'm trying to read chapter XIII of SGA1, and I'd appreciate some help about a few issues I'm having.
1. Definition 2.1.1. is of tamely ramified sheaves. The definition is as such: if $U$ is an open subscheme of $X$ (which is a scheme over $S$), $F$ a sheaf of sets on $U$, and $Y$ the reduced induced closed subscheme $X - U$, which we assume to be a normal crossings divisor. $F$ is said to be tamely ramified over $X$ along $Y$ relatively to $S$, if for every geometric point $\bar s$ of $S$, and for every closed point $y$ of $Y\_{\bar s}$, the restriction of $F$ to the field of fractions $K$ of $\mathcal{O}\_ {X\_{\bar s}}$ is representable by the spec of an etale $K$-algebra $L$, tamely ramified over $\mathcal{O}\_ {X\_ {\bar s, y}}$.
I can't say that I fully understand this. I can think of several interpretations of this, but I'd feel more comfortable if someone will corroborate one of them. I'm guessing $\mathcal{O}\_ {X\_{\bar s}}$ means the stalk of the generic point of $X\_{\bar s}$ in $X$, rather than the structure sheaf of $X\_{\bar s}$. Still, I feel this definition is lacking. For example, $L$ is an etale $K$-algebra, but it makes sense to ask that it be tamely ramified over $\mathcal{O}\_ {X\_ {\bar s, y}}$? Also, by sheaf in sets I'm guessing they mean a stack in the small etale site fibered in sets such that it's a functor. I guess what I'm really asking is how to think of this. How does this definition relate to tamely ramified covers? How do these "tamely ramified" sheaves have to do with the tame $\pi\_1$'s?
2. This one is mostly me not knowing French well enough: In the second paragraph of 2.1.5. it says "Si maintenant $F$ est un faisceau en groupes sur $U$, l'image inverse sur $S'$ d'un torseur sous $F$ moderement ramifie sur $X$ relativement a $S$ est un torseur sous $F'$ moderement ramifie sur $X'$ relativement a $S'$."
I don't understand this grammatically. Is it "If now $F$ is a sheaf of groups of $U$, the inverse image of $S'$ of a sub-torsor of $F$ ..." or "of $F$-torsors", or "sub-sheaf of $F$ that is made up of torsors". None of these makes much sense to me. Do you understand the statement and how it would make sense?
| https://mathoverflow.net/users/2665 | SGA1 Chapter XIII (tamely ramified sheaves) | If you look in the text, $y$ is not a closed point but a *maximal* point, meaning a maximally generic point. Then $O\_{X\_{\bar{s}y}}$ is a discrete valuation ring, and so it makes sense to talk about tameness of extensions. Also, surely she's working in the etale topology, otherwise it would be kind of silly to try and represent functors by etale covers, but I didn't actually see where she says that. It's got to be written somewhere, though!
I suppose you're right about a sheaf of sets being a stack with no non-identity maps, but taking this as a definition is exactly as sensible as defining a set to be a category with no non-identity maps.
"torseur sous F" = "torsor under F" = "F-torsor"
| 2 | https://mathoverflow.net/users/1114 | 22787 | 15,019 |
https://mathoverflow.net/questions/22771 | 4 | Suppose $F/\mathbb{Q}$ is a totally real field of degree $d$ and class number one. Fix an ordering $\sigma\_1, \dots, \sigma\_d$ on the embeddings of $F$. Is
$\sum\_{\alpha \in \mathcal{O}\_F}e^{2\pi i (z\_1 \sigma\_1 (\alpha^2)+ \dots +z\_d \sigma\_d(\alpha^2))}$
a Hilbert modular form of parallel weight $1/2$?
| https://mathoverflow.net/users/1464 | Jacobi's theta function over totally real fields | Yep -- though I have never thought through any technicalities regarding definition of half-integral weight Hilbert modular forms; I'm comfortable saying, at least, that the square of that theta function is a Hilbert modular form of weight 1. Harvey Cohn wrote several papers about this: see e.g.
MR0113855 (22 #4686)
Cohn, Harvey
Decomposition into four integral squares in the fields of $2^{1/2}$ and $3^{1/2}$.
Amer. J. Math. 82 1960 301--322.
I advised a senior thesis student at Princeton, Jorge Cisneros, who wrote a very nice thesis working out representations by sums of four squares for Q(sqrt(7)); in this case, there's a cusp form in the relevant space of Hilbert modular forms so the "error" betweeen the number of representations and the relevant divisor sum forms a very nice distribution...
| 5 | https://mathoverflow.net/users/431 | 22793 | 15,023 |
https://mathoverflow.net/questions/22719 | 11 | Given a real manifold $M$ with symplectic $2$-form $\omega$,
one can ask whether the cohomology class $[\omega] \in H^2(M;{\mathbb R})$ lies in the image of
$H^2(M;{\mathbb Z})$. If so, one can ask for a line bundle ${\mathcal L}$
with $c\_1({\mathcal L}) = [\omega]$ (or even better,
a connection $\alpha$ on $\mathcal L$ whose curvature $curv(\alpha)$ is $\omega$).
In the weakest definition of "geometric quantization", one puts a compatible
almost complex structure on $M$ and uses it to define the pushforward
of $\mathcal L$ to a point in $K$-theory. Call this $Q(M)$.
Are there examples worked out somewhere in which $H^2(M;{\mathbb Z})$
has torsion, so that $\mathcal L$ is not uniquely determined by $\omega$?
Can $Q(M)$ depend on the choice of $\mathcal L$?
I don't have any very good reason for asking this, other than I've felt
it to be a hole in my understanding of geometric quantization. The spaces I care about quantizing never seem to have torsion in $H^2$.
| https://mathoverflow.net/users/391 | What are the implications of torsion in H^2 for geometric quantization? | If L\_1 and L\_2 are two line bundles on a manifold $M$ that differ by torsion, then their Chern characters
$$ch(L\_1) = 1 + c\_1(L\_1) + \frac{1}{2}c\_1(L\_1)^2 + \cdots$$
$$ch(L\_2) = 1 + c\_1(L\_2) + \frac{1}{2}c\_1(L\_2)^2 + \cdots$$
agree, if only because the right-hand sides of these formulas are taking place in $H^\*(M;\mathbb{Q})$ and there $c\_1(L\_1) = c\_1(L\_2)$. If $L\_1$ and $L\_2$ are prequantizations of a symplectic structure on M, and if we can give M a compatible complex structure, then we have Q(L\_1) = Q(L\_2) by the Riemann-Roch formula
$$Q(L\_1) = \int ch(L\_1) td(M) = \int ch(L\_2) td(M) = Q(L\_2)$$
We do not get a finer equation than $Q(L\_1) = Q(L\_2)$. Below the line is an example of $M$, $L\_1$, and $L\_2$ for which $H^\*(M;L\_1)$ and $H^\*(M;L\_2)$ are not isomorphic, that I thought of first. I'll integrate these answers later.
---
I had written an answer with a bogus example of a Kahler manifold with two prequantum line bundles $L\_1$ and $L\_2$ for which $Q(L\_1)$ was different from $Q(L\_2)$. Here's my attempt to repair it--the best I can do is a symplectic manifold with two prequantum line bundles $L\_1$ and $L\_2$ for which $H^0(L\_1) \neq H^0(L\_2)$. But there is definitely some higher cohomology that might cancel this difference when you compute $Q$. I don't have a guess for what happens.
I am using "prequantum line bundle" to mean a line bundle whose real chern class (times $2\pi i$?) is equal to the class of the symplectic form. The premise of my example is that if we have two such line bundles $L\_1$ and $L\_2$ whose integral Chern classes differ by $n$-torsion, then $L\_1$ and $L\_2$ will become isomorphic over some $\mathbb{Z}/n$-cover $Y$ of $M$. (One can see this by thinking about the bundle-theoretic meaning of the Bockstein map $H^1(M;\mathbb{Z}/n) \to H^2(M;\mathbb{Z})$ and the long exact sequence it fits into.).
We can organize the data this way: $f:Y \to M$ is a $\mathbb{Z}/n$-bundle over $M$ with a $\mathbb{Z}/n$-equivariant prequantum line bundle $L$, and $L\_1$ and $L\_2$ are built from $L$ by the formula
$$H^0(U;L\_i) = \alpha\_i\text{-eigenspace of }H^0(f^{-1}(U);L)$$
where $\alpha\_1$ and $\alpha\_2$ are two different characters of $\mathbb{Z}/n$.
So to find such an $M$, we should find a prequantized symplectic manifold $Y$ equipped with a free $\mathbb{Z}/n$-action, such that $\mathbb{Z}/n$ does not act so uniformly on $H^0(Y;L)$ (i.e. so that the different eigenspaces of a generator for $\mathbb{Z}/n$ have different dimensions). The only source of prequantized symplectic manifolds available to me are projective varieties, and the only source of free $\mathbb{Z}/n$-actions are those on projective hypersurfaces.
So, let $Y$ be the degree 5 hypersurface in $\mathbb{P}^3$ given by the equation $x^5 + y^5 + z^5 + w^5 = 0$. If $\eta$ is a 5th root of unity, then the $\mathbb{Z}/5$-action given by $x \mapsto x$, $y \mapsto \eta y$, $z \mapsto \eta^2 z$ and $w \mapsto \eta^3 w$ is free (to check the absence of fixed points, it's important that the eigenvalues of the different letters are distinct). If $L$ is the restriction of $O(1)$ on $P^3$ to $Y$ then $H^0(Y;L) = H^0(P^3;O(1))$, which is the vector space spanned by $x,y,z,$ and $w$, and the generator of $\mathbb{Z}/5$ acts by $x \mapsto x$, $y \mapsto \eta^{-1} y$, $z \mapsto \eta^{-2} z$, and $w \mapsto \eta^{-3}w$.
The conclusion is that if $L\_1$ is the line bundle on $M = Y/(\mathbb{Z}/5)$ corresponding to the eigenvalue 1 (or $\eta^{-1}$ or $\eta^{-2}$ or $\eta^{-3}$) then $H^0(M;L\_1)$ is one-dimensional, and if $L\_2$ is the line bundle on $M$ corresponding to the eigenvalue $\eta^{-4} = \eta$ then $H^0(M;L\_2)$ is zero-dimensional.
More generally one can take (degree n hypersurface in $\mathbb{P}^r$)/(action of $\mathbb{Z}/n$), but to make the action free one needs $n > r+1$ (and maybe $n$ prime). Taking large $n$ like this creates cohomology in $H^{r-1}(M;L\_1)$ and $H^{r-1}(M;L\_2)$.
| 8 | https://mathoverflow.net/users/1048 | 22794 | 15,024 |
https://mathoverflow.net/questions/22800 | 7 | Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?
| https://mathoverflow.net/users/1148 | Must a locally compact group be Hausdorff in order to possess a Haar measure? | No. Simon Rubinstein-Salzedo's ["On the existence and uniqueness of invariant measures on locally-compact groups"](http://simonrs.com/HaarMeasure.pdf) presents a proof of existence (and uniqueness up to a multiplicative strictly positive constant) of a left Haar measure given a locally-compact, not necessarily Hausdorff, group.
| 7 | https://mathoverflow.net/users/441 | 22802 | 15,026 |
https://mathoverflow.net/questions/22801 | 10 | Let $X$ be a closed, orientable surface of genus at least 2, and let $\phi: \pi\_1(X) \to \pi\_1(X)$ be a surjective homomorphism. Is $\phi$ necessarily injective?
| https://mathoverflow.net/users/5499 | Self-homomorphisms of surface groups | Yes. Surface groups are [Hopfian](http://en.wikipedia.org/wiki/Hopfian_group). More generally, all [residually finite](http://en.wikipedia.org/wiki/Residually_finite) groups are Hopfian -- see Theorem IV.4.10 in Lyndon and Schupp's book "Combinatorial Group Theory".
| 10 | https://mathoverflow.net/users/317 | 22803 | 15,027 |
https://mathoverflow.net/questions/13291 | 10 | Let $X$ be a smooth projective variety. If I've understood correctly, the Weil conjectures imply that it is possible to compute the Betti numbers of $X(\mathbb{C})$ by computing the local zeta function associated to $X(\overline{\mathbb{F}\_p})$ for $p$ a prime of good reduction. This is because one can identify the factors coming from different $\ell$-adic cohomology groups by the absolute value of their roots (the "Riemann hypothesis," by analogy with the case of curves).
Unfortunately, this doesn't seem to be true for the analogous situation with compact triangulable spaces $Y$ and the Lefschetz zeta function $\zeta\_f$, at least not for an arbitrary choice of function $f : Y \to Y$. For example, if $f$ is homotopic to the identity, then $\zeta\_f$ can only see the Euler characteristic of $Y$. Even if $f$ acts "generically" (i.e. none of the factors of $\zeta\_f$ cancel), there doesn't seem to be a way to distinguish which factors are associated to which degree. (Of course, I would love if I were wrong about this.)
**Question 1:** When is there an analogue of the geometric Frobenius for compact triangulable spaces $Y$? By this I mean a more-or-less canonical function $f : Y \to Y$ such that some analogue of the Riemann hypothesis holds for $\zeta\_f$.
**Question 2:** Regardless of the answer to Question 1, is it always possible to choose $f$ such that $\zeta\_f$ can tell you which of its factors are associated to which homology groups?
(Side question: is it true that given any $f : Y \to Y$ there is always some $f'$ homotopic to $f$ which has finitely many fixed points?)
| https://mathoverflow.net/users/290 | Is it always possible to compute the Betti numbers of a nice space with a well-chosen Lefschetz zeta function? | Having resolved my [ignorance](https://mathoverflow.net/questions/22801/self-homomorphisms-of-surface-groups) concerning surface groups I can now answer question 1 negatively (or at least some formulation thereof). It is impossible if $Y$ is an oriented surface of genus at least $2$.
Suppose that $f: Y \to Y$ is a self map of the surface such that the eigenvalues of $f^\*$ acting on each $H^i(Y)$ are all nonzero (otherwise we can't "detect" the betti numbers), and such that $H^i(Y)$ and $H^j(Y)$ do not have eigenvalues of common magnitude for $i \neq j$. Then in particular $f^\*$ acts on $H^2(Y)$ nontrivially, say by multiplication by some integer $d$. This integer cannot be $\pm 1$ since then $H^0(Y)$ and $H^2(Y)$ would contain eigenvectors with eigenvalues of equal magnitude.
Consider the subgroup $H = f\_\*(\pi\_1(Y))$ inside $G = \pi\_1(Y)$. If this had infinite index, then $f$ would lift to a map to some infinite covering of $Y$, so it would induce a trivial map of $H^2$. So $H$ has finite index in $G$. Let $X \to Y$ be the corresponding covering space. Then $\pi\_1(X)$ is a quotient of $\pi\_1(Y)$, hence its abelianization has rank $\leq 2g$ where $g$ is the genus of $Y$. This implies that $X$ is a closed surface of genus at most $g$. But its Euler characteristic is precisely $[G:H]$ times the Euler characteristic of $Y$, so $X = Y$. Thus $f$ induces a surjection on $\pi\_1(Y)$. By the post cited above, $f$ actually induces an isomorphism on $\pi\_1(Y)$, so it is a homotopy equivalence. In particular, $d = \pm 1$, contrary to assumption.
After writing this it occurs to me that you might object to me ruling out the case $d = -1$... At any rate, this shows that the eigenvalues can't ever look like they do in the case of the Riemann hypothesis, with magnitude $q^{i/2}$ on $H^i$ for some $q>1$.
| 4 | https://mathoverflow.net/users/5499 | 22804 | 15,028 |
https://mathoverflow.net/questions/22806 | 3 | All varieties are defined over $\mathbb{C}$. Let $\pi : X \to C$ be an elliptic surface with $X$ and $C$ smooth. Then there is a Jacobian surface $\overline{\pi}: J(X) \to C$ (with a section) associated to $X$. Do we also have a morphism $\varphi : X \to J(X)$ such that $\overline{\pi}\circ \varphi = \pi$? (If yes, a precise reference for the construction of that morphism would be greatly appreciated!)
| https://mathoverflow.net/users/5197 | Jacobian surface associated to an elliptic surface | I think the following at least gives a rational map: Choose a curve $D$ in $X$ which maps dominantly to $C$ with degree $n$. For a smooth fibre $F$ of $\pi$ define a map to the correspoding fibre of $\bar{\pi}$ by sending a point $p$ to the class of $n[p] - [F\cdot D]$. (This is a cycle of degree 0.) With a little work one can check that this gives a well defined rational map which is a morphism over the smooth locus.
It seems unlikely though that the above actually gives a morphism in general.
| 3 | https://mathoverflow.net/users/519 | 22810 | 15,031 |
https://mathoverflow.net/questions/22799 | 5 | Let $E$ be an arbitrary Banach space and let $T:E^{\*}\rightarrow\ell^{2}$
be a linear continuous operator. Is it true that $T$ must be the
$so$-limit (i.e., limit w.r.t. the strong operator topology) of a
net $(S\_{d})^{\*}$ $\left(d\in\mathcal{D}\right)$ of adjoint operators,
with $S\_{d}$:$\ell^{2}$ $\rightarrow$ $E$ and $||S\_{d}||\leq||T||$
$\left(d\in\mathcal{D}\right)$?
I guess not, say $E=c\_{0}(I)$ and $T$ is a surjection, where $I$
has a "big" cardinality. But maybe I'm
wrong.
Any help will be highly appreciated.
| https://mathoverflow.net/users/2508 | Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? | You have that $T^\*:(\ell^2)^\* \rightarrow E^{\*\*}$, so using that $(\ell^2)^\*$ is isomorphic to $\ell^2$ (just in the linear sense, as we already have a co-ordinate system), we can regard $T^\*$ as a map $\ell^2\rightarrow E^{\*\*}$.
By the Principle of Local Reflexivity (I've used a paper of Behrends in the past, which is overkill, but is freely available: <http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf> Or look in a book on Banach space theory) for each triple $i=(M,N,\epsilon)$, where $M\subseteq E^{\*\*}$ and $N\subseteq E^\*$ are finite-dimensional, and $\epsilon>0$, we can find an operator $S\_i:M\rightarrow E$ such that $(1-\epsilon)\|x\| \leq \|S\_i(x)\| \leq (1+\epsilon)\|x\|$ for $x\in M$, and with $\phi(S\_i(x)) = x(\phi)$ for $x\in M$ and $\phi\in N$.
So, let $P\_n:\ell^2 \rightarrow \ell^2$ be the projection onto the first $n$ co-ords, let $M\supseteq T^\*(P\_n(\ell^2))$ and let $i=(M,N,\epsilon)$, so $S=S\_i T^\* P\_n$ makes sense, and is a map $\ell^2\rightarrow E$. Then, for $a\in\ell^2$ and $\phi\in E^\*$, $$S^\*(\phi)(a) = \phi(S\_i T^\* P\_n(a)) \rightarrow T^\*(a)(\phi) = T(\phi)(a),$$ as $i$ and $n$ increase.
So we have found a bounded net $(S\_d)$ (we can even choose it bounded by $\|T\|$ be rescaling a little) with $S\_d^\* \rightarrow T$ in the weak operator topology. But now a standard trick (take convex combinations, as the closure of a convex set is the same in the weak and norm topologies) allows us to find a net which converges SOT.
I think the proof would work for any Banach space F replacing $\ell^2$, as long as we can find a bounded net of finite rank operators $(F\_\alpha)$ with $F\_\alpha\rightarrow 1$ SOT. That is, F should have the bounded approximation property.
| 7 | https://mathoverflow.net/users/406 | 22822 | 15,038 |
https://mathoverflow.net/questions/22814 | 18 | A group $G$ is said to be linear if there exists a field $k$, an integer $n$ and an injective homomorphism $\varphi: G \to \text{GL}\_n(k).$
Given a short exact sequence
$1 \to K \to G \to Q \to 1$ of groups where $K$ and $Q$ are linear (over the same field), is it true that $G$ is linear too?
Background: [Arithmetic groups](http://en.wikipedia.org/wiki/Arithmetic_group) are by definition commensurable with a certain linear group, so they are finite extensions of a linear group, and finite groups clearly are linear (over any field).
| https://mathoverflow.net/users/3380 | Are extensions of linear groups linear? | The universal central extension $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$ is the preimage of $\text{Sp}\_{2n}\mathbb{Z}$ in the universal cover of $\text{Sp}\_{2n}\mathbb{R}$, and fits into the sequence
$$1\to \mathbb{Z}\to \widetilde{\text{Sp}\_{2n}}\mathbb{Z}\to \text{Sp}\_{2n}\mathbb{Z}\to 1.$$
Deligne proved that $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$
is not residually finite; the intersection of all finite-index subgroups of is $2\mathbb{Z}<\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$. In particular, this implies that $\widetilde{\text{Sp}\_{2n}}\mathbb{Z}$ is not linear. But certainly $\mathbb{Z}$ and $\text{Sp}\_{2n}\mathbb{Z}$ are. If you want an arithmetic group, you can take the corresponding $\mathbb{Z}/k\mathbb{Z}$-extension of $\text{Sp}\_{2n}\mathbb{Z}$, which will not be linear as long as $k\neq 2$.
I learned the proof of this theorem from Dave Witte Morris, who has written up his fairly-accessible notes as "[A lattice with no torsion-free subgroup of finite index](http://people.uleth.ca/~dave.morris/talks.shtml) (after P. Deligne)" ([PDF](http://people.uleth.ca/~dave.morris/talks/deligne-torsion.pdf) link).
| 25 | https://mathoverflow.net/users/250 | 22823 | 15,039 |
https://mathoverflow.net/questions/19410 | 26 | This is a more focused version of [Summation methods for divergent series](https://mathoverflow.net/questions/19201/summation-methods-for-divergent-series).
>
> Let $a\_n$ be a sequence of real
> numbers such that $\lim\_{x \to 1^{-}}
> > \sum a\_n x^n$ and $\lim\_{s \to 0^{+}}
> > \sum a\_n n^{-s}$ both exist. (In
> particular, we assume that both of the
> sums in question converge in the
> appropriate region.) Need the limits
> be equal?
>
>
>
---
I've thought about this before, and am doing so again thanks to the linked question. Here are some things I've tried:
It is true if $a\_n$ is periodic with average $0$.
If it is true for $a\_n$, then it is true for the sequence $b\_{kn+r}=a\_n$, $b\_m=0$ if $m \not \equiv r \mod k$.
It is true for $a\_n=1$ if $n$ is an even square, $-1$ if $n$ is an odd square and $0$ otherwise. I tried to prove in general that, if it is true for $a\_n$ then it is true for $b\_{n^2}=a\_n$, $b\_m=0$ for $m$ not a square, but couldn't.
It appears to be true for $a\_n = (-1)^n \log n$, although I didn't check all the details.
I do not know any explicit sequence $a\_n$ which obeys the hypotheses of the question and is not $(C, \alpha)$-[summable](http://en.wikipedia.org/wiki/Cesaro_summation) for some $\alpha$. So it is possible that this is really a theorem about higher Cesaro summability. But I suspect such sequences do exist.
A natural generalization is: Let $a\_n$, $\lambda\_n$ and $\mu\_n$ be three sequences of real numbers, with $\lambda\_n$ and $\mu\_n$ approaching $\infty$. If
$\lim\_{s \to 0^{+}} \sum a\_n e^{-\lambda\_n s}$ and $\lim\_{s \to 0^{+}} \sum a\_n e^{-\mu\_n s}$ both exist, need they be equal?
As far as I can tell, Wiener's generalized Tauberian theorem does not apply.
| https://mathoverflow.net/users/297 | Do Abel summation and zeta summation always coincide? | I think the answer is 'yes.' I don't have a suitably general reason why this is the case, although surely one exists and is in the literature somewhere.
At any rate, for the problem at hand, we have for $s > 0$
$$\sum \frac{a\_n}{n^s} = \frac{1}{\Gamma(s)}\int\_0^\infty \sum a\_n e^{-nt} t^{s-1} dt.$$
*Edit: the interchange of limit and sum used here requires justification, and this is done below.* Supposing that $\sum a\_n x^n \rightarrow \sigma$, we may write $\sum a\_n e^{-nt} = (\sigma + \epsilon(t))\cdot e^{-t}$ where $\epsilon(t) \rightarrow 0$ as $t \rightarrow 0$, and $\epsilon(t)$ is bounded for all t. In this case
$$\sum \frac{a\_n}{n^s} = \sigma + O\left(s\int\_0^\infty \epsilon(t) e^{-t} t^{s-1} dt\right)$$
Showing that error term tends to $0$ is just a matter of epsilontics; for any $\epsilon > 0$, there is $\Delta$ so that $|\epsilon(t)| < \epsilon$ for $t < \Delta$. Hence
$$\left|s\int\_0^\infty \epsilon(t) e^{-t} t^{s-1} dt \right| < s\epsilon \int\_0^\Delta t^{s-1} dt + s\int\_\Delta^\infty e^{-t}t^{s-1}dt < \epsilon \Delta^s + s\int\_\Delta^\infty e^{-t} t^{-1} dt.$$
Letting $s \rightarrow 0$, our error term is bounded by $\epsilon$, but $\epsilon$ of course is arbitrary.
*Edit:* Justifying the interchange of limit and sum above is surprisingly difficult. We will require
*Lemma:* If for fixed $\epsilon > 0$, the partial sums $D\_{\epsilon}(N) = \sum\_{n=1}^N a\_n/n^\epsilon = O(1),$ then
(a) $A(N) = \sum\_{n \leq N} a\_n = O(n^\epsilon)$, and
(b) $\sum\_{n \leq N} a\_n e^{-nt} = O(t^{-\epsilon})$,
where the O-constants depend on $\epsilon.$
This, with the hypothesis that $\sum a\_n/n^s$ converges for all $s > 0$, imply the conclusions a) and b) for all positive $\epsilon$.
To prove part a), note that
$$\sum\_{n \leq N} a\_n = \sum\_{n \leq N} a\_n n^{-\epsilon}n^\epsilon = \sum\_{n \leq N-1} D\_{\epsilon}(n) (n^\epsilon - (n+1)^\epsilon) + D\_\epsilon(N)N^\epsilon,$$
which is seen to be $O(N^\epsilon)$ upon taking absolute values inside the sum.
To prove part b), note that
$$t^\epsilon \sum\_{n \leq N} a\_n e^{-nt} = t^\epsilon \sum\_{n=1}^{N-1} A(n)(e^{-nt} - e^{-(n+1)t}) + t^\epsilon A(N) e^{-Nt} = O \left( \sum\_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) + (tN)^\epsilon e^{-Nt}\right).$$
Now, $(tN)^\epsilon e^{-Nt} = O(1)$, and
$$\sum\_{n\leq N} (tn)^\epsilon e^{-nt}(1-e^{-t}) = 2^\epsilon(1-e^{-t}) \sum\_{n\leq N} (tn/2)^\epsilon e^{-nt/2} e^{-nt/2} = O\left(\frac{1-e^{-t}}{1-e^{-t/2}}\right) = O\left(\frac{1}{1+e^{t/2}}\right) = O(1),$$
and this proves b).
We use this to justify interchanging sum and integral as follows: note that
$$\sum\_{n=1}^N \frac{a\_n}{n^s} = \frac{1}{\Gamma(s)}\int\_0^\infty \sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt,$$
and therefore
$$\frac{1}{\Gamma(s)}\int\_0^\infty \lim\_{N\rightarrow\infty}\sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt = \frac{1}{\Gamma(s)}\int\_0^1 \lim\_{N\rightarrow\infty}\sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt + \frac{1}{\Gamma(s)}\int\_1^\infty \lim\_{N\rightarrow\infty}\sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt.$$
In the first integral, note that for $\epsilon < s$, $\sum\_{n \leq N} a\_n e^{-nt} t^{s-1} = O(t^{s-\epsilon -1})$ for all $N$. So by dominated convergence in the first integral, and uniform convergence of $e^t \sum\_{n=1}^N a\_n e^{-nt}$ for $t \geq 1$ in the second, this is limit is
$$\lim\_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int\_0^1 \sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt + \lim\_{N\rightarrow\infty}\frac{1}{\Gamma(s)}\int\_1^\infty \sum\_{n=1}^N a\_n e^{-nt} t^{s-1} dt = \lim\_{N\rightarrow\infty} \sum\_{n=1}^N a\_n \frac{1}{\Gamma(s)}\int\_0^\infty e^{-nt}t^{s-1} dt.$$
This is just $\sum\_{n=1}^\infty \frac{a\_n}{n^s}$.
Note then that we do not need to assume from the start that the infinite Dirichlet sum tends to anything as $s \rightarrow 0$; once it converges for each fixed $s$, that is implied by the behavior of the power series.
| 11 | https://mathoverflow.net/users/5621 | 22825 | 15,041 |
https://mathoverflow.net/questions/22821 | 8 | I'm studying the deformation theory of compact complex manifolds as developed by Kodaira and Spencer. On the side I'm reading as much about deformation theory in general as I can get my hands on (and understand), and I've been wondering about the relationship between the basic definitions in the analytic and algebraic categories. To summarize:
*Analytic side*: A complex analytic family of smooth compact manifolds is a holomorphic map $\pi : \mathcal X \to S$ of smooth complex manifolds $\mathcal X$ and $S$ such that $\pi$ is a proper submersion and each fiber $X\_t = \pi^{-1}(t)$ is a compact complex manifold. This implies some other conditions, like that $\mathcal X$ is locally trivial over $S$.
*Algebraic side*: A family of schemes is a proper flat morphism $\pi : X \to Y$ of schemes.
I've been asking myself what the relationship between these definitions is. To get something like the algebraic definition in the analytic category we just replace "scheme" by "complex space".
Now, a complex manifold is a smooth complex space, and ~~local triviality of $\mathcal X$ along with compactness of the fibers implies that $\pi : \mathcal X \to S$ is proper~~ (edit: unnecessary). I'm also fairly certain that $\pi$ is flat (my algebraic side is weak), so $\pi : \mathcal X \to S$ will be a family of complex spaces in the algebraic sense.
My question is: what conditions do we need on $\pi : X \to Y$ to pass in the other direction? Is it enough that the complex spaces $X$ and $Y$ be smooth? I've been thinking about this and I've got this vague idea that flatness of $\pi$ and coherence of the structure sheaves will lead to local triviality, but I haven't been able figure out how.
| https://mathoverflow.net/users/4054 | Complex analytic vs algebraic families of manifolds | The standard situation in Kodaira-Spencer's work is the following:
If you're on the algebraic side and you have a smooth ("smooth" in the sense of algebraic geometry) and proper (proper in the sense of algebraic geometry) map $\pi: X \to Y$ , then when you translate this to the analytic side, "smooth" turns into "submersion" (in the sense of: pushforward of vector fields is surjective), and "proper" turns into "proper" (in the sense of: inverse image of compact set is compact). And "map" turns into "holomorphic map". Then you can use, for instance, the "[preimage theorem](http://en.wikipedia.org/wiki/Preimage_theorem)" (be careful to not get confused by the usage of "smooth" in that article --- there smooth means $C^\infty$) to deduce that the fibers are holomorphic complex manifolds. Strictly speaking we must use the holomorphic version of the "preimage theorem". But the holomorphic version does hold, as do holomorphic versions of other standard theorems like implicit function theorem and inverse function theorem. Perhaps this is in Chapter 0 of Griffiths-Harris, or Chapter 1 of Huybrechts.
The fibers are compact because a point is compact. :)
| 4 | https://mathoverflow.net/users/83 | 22829 | 15,044 |
https://mathoverflow.net/questions/22830 | 5 | The Baire category theorem implies that a nonempty complete metric space without isolated points must be uncountable. In many situations I have encountered, the "natural examples" of complete metric spaces without isolated points (of a certain type, or possibly with some additional structure) in fact have at least continuum cardinality. This is not so surprising, since if Cantor's continuum hypothesis holds, then uncountable is equivalent to at least continuum cardinality.
However, if we do not wish to assume CH -- and, ever since Godel and Cohen proved that (G)CH is independent of ZFC set theory, this seems to be the prevalent attitude -- what can be said about the existence of such spaces of uncountable cardinality less than the continuum?
I asked this question to someone before, and I seem to remember that it is known that one **cannot** unconditionally improve the conclusion of this application of Baire category to say "continuum cardinality". But could someone say a little bit about how this goes? Preferably in words that are comprehensible to a non-set theorist like myself?
**Addendum**: Thanks to Sergei Ivanov for a quick and convincing answer: evidently I was making things much more complicated than I needed to. Just to get myself reoriented properly, I would like to try to remember where the set-theoretic subtleties come in. Suppose I ask about the conclusion of BCT itself, rather than this particular corollary: not assuming CH, what can we say about the minimal cardinality of a covering family of nowhere dense subsets in a complete metric space?
**Second Addendum**: I was even more turned around than I had realized: I was (i) worrying needlessly about uncountable cardinals smaller than the continuum and (ii) not worrying enough about cardinals greater than the continuum! In particular, I was under the misimpression that for any cardinal $\kappa \geq 2^{\aleph\_0}$, $\kappa^{\aleph\_0} = \kappa$. This led me to incorrectly guess the strong form a classical theorem of F.K. Schmidt. I think I have it right now: if you are interested, see pp. 13-16 of
[http://alpha.math.uga.edu/~pete/8410Chapter3.pdf](http://alpha.math.uga.edu/%7Epete/8410Chapter3.pdf)
| https://mathoverflow.net/users/1149 | Improvements of the Baire Category Theorem under (not CH)? | A complete space without isolated points has at least continuum cardinality. At least if you agree to use (some form of) Axiom of Choice.
Choose two disjoint closed balls $B\_1$ and $B\_2$. Inside $B\_1$, choose disjoint closed balls $B\_{11}$ and $B\_{12}$. Inside $B\_2$, choose disjoint closed balls $B\_{21}$ and $B\_{22}$. And so on. At $n$th step, you have $2^n$ disjoint balls indexed by binary words of length $n$, and you choose two disjoint balls of level $n+1$ inside each ball of level $n$. This is possible because the balls are not single points. Make sure that radii go to zero. Now you have continuum of sequences of nested balls each having a common point.
| 15 | https://mathoverflow.net/users/4354 | 22831 | 15,045 |
https://mathoverflow.net/questions/22839 | 3 | Let X be locally compact and Hausdorff, and let $f:X\rightarrow\mathbb R$ be a function. Suppose that for all finite regular (positive) Borel measures $\mu$, we know that $f$ is $\mu$-measurable. Does it follow that $f$ is Borel? If not, what's a good counter-example?
Definitions: The Borel sigma-algebra is generated by the open sets. So $f$ is Borel if $f^{-1}(U)$ is Borel for each open $U\subseteq\mathbb R$.
I think the definition of $\mu$-measurable is that for each open $U\subseteq\mathbb R$, we have that $f^{-1}(U)$ is in the completed sigma-algebra for $\mu$. That is, we can find Borel sets $A$ and $B$ with $A \subseteq f^{-1}(U) \subseteq B$ with $\mu(B\setminus A)=0$.
Remark: The "obvious" measures are the point mass measures, but then the completed sigma-algebra is $2^X$, so all functions are measurable! So the question is, in some sense, whether $X$ supports "enough" finite regular Borel measures.
Vague motivation: This old paper of Barry Johnson: Separate continuity and measurability.
Proc. Amer. Math. Soc. 20 1969 420--422 see <http://www.jstor.org/stable/2035668> But Barry's paper clearly gives me enough for what I want, so really this question is out of curiosity, not an attempt to understand the paper better!
| https://mathoverflow.net/users/406 | Borel vs measure for all Borel measures | Let $f: \mathbb{R} \to \mathbb{R}$ be the characteristic function of a subset $A \subseteq \mathbb{R}$ which is analytic but not Borel. Then $f$ is universally measurable but not Borel.
| 11 | https://mathoverflow.net/users/4706 | 22840 | 15,050 |
https://mathoverflow.net/questions/22838 | 89 | I am interested in how to select interesting yet reasonable problems for students to work on, either at Honours (that is, a research-based single year immediately after a degree) or PhD.
By this I mean a problem that is unsolved but for which there is a good chance that a student can solve it either completely or partially and come out with a thesis either way.
There are a number of possible strategies that I see some of my colleagues use, but which are sadly not available to me:
(1) Be sufficiently brilliant that you already know roughly how something unsolved *should* be solved and guide the student accordingly, modifying the strategy on the fly.
(2) Have a major project, say classifying a big class of structures, that is amenable to attack with a big general theorem with many well-defined sub-cases that can be assigned individually to students.
Personally I often work on problems that lead nowhere - I don't solve it, it's too hard, I only rediscover known examples etc - but provided at least *some* of the problems work out - it doesn't matter.
But for students, its more of a "one-shot" affair - they can't afford to work for a year or, worse, three years, and get nothing.
Are there any general principles that will help in the selection of problems? Or is just a situation where you've either "got the knack" or you haven't?
| https://mathoverflow.net/users/1492 | How do you select an interesting and reasonable problem for a student? | Let me first answer a slightly different question, how to organize one's thoughts about such problems. I simply maintain a list of suitable projects, with ideas on how to approach them, and put them in a file "Dissertation Problems.tex". Some of these are projects that I might like to carry out myself, but many are projects that would be suitable for a math PhD dissertation. I have another similar file called "Math Ideas" that I maintain on my mobile phone, so that I can write into it when I am traveling.
It often happens that when I am working on one problem, I have an idea for a related project, or a side question, or a generalization to a method that arises, or an idea for a counterexample to such a generalization. Sometimes, of course, these side problems can be solved or incorporated into the original project. But just as often, they are not directly necessary for the original project, but still interesting, and so I save them for a later project or for a student. Of course, not all ideas pan out, and in many cases, these ideas turn out to be uninteresting or wrong in some way. But often, they have turned into very interesting questions whose resolution forms a dissertation or a paper or joint paper.
In this way, in time I accumulate more problems and questions than I can work on myself, and when my PhD students are ready for a problem, I can suggest several that may be to their liking. Often the student already has ideas for a problem, and in these cases I help them to focus their questions and efforts. These files also work for joint projects with colleagues looking to do a joint project with me.
The most important thing, however, in this method, is **to remember to write the idea down**. Surely we all have great ideas from time to time, but then, after we become absorbed in another task or project, the fleeting idea is regrettably forgotten. So it is important to be systematic about recording it. Create a file on your laptop (and on your mobile phone!) to which you add suitable mathematical ideas when they occur to you. Perhaps you want two files, one for student projects, and one for projects you want to reserve for yourself.
This question I have answered---how to organize one's thoughts about mathematical ideas--- may seem to be a trival matter, but I believe that many mathematical ideas are held in mind only briefly before being forgotton forever, and so I find it to be a critically important issue, an important key to successful mathematical practice.
Let me now turn to the question you actually asked, namely, how to select problems in the first place. This I find to be an intensely personal matter, having to do with one's style of mathematics. We all likely have different mathematical styles, with some of us interested in easy-to-state or sweeping questions about fundamental issues, others preferring to understand motivating principal examples very deeply, and others interested in aspects of some big-machinery construction method, and so on. These mathematical styles will lead to very different sorts of questions and problems, and will naturally affect the kinds of problems that we would find suitable for students. It is of course much easier to find suitable problems for students that arise as part of a big program with many examples or cases that need to be worked out, and these kinds of problems often carry with them the opportunity to learn a part of that machinery. (My own style surely tends more toward quirky but fundamental questions, perhaps underlying or alongside the well-beaten path but not directly on it, and less towards big machinery, but in others eyes I am likely merely presupposing a certain amount of big machinery, such as forcing or large cardinals.) The sweeping-questions problems, however, are more dangerous for students because they can often be harder or of unknown difficulty. But occasionally, I find that an interesting special case or aspect of an interesting new sweeping question, which arises during my own investigations, and when these occur I have found these smaller projects very satisfying mathematically, both for myself and my students. I would never give a problem to a PhD student that I didn't myself find compelling and interesting.
| 59 | https://mathoverflow.net/users/1946 | 22844 | 15,053 |
https://mathoverflow.net/questions/22818 | 8 | The following is a somewhat vague question concerning logic, but with ideas from algebraic geometry (see in particular the example at the end). The vagueness is in the notion of "language".
Let $A$ be a set and fix some language in which to write propositions which assign a value of true or false to each element of A. Let $R$ be the set of such propositions about elements of $A$, and suppose that $R$ contains "True" and "False" and is closed under "and" ($\wedge$) and "or" ($\vee$).
Now let $V\subseteq A$ be a subset and consider the set $I(V)\subseteq R$ such that $r\in I(V)$ iff $r$ holds of every element $v\in V$. This set $I(V)$ has the following properties:
1. "True" is in $I(V)$,
2. If $i$ and $j$ are in $I(V)$ then $i\wedge j$ is in $I(V)$, and
3. If $i$ is in $I(V)$ and $r\in R$ then $r\vee i$ is in $I(V)$.
Thus $I(V)$ satisfies the axioms of an "ideal" in the "ring" $R$ if we take the following "strange correspondence" between statements and algebraic operations: $T\mapsto 0$, $F\mapsto 1$, $\wedge\mapsto +$ and $\vee\mapsto \times$. Note that in fact $R$ is a commutative "Rig" (ring without negatives) under this "strange correspondence." The correspondence is "strange" because we usually think of "or" as $+$, "and" as $\times$, "true" as $1$, and "false" as $0$. But we can find intuition for this correspondence via Example 2 below.
Suppose we define an ideal in $R$ to be a subset $I\subseteq R$ satisfying conditions 1,2,3. Given an ideal $I$ we can consider the set $Z(I)\subseteq A$ of all elements $a\in A$ such that $i$ holds of $a$ for each $i\in I$. Call such subsets "closed".
Observation 1: There is an (order-reversing) correspondence between the ideals of $R$ and the closed subsets of $A$.
Example 2: Suppose we take the language of commutative rings. Say $A$ is the set ${\mathbb R}^n$ and $R$ is the set of statements of the form $f(x\_1,\ldots,x\_n)=0$ where $f$ is a polynomial with real coefficients. The set of such statements forms a ring, by operating on the polynomials; that is $R\cong {\mathbb R}[x\_1,\ldots,x\_n]$.
Given a subset $V\subseteq A$ we can consider those equations that are true of all points in $V$. Such a subset will satisfy conditions 1,2,3 above (where "True" is 0=0). We see that the "strange correspondence" discussed above does make sense: if $f=0$ and $g=0$ hold of every point in $V$ then so does $f+g=0$; for any equation $r=0$ in $R$, if $f=0$ holds of every point in $V$ then so does $rf=0$. The empty subset of $A$ is satisfied by $1=0$ (or "false"). In fact, Observation 1 is the basic observation of algebraic geometry in this context.
Question 3: Has anyone looked at this correspondence? Can it be made more rigorous? Can algebro-geometric notions (like schemes) be applied to other "languages" in an interesting way?
| https://mathoverflow.net/users/2811 | Ideals of statements? | Yes, of course, the algebraic aspects of logic have been very well studied. There is a lot to say about this, but since I am supposed to be on a voluntary MO hiatus until the end of the semester, I will only mention a few things.
You might want to ask for your "ideals" to be closed under logical equivalence too. Otherwise, statements of length 17 or more form a rather silly ideal. With this change, your "ideals" are called *deductively closed theories*.
The space you're describing is basically the [Stone space](http://en.wikipedia.org/wiki/Stone%27s_representation_theorem_for_Boolean_algebras#Stone_spaces) of the [Lindenbaum algebra](http://en.wikipedia.org/wiki/Lindenbaum%E2%80%93Tarski_algebra) of your language. The Lindenbaum algebra is the Boolean algebra which consists of all sentences of the language modulo logical equivalence. This construction also makes sense over a nontrivial base theory T, where logical equivalence is replaced by T-provable equivalence.
Another name for this Stone space is the space of (complete) 0-types over the theory T. The more general space of (complete) [n-types](http://en.wikipedia.org/wiki/Type_%28model_theory%29) is obtained in the same way by using formulas in n fixed variable symbols x1,...,xn (instead of sentences, which have no free variables). These spaces of types and their topology are a central concept in model theory. One usually takes T to be a the complete theory of a structure M. Then the space of 0-types has only one point since T is complete and n-types correspond to coherent things that one could say about a n-tuple of elements of M.
Several theorems of model theory have interesting meanings in this context. For example, the Compactness Theorem corresponds to the fact that Stone spaces are compact, and the Omitting Types Theorem corresponds to the Baire Category Theorem (for compact Hausdorff spaces). Some leading model theorists have supported the view that model theory should become more and more algebraic/geometric, which is indeed a current trend.
| 9 | https://mathoverflow.net/users/2000 | 22854 | 15,062 |
https://mathoverflow.net/questions/22851 | 2 | Is there a neat way to use something like the Brier score to score an infinite set of forecasts/outcomes?
| https://mathoverflow.net/users/4076 | How would one extend the Brier score to an infinite number of forecasts? | A general form of the Brier score, for an essentially arbitrary outcome space $\cal X$, is as follows.
Let $q(\cdot)$ be your quoted density for a random quantity $X$, with respect to a dominating measure $\mu$ over $\cal X$. Then your score, when outcome $X=x$ is realised, is
$$S(x, q(\cdot)) = \int q(t)^2 d\mu(t) - 2q(x).$$
The Brier score is just one of an infinity of proper scoring rules. For some alternatives, see e.g.:
Gneiting, T. and Raftery, A. E. (2007). Strictly proper scoring rules, prediction, and estimation. Journal of the American Statistical Association, 102, 359-378.
Dawid, A. P. (2007). The geometry of proper scoring rules. Annals of the Institute of Statistical
Mathematics 59, 77-93.
doi:10.1007/s10463-006-0099-8
| 5 | https://mathoverflow.net/users/5689 | 22856 | 15,063 |
https://mathoverflow.net/questions/22828 | 8 | I am confused and don't get the big picture concerning the connection between
* [Ito integral](http://en.wikipedia.org/wiki/Ito_integral)
* [Stratonovich integral](http://en.wikipedia.org/wiki/Stratonovich_integral)
* Standard results in probability theory concerning skewed distributions.
**Example:** Take e.g. the mean of the [log-normal distribution](http://en.wikipedia.org/wiki/Log-normal_distribution) $e^{\mu+{\sigma^2\over 2}}$. The extra term $\sigma^2\over 2$ is a result of [Jensen's inequality](http://en.wikipedia.org/wiki/Jensen%27s_inequality#Proofs). The same result can be produced by Ito's lemma, it is the result of the famous extra term. Therefore this term is sometimes also called the Ito correction term.
**Question:** What I don't understand is why we need a strange integral which can't be used in the classical way (e.g. the standard chain rule doesn't hold any more) to come to a result which could as well be derived by standard techniques. On the other hand if we use an integral where the classical rules still apply (Stratonovich integral) we miss this additional term.
Why can't we just transform the stochastic process into an equivalent probability distribution and use the classical integrals to arrive at the right results (and, too, stick to the classical calculus rules). In a way we then would integrate not with respect to a stochastic process but with respect to the resulting probability distribution. It all seems like outsmarting ourselves and overcomplicating matters?!?
**Addendum:** Don't get me wrong: I think I understand (most) of these results piece by piece, what I miss is the bigger picture how everything fits together - perhaps somebody can enlighten me - thank you!
| https://mathoverflow.net/users/1047 | Big picture concerning Ito integral, Stratonovich integral and standard results in probability theory | It's not quite clear where exactly you have a difficulty. Of course, using stochastic calculus and Ito's integral (which is central to the modern theory of stochastic processes) to derive properties of the log-normal distribution is an overkill, but it might be a nice exercise.
Some random quick points on why Ito's integral is important:
1. Ito introduced his integrals to describe continuous Markov processes. This was successfully carried out so that now we have a nice view of Markov processes as solutions of stochastic differential equations.
2. Ito's integral seen as a process is a martingale which is very convenient since martingales "have a lot of structure", and there are lots of martingale tools. As for Stratonovich's integral, the change of variables formula for it is the same as in classical calculus, so it might appear more natural, but it is harder to work with due to the lack of martingale structure.
| 10 | https://mathoverflow.net/users/2968 | 22858 | 15,064 |
https://mathoverflow.net/questions/22857 | 2 | Is there any introduction to abelian varieties of CM type?any reference?Like how to construct a abelian varieties given a CM field E?What is the properites of the Mumford Tate group of the abelian varieties of CM type?
| https://mathoverflow.net/users/3945 | Abelian varieties of CM type? | IIRC I learnt a lot from Katz' papers from the 1970s. Of course the basic construction is the same as the elliptic curve case: you take C^g, quotient out by the lattice coming from E via its g embeddings into C, and then you have to prove that the quotient is an abelian variety, which involves writing down a non-degenerate Riemann form. This isn't hard, but I think I first saw it in one of Katz' papers. Oh---I should say that before I read Katz I read the section on abelian varieties over C in Cornell-Silverman (although there will be other references for this stuff).
| 3 | https://mathoverflow.net/users/1384 | 22863 | 15,067 |
https://mathoverflow.net/questions/22855 | 7 | A $\Delta$-set is a contravariant functor from the category $\Delta'$ of order-preserving injections to the category of sets (this is essentially what Allen Hatcher calls a $\Delta$-complex).
A main reason for working with simplicial sets instead of $\Delta$-sets should be that they allow quotients (see e.g. Allen Hatcher's nice appendix "CW complexes with simplicial structure" to his Algebraic Topology book: "A major disadvantage of $\Delta$-complexes is that they do not allow quotient constructions"), How does this go well with the fact that the category of functors $\Delta'op\to Sets$ **has** colimits?
(This question was already asked in a comment on Allen Hatcher's answer to [this](https://mathoverflow.net/questions/6281/definition-of-simplicial-complex/6302#6302) question on the definition of simplicial complexes. I apologize for asking it twice but there has been no answer given and I am afraid that the reason is - if it's not the silliness of my question - that the comment appears only after pressing the "more comments" button. However, I apologize.)
| https://mathoverflow.net/users/4676 | Why do Delta-sets not allow quotients? | The basic issue is that not every function that we would like to describe between $\Delta$-complexes can be realized by a natural transformation between functors. The lack of degeneracy maps means that no map $X \to Y$ of $\Delta$-complexes that sends any simplex down to a degenerate simplex can be realized by a natural transformation of functors. For example, if $X$ is a $\Delta$-complex interval realizing $[0,1]$ and $Y$ is a $\Delta$-complex realizing $[0,1]^2$, then there is no natural transformation of functors realizing the projection maps $p\_i:[0,1]^2 \to [0,1]$.
As a consequence, the category of $\Delta$-complexes does not have enough immediately-available maps between objects to construct the kinds of colimit diagrams one would like to realize.
| 6 | https://mathoverflow.net/users/360 | 22864 | 15,068 |
https://mathoverflow.net/questions/22865 | 4 | Does there exist a two variable analogue of the Weil conjecture?
What I mean is that the usual Weil involves a one-variable zeta-function which you get by using numbers $V\_n = V ( GF(p^n))$ of points of a smooth algebraic variety over finite fields of characteristic $p$. Is it possible to have a sensible two-parameter family of finite rings instead? Any references?
For instance, one can consider finite quotients of Witt vectors, and form a two-parameter family of numbers $V\_{n,m} = V ( Witt(GF(p^n))/I^m)$ (where $I$ is the maximal ideal of the Witt vectors) from a variety $V$ (smooth, projective over $\mathbb Z$). Is there a sensible two variable zeta-function cooked with these numbers?
| https://mathoverflow.net/users/5301 | 2d Weil conjecture | The theory of arithmetic motivic integration (see, for example, the paper by Denef and Loeser in the proceedings of the 2002 ICM) takes into account the numbers you wish to encode in your two variable zeta function.
As Kevin Buzzard rightly points out, if the scheme is smooth, then the counts along powers of primes do not give much additional information. This motivic theory is useful only in so far as it allows for possible singularities.
| 6 | https://mathoverflow.net/users/5147 | 22875 | 15,072 |
https://mathoverflow.net/questions/22860 | 22 | I need to give a lot of quite basic background to this question because I think (at least from conversing with fellow graduate students) that most mathematicians have not *really* thought about fractions for a long time. I think that there is an interesting germ of an idea in here somewhere, but I cannot exactly pinpoint it. Essentially there seems to be two canonical ways to solve division problems and there does not seem to be a "natural isomorphism" relating the two ways. I am interested in framing this duality formally: is there a "categorification" of the rational numbers where this duality can be precisely framed?
I TA a class for future elementary school teachers. The idea is to go back and really understand elementary school mathematics at a deep level. Hopefully this understanding gets passed on to the next generation.
We were discussing division of fractions. Rather than say "Yours is not to wonder why, just invert and multiply", we try to make sense of this question physically, and then use reasoning to solve the problem. Take (3/4) / (2/3). When doing this there seems to be two reasonable interpretations:
1) 3/4 of a cup of milk fills 2/3 of a container. How much milk (in cups) does it take to fill the entire container?
This is a "How many in each group" division problem, analogous to converting 6/3 into the question "If I have six objects divided into three equal groups, how many objects will be in each group?"
The solution that stares you in the face if you draw a picture of this situation is the following: 3/4 of a cup fills 2 thirds of a container. That means there must be 3/8 cups of milk in each third of the container. One container must have 9/8 cups of milk then, because this is 3 of these thirds. Note that the solution involved first dividing and then multiplying.
2) I have 3/4 cups of milk, and I have bottles which each hold 2/3 of a cup. How many bottles can I fill?
This is a "How many groups" division problem, analogous to converting 6/3 into the question "If I have six objects divided into groups of two, how many groups do I have?"
The solution suggested by this situation is the following: 3/4 of a cup of milk is actually 9 twelfths of a cup . Each twelfth is an eighth of a bottle. So I have 9/8 of a bottle. This solution involved first multiplying and then dividing.
Now I come to my question. This pattern persists! Every real world example of a "how many in each group" division problem suggests a solution by first dividing and then multiplying, whereas each "How many groups" division problem involves first multiplying and then dividing. It seems that solving the problem in the other order does not admit a conceptual realization in terms of the original problem. This is interesting to me! It suggests that the two solution methods are fundamentally different somehow. The standard approach to rational numbers (natural numbers get grothendieck grouped into integers, which get ring of fractioned into rational numbers) ignores this kind of distinction. Is there a "categorification" of the rational numbers which preserves the duality between these two types of question?
UPDATE 1: In the category of sets, if you wanted to express $(\frac{6}{2})(3) = \frac{(6)(3)}{2}$ you would have to do something like this:
Let $A$ be a set with 6 elements, $B$ a set with 3 elements, $\sim$ an equivalence relation on A where each equivalence class has 2 elements, and $\cong$ an equivalence relation on $A \times B$ where each equivalence class has 2 elements. Then there is no canonical morphism from $(A/ \sim) \times B \to (A \times B)/ \cong$. This seems to explain things somewhat on the level of integers, but we are talking fractions here.
UPDATE 2: Qiaochu points out in a comment to his answer that the order of operations is not the most essential thing here. You can solve the first problem by observing that 9/4 cups of milk fill 2 containers, so 9/8 must fill one. Torsors give a formal distinction between the two situations, but it still feels like UPDATE 1 should go through in a suitable category of "fractional sets".
UPDATE 3: For a very nice discussion of the ideas related to Theo's answer see <http://golem.ph.utexas.edu/category/2008/12/groupoidification_made_easy.html>
| https://mathoverflow.net/users/1106 | Do rational numbers admit a categorification which respects the following "duality"? | I don't claim to have a full answer, but I wanted to put it here so that others might help elaborate it.
I think that the "classical" (aside: at a recent seminar on Heegaard Floer homology, the word "classical" was defined as "posted on the arXiv") categorification of the positive rationals is to the world of finite groupoids. This is, in any case, the approach pushed by John Baez and collaborators in their HDA and Groupoidification series. You may know this already, but I'll review the basic structure for other readers. So recall that a finite groupoid is precisely: a finite set of objects, and for each pair of objects a finite set of ways that they are isomorphic, and some composition rules so that isomorphism is transitive. The simplest examples of groupoids are: discrete groupoids, aka "sets", in which each object is isomorphic to itself in exactly one way, and not isomorphic to any other object; and "point mod G", in which there is a single object, and a finite group G worth of ways that it is isomorphic to itself (G should be a group so that isomorphism is correctly transitive). There is a good notion of "equivalence" of groupoids, in which, for example, the groupoid with only one object, which is isomorphic to itself in only one way, is equivalent to the groupoid with two objects which are each isomorphic to themselves in precisely one way and which are isomorphic to each other in precisely one way. Groupoids have a good notion of "disjoint union", and in fact every finite groupoid is equivalent to a groupoid that is a disjoint union of various "point mod G"s for various groups G, just as every finite set is a disjoint union of points.
What Baez and Dolan described was the correct notion of "cardinality" of a groupoid. Namely, cardinality is additive under disjoint union, preserved under equivalence of groupoids, and the cardinality of "point mod G" is 1/|G|. (They prove that this uniquely defines a cardinality.) There is also a notion of cartesian product of groupoids, and this notion of cardinality is multiplicative under products. Moreover, if a group G acts on another groupoid X, then there is a "quotient groupoid" X//G, which has the same objects as X and extra isomorphisms for the G action. If the action is free, then X//G is equivalent to the usual "coarse" quotient X/G. Moreover, the cardinality of X//G is the cardinality of X divided by the number of group elements of G.
Since there are groups of arbitrary positive-integer cardinality, there are groupoids of arbitrary nonnegative-rational cardinality. In this sense, {Groupoids} categorifies {Rationals}.
But there are at least two reasons to be suspicious of this proposed categorification. First of all, although addition and multiplication categorify as we expect them to --- to disjoint union and cartesian product --- division is a bit strange. It is not inverse to multiplication except on one side: if you take a groupoid A, multiply it by a set B, and then divide by a group that acts freely and transitively on B, you get a groupoid equivalent to A; but if you divide first, you cannot multiply back. Similarly, it is not generally true that X//G is equivalent to X times pt//G. Maybe there is some sort of "semidirect product" or "crossed product" of groupoids that really does unify it all --- I don't know. In any case, when Baez explains his division, he treats numerator and denominator very differently: the numerator is something like a set, whereas the denominator is something like a group.
Second, it is not true that {Rationals} are the "grothendieck group" of {Groupoids}, at least not without some heavy duty groupifying. At the very least, there are many inequivalent groupoids with the same cardinality. Maybe this is a plus from the point of view of your question: {Groupoids} may be able to tell apart the two kinds of division that you sketch.
| 12 | https://mathoverflow.net/users/78 | 22879 | 15,075 |
https://mathoverflow.net/questions/22884 | 1 | I have an graph with the following attributes:
* Undirected
* Not weighted
* Each vertex has a minimum of 2 and maximum of 6 edges connected to it.
* Vertex count will be < 100
* Graph is static and no vertices/edges can be added/removed or edited.
I'm looking for **all** subgraphs between a random subset of the vertices (at least 2).
I've created a (warning! programmer art) animated gif to illustrate what i'm trying to achieve: <https://i.stack.imgur.com/Hgaq5.gif>
My end goal is to have a set of subgraphs that allow moving from one of the subset vertices (blue nodes) and reach **any** of the other subset vertices (blue nodes).
| https://mathoverflow.net/users/5692 | Graph algorithm to find all subgraphs that connect N arbitrary vertices | It looks like the paper
*Generating all the Steiner trees and computing Steiner intervals for a fixed number of terminals*
by Costa Dourado, de Oliveira, and Protti is what you want (available from ScienceDirect). I think the paper gives an algorithm for generating all the *minimal* (under subgraph inclusion) subgraphs connecting the blue vertices (from which it is easy to obtain all such subgraphs).
| 4 | https://mathoverflow.net/users/2233 | 22888 | 15,079 |
https://mathoverflow.net/questions/22881 | 7 | Let $S$ be a locally noetherian scheme, $C$ the category of proper smooth $S$-schemes with geometrical connected fibres and $C\_\*$ the category of pointed objects of $S$, i.e. objects of $C$ together with a morphism $S \to C$. Also denote $A$ the category of abelian schemes over $S$. There is a well-known rigidity result stating that a pointed morphism between $X,Y \in A$ is already a group morphism. In other words, the inclusion functor
$A \to C\_\*$
is fully faithful. Is there a nice description for the image? In other words, which purely scheme-theoretic properties do abelian schemes have and are there enough to characterize them? For example, $X \in A$ is "homogeneous".
| https://mathoverflow.net/users/2841 | scheme-theoretic description of abelian schemes | How about smooth proper morphisms $X \to S$ with connected fibers, a section $S \to X$, such that the sheaf of Kähler differentials $\Omega\_{X/S}$ is a pullback from $S$, and such that the group scheme $\underline{\rm Aut}\_S X$ acts transitively on the fibers? The essential point is that the hypothesis on the differentials insures that no geometric fiber can contain a rational curve, so no affine algebraic group can act non-trivially. The result should follow from Chevalley's structure theorem for algebraic groups, with some fairly standard arguments (I haven't checked the details, though).
| 6 | https://mathoverflow.net/users/4790 | 22892 | 15,082 |
https://mathoverflow.net/questions/22883 | 14 | Let $U$ be an open subscheme of $\textrm{Spec} \ \mathbf{Z}$. The complement of $U$ is a divisor $D$ of $\textrm{Spec} \ \mathbf{Z}$.
**Q**. Can we classify the etale coverings of $U$ of a given degree?
Given a finite etale morphism $\pi:V\longrightarrow U$, the normalization of $X$ in the function field $L$ of $V$ is $\textrm{Spec} \ O\_L$. Can we also say what $V$ itself should be?
Of course, we can complicate things by replacing $\textrm{Spec} \ \mathbf{Z}$ by $\textrm{Spec} \ O\_K$.
**Example**. Take $U= \textrm{Spec} \ \mathbf{Z} -\{ (2)\}$ and let $V\rightarrow U$ be an finite etale morphism. Suppose that $V$ is connected and that let $K$ be its function field. The normalization of $\textrm{Spec} \ \mathbf{Z}$ in $K$ is of course $O\_K$. The extension $\mathbf{Z}\subset O\_K$ is unramified outside $(2)$ and (possibly) ramified at $(2)$. Can one give a description of $V$ here?
**EDIT**. I just realized one can also ask themselves a similar question for $\mathbf{P}^1\_{\mathbf{C}}$. Or even better, for any Riemann surface $X$.
| https://mathoverflow.net/users/4333 | Etale coverings of certain open subschemes in Spec O_K | As Kevin points out, $V$ is indeed $\mathcal{O}\_K[\frac{1}{2}]$ in your example. Your link to the fundamental group is also correct. $\pi\_1(U)$ is the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside 2 (since you can restrict attention to the *connected* etale $\mathbb{Q}$-algebras)\*. More generally, in your original question, these are replaced by $\mathcal{O}\_L$ with the primes above the support of $D$ inverted, and the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside the support of $D$. These groups can get pretty horrendous, and so number-theorists tend to (at least in my view) study the more well-behaved but still very mysterious maximal pro-$p$-quotients of these groups. Now these pro-$p$ groups are not only finitely generated by work of Shafarevich ("Extensions with prescribed ramification points"), they are $d$-generated, where $d$ is the cardinality of the support of $D$ (for "tame" and not silly $D$). More impressively, the relation rank is also calculated/bounded (in this case, it's also $d$!!!), frequently leading to conclusions about when these groups are finite or infinite.
The "more complicated" starting point of $\mathcal{O}\_K$ is in a sense not actually much more complicated. You're still asking for the Galois group of the maximal extension of $K$ unramified outside a finite set of primes, and Shafarevich's work gives formulas for the relation rank and generator rank for the pro-$p$-quotients.
The answer to your overall classification question is thus pretty difficult, and consume entire subdisciplines of number theory. As an example, Christian Maire has constructed number fields with a trivial class group but infinite unramified extensions -- you'd have to have a complete understanding of when this could happen before you could hope to prescribe all unramified extensions of a given degree, with or without certain primes inverted. There are certain cases where this can be done via, e.g., root discriminant bounds, but the story is far from being complete at this point.
As in Lars's answer, the situation is much much better understood for $\mathbb{P}\_\mathbb{C}^1$ and Riemann surfaces than it is for number fields.
\*: Actually, you have to be a little careful about 2-extensions. Etaleness doesn't pick up on whether or not infinite primes ramify. Everything's fine if you start with an totally imaginary field.
| 5 | https://mathoverflow.net/users/35575 | 22893 | 15,083 |
https://mathoverflow.net/questions/22891 | 5 | For a projective variety $X$, Serre's vanishing theorem says that $H^i(X, \mathcal{F}(n))=0$ for any coherent sheave, $i>1$ and sufficiently large $n$. I am wondering, is there a similar type of vanishing theorem on quasi-projective varieties, namely, let $Y$ be a quasi-projective variety, what can we say about the vanishing of $H^i(Y, \mathcal{F}(n))$ under the same setting as projective case.
Or is there a similar type of theorem for local cohomology, say, when is $H^i\_{pt}(X, \mathcal{F}(n))$ vanishing?
| https://mathoverflow.net/users/2348 | Serre type vanishing theorem of coherent sheaves on quasi-projective variety? | Even for quasi-affine variety you don't have a vanishing theorem except Grothendieck's vanishing theorem on a noetherian topological space of finite dimension.
Consider, for example, affine plane without point $\mathbf{A}^2\backslash 0.$ Then the structure sheaf is ample but the first cohomology $H^1(\mathbf{A}^2\backslash 0, \mathcal{O})$
is not trivial and even it is infinite dimensional.
| 11 | https://mathoverflow.net/users/2464 | 22894 | 15,084 |
https://mathoverflow.net/questions/22897 | 42 | Is there a nice characterization of fields whose automorphism group is trivial? Here are the facts I know.
1. Every prime field has trivial automorphism group.
2. Suppose *L* is a separable finite extension of a field *K* such that *K* has trivial automorphism group. Then, if *E* is a finite Galois extension of *K* containing *L*, the subgroup $Gal(E/L)$ in $Gal(E/K)$ is self-normalizing if and only if *L* has trivial automorphism group. (As pointed out in the comments, a field extension obtained by adjoining one root of a generic polynomial whose Galois group is the full symmetric group satisfies this property).
3. The field of real numbers has trivial automorphism group, because squares go to squares and hence positivity is preserved, and we can then use the fact that rationals are fixed. Similarly, the field of algebraic real numbers has trivial automorphism group, and any subfield of the reals that is closed under taking squareroots of positive numbers has trivial automorphism group.
My questions:
1. Are there other families of examples of fields that have trivial automorphism group? For instance, are there families involving the p-adics? [EDIT: One of the answers below indicates that the p-adics also have trivial automorphism group.]
2. For what fields is it true that the field cannot be embedded inside any field with trivial automorphism group? I think that any automorphism of an algebraically closed field can be extended to any field containing it, though I don't have a proof) [EDIT: One of the answers below disproves the parenthetical claim, though it doesn't construct a field containing an algebraically closed field with trivial automorphism group]. I suspect that $\mathbb{Q}(i)$ cannot be embedded inside any field with trivial automorphism group, but I am not able to come up with a proof for this either. [EDIT: Again, I'm disproved in one of the answers below]. I'm not even able to come up with a conceptual reason why $\mathbb{Q}(i)$ differs from $\mathbb{Q}(\sqrt{2})$, which can be embedded in the real numbers.
**ADDED SEP 26**: All the questions above have been answered, but the one question that remains is: can every field be embedded in a field with trivial automorphism group? Answering the question in general is equivalent to answering it for algebraically closed fields.
| https://mathoverflow.net/users/3040 | Fields with trivial automorphism group | As Robin as pointed out, for all primes $p$, $\mathbb{Q}\_p$ is **rigid**, i.e., has no nontrivial automorphisms. It is sort of a coincidence that you ask, since I spent much of the last $12$ hours writing up some material on multiply complete fields which has applications here:
Theorem (Schmidt): Let $K$ be a field which is complete with respect to two inequivalent nontrivial norms (i.e., the two norms induce distinct nondiscrete topologies). Then $K$ is algebraically closed.
Corollary: Let $K$ be a field which is complete with respect to a nontrivial norm and *not* algebraically closed. Then every automorphism of $K$ is continuous with respect to the norm topology. (Proof: To say that $\sigma$ is a discontinuous automorphism is to say that the pulled back norm $\sigma^\*|| \ ||: x \mapsto ||\sigma(x)||$ is inequivalent to $|| \ ||$. Thus Schmidt's theorem applies.
In particular this applies to show that $\mathbb{Q}\_p$ and $\mathbb{R}$ are rigid, since every continuous automorphism is determined by its values on the dense subspace $\mathbb{Q}$, hence the identity is the only possibility. (It is possible to give a much more elementary proof of these facts, e.g. using the Ostrowski classification of absolute values on $\mathbb{Q}$.)
At the other extreme, each algebraically closed field $K$ has the largest conceivable automorphism group: $\# \operatorname{Aut}(K) = 2^{\# K}$: e.g. Theorem 80 of
[http://alpha.math.uga.edu/~pete/FieldTheory.pdf](http://alpha.math.uga.edu/%7Epete/FieldTheory.pdf).
There is a very nice theorem of Bjorn Poonen which is reminiscent, though does not directly answer, your other question. For any field $K$ whatsoever, and any $g \geq 3$, there exists a genus $g$ function field $K(C)$ over $K$ such that $\operatorname{Aut}(K(C)/K)$ is trivial. However there may be other automorphisms which do not fix $K$ pointwise.
There is also a sense in which for each $d \geq 3$, if you pick a degree $d$ polynomial $P$ with $\mathbb{Q}$-coefficients at random, then with probability $1$ it is irreducible and $\mathbb{Q}[t]/(P)$ is rigid. By Galois theory this happens whenever $P$ is irreducible with Galois group $S\_d$, and by Hilbert Irreducibility the complement of this set is small: e.g. it is "thin" in the sense of Serre.
**Addendum**: Recall also Cassels' embedding theorem (J.W.S. Cassels, *An embedding theorem for fields*, Bull. Austral. Math. Soc. 14 (1976), 193-198): every finitely generated field of characteristic $0$ can be embedded in $\mathbb{Q}\_p$ for infinitely many primes $p$. It would be nice to know some positive characteristic analogue that would allow us to deduce that a finitely generated field of positive characteristic can be embedded in a rigid field (so far as I know it is conceivable that every finitely generated field of positive characteristic can be embedded in some Laurent series field
$\mathbb{F}\_q((t))$, but even if this is true it does not have the same consequence, since Laurent series fields certainly have nontrivial automorphisms).
| 32 | https://mathoverflow.net/users/1149 | 22903 | 15,090 |
https://mathoverflow.net/questions/22906 | 1 | Equivalently, is there a graph that contains an infinite simple path that has a start and an end point? My intuition is that there is no such graph, but I'm finding it hard to articulate why.
| https://mathoverflow.net/users/800 | Is it possible to define a graph that has two vertices that are infinitely far apart? | One reasonable definition of "simple path" is a connected acyclic subgraph with vertex degree at most two. But, to avoid circularity, we need a definition for being connected. I think the standard definition is that a subgraph is connected iff every two vertices are at a finite distance from each other, so with this definition the answer is trivially no.
On the other hand, if you define a "simple path" to be an acyclic graph in which all vertices have degree one or two and at most two of them have degree one (not the definition I'd choose, but it works for finite graphs) then with this definition the answer is yes: remove a single vertex from the middle of the double-ended infinite path indexed by the integers.
So the answer to your question depends on the details of your definitions.
| 8 | https://mathoverflow.net/users/440 | 22912 | 15,094 |
https://mathoverflow.net/questions/22868 | 5 | Let $f(x)$ be a polynomial with real coefficients, and let $||\cdot||$ be the distance-from-the-nearest-integer function. It is known that for any $ \epsilon > 0 $, the set $S$ of positive integer solutions of the inequality $||f(x)|| < \epsilon$ has bounded gaps. This means that if $x\_1 < x\_2 < \ldots$ are the elements of $S$ in increasing order, then the set of differences $x\_{n+1}-x\_n$ is bounded. This is proved in "Simultaneous Diophantine Approximation and IP Sets" by Furstenberg and Weiss, Acta Arith. 1988.
I would like to know how to compute the maximum gap, i.e. the maximum of the values $ x\_{n+1}- x\_n $, given a specific polynomial and a specific value of $\epsilon$. I can do this for linear polynomials, but not for quadratics. The proof of Furstenberg and Weiss appears to be non-constructive.
To take a particular example, what is the maximum possible gap between successive solutions of the inequality $||2^{1/2}x^2|| < .01$? Can anyone suggest a method, no matter how impractical, that would eventually lead to an answer? Could it be that, even in this particular case, no one knows how to find the maximum gap?
Note: Experimentation suggests that the maximum gap is 627, which occurs for the first time following the solution $x=1115714$.
| https://mathoverflow.net/users/5229 | Wanted: A constructive version of a theorem of Furstenberg and Weiss | Dear RJS,
I think Tim Gowers is right - the problem seems too hard. Reasonably good bounds are known on (for example) the *least* $n \geq 1$ for which $\Vert n^2 \sqrt{2} \Vert \leq \epsilon$; one can find such an $n$ with $n \leq \epsilon^{-7/4 + o(1)}$. This is a result of Zaharescu [Zaharescu, A; Small values of $n^2\alpha\pmod 1$. Invent. Math. 121 (1995), no. 2, 379--388.] Zaharescu in fact obtains this result for any $\theta$ in place of $\sqrt{2}$. From a cursory glance at the paper I see that he uses the continued fraction expansion for $\theta$ and so it may be that one can slightly improve his bound in the particular case of $\sqrt{2}$.
It is an old conjecture of Heilbronn that the right bound here should be $\epsilon^{-1 + o(1)}$. I don't know off the top of my head whether any more precise conjectures have been made based on sensible heuristics either for this or for your original problem.
To get *an* explicit upper bound for your problem one can proceed quite straightforwardly using arguments due to Weyl. I don't think this is the right place to describe an argument in detail: there are several variants, and I first learnt this from a Tim Gowers course at Cambridge. See Theorem 3.10 of these notes:
<http://www.math.cmu.edu/~af1p/Teaching/AdditiveCombinatorics/notes-acnt.pdf>
If you really had to show that 627 is the answer to your specific problem, probably the best bet would be to inspect all the quadratics $n^2\sqrt{2} + \theta n + \theta'$ for $\theta,\theta'$ in some rather dense finite subset of $[0,1]^2$ and show using a computer that each takes (mod 1) a value less than 0.009999 for some $n \leq 627$. Painful!
The argument of Furstenberg and Weiss uses ergodic theory and so will not directly lead to an effective bound.
There are quite detailed conjectures about the fractional parts of $n^2\sqrt{2}$ (and other similar sequences) due to Rudnick, Sarnak and Zaharescu, essentially encoding the fact that this sequence of fractional parts is expected to behave like a Poisson process. I don't think those conjectures are likely to be helpful in your context since, taken too literally, they would seem to suggest that there are arbitrarily long intervals without a number such that $\Vert n^2 \sqrt{2} \Vert < 0.01$ - contrary to Furstenberg-Weiss.
Nonetheless let me point out a recent paper of Heath-Brown which is very interesting in connection with these matters:
<http://arxiv.org/pdf/0904.0714v1>.
One more point perhaps worthy of mention: sequences with bounded gaps are usually known as *syndetic*.
| 5 | https://mathoverflow.net/users/5575 | 22913 | 15,095 |
https://mathoverflow.net/questions/22914 | 13 | Let A be an Abelian category.
From this category, we can form the chain complex category Ch(A). The objects of Ch(A) are chain complexes of objects of A. The morphisms of Ch(A) are chain maps. Ch(A) is an Abelian category for every Abelian category A.
Now from Ch(A), we can form the chain homotopy category K(A). The objects of K(A) are just objects of Ch(A), but the morphisms of K(A) are chain homotopy classes of chain maps. Thus, K(A) is a quotient of Ch(A).
It turns out that K(A) is an additive category for any Abelian category A. From asking people, I seem to get the impression that K(A) is not always abelian, but I've had a hard time showing this. If all objects of A are projective (e.g. if A is the category of vector spaces over some field k), then K(A) will be Abelian.
I've been trying to show that K(**Ab**) is not Abelian (where **Ab** is the category of Abelian groups). More specifically, I've been trying to show this by showing the following (which may or may not be true):
Let X be a chain complex with the group of integers in dimension 0, and zero in every other dimension. Let f be the chain map from X to X, that sends each integer x to 2x. I've been trying to show that in K(**Ab**), the homotopy class of this chain map does not have a cokernel.
Any answers, suggestions, hints, or comments would be greatly appreciated!
| https://mathoverflow.net/users/5698 | Is the chain homotopy category, K(Ab), an Abelian category? By Ab, I mean the category of Abelian groups. | You might want to have a look at [my answer](https://mathoverflow.net/questions/15658/how-do-i-know-the-derived-category-is-not-abelian/15662#15662) to this question.
| 11 | https://mathoverflow.net/users/310 | 22916 | 15,097 |
https://mathoverflow.net/questions/22907 | 9 | This question is inspired by, but is independent of: [Sheaf Description of G-Bundles](https://mathoverflow.net/questions/2414/sheaf-description-of-g-bundles)
Line bundles are classified by $H^1(X,\mathcal{O}^\times\_X)$. We also know that in general that $H^1(X,G)$, where $G$ is a sheaf from open sets in $X$ to $Grps$, classifies $G$-torsors over X.
With this insight in mind: $\mathcal{O}^\times\_X$-torsors should correspond to line bundles. Indeed, if $P$ is one, then $\mathcal{O} \_X \times \_{\mathcal{O} \_X^\times}P$ gives the desired line bundle, and all line bundles are achieved this way (see the question I linked to).
My question is about the more mundane $H^1(X,\mathbb{C})$, which can be thought of as $H^1(X,\underline{\mathbb{C}})$ where $\underline{\mathbb{C}}$ is the constant sheaf $\mathbb{C}$ (which I think of as a sheaf going to $Grps$). These should supposedly correspond to $\mathbb{C}$-bundles over $X$. Which appears to be line bundles. But of course there's no reason for $H^1(X,\mathbb{C})$ to equal $H^1(X,\mathcal{O}^\times\_X)$... My intuition is that this should correspond to the more naive version of fiber bundles that doesn't involve a structure group. Do you have any thoughts?
| https://mathoverflow.net/users/5309 | Confusion about how the first cohomology classifies torsors | The general principle is: if you have some objects which are locally trivial but globally possibly not trivial then the isomorphism classes of such objects are classified by $H^1(X,\underline{Aut})$, where $\underline{Aut}$ is the sheaf of automorphisms of your objects.
So, if your objects locally are $U\times \mathbb A^n$ (i.e. vector bundles) or they are $\mathcal O\_U^{\oplus n}$ (i.e. locally free sheaf) then either of these are classifed by $H^1(X, GL(n,\mathcal O))$. For $n=1$, you get $GL(1,\mathcal O)=\mathcal O^\*$.
Now what is $\mathbb C$ an automorphism group of? Certainly not of line bundles (zero has to go to zero).
| 14 | https://mathoverflow.net/users/1784 | 22921 | 15,101 |
https://mathoverflow.net/questions/22929 | 12 | A generic elliptic curves over C has automorphism group of order 2. The elliptic curves with extra automorphisms are C/Z[i] (automorphism group of order 4) and C/Z[w] where w is a primitive 3rd root of unity (automorphism group of order 6). One can use their extra automorphisms to prove that they can be defined over Q
Similarly, the Klein quartic (a genus 3 curve with 168 automorphisms, the maximum possible number for genus 3 curves) can be defined over Q.
>
> Suppose X is a genus g curve over C.
> Is there a condition on the group
> Aut(X) that guarantees that X is
> defined over a number field?
>
>
>
I feel like maybe one could try to approach this question by passing to the Jacobian of X and arguing that extra automorphisms of X give extra automorphisms of J(X) which means that J(X) has "large" endomorphism ring and hence must be defined over a number field (just as CM elliptic curves are defined over number fields), but I don't know whether "large" is large enough.
| https://mathoverflow.net/users/683 | Extra automorphisms of curves and definability over \bar Q | If the group has order 84(g-1), then the curve is a Galois cover of the (2,3,7) orbifold. By Belyi's theorem, it is defined over a number field.
| 9 | https://mathoverflow.net/users/121 | 22933 | 15,109 |
https://mathoverflow.net/questions/22927 | 242 | As I understand it, it has been proven that the axiom of choice is independent of the other axioms of set theory. Yet I still see people fuss about whether or not theorem X depends on it, and I don't see the point. Yes, one can prove some pretty disturbing things, but I just don't feel like losing any sleep over it if none of these disturbing things are in conflict with the rest of mathematics. The discussion seems even more moot in light of the fact that virtually none of the weird phenomena can occur in the presence of even mild regularity assumptions, such as "measurable" or "finitely generated".
So let me turn to two specific questions:
>
> If I am working on a problem which is not directly related to logic or set theory, can important mathematical insight be gained by understanding its dependence on the axiom of choice?
>
>
> If I am working on a problem and I find a two page proof which uses the fact that every commutative ring has a maximal ideal but I can envision a ten page proof which circumvents the axiom of choice, is there any sense in which my two page proof is "worse" or less useful?
>
>
>
The only answer to these questions that I can think of is that an object whose existence genuinely depends on the axiom of choice do not admit an explicit construction, and this might be worth knowing. But even this is largely unsatisfying, because often these results take the form "for every topological space there exists X..." and an X associated to a specific topological space is generally no more pathological than the topological space you started with.
Thanks in advance!
| https://mathoverflow.net/users/4362 | Why worry about the axiom of choice? | The best answer I've ever heard --- and I think I heard it here on MathOverflow from Mike Shulman, which suggests that this question is roughly duplicated somewhere else --- is that you should care about constructions "internal" to other categories:
1. For many, many applications, one wants "topological" objects: topological vector spaces, topological rings, topological groups, etc. In general, for any algebraic gadget, there's a corresponding topological gadget, by writing the original definition (a la Bourbaki) entirely in terms of sets and functions, and then replacing every set by a topological space and requiring that every function be continuous.
2. A closely related example is that you might want "Lie" objects: sets are replaced by smooth manifolds and functions by smooth maps.
3. Another closely related example is to work entirely within the "algebraic" category.
In all of these cases, the "axiom of choice" fails. In fact, from the internal-category perspective, the axiom of choice is the following simple statement: every surjection ("epimorphism") splits, i.e. if $f: X\to Y$ is a surjection, then there exists $g: Y \to X$ so that $f\circ g = {\rm id}\_Y$. But this is simply false in the topological, Lie, and algebraic categories.
This leads to all sorts of extra rich structure if you do algebra internal to these categories. You have to start thinking about bundles rather than products, there can be "anomalies", etc.
**Update:**
In the comments, there was a request for a totally explicit example, where Axiom of Choice is commonly used but not necessary. Here's one that I needed recently. Let $\mathcal C$ be an abelian tensor category, by which I mean that it is [abelian](https://en.wikipedia.org/wiki/Abelian_category), has a monoidal structure $\otimes$ that is biadditive on hom-sets, and that has a distinguished natural isomorphism $\text{flip}: X\otimes Y \overset\sim\to Y\otimes X$ which is a "symmetry" in the sense that $\text{flip}^2 = \text{id}$. Then in $\mathcal C$ is makes sense to talk about "Lie algebra objects" and "associative algebra objects", and given an associative algebra $A$ you can define a Lie algebra by "$[x,y] = xy - yx$", where this is short-hand for $[,] = (\cdot) - (\cdot \circ \text{flip})$ — $x,y$ should not be read as elements, but as some sort of generalization. So we can makes sense of the categories of $\text{LIE}\_{\mathcal C} = $"Lie algebras in $\mathcal C$" and $\text{ASSOC}\_{\mathcal C} = $"associative algebras in $\mathcal C$", and we have a forgetful functor $\text{Forget}: \text{ASSOC}\_{\mathcal C} \to \text{LIE}\_{\mathcal C}$.
Then one can ask whether $\text{Forget}$ has a left adjoint $U: \text{LIE}\_{\mathcal C} \to \text{ASSOC}\_{\mathcal C}$. If $\mathcal C$ admits arbitrary countable direct sums, then the answer is yes: the tensor algebra is thence well-defined, and so just form the quotient as you normally would do, being careful to write everything in terms of objects and morphisms rather than elements. In particular, if $\mathfrak g \in \text{LIE}\_{\mathcal C}$, then $U\mathfrak g \in \text{ASSOC}\_{\mathcal C}$ and it is universal with respect to the property that there is a Lie algebra homomorphism $\mathfrak g \to U\mathfrak g$.
Let's say that $\mathfrak g$ is **representable** if the map $\mathfrak g \to U\mathfrak g$ is a monomorphism in $\text{LIE}\_{\mathcal C}$. By universality, if there is any associative algebra $A$ and a monomorphism $\mathfrak g \to A$, then $\mathfrak g \to U\mathfrak g$ is mono, so this really is the condition that $\mathfrak g$ has some faithful representation. The statement that "Every Lie algebra is representable" is normally known as the Poincare-Birkoff-Witt theorem.
The important point is that the usual proof — the one that Birkoff and Witt gave — requires the Axiom of Choice, because it requires picking a vector-space basis, and so it works only when $\mathcal C$ is the category of $\mathbb K$ vector spaces for $\mathbb K$ a field, or more generally when $\mathcal C$ is the category of $R$-modules for $R$ a commutative ring and $\mathfrak g$ is a *free* $R$-module, or actually the proof can be made to work for arbitrary *Dedekind domains* $R$. But in many abelian categories of interest this approach is untenable: not every abelian category is semisimple, and even those that are you often don't have access to bases. So you need other proofs. Provided that $\mathcal C$ is "over $\mathbb Q$" (hom sets are $\mathbb Q$-vector spaces, etc.), a proof that works constructively with no other restrictions on $\mathcal C$ is available in
* Deligne, Pierre; Morgan, John W.
Notes on supersymmetry (following Joseph Bernstein). *Quantum fields and strings: a course for mathematicians*, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. [MR1701597](https://mathscinet.ams.org/mathscinet-getitem?mr=1701597).
They give a reference to
* Corwin, L.; Ne'eman, Y.; Sternberg, S.
Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry).
*Rev. Modern Phys*. 47 (1975), 573--603. [MR0438925](https://mathscinet.ams.org/mathscinet-getitem?mr=0438925).
in which the proof is given when $\mathcal C$ is the category of modules of a (super)commutative ring $R$, with $\otimes = \otimes\_R$, and, importantly, $2$ and $3$ are both invertible in $R$. [Edit: I left a comment July 28, 2011, below, but should have included explicitly, that Corwin--Ne'eman--Sternberg require more conditions on $\mathcal C$ than just that $2$ and $3$ are invertible. Certainly as stated "PBW holds when $6$ is invertible" is inconsistent with the examples of Cohn below.]
Finally, with $R$ an arbitrary commutative ring and $\mathcal C$ the category of $R$-modules, if $\mathfrak g$ is torsion-free as a $\mathbb Z$-module then it is representable. This is proved in:
* Cohn, P. M.
A remark on the Birkhoff-Witt theorem.
*J. London Math. Soc*. 38 1963 197--203. [MR0148717](https://mathscinet.ams.org/mathscinet-getitem?mr=0148717)
So it seems that almost all Lie algebras are representable. But notably Cohn gives examples in characteristic $p$ for which PBW fails. His example is as follows. Let $\mathbb K$ be some field of characteristic $p\neq 0$; then in the free associative algebra $\mathbb K\langle x,y\rangle$ on two generators we have $(x+y)^p - x^p - y^p = \Lambda\_p(x,y)$ is some non-zero Lie series. Let $R = \mathbb K[\alpha,\beta,\gamma] / (\alpha^p,\beta^p,\gamma^p)$ be a commutative ring, and define $\mathfrak g$ the Lie algebra over $R$ to be generated by $x,y,z$ with the only defining relation being that $\alpha x = \beta y + \gamma z$. Then $\mathfrak g$ is not representable in the category of $R$-modules: $\Lambda\_p(\beta y,\gamma z)\neq 0$ in $\mathfrak g$, but $\Lambda\_p(\beta y,\gamma z)= 0$ in $U\mathfrak g$.
| 151 | https://mathoverflow.net/users/78 | 22938 | 15,113 |
https://mathoverflow.net/questions/22923 | 39 | Does there exist an algorithm which computes the Galois group of a polynomial
$p(x) \in \mathbb{Z}[x]$? Feel free to interpret this question in any reasonable manner. For example, if the degree of $p(x)$ is $n$, then the algorithm could give a set of permutations $\pi \in Sym(n)$ which generate the Galois group.
| https://mathoverflow.net/users/4706 | Computing the Galois group of a polynomial | There is an algorithm described in an ancient and interesting book on Galois Theory by Leonard Eugene Dickson. Here is a brief sketch in the case of an irreducible polynomial $f\in \mathbb{Q}[x]$.
Suppose that $z\_1\ldots z\_n$ are the roots of $f$ in some splitting field of $f$ over $\mathbb{Q}$. (We don't need to construct the splitting field. The $z\_i$ are mentioned here for the sake of explanation.) Let $x\_1\ldots x\_n$ be indeterminates. For a permutation $\sigma\in S\_n$, let
$$E\_\sigma=x\_1z\_{\sigma(1)}+\ldots+ x\_n z\_{\sigma(n)}.$$
Let $g(x):=\prod \_{\sigma} (x-E\_\sigma)$,
where $\sigma$ runs through all permutations in $S\_n$. Each coefficient $c\_i$ of $x^i$ in $g$ is symmetric in $z\_1 \ldots z\_n$, so (using the theorem on symmetric functions) we can write $c\_i$ as a polynomial in $x\_1\dots x\_n$ with rational coefficients.
Assuming that this has been done, factor $g$ into irreducibles over the ring $\mathbb{Q}[x\_1 \ldots x\_n]$. Let $g\_0$ be the irreducible factor of $g$ that is satisfied by $E\_{Id}$, where $Id$ is the identity permuation. Then the galois group of $f$ consists of all permutations of $x\_1\ldots x\_n$ that fix $g\_0$.
The point is that the computation of $g\_0$ is effective (albeit horrendous) and so is the determination of the permutations that fix $g\_0$.
| 35 | https://mathoverflow.net/users/5229 | 22946 | 15,118 |
https://mathoverflow.net/questions/22941 | 4 | For example, deciding whether or not the following is a category seems to depend on the above question (from Awodey's Category Theory, pg. 6):
> "What if we take sets as objects and as arrows, those $f : A \rightarrow B$ such that for all $b \in B$, the subset $f$-1$(b) \subseteq A$ is finite?"
Define for each $n \in N$ the function $f$n: $N \rightarrow N$ where $f$n$(x) = max(0, x - n)$.
Then any $f$n or any finite composition thereof has finite inverse images. Yet the "infinite composition" $... f$1$f$2$f$0 has an infinite inverse image for 0, and so the above does not meet the definition of a category.
If this "infinite composition" is legit, does it follow from the basic definition of a category, or must the definition be made more flexible or precise to accommodate it?
For reference, this is Awodey's definition concerning composition:
> "Given arrows $f : A \rightarrow B$ and $g : B \rightarrow C$, [...] there is given an arrow: $g$ o $f : A \rightarrow C$ called the composite of $f$ and $g$."
Thank you for your insight.
| https://mathoverflow.net/users/5700 | Is an "infinite compositions of arrows" meaningful? | This is probably not what the OP is looking for, but there is a notion of "infinite composition of arrows" which often appears for example in categorical homotopy theory:
If $f\_0 : X\_0 \to X\_1$, $f\_1 : X\_1 \to X\_2$, $\ldots$ are morphisms in a category $C$ and the colimit of the diagram $X\_0 \to X\_1 \to X\_2 \to \cdots$ exists (call it $X$) then $X$ is equipped in particular with a canonical map $X\_0 \to X$ which is called the *transfinite composition* of the maps $f\_i$. Of course, technically this morphism of $C$ is specified only up to canonical isomorphism because $X$ may be replaced by a (uniquely isomorphic) different colimit of the $X\_i$. More generally given an ordinal $\alpha$ which we may view as a category (poset) and a colimit-preserving functor $X\_\cdot : \alpha \to C$ (so that for each limit ordinal $\beta \in \alpha$, we have $X\_\beta = \operatorname{colim}\_{\gamma < \beta} X\_\gamma$), we may form the transfinite composition $X\_0 \to \operatorname{colim}\_\alpha X\_\alpha$. One is often concerned with questions such as whether a given class of maps is closed under transfinite compositions (for example, the class of cofibrations in a model category has this property).
| 10 | https://mathoverflow.net/users/126667 | 22947 | 15,119 |
https://mathoverflow.net/questions/22950 | 17 | When I hear the phrase "line bundle" the first thing that pops into my head is a mobius band. But this is a bad picture from an algebraic point of view since any line bundle on an affine variety is trivial. Anyway, my question is: is there a way of seeing more concretely what "goes wrong" when you try to construct the mobius band as an algebraic line bundle over ℝ, and what changes when you move to analytic line bundles?
| https://mathoverflow.net/users/2615 | why isn't the mobius band an algebraic line bundle? | Consider the real algebraic line bundle $\mathcal{O}(-1)$ over the real algebraic variety $\mathbb{R}\mathbb{P}^1$. It is nontrivial hence continuously isomorphic to the "Moebius" line bundle (there are only 2 line bundles on the circle, up to continuous isomorphism), so its total space is homeomorphic o the "Moebius strip".
By the way, it is false that line bundles on affine algebraic varieties are trivial. One example is the above universal line bundle over $\mathbb{R}\mathbb{P}^1$. If you want an example over $\mathbb{C}$, consider the complement of a point in an elliptic curve: it's an affine variety but its Picard group is far from being trivial.
| 27 | https://mathoverflow.net/users/4721 | 22954 | 15,121 |
https://mathoverflow.net/questions/22078 | 48 | It is a standard and important fact that any smooth affine group scheme $G$ over a field $k$ is a closed $k$-subgroup of ${\rm{GL}}\_n$ for some $n > 0$. (Smoothness can be relaxed to finite type, but assume smoothness for what follows.) The proof makes essential use of $k$ being a field, insofar as it uses freeness of finitely generated $k$-submodules of the coordinate ring of $G$. The same argument (appropriately formulated) then works when $k$ is a PID. (edit: I originally mentioned that I didn't know if this is also true over any Dedekind domain, and wasn't asking about it; nonetheless, comments from George and Kevin below give proof for Dedekind base case.)
The question is this: is the above result true for all artin local rings $k$, or even just the ring of dual numbers over a field? Or can one give a counterexample? Since monic homomorphisms between finite type groups over an artin ring are closed immersions, an equivalent formulation which may be more vivid is: does $G$ admit a (functorially) faithful linear representation on a finite free $k$-module?
(I originally thought I needed an affirmative such result over artin local rings to prove a certain general fact for smooth affine group schemes over noetherian rings, but eventually that motivation got settled in another way. So for me it is now an idle question, though I think a very natural one from the viewpoint of deformation theory of smooth linear algebraic groups.)
It sounds like the sort of thing which must have been thought about back in the 1960's when SGA3 was being written, so I mentioned the question to a couple of the SGA3 collaborators as well as some other experts in these matters. Unfortunately nobody whom I have asked knew one way or the other, even for the dual numbers. One of them suggested a couple of days ago that I should "advertise this problem; it is very provocative." Fair enough; I suppose this kind of advertising on MO is OK.
| https://mathoverflow.net/users/3927 | Smooth linear algebraic groups over the dual numbers | This is not a direct answer to the question for a general group scheme $G \to S$ and I am not an expert in this area. However, I would like to point out that the resolution property of stacks is a natural condition that appears in this context of Hilbert's 14th problem by work of R. W. Thomason:
*Equivariant resolution, linearization, and Hilbert's fourteenth problem over arbitrary base schemes
Advances in Mathematics 65, 16-34 (1987)*
Once and for all let $ \pi \colon G \to S$ be an affine, flat, finite type group scheme over a noetherian and separated base scheme $S$.
Recall, that a noetherian algebraic stack has the *resolution property* if every coherent sheaf is a quotient of a vector bundle (a locally free sheaf, which will be always assumed to be of finite and constant rank).
Therefore, the classifying stack $B\_S G$ has the resolution property if and only if every coherent $G$-comodule on $S$ is the equivariant quotient of some locally free $G$-comodule. The latter is the definition of the *$G$-equivariant resolution property* of $S$.
What we need is his Theorem 3.1:
$G \to S$ can be embedded as a closed subgroup scheme of $GL(V)$ for some vector bundle $V$ on $S$ if $B\_S G$ has the resolution property. If $S$ is affine, $V$ can be taken to be free.
Thomason does not say that the converse to Theorem 3.1. also holds. I guess that this is true if $S$ is affine, but as I am always getting confused while working with comodules, I cannot give a rigorous proof at the moment.
Nevertheless, it is worth to ask when $B\_S G$ has the resolution property. Thomason proved this in the following cases:
1. $S$ regular and dim $S \leq 1$,
2. $S$ regular; dim $S \leq 2$; $\pi\_\* O\_G$ is a locally projective $O\_S$-module, e.g, if $\pi \colon G \to S$ is smooth and with connected fibres.
3. $S$ regular or **affine** or has an ample family of line bundles; $G$ a reductive group scheme which is either split reductive, or semisimple, or with isotrivial radical and coradical, or over a normal base $S$.
In particular, if $S$ is the the spectrum of the ring of dual numbers, then this provides an affirmative answer to the posted question if $G \to S$ satisfies the conditions in (3).
Even for $G \to S$ arbitrary with reduction $G\_0 \to S\_0$, we know that the reduction $X\_0=B\_{S\_0}G\_0$ of $X= B\_S G$ has the resolution property by (1).
So we may reformulate the original question as follows:
(Q2) Is the resolution property preserved under the first order deformation $X\_0 \to X$?
Lifting of various locally free resolutions from $X\_0$ to $X$ is probably not the best approach. However, it suffices to lift a *single* locally free sheaf.
Let us see, why this is true.
A noetherian algebraic stack with affine diagonal has the resolution property if and only if there exists a vector bundle $V$ whose associated frame bundle has quasi-affine total space.
The normal case was proven by Totaro in
*The resolution property for schemes and stacks.
J. Reine Angew. Math. 577 (2004), 1--22.
14A20 (14C35)*
and in my thesis, I am currently working on, I show that this really holds for non-reduced stacks too.
Therefore if we can lift $V\_0$ from $X\_0$ to a vector bundle $V$ on $X$, then $V$ has still quasi-affine frame bundle as its reduction is quasi-affine.
The obstruction for this lies in $H^2(X\_0, I \otimes V\_0^\vee \otimes V\_0)$ where $I$ is the coherent ideal of order two defining the deformation $X\_0 \to X$. Probably, the ideal can be removed here with some tricks.
In our case this cohomology boils down to the second group cohomology of the $G\_0$-representation $I \otimes V\_0^\vee \otimes V\_0$. In particular, if $G\_0 \to S\_0$ is linearly reductive, the obstruction is zero.
Therefore we have proven:
If $G \to S$ is a group scheme over an artinian base with linearly reductive special fibre, then $G \to S$ can be embedded into some $GL\_{n,S}$ as a closed subgroup scheme.
Clearly this still leaves out interesting cases and probably this can be proven more directly avoiding stack theory.
| 11 | https://mathoverflow.net/users/4101 | 22955 | 15,122 |
https://mathoverflow.net/questions/22953 | 12 | Let $0 < a < 1$ be fixed, and integer $n$ tends to infinity. It is not hard to show that the number of integers $k$ coprime to $n$ such that $1\leq k\leq an$ asymtotically equals $(a+o(1))\varphi(n)$.
The question is: what are the best known estimates for the remainder and where are they written?
Many thanks!
| https://mathoverflow.net/users/4312 | distribution of coprime integers | Vinogradov, I. M. An introduction to the theory of numbers, Ch. 2, problem N 19. It gives error term $O(\tau(n))$. But direct application of inclusion-exclusion principle gives $O(2^{\omega(n)})$ (where $\tau$ is the number of divisors, and $\omega$ is the number of prime divisors. Standart solution with Mobius function also gives $O(2^{\omega(n)})$ if one take into account that $\mu(p^2)=0.$
| 10 | https://mathoverflow.net/users/5712 | 22959 | 15,125 |
https://mathoverflow.net/questions/22943 | 3 | This question comes from reading Washington's proof of Iwasawa's theorem, and wanting to learn the commutative algebra version of the classification of finitely-generated $\Lambda$-modules. I went to the reference in Serre, and am now hung up on a point in his classification.
---
Let $A$ be a regular local ring of dimension $2$, and let $M$ be a finitely-generated, torsion-free $A$-module. $M$ is reflexive if $M = M^{\*\*}$, or equivalently, if $M = \cap\_P M\_P$, where the intersection ranges over all prime ideals of height $1$. The detail that I am hung up on is the following:
>
> $M$ reflexive implies that it is free.
>
>
>
---
Serre is kind enough to give a proof of this, however, he uses confusing notation which is contained in the lecture notes from a 1955 lecture in Tokyo. For completeness, his proof is as follows (I will try to stay close to the original French, odd word choices are due to the French, and might be insightful): One can plunge $M$ into a free module $N$ of the same rank. $M$ being reflexive signifies that that, in the primary decomposition of $M$ in $N$, the prime ideals that intervene are of height $1$. One then has that $\text{codh}(N/M) \geq 1$, then $\text{dh}(N/M) = \text{dim}(A) - \text{codh}(N/M) \leq 1$; since $N$ is free, one has $\text{dh}(M) = \text{dh}(N/M) - 1 \leq 0$, which signifies that $M$ is free.
My guess on the dh is that it denotes projective dimension (I am sure that it stands for some form of homological dimension, based on the name of the paper). That would enable concluding that $M$ is free if the projective dimension is $0$. The prime decomposition fact is proven in Bourbaki (Commutative Algebra, Chapter 7, Section 4, Subsection 2, Proposition 7 i)). The confusing thing about that interpretation is that it seems to imply that the co-projective dimension is easier to get than the projective dimension. Also, where does the equality between the dimension of $M$ and $N/M$ come from (idea: the projective dimension of $N/M$ is at most one, and this is a start of a projective resolution of $N/M$, so the completion indicates that $M$ has dimension one fewer than $N/M$. However, why does it follow that every projective resolution has the same length, or if that isn't true, why is this a start of a minimal projective resolution?).
Also, I will give a proof that a friend gave me. Use the embedding of $M$ into $N$, and look at a regular sequence for $N$. This gives rise to a regular sequence of $M$. But then, since $N$ is free, the regular sequence for $M$ is of length at least $2$, so the projective dimension of $M$ is at most $0$, which is again what we wanted.
However, that only uses $M$ is torsion-free, and not reflexive. If there is no error in the above proof, then one can easily show that the maximal ideal is free for a sample counter-example to that statement.
---
Finally, some remarks are in order for how I am using this. I am using this in the case that $A = \mathbb{Z}\_p[[T]]$, so a proof specific to that ring wouldn't be too troubling (while it is true in general, I would only be generally disturbed without a complete proof). Also, I only really need that if $M$ is torsion-free, and $M/f$ is finite, then $M$ is trivial (applied to the case $f(T) = ((1+T)^{p^n} - 1)/T$). Finally, I am trying to get around having to understand a long matrix bash (on the order of 5 pages) in order to understand this theorem.
| https://mathoverflow.net/users/5473 | Reflexive modules over a 2-dimensional regular local ring | Hello,
I guess that $\mathrm{codh}$ actually means $\mathrm{depth}$, that is the length of a maximal regular
sequence on a module.
Then $\mathrm{codh}(N/M)\geq 1$ is a consequence of the fact that the maximal ideal of $A$ does not annihilate the module due to reflexivity.
The next inequality is Auslander-Buchsbaum.
Finally $\mathrm{dh}(M)=\mathrm{dh}(N/M)-1$ is a standard fact from homological algebra: if for some module $Q$ $\mathrm{dh}(Q)\leq d$ and $0\rightarrow S\rightarrow P\_{d-1}\rightarrow\ldots\rightarrow P\_0\rightarrow Q\rightarrow 0$ is exact with projective $P\_k$, then $S$ is projective too. Apply this to $0\rightarrow M\rightarrow N\rightarrow N/M\rightarrow 0$.
H
| 2 | https://mathoverflow.net/users/3556 | 22961 | 15,127 |
https://mathoverflow.net/questions/22908 | 36 | A few months ago, I was curious about some properties of Maass cusp forms, of nonabelian arithmetic origin. As a result, I went through a somewhat predictable process of finding a totally real $A\_4$ extension of $Q$, lifting the resulting projective Galois representation to an honest Galois representation, and writing a short program to compute as many coefficients of the Artin L-function (thus coefficients of the Maass form) as needed.
Well, as often happens, I didn't find anything particularly surprising in the end.
But now I "have a Maass form". Its a pretty Maass form -- the simplest one of eigenvalue 1/4, of "nonabelian" origin (not arising from a dihedral Galois representation). Its conductor is 163 -- a very attractive prime number (though its appearance here seems coincidental). Some class number 1 coincidences make the computation of its coefficients extremely quick and simple.
So, does anyone want the Maass form (i.e. code to output coefficients quickly)? It's fun to play with, and doesn't take up too much space. I guarantee its modularity. If not, any suggestions where to put it (a little journal that publishes such cute examples)?
| https://mathoverflow.net/users/3545 | Does anyone want a pretty Maass form? | [these are comments, not an answer, but there were too many for the comments box]
Hey---I wrote that code too! I did it to teach myself "practical Maass forms". I wrote in pari, not sage. I didn't do the example you did. Here's what I did, for what it's worth. First I tried a dihedral example. I used the Hilbert class field of $\mathbf{Q}(\sqrt{145})$; the class group is cyclic of order 4, giving a $D\_8$ extension of $\mathbf{Q}$ with a faithful 2-dimensional representation. It's an easy exercise in factoring polynomials mod $p$ to compute traces of Frobenius, and I got the Hecke eigenvalues with little trouble.
But here's the big question: how do you know you got them right? Here's how I did it. I computed the first 200,000 Hecke eigenvalues, created the formal power series defining a function on the upper half plane as per usual, and then I evaluated it to many decimal places at lots of random points $z$ and $\gamma.z$ with $\gamma$ in the level. In all the cases I tried, the answers were the same to within experimental error. I concluded that probably I'd got everything working.
I also did an $S\_3$ extension (the class group of $\mathbf{Q}(\sqrt{79})$) and a non-algebraic example coming from a Grossencharacter---this one had conductor 8 but eigenvalue not $1/4$.
I also tried to do an $A\_5$ example! As you probably know from the $A\_4$ example, an issue that needs resolving here is that the image of Galois in $GL\_2(\mathbf{C}$) isn't $A\_4$, it's the central extension, so you need to know how primes split in that last extension, which is computationally more expensive. Bjorn Poonen showed me a wonderful trick though, so I could do it. I computed $a\_n$ for hundreds and thousands of $n$, built the function on the upper half plane, and checked to see if it was invariant by the level group. It wasn't :-( I concluded that either the Langlands program was wrong or my code was wrong, and I had a good idea which. [EDIT May 2012: for what it's worth I did actually get the code working in the end (in April 2011 in fact) -- my code *was* wrong (stupid error: all the "hard" code was fine but I had miscalculated $a\_{1951}$!), and Langlands' programme still looks fine. AFAIK one still cannot prove that the candidate Maass form whose power series expansion I can compute a very long way is actually a Maass form.]
Here is the pari script for the 145 example, by the way:
```
N=200000;
f=x^4 - x^3 - 3*x^2 + x + 1;
ap(p)=if(p==5,-1,if(p==29,-1,if(issquare(Mod(p,5))&&issquare(Mod(p,29)),2*(matsize(factormod(f,p))[1]-3),0)));
chi(n)=kronecker(n,145);
v=vector(N,i,0);
v[1]=1;
for(i=2,N,fac=factor(i);k=matsize(fac)[1];\
if(k>1,v[i]=prod(j=1,k,v[fac[j,1]^fac[j,2]]),\
if(fac[1,2]==1,v[i]=ap(i),\
p=fac[1,1];e=fac[1,2];v[i]=v[p]*v[p^(e-1)]-chi(p)*v[p^(e-2)]))\
);
F(z)=local(x,y,M);x=real(z);y=imag(z);M=ceil(11/y);if(M>N,error("y too small."));sqrt(y)*sum(n=1,M,if(v[n]==0,0,v[n]*besselk(1e-30*I,2*Pi*n*y)*cos(2*Pi*n*x)))
```
That's it! It's pretty self-explanatory. ap(p) returns the coefficient $a\_p$ of the form. chi is the character of the form. If you run this the computer will pause for a few seconds while it computes the first 200,000 coefficients of the Maass form. After that it will give you a function $F$ on the upper half plane, which is defined by a Fourier expansion, and the miracle will be that it will be $\Gamma\_1(145)$-invariant. For example, after running the code above, you can try this:
```
gp > z=-0.007+0.08*I
%5 = -0.007000000000000000000000000000 + 0.08000000000000000000000000000*I
gp > F(z)
%6 = 0.2101332751524672135753981488 + 0.E-30*I
gp > F(z/(145*z+1))
%7 = 0.2101332751524672135753981489 + 0.E-30*I
gp > %6-%7
%8 = -5.67979851 E-29 + 0.E-30*I
```
What this says is that for $z$ a random element of the upper half plane such that $z$ and $\gamma.z$ both have imaginary part which is not too small (if the im part is too small you need more Fourier coeffts), where here $\gamma=(1,0;145,1)$, $F$ evaluates, to within experimental error, to the same value at $z$ and $\gamma.z$.
Note that Marty's example is of $A\_4$ type, so more interesting than this example, but a theorem of Langlands tells us that Marty's example really will be a cusp form.
| 13 | https://mathoverflow.net/users/1384 | 22963 | 15,128 |
https://mathoverflow.net/questions/22975 | 14 | A quote from Wikipedia's article on the [Rotation group](http://en.wikipedia.org/wiki/Rotation_group):
>
> Consider the solid ball in $\mathbb{R}^3$ of
> radius $\pi$ [...].
> Given the above, for every point in
> this ball there is a rotation, with
> axis through the point and the origin,
> and rotation angle equal to the
> distance of the point from the origin.
> The identity rotation corresponds to
> the point at the center of the ball.
> Rotation through angles between $0$ and
> $-\pi$ correspond to the point on the same axis and distance from the origin but
> on the opposite side of the origin.
> The one remaining issue is that the
> two rotations through $\pi$ and through $-\pi$
> are the same. So we identify [...] antipodal points on the
> surface of the ball. After this
> identification, we arrive at a
> topological space homeomorphic to the
> rotation group.
>
>
>
So far, so good. This illustrates $SO(3)\cong \mathbb{RP}^3$.
>
> These identifications illustrate that
> $SO(3)$ is connected but not simply
> connected. As to the latter, in the
> ball with antipodal surface points
> identified, consider the path running
> from the "north pole" straight through
> the center down to the south pole.
> This is a closed loop, since the north
> pole and the south pole are
> identified. This loop cannot be shrunk
> to a point, since no matter how you
> deform the loop, the start and end
> point have to remain antipodal, or
> else the loop will "break open".
>
>
>
I believe that $SO(3)$ is connected but the "intuitive argument" for $\pi\_1(SO(3))\neq 0$ is not clear to me: The starting point at the "north pole" is a rotation of $\pi$ counterclockwise around the $z$ axis. This agrees with the "south pole", a rotation of $\pi$ clockwise around the $z$ axis. So the described path is a full $2\pi$ rotation counterclockwise around the $z$ axis, stating not in the identity position. Why isn't this homotopic to the trivial path? Antipodal points are identified, so what does "start and end point have to remain antipodal, or else the loop will "break open"" mean?
| https://mathoverflow.net/users/5716 | How to demonstrate $SO(3)$ is not simply connected? | A loop is homotopically trivial if it can be continuously deformed to the constant loop. This means that at every step of the deformation (every "instant in time") you still have a loop. There are two kinds of loops on the unit ball with antipodal identifications in the boundary: either it's also a loop in the ball (without identifications) or else it starts and ends at antipodal points. The example curve in the question is of the latter kind. It seems intuitive that any continuous deformation of this curve which remains closed has to still connect antipodal points, since you cannot move the ends closer to each other -- which is what you'd have to do in order to get a contractible loop -- while keeping it a closed curve.
| 12 | https://mathoverflow.net/users/394 | 22977 | 15,138 |
https://mathoverflow.net/questions/22985 | 0 | The question looks like an exercise in elementary algebraic topology, but I didn't manage to solve it. I am considering this question because it is a toy example in a problem I'm thinking about.
Let's consider the natural embedding $GL\_n(\mathbb C) \to \mathbb C^{n^2} \backslash \{0\}$. As was discussed [in this question](https://mathoverflow.net/questions/18677/cohomology-rings-of-gl-nc-sl-nc), cohomology with rational coefficients of $GL\_n(\mathbb C)$ is an exterior algebra on generators in degrees 1, 3, ..., 2n-1 (one generator in each degree), whereas $\mathbb C^{n^2} \backslash \{0\}$ is homotopy equivalent to a sphere $S^{n^2-1}$.
I'd like to prove that the map induced on cohomology of degree $n^2 - 1$ is a zero map.
Any ideas?
Thanks.
| https://mathoverflow.net/users/2260 | Natural embedding GL_n(C) -> C^{n^2} \ {0} induces zero on cohomology | $\mathbb C^{n^2} \smallsetminus {0}$ is homotopy equivalent to $S^{2n^2-1}$, not $S^{n^2-1}$, and $\mathrm H^{2n^2-1}(GL\_n(\mathbb C)) = 0$.
| 11 | https://mathoverflow.net/users/4790 | 22988 | 15,144 |
https://mathoverflow.net/questions/22984 | 18 | Let $K$ be a compact oriented 3-dimensional handlebody of genus $g$. The group $H\_g$ of isotopy classes of diffeomorphisms of $K$ is called the *handlebody group*. (It embeds as a subgroup of the mapping class group of the genus $g$ surface $\partial K$.) The fundamental group of $K$ is a free group of rank $g$, so there is a homomorphism
$H\_g \to Out(F\_g).$
I've been thinking about this homomorphism and its kernel, and I've come to suspect that the kernel is generated by Dehn twists around curves in $\partial K$ that bound discs in $K$. These elements are all clearly contained in the kernel, but do they generate the entire kernel?
Does anyone know of a reference, proof, or counter example?
| https://mathoverflow.net/users/4910 | The kernel of the map from the handlebody group to Outer automorphisms of a free group | This result is due to Luft; see "Actions of the homeotopy group of an orientable 3-dimensional handlebody".
McCullough, in "Twist groups of compact 3-manifolds", proves that the twist group is not finitely generated and gives further references.
EDIT. Ninja'ed (well, at least the first sentence).
| 12 | https://mathoverflow.net/users/1650 | 23006 | 15,152 |
https://mathoverflow.net/questions/23003 | 5 | These are parametrized by $H^1(Gal(\mathbb{Q}), Aut X)$, where X is *some* $\mathbb{Q}$-model of the curve.
It was established in [Confusion about how the first cohomology classifies torsors](https://mathoverflow.net/questions/22907/confusion-about-how-the-first-cohomology-classifies-torsors)
that fiber bundles over $B$ with fiber $F$, structure group $G$ and transition maps with property $P$ are classified by $T$-torsors, where $T$ is the sheaf on $B$ of functions to $G$ with property $P$. $T$-torsors, in turn, are classified by $H^1(B, T)$.
Is there a way to interpret the aforementioned classification of $\mathbb{Q}$-models of a curve in these terms?
| https://mathoverflow.net/users/5309 | Understanding different Q-models of a curve over C | Yes.
Roughly, the idea is that once your base object $X/B$ is fixed then for any object $Y$ you can consider the sheaf $Iso(X,Y)$ of isomorphisms from $X$ to $Y$. This has an action of the sheaf $Aut(X)$ and on any open cover $U$ where $Y|\_U \cong X|\_U$, making a choice of such an isomorphism gives an isomorphism of $Aut(X)(U)$-sets $Iso(X,Y)(U) \cong Aut(X)(U)$.
The difference from the standard argument in this case is that you need to use the etale topology instead of the Zariski topology. The etale covers of $\mathbb{Q}$ are finite field extensions $F$ and so the statement that $Y$ is "locally" equivalent to $X$ means that $Y\_F \cong X\_F$ for a large enough field extension $F$.
| 4 | https://mathoverflow.net/users/360 | 23015 | 15,158 |
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