mg.metric geometry - Best fit for multiple shapes inside an area

Where is everybody? Only one answer, which hasn't received any upvotes, nor any downvotes. I don't know any way to put this problem back on the radar screen other than posting an answer to it, so here's an extended version of the answer I already posted.

Garey and Johnson, Computers and Intractability, has a list of problems that are known to be NP-complete. Subset Sum is problem SP13 in that list, on page 223 of the book. I quote:

Instance: Finite set $A$, size $s(a)in{bf Z}^+$ for each $ain A$, positive integer $B$.

Question: Is there a set $A'subseteq A$ such that the sum of the sizes of the elements in $A'$ is exactly $B$?

Now let's look at a special case of the current MO question. Suppose that the multiple shapes are rectangles with base 1 and various integer heights, and the (big) rectangular area also has base 1 and integer height B. Suppose that there is an efficient way to determine the best fit of the shapes in the big rectangle (I am reinterpreting the original request for a "formula" as a request for an efficient method). Then you have an efficient way to solve Subset Sum. Namely, apply the alleged efficient best fit method to the data. If it finds a way to completely fill the big rectangle, then it has found a subset summing to $B$, and if it doesn't find a way to completely fill the big rectangle, then it has proved that there is no subset summing to $B$.

It follows that there is no efficient method for finding a best fit in this special case of the original problem, unless P = NP.

ag.algebraic geometry - Existence of (smooth) models

I'm happy to present my example of a smooth projective surface $X$ over
$K=mathbb{Q}_p$ ($p$ prime) such that $X(K)neqemptyset$, whose
$l$-adic cohomology groups are unramified (for all primes $l$) and which still
has bad reduction : there is no smooth $mathbb{Z}_p$-scheme whose generic
fibre is $X$. (The method works for any finite extension of $mathbb{Q}_p$ and was worked out a few years ago.)

The surface $X$ is going to be a conic bundle over $mathbb{P}_1$ with four
degenerate fibres, so it is a rational surface in the sense of being $bar
K$-birational to $mathbb{P}_2$. It will be clear that the example is not
isolated.

If $p$ is odd, let $dinmathbb{Z}_p^times$ be a unit which is
not a square, and take $d=5$ if $p=2$, so that $K(sqrt{d})|K$ is the unramified quadratic extension.

Let $e_1, e_2$ be two distinct units of $K$. We take $X$ to be the surface in
$mathbb{P}({cal O}(2)oplus{cal O}(2)oplus{cal O})$ (coordinates $y:z:t$)
over $mathbb{P}_1$ (coordinates $x:x'$) defined by the equation
$$
y^2-dz^2=xx'(x-e_1x')(x-e_2x')t^2.
$$
I claim that this $X$ has all the properties stated above, if $v_p(e_1-e_2)>0$.

First, $X(K)neqemptyset$ because each degenerare fibre is a pair of
intersecting lines conjugated by $mathrm{Gal}(bar K|K)$.

Secondly, the $l$-adic cohomology is unramified because the action of
$mathrm{Gal}(bar K|K)$ on the Picard group $mathrm{Pic}(bar{X})$ of $bar
X=Xtimes_Kbar K$ factors via the quotient $mathrm{Gal}(K(sqrt{d})|K)$.

Finally, $X$ has bad reduction because its Chow group $A_0(X)_0$ of $0$-cycles of degree
$0$ is $mathbb{Z}/2mathbb{Z}$ (cf. prop. 1 of arXiv:math/0302156), and a theorem of Bloch (th. 0.4, On the Chow groups of certain
rational surfaces, Annales scientifiques de l'École Normale Supérieure, Sér.
4, 14 no. 1 (1981), p. 41-59, available at Numdam) asserts that if a conic bundle has good
reduction, then its Chow group of $0$-cycles of degree $0$ is $0$.

Addendum (in response to a question in an email I received). One can show moreover that no smooth projective surface $Y$ over $mathbf{Q}_p$ which is $mathbf{Q}_p$-birational to $X$ can have good reduction. This follows from the facts recalled above and the theorem of Colliot-Thélène and Coray (which can be found in Fulton's Intersection theory) : $A_0(Y)_0$ is isomorphic to $A_0(X)_0$.

ac.commutative algebra - sum of radical ideals

Let $A$ be a commutative ring and endow the closed subsets of $mathrm{Spec}(A)$ with the Grothendieck topology of finite covers. One may ask if the presheaf $V mapsto A/I(V)$ is a sheaf. This is not true in general and is related (but not equivalent) to the following pure algebraic question:

In which commutative rings $A$ are the radical ideals closed under sum?

The property can be checked locally. It holds in dimension $0$, and also for integral domains of dimension 1. It doesn't hold for the $2$-dimensional ring $k[x,y]$ (consider $(x^2 + y)+(y) = (x^2,y)$), nor for the 1-dimensional ring $bigl(k[x,y]/(x^2 y + y^2)bigr)_{(x,y)}$.

Are there other interesting examples/counterexamples or approaches for a general classification? I think the property has some algebro-geometric interpretation: All intersections of closed subschemes are transversal. See also SE/322872.

gt.geometric topology - Presentation of the monoid of surfaces

I agree with Tom and Ryan that it is worthwhile to learn proofs of the classification of surfaces. I think that the result get a bit of a bad rap since the "standard" combinatorial proof that everyone used to learn (which appears in Seifert-Threlfell's book and Massey's book) is complicated and unenlightening. However, there are now a number of nicer proofs available. Here are a few of my favorites.

1) If you like Morse theory, there is a nice proof in Hirsch's book on differential topology.

2) There is a slick combinatorial proof in Armstrong's book "Basic Topology". I believe that this is the source for the proof mentioned above by Tom Church that Benson Farb likes to give.

3) In Fomenko-Matveev's book "Algorithmic and Computer Methods for Three-Manifolds", there is a nice proof using handle decompositions.

4) There is finally John Conway's "ZIP proof", which was written up by Francis and Weeks in their paper "Conway's ZIP Proof".

All of these proofs assume that the surface has been equipped with either a triangulation (for numbers 2-4) or a smooth structure (for 1). For nice approaches to this, see the answers to my question here. However, when you are first approaching these types of results, I would recommend just assuming that the surfaces can be triangulated or smoothed.

Why fourier transform tell us energy of any frequency of f(t)

In signal processing, the energy of a continuous-time signal is defined as the square of its $L^2$ norm. Hence, the spectral energy of this signal is the square of the $L^2$ norm of this signal in the spectral domain, i.e. of its Fourer transform. By Parceval's theorem, these two energies are equal. But as a consequence, $|x(t)|^2$ is the energy density of the signal at the moment $t$, and $|F(x)(tau)|^2$ is the spectral energy density at the frequency $tau$.

This is in complete analogue with the discrete case: in your notation, $a_n^2 + b_n^2$ is equal (up to a multiplicative constant) to the square of the absolute value of $int f(t) e^{-i theta t} dt$ for $theta=n$ (for the real-valued $f$).

gn.general topology - Freeing a sphere from within a sphere

For your first question, look at complements (a "fundamental" technique in analyzing ambient isotopies/ambient homeomorphisms; e.g. the "knot group"):

Let the union of the initial two spheres be $S$, and the union of the final two spheres be $T$. An isotopy on $mathbb{R}^3$ taking $S$ to $T$ in particular ends with a homeomorphism from $mathbb{R}^3$ to itself taking $S$ to $T$, hence producing a homeomorphism from $U=mathbb{R}^3setminus S$ to $V=mathbb{R}^3setminus T$. But $U$ has only one of its three connected components contractible (the inside of the inner sphere), whereas $V$ has two of its three components contractible (the insides of the two spheres), a contradicting that they be homeomorphic.

$Big($One way to prove the other components are not contractible is that they have non-trivial second homotopy and homology groups, as exhibited by the elements represented by the spheres themselves.$Big)$

For your third (and second) question, in generality, here's an argument that there is no embedding from $M=S^{n-1}times I$ to $mathbb{R}^n$ with boundary ambient isotopic to the "side-by-side" embedding of
$S_0sqcup S_1$. By any letter $S$ I denote a copy of $S^{n-1}$. For comfort of the imagination, think $n=2$:

(1) Up to ambient isotopy, there are only three embeddings of $S_0sqcup S_1$ in $mathbb{R}^n$ : the "avocado$^+$ type" embeddings $A^+$ with $S_0$ inside $S_1$, the "avocado$^-$ type" embeddings $A^-$ with $S_1$ inside $S_0$, and the "side-by-side type" embeddings $B$ with $S_0$ and $S_1$ "next to" each other.

$Big($That they fall into these three types is just a case analysis; that any two embeddings in the same "type" are ambient isotopic is also easy: (a) first translate and dilate until the $S_1$ embeddings match up, and then being in the same case means the two $S_2's$ are embedding in the same component of the compliment of $S_1$, so (b) you can move them around in that component until they match up, too.$Big)$

(2) $A^pm$ are not ambient isotopic to $B$, by counting contractible components of their complements, as in the answer to your first question: $A^pm$ has one, $B$ has two.

(3) An embedding from $M=Stimes I$ to $R^n$ cannot have $partial M = B$ up to ambient isotopy.

$Big($ Why? Here's one proof, which I found kind of fun: Suppose $partial M = B$, i.e. is in the "side-by-side" arrangement. Since $M$ is connected, $int(M)=Msetminus partial M$ must lie outside both spheres of $partial M$, since if it lies inside one, it can't touch the other. Let $x$ be the center of of the $S_1$ component of $partial M$. Now $M$ is compact, so it has a point $p$ which is of maximal distance from $x$. Such a $p$ must be a boundary point of
$M$, since at an interior point, we could move slightly in a direction away from $x$ and stay inside $M$. But $p$ can't be on $S_1$ or $S_2$ either, since their outer sides are "padded" by the open set $int(M)$... this a Euclidean geometry argument which I'll stop rigorizing here. So $p$ is nowhere, a contradiction.$Big)$

(4) By process of elimination, $partial M$ must be ambient isotopic to $A^pm$, which by (2) is not ambient isotopic to the "side-by-side" embedding $B$, which is what you wanted.

ag.algebraic geometry - Examples of divisors on an analytical manifold

Hartshorne is the reference where you can find the following example which might be useful.
I what follow everything is with multiplicity. Now Alberto pointed out above the case of the divisor over $mathbb{P}^1$ associated to its "tangent bundle": Two points over the sphere counted with multiplicity (from here though, it is not hard to believe that the Chern class of such a bundle is going to be 2). Notice that these two points are given by zeros of polynomials of degree two defined over the sphere. I think nothing stops you taking now polynomial of degree 3, 4 and so on. Then what we get are nothing but 3, 4 points over the sphere: Divisors of degree 3, 4 and so on. We can do something similar over all the curves (Riemann Surfaces) and what we get are divisors: points with labels. Such labels are the multiplicities. Chapter IV Hartshorne. or Klaus-Hulek: Elementary Algebraic Geometry.

Now, let's take a look at divisors over the surface $mathbb{P}^2$: they are algebraic curves (Riemann Surfaces). Do not get confused please by the name Surface here. Applying the same argument as before, a divisor of degree two is going to be the zero locus of polynomials of degree 2: conics. Same for degree three (cubics), four (quartics), and so on and so forth. For instance, in degree two we might have the divisor $C=([x:y:z]in mathbb{P}^2| x^2+y^2=z^2)$. Deshomogenizing with $H=[z=1]$ you get a perfect polynomial $x^2+y^2=1$ which defines the intersection $Hcap C$. This is how your global divisor $C$ looks like locally.

Now taking a family of divisors of degree two, the conics, it is well known that the space of embeddings of conics in $mathbb{P}^2$ is (the linear system) $mathbb{P}^5$. We get this by considering the coefficients in the equation $ax^2+by^2+cz^2+dxy+exz+fyz=0$ as coordinates in $mathbb{P}^5$. Notice that we get the following map out of the previous considerations, $$phi:mathbb{P}^2rightarrow mathbb{P}^5$$ given by $[x:y:z]mapsto [x^2:y^2:z^2:xy:xz:yz]$. Here pencils are a subfamily of conics in the complete linear system given above with a certain property (find out which one). However, we can consider the following subfamily of conics: all those conics passing through a fixed point in $mathbb{P}^2$. This is nothing but a hyperplane $H$ in $mathbb{P}^5$. We can even consider $phi(mathbb{P}^2)cap H$. This is going to be a divisor on $mathbb{P}^2cong phi(mathbb{P}^2)$. Guess which one?. Hartshorne II section 7.

One can apply the the ideas with zero locus of polynomials of degree three: Divisors of degree 3 in $mathbb{P}^2$. These were given the name of elliptic curves. (did someone say that in considering such curves, we find the divisor associated to the canonical bundle of $mathbb{P}^2$?). We can go on with the degree and getting divisors on the projective plane of higher degree. These were only examples of divisors on $mathbb{P}^2$. Notice that all of them have a nontrivial topology and geometry. This fact is not a coincidence and the book of HG argues in this direction in Chapter zero.

fa.functional analysis - Finite Rank Commutators

My former student Detelin Dosev and I are interested in classifying the commutators in $L(X)$, the bounded linear operators on the Banach space $X$ (see our joint paper on my home page or the ArXiv and the references therein, and recall that $T$ is a commutator means that there are $A$ and $B$ in $L(X)$ so that $T=AB-BA$).

If $T$ is a finite rank operator in $L(X)$ with zero trace then it is classical that $T$ is a commutator. The converse is obviously true when $X$ is finite dimensional.

Conjecture. There is an infinite dimensional Banach space $X$ such that all finite rank commutators in $L(X)$ have zero trace.

If true, this conjecture is very hard to prove, but maybe it is known to be false?

We are also interested in conditions on $X$ that guarantee that all finite rank operators in $L(X)$ are commutators. If $X$ is isomorphic (meaning linearly homeomorphic) to its closed subspaces of codimension one, then rank one projections in $L(X)$ are commutators and hence every finite rank operator in $L(X)$ is the sum of two commutators, but we do not know whether every finite rank operator in $L(X)$ is a commutator. If also $X$ contains a complemented subspace isomorphic to $X oplus X$, then every diagonalizable finite rank operator in $L(X)$ is a commutator, but we still don't know whether all finite rank operators in $L(X)$ are commutators. Are we missing something easy?

A Banach space $X$ is said to have a Pelczynski decomposition provided $X$ is isomorphic to $(X oplus X oplus X oplus ...)_p$ for some $1 le p le infty$ or $p = 0$. A general result is that if $X$ admits a Pelczynski decomposition, then every compact operator on $X$ is a commutator, so standard spaces do not give examples of finite rank non commutators.

ho.history overview - How to find ICM talks?

Just recently the International Mathematical Union (IMU) put online all the previous proceedings on the ICM's!!! Here: http://www.mathunion.org/ICM/ . This webpage is based on joint work by R. Keith Dennis (Ithaca) and Ulf Rehmann (Bielefeld). So you can read for free all the articles (Including Hilbert's famous problem paper; Martin Grötschel
demonstrated it at the opening ceremony of ICM 2010). I dont know when the proceedings of ICM2010 will be added. UPDATE: the 2010 articles have now been added!

As for the talks themselves there is a page with all the ICM 2010 plenary talks here http://www.icm2010.in/from-the-venue/online-streaming-archives links for videos from earlier ICM's (plenary talks and other events) can be found here http://www.mathunion.org/activities/icm

Update (August, 6 2014) Many (46 for now) of the ICM 2014 proceedings contributions are already available on arXiv, via this search. (I got it from Peter Woit's blog.) Videos of lectures can be found here.

Update (May 2014): Starting 1992 there is also every four years the European Congress of Mathematics (ECM) that the European Mathematical Society (EMS) is running.
The proceedings of the first three ECMs are now freely open. These volumes are available at
http://ada00.math.uni-bielefeld.de/ECM/.

(Digitising the proceedings of the first three ECMs, published by Birkhäuser was a task carried out by the EMS Electronic Publishing Committee.)

Starting with the 4ECM (2004), the Proceedings are published by the EMS Publishing House. The EMS decided to make them freely available online too. I expect that this will happen soon and I will keep you posted. Further update (June 2014) The ECM Proceedings are now available on http://www.euro-math-soc.eu/ECM/! I was told that in a few months, the EMS will put also the 6ECM volume.

ag.algebraic geometry - In what topology DM stacks are stacks

Background/motivation

One of the main reason to introduce (algebraic) stacks is build "fine moduli spaces" for functors which, strictly speaking, are not representable. The yoga is more or less as follows.

One notices that a representable functor on the category of schemes is a sheaf in the fpqc topology. In particular it is a sheaf in coarser topologies, like the fppf or étale topologies. Now some naturally defined functors (for instance the functor $mathcal{M}_{1,1}$ of elliptic curves) are not sheaves in the fpqc topology

(actually $mathcal{M}_{1,1}$ is not even an étale sheaf) so there is no hope to represent them.

Enters the $2$-categorical world and we introduce fibered categories and stacks. Many functors which are not sheaves arise by collapsing fibered categories which ARE stacks, so not all hope is lost. But, as not every fpqc sheaf is representable, we should not expect that every fpqc stack is in some sense "represented by a generalized space", so we make a definition of what we mean by an algebraic stack.

Let me stick with the Deligne-Mumford case. Then a DM stack is a fibered category (in groupoids) over the category of schemes, which

1) is a stack in the étale topology

2) has a "nice" diagonal

3) is in some sense étale locally similar to a scheme.

I don't need to make precise what 2) and 3) mean.

By the preceding philosophy we should expect that DM stacks generalize schemes in the same way that stacks generalize sheaves. In particular I would expect that DM stacks turn out to be stacks in finer topologies, just as schemes are sheaves not only in the Zariski topology (which is trivial) but also in the fpqc topology (which is a theorem of Grothendieck).

Question

Is it true that DM stacks are actually stacks in the fpqc topology? And if not, did someone propose a notion of "generalized space" in the context of stacks, so that this result holds?

galois representations - One dimensional (phi,Gamma)-modules in char p

OK...I think I see how to do this now. In the end, I am seeing $(p-1)^2$ distinct $(phi,Gamma)$-modules which matches well with the Galois side.

To do this, let $D$ be any 1-dimensional etale $(phi,Gamma)$-module. Let $e$ be a basis, and set $phi(e)=h(T)e$ with $h(T) in F_p((T))^times$. Write $h(T) = h_0 T^a f(T)$ with $h_0 in F_p^times$ and $f(T) in F_p[[T]]$ with $f(0)=1$.

Changing basis from $e$ to $u(T)e$ with $u(T) in F_p((T))^times$ gives
$$
phi(u(T)e) = u(T^p)h(T)e = frac{u(T^p)}{u(T)} h(T) (u(T)e).
$$
I claim one can find $u(T)$ such that $u(T)/u(T^p)$ equals any element of $1+TF_p[[T]]$. Indeed, for such an element $g(T)$, the infinite product $prod_{j=1}^infty phi^j(g(T))$ (which hopefully converges since $g(0)=1$) works.

Thus, we can change basis so that $phi$ has the form $phi(e) = h_0 T^a e$ -- i.e. we can kill off the $f(T)$ term. Further, by making a change of basis of the form $e$ goes to $T^b e$, we may assume that $0 leq a < p-1$.

Now, we use the fact that the $phi$ and $Gamma$ actions commute (which is a strong condition even in dimension 1). Namely, let $gamma$ be a generator of $Gamma$, and set $gamma e = g(T) e$. Then $gamma phi e = phi gamma e$ implies
$$
((1+T)^{chi(gamma)}-1)^a g(T) = g(T^p) T^a.
$$
Comparing leading coefficients, we see this is only possible if $a=0$ and $g(T)$ is a constant.

Thus, $D$ has a basis $e$ so that $phi(e) = h_0 e$ and $gamma(e) = g_0 e$ with $h_0,g_0 in F_p^times$ as desired.

Does this look okay? Any takers for the 2-dimensional case?

dg.differential geometry - Where does the generic triangle live?

The generic triangle lives in a subset of this function space $Axioms^{Languages}$.

What do I mean? First, let's us what is generic addition? Immediately, you would ask me, addition on which number system? The natural numbers, integers or what? Properly then, one can only say that for each (semi)ring $R$, addition is specified as some subset of $R^{Rtimes R}$.

Now I invoke this duality principle: Operations in number theory are analogous to objects in geometry. Objects in number theory are analogous to operations in geometry.

One of the reasons why this question by Six Winged Seraph is so difficult (and so interesting) is that we should think of the object "triangle" in geometry as analogous to the operation "addition" in number theory. We should not think of the "triangle" as analogous to the number "$n$". It is easy to say where the generic number "$n$" lives in: namely the set of natural numbers ${mathbb{N}}$! (or whatever number system you might be considering) This is an easy question just as where does the generic rotations, dilations and reflections live in: namely the Lie group $Isom({mathbb{R}}^2)$? (or ${mathbb{H}}^2$ or ${mathbb{S}}^2$ depending on which constant curvature space you are considering.)

The "triangle" question is hard because what you can pin down about a generic triangle is not what it is, but what you can do with it. That is, axioms. Using the duality principle again, we pin down what addition is by the axioms, not just axioms for addition, but also how multiplication interacts with it. Are you interested in statements like $23+34=57$, $n+n=2n$ etc? Only to the extent that these statements form part of the completed infinity of theorems from which can be generated by the axioms.

In a similar way, for each constant curvature plane, axioms that govern reflection, axioms that provide congruence criteria, Pappus' proof is generated from these axioms.

As a final refinement, one might also consider triangles in other non-constant curvature spaces. Just as addition can be considered for vector spaces or groups, not just rings. In this case, the full answer is that for any language in which triangles can be talked about, there is a corresponding axiomatization of what can be done with triangles.

Eg, the Pappus proof is for the first-order axiomatization of (non)-Euclidean geometry. There are different proofs for the plane as a Riemannian manifold (via integration) etc. For each of these axiomatizations, we have a different notion of triangle.

lie algebras - In what way do exact sequences of Lie ideals integrate to the category of groups?

Please excuse, very naive question:

Suppose $g$ is a topological Lie algebra over Q and $G$ = $exp(g)$ the associated group

(take free group on formal symbols $exp(X)$, X $in$ $G$, and impose all relations
formally coming from the BCH-formula).

Suppose I have a short exact sequence

$0longrightarrow b_{1}cap b_{2}longrightarrow b_{1}oplus b_{2}longrightarrow glongrightarrow 0$

of g-modules, but special in the sense that

• $g$ is the full Lie algebra,


• $b_1$, $b_2$ (and then their intersection) are supposed to be proper Lie ideals (and not just any kind of g-module)


• and $b_1 + b_2 = g$ ; this can happen if $g$ is weird enough

It seems to me (in some cases)/(always)/(never? ;-) such a sequence should induce
something like an exact sequence

1 -> A -> B -> G -> 1

(exact in the obvious classical sense, clearly groups are not an abelian category...)

where B could be something like the coproduct/free product of the normal
subgroups associated to the Lie ideals b1, b2; and A the subgroup associated
to the intersection of b1 and b2.

Is that true or is it complete nonsense? Is it trivially totally ridiculously false?

[please note, even though it may sound so, I do not want to go in
the direction of 'integrating' g-modules to G-modules, I would like to
transfer Lie algebra decompositions to 'nonlinear' group 'decompositions',
whatever that means....]

ac.commutative algebra - Which rings are subrings of matrix rings?

Kevin Buzzard informs me, by e-mail and in the above comment, that what he cares about is the case where $A$ is regular. In particular, it would be good to know whether every $T$ is a matrix ring in this case. I thought about it but didn't solve this, so here is a record of my ideas. Let $n$ be the dimension of $A$.

(1) A subring of a matrix is obviously a matrix ring. So, if we knew that $T$ was a matrix ring whenever $T$ was normal, this would establish that $T$ was always a matrix ring.

(2) As explained in my previous answer, if $T$ is free as an $A$-module, then $T$ is a matrix ring. We have the following implications: if $T$ is Cohen-Macaulay then $T$ torsion-free and finite over $A$ implies $T$ flat over $A$; if $A$ is a polynomial ring then $T$ flat over $A$ implies $T$ free over $A$. So, if $A$ is a polynomial ring and $T$ is Cohen-Macaulay, then $T$ is a matrix ring.

In particular, if $n=1$ or $2$, then $T$ normal implies $T$ Cohen-Macaulay. So, in these cases, and with $A$ a polynomial ring, $T$ is a matrix ring.

(3) As explained above, if $T^*$ is free over $A$, we also get to conclude that $T$ is a matrix ring. Unfortunately, this can fail when $n geq 3$.

(4) If there is some $T$-module $M$ on which $T$ acts without kernel, and $T$ is free as an $A$-module, then $T$ is a matrix algebra. Restated geometrically, if there is any coherent sheaf on $mathrm{Spec}(T)$, with support on the whole of $mathrm{Spec}(T)$, whose pushforward to $mathrm{Spec}(A)$ is a trivial vector bundle, then $T$ is a matrix algebra. If we restrict our attention to $A$ a local ring, or a polynomial ring, then the adjective "trivial" comes for free.

So, if there were to be a counter-example, we would want $n geq 3$ and we would want $T$ to be normal but not Cohen-Macaulay. Moreover, we would want that basically any $T$-module is not free as an $A$-module.

Any ideas?

at.algebraic topology - Homotopy type of set of self homotopy-equivalences of a surface

Hi Andy,

Here is a proof for the case with marked points (see below for some ideas for the case of closed surfaces).

Proof: straight-line homotopy.

Less tersely: let $HE_0(Sigma,ast)$ be the identity component of the monoid of self-homotopy equivalences of $Sigma$ fixing the basepoint; in particular, each $fin HE_0(Sigma,ast)$ is homotopic rel $ast$ to the identity. Fix a hyperbolic metric on $Sigma$ and thus an identification of the universal cover $widetilde{Sigma}$ with the hyperbolic plane $mathbb{H}^2$, and a basepoint $ast$ in $mathbb{H}^2$.

Each $fcolon Sigmato Sigma$ has a unique lift to $mathbb{H}^2$ fixing the basepoint. because $f$ acts trivially on $pi_1(Sigma)$, $f$ commutes with the deck transformations. Thus we may take the straight-line homotopy $f_t(x)=tx-(1-t)f(x)$, where by this convex combination I mean to move with unit speed along the geodesic from $f(x)$ to $x$. Since the deck transformations act by isometries on $mathbb{H}^2$, this homotopy descends to $Sigma$; each $f_t$ is still a homotopy equivalence. We can perform this straight-line homotopy for all $f$ simultaneously; since the lifts of $f$ are uniformly continuous, this homotopy is continuous on $HE_0(Sigma,ast)$ and gives a contraction to the identity.

There must be some work needed to get from this to the case for closed surfaces, because this proof works for a genus 1 surface with marked point, and of course $HE_0(T^2)$ is homotopy equivalent to $T^2$ itself. But it seems to me like a reduction should be possible; I think the important thing is that $pi_1(Sigma)$ is centerless.

Acknowledgement: I learned this idea from Rita Jimenez Rolland, based on conversations she had with Mladen Bestvina about the related case of $text{Aut}(F_n)$.

knot theory - What is the Alexander polynomial of a point?

According to the Baez-Dolan cobordism hypothesis, an extended TQFT is determined by its value on a single point. This value a fully dualizable object of a symmetric monoidal $n$ category (a fully dualizable object is a higher categorical analogue of a finite dimensional vector space). The Alexander polynomial is a quantum invariant, and comes from a TQFT.

How can an "Alexander polynomial" TQFT be put into an extended TQFT, and what is its value at a single point?

The question I just asked is closely related to this question. I also asked the question on the ldtopology blog here, and Theo Johnson-Freyd suggested that MO might be the place to ask it.

---

Briefly, I will summarize what an extended TQFT is. A TQFT as a symmetric monoidal functor Z:Cob(n)-> Vect(k) from the tensor category of $n-1$ dimensional manifolds and cobordisms between them to the tensor category of vector spaces over a field k. An extended TQFT is a symmetric monoidal functor Z:Cob_k(n)->C from the n-category of cobordisms to a symmetric monoidal n-category C. I vote for the introduction to Lurie's expository account of his work proving the Baez-Dolan cobordism hypothesis as the best place to read about why extended TQFT's are natural objects, to understand their motivation, and to understand why people are so excited about them. An extended TQFT assigns a fully dualizable object to a point, and a higher “trace” on this object to a closed n-dimensional manifolds.

cv.complex variables - Minimizing the modulus of a polynomial around a circle

There is another way.
Every non-negative trigonometric polynomial $f$ on the circle
is of the form $|q|^2$, where $q$ is an analytic polynomial.
(I mean by this that $f$ is of form $sum_{-N}^N a_n z^n$ and
$q(z) = sum_0^N b_n z^n$).
This is called the Fejer-Riesz theorem.

So, you guess a minimum for $|p|^2$, call it $m$, and then see
whether $f = |p|^2 - m$ is the modulus squared of a polynomial
(an algebraic identity).
If it is, try again with larger $m$; if not, reduce $m$.

For a fuller account, see the survey article by Helton and Putinar:

@incollection {MR2389626,
AUTHOR = {Helton, J. William and Putinar, Mihai},
TITLE = {Positive polynomials in scalar and matrix variables, the
spectral theorem, and optimization},
BOOKTITLE = {Operator theory, structured matrices, and dilations},
SERIES = {Theta Ser. Adv. Math.},
VOLUME = {7},
PAGES = {229--306},
PUBLISHER = {Theta, Bucharest},
YEAR = {2007},
MRCLASS = {47-02 (14P10 47A13 47A57 47A63 90C22)},
MRNUMBER = {MR2389626 (2009i:47001)},
MRREVIEWER = {Joseph A. Ball},
}

-John E. McCarthy

ag.algebraic geometry - Expressing fiber product of affines via an ideal

Let $X$ (resp. $Y$) be the affine $k$-scheme defined by the ideal $I$ (resp. $J$) in the polynomial ring $k[x_1,...x_n]$ (resp. $k[y_1,...,y_m]$).
Let $Z$ be the affine scheme defined by the ideal $L$ in $k[z_1,...z_s]$, and let $f^*:k[z]/Lrightarrow k[x]/I$ (resp. $g^*:k[z]/Lrightarrow k[y]/J$) be $k$-homomorphisms, where $x=(x_1,...,x_n)$ and so forth, corresponding to scheme morphisms $f:Xrightarrow Z$ (resp. $Yrightarrow Z$).

Then it should be possible to express the fiber product $Xtimes_{f,Z,g}Y$ via an ideal $W$ in the polinomial ring $k[x,y,z]$ [edit: actually, $W$ should be an ideal in $k[x,y]$] (where $x$ stands for the string of variables $x_1,...,x_n$, and so on).

Question: how to express $Wsubseteq k[x,y,z]$ explicitely in terms of $I$, $J$, $L$, $f^*$ and $g^*$?

Edit:
You can express things explicitely in terms of some polynomials $F_i$, $G_i$ and $H_i$ such that $I=(F_1,...,F_N)$, $J=(G_1,...,G_M)$ and $L=(H_1,...,H_S)$, and in terms of the components $(f_1,...,f_s)$ (resp. $(g_1,...,g_s)$) of $f$ (resp. $g$).

ca.analysis and odes - How to estimate the growth of a recurrence sequence

If we have a linear recurrence sequence where each term depends on all previous terms, say

$a_n = sum_{k=0}^{n-1} binom{n}{k} a_k, quad a_0 = 1$

is there any way to estimate the growth of a_n in terms of a Big-O notation?

I suppose the growth must be super-exponential, because if $a_1, ldots, a_{n-1}$ grows exponentially, say, $q^i$, then we have $a_n = (q+1)^n - q^n$. Hence The exponent grows from $q$ to $q+1$. But I am not sure if this serves as an argument.

Thanks!

ag.algebraic geometry - Reference Request for Drinfeld and Laumon Compactifications

Background

Let $X$ denote a smooth projective curve over $mathbb{C}$ and let $G$ denote a semi-simple simply connected algebraic group over $mathbb{C},$ which has associated flag variety $G/B.$

Then we can consider the variety $Maps^d(X, G/B)$ of maps from $X$ to $G/B$ of fixed degree $d$ where $d$ is an $mathbb{N}$-linear combination of coroots of $G.$ See the top of page 2 of this paper by Alexander Kuznetsov Kuznetsov for the definition of degree. The Plucker embedding of the flag variety into projective space gives an alternative formulation of $Maps^d(X, G/B)$ which can be found in section 1.2 of Kuznetsov or in this survey article of Alexander Braverman Braverman.

In general, $Maps^d(X, G/B)$ is not compact, but there is a compactification due to Drinfeld, which is referred to as the variety of quasi-maps and denoted $QMaps^d(X, G/B).$ See Kuznetsov or Braverman.

On the other hand, when $G = SL_n,$ there is a second compactfication due to Laumon. This is because when $G = SL_n,$ we have both the Plucker embedding description of the flag variety, but also the description of the flag variety as flags of vector spaces. This latter description gives another formulation of $Maps^d(X, G/B)$ but leads to a compactification known as quasi-flags. Once again, see Kuznetsov. When $n>2,$ varieties of quasi-maps and of quasi-flags are different. It turns out that quasi-flags are always smooth, while quasi-maps have singularities.

Broadening our focus somewhat, we could instead consider the representable map of stacks $Bun_B(X) to Bun_G(X),$ and note that the fiber over the trivial $G$-bundle is the union of all the $Maps^d(X, G/B)$ for all possible degrees (note that the degree just tells us which connected component of $Bun_B$ we live in).

Just as the variety of maps above was not compact, the map $Bun_B to Bun_G$ is not proper. But there exists a relative compactification of $Bun_B,$ also referred to as the Drinfeld compactification, which I will denote $Bun_B^D.$ This compactification still maps to $Bun_G,$ but the map is now proper. The fiber over the trivial bundle of this map coincides with the union of all $QMaps^d(X, G/B).$

As before, when $G = SL_n,$ there is a second compactification of $Bun_B$ which I will denote $Bun_B^L$ whose fiber over the trivial bundle coincides with the union of all the quasi-flags varieties. See this paper by Braverman and Gaitsgory BG or this follow-up paper by Braverman, Gaitsgorgy, Finkelberg, and Mirkovic BGFK for more details.

Question

In Kuznetsov, Kuznetsov proves that when $X = mathbb{P}^1$ and $G = SL_n,$ there is a map from the space of quasi-flags of degree $d$ to the space of quasi-maps of degree $d$ which is a small resolution of singularities.

Later, in BG, it is asserted that Kuznetsov proved that $Bun_B^L(X)$ is a small resolution of singularities of $Bun_B^D(X)$ for any smooth projective curve $X.$

It seems to me that there are two discrepancies here. One has to do with an arbitrary smooth projective curve versus $mathbb{P}^1.$ The second has to do with moving from the varieties of quasi-maps and quasi-flags to the stacks $Bun_B^D$ and $Bun_B^L.$

Does anyone know a reference which explains the bridge between Kuznetsov and the assertions of BG? Or perhaps this was just something clear to the experts which never warranted an explanation?

nt.number theory - about Hilbert and Siegel modular varieties (forms)

It seems to me that Hilbert modular varieties (forms) are generalization from Q to totally real fields. While Siegel modular varieties (forms) are generalization from 1 dimensional to higher dimensional abelian varieties. But they should both be some kind of Shimura variety (automorphic forms), right?

According to Milne's note of Shimura varieties, Siegel modular varieties are Shimura varieties coming from the Shimura datum (G, X) where G is the symplectic similitude group of a symplectic space (V, phi). So what is the corresponding Shimura datum for the Hilbert modular variety? Or am I asking a wrong question?

Also, in the definition of Shimura varieties G(Q)G(A_f)X/K, why we only consider Q and its adele group? why not general number fields and their adeles? (again, maybe a wrong question)

Thank you.

ag.algebraic geometry - What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem?

Added 2 June:

Since the summary below is already a bit long, I thought I'd add a few lines at the beginning as a guide. The proofs all proceed as follows:

1. Identify the quantity of interest (like the Euler characteristic) as the index of an operator going from an 'even' bundle to an 'odd' bundle.




2. Use Hodge theory to write the index in terms of the dimensions of harmonic sections, i.e., kernels of Laplacians.




3. Use the heat evolution operator for the Laplacians and 'supersymmetry' to rewrite this as a 'supertrace.'




4. Write the heat evolution operator in terms of the heat kernel to express the supertrace as the integral of a local density.




5. Use the eigenfunction expansion of the heat kernel to identify the constant (in time) part of the local density.

Most of this is general nonsense, and the difficult step is 5. By and large, the advances made after the seventies all had to do with finding interpretations of this last step that employed intuition arising from physics.

---

I suffered over this proof quite a bit in my pre-arithmetic youth and wrote up
a number of summaries. A condensed and extremely superficial version is given here, mostly for my own review.
If by chance someone finds it at all useful, of course I will be delighted. I apologize that I don't say anything about physical
intuition (because I have none), and for repeating parts of the previous nice answers.
It's been years since I've thought about these matters, so I will forgo
all attempts at even a semblance of analytic rigor. In fact, the main pedagogical reason for posting is that a basic outline of the proof is possible to understand with almost no analysis.

The usual setting has a compact Riemannian manifold $M$, two hermitian bundles $E^+$ and $E^-$, and a linear
operator
$$P:H^+rightarrow H^-,$$
where $H^{pm}:=L^2(E^{pm})$.
With suitable assumptions (ellipticity), $ker(P)$ and $coker(P)$
have finite dimension, and the number of interest is the index:
$$Ind(P)=dim(ker(P))-dim(coker(P)).$$
This can also be expressed as
$$dim(ker(P))-dim(ker(P^{*})),$$ where
$$P^{*}:H^-rightarrow H^+$$
is the Hilbert space adjoint. A straightforward generalization of the Hodge theorem allows us also to write this in terms of Laplacians
$Delta^+=P^* P$ and $Delta^-=PP^*$
as
$$dim(ker(Delta^+))-dim(ker(Delta^-)).$$
Things get a bit more tricky when we try to identify the index with the expression ('supertrace,' so-called)
$$Tr(e^{-tDelta^+})-Tr(e^{-tDelta^-}).$$
The operator
$$e^{-tDelta^{pm}}:H^{pm}rightarrow H^{pm}$$
sends a section $f$ to the solution of the heat equation
$$frac{partial}{partial t} F(t,x)+Delta^{pm}F(t,x)=0$$
($x$ denoting a point of $M$) at time $t$ with intial condition $F(0,x)=f(x).$
One important part of this is that there are discrete Hilbert direct sum decompositions
$$H^+=oplus_{lambda} H^+(lambda)$$
and $$H^-=oplus_{mu} H^-(mu)$$
in terms of finite-dimensional eigenspaces for the Laplacians with non-negative eigenvalues. And then, the identities
$$Delta^-P=PP^{*}P=PDelta^+$$
and
$$Delta^+P^{*}=P^{*}PP^{*}=P^{*}Delta^-$$
show that the (supersymmetry) operators $P$ and $P^{*}$ can be used to define isomorphisms between all non-zero eigenspaces of the two Laplacians with
a correspondence for eigenvalues as well.
Thus, once you believe that the exponential operators are trace class,
it's easy to see that the only contributions to the trace are from the kernels of the plus and minus Laplacians. This is the 'easy cancellation' that occurs in this proof.
But on the zero eigenspaces, the heat evolution operators are clearly the identity, allowing us to identify the supertrace with the index.
To summarize up to here, we have
$$Ind(P)=Tr(e^{-tDelta^+})-Tr(e^{-tDelta^-}).$$
This identity also makes it obvious that the supertrace is in fact independent of $t>0$.

The proofs under discussion all have to do with identifying this supertrace in terms of local expressions that
relate naturally to characteristic classes. The beginning of this process involves first writing the operator
$e^{-tDelta^+}$ in terms of an integral kernel
$$K^+_t(x,y)=sum_i e^{-tlambda_i } phi^+_i(x)otimes phi^+_i(y)$$
where the $phi^+_i$ make up an orthonormal basis of eigenvectors for the Laplacian.
That is,
$$[e^{-tDelta^+}f](x)=int_M K^+_t(x,y)f(y)dvol(y)=sum_i e^{-tlambda_i } int_M phi^+_i(x) langle phi^+_i(y),f(y)rangle dvol(y).$$
Formally, this identity is obvious, and the real work consists of the global analysis necessary to justify the formal computation.
Obviously, there is a parallel discussion for $Delta^-$. Now, by an infinite-dimensional version of the formula
that expresses the trace of a matrix as a sum of diagonals, we get that
$$Tr(e^{-tDelta^+})=int_M Tr(K^+_t(x,x))dvol(x)=int_M sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x),$$
an integral of local (point-wise) traces, and similarly for $Tr(e^{-tDelta^-})$. One needs therefore, techniques to evaluate the density

$$sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x).$$

More analysis gives an asymptotic expansion for the plus and minus densities of the form
$$ a^{ pm }_{-d/2}(x) t^{-d/2}+a^{ pm }_{d/2+1}(x) t^{-d/2+1}+cdots $$
where $d$ is the dimension of $M$.

Up to here the discussion was completely general, but then the proof begins to involve special cases, or
at least, broad division into classes of cases. But note that even for the special cases mentioned in the original question,
one would essentially carry out the procedure outlined above for a specific operator $P$.

The breakthrough in this line of thinking
came from Patodi's incredibly complicated computations for the operator $d+d^*$
going from even to odd differential forms,
where one saw that the
$$a^{+}_i(x)$$
and
$$a^{-}_i(x)$$
canceled each other out locally, that is, for each point $x$, for all the
terms with negative $i$. I think it was fashionable to refer to this cancellation as 'miraculous,' which it is, compared to the easy cancellation above.
At this point, Patodi could take a limit
$$lim_{trightarrow 0}[sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x)],$$
that he identified with the Euler form. This important calculation set a pattern that recurred in all other versions of
the heat kernel approach to index theorems. One proves the existence of an analogous
limit as $trightarrow 0$ and identifies it. The identification
as a precise differential form representative for a characteristic class is referred to sometimes as a local index theorem, a statement
more refined than the topological formula for the global index. There is even a beautiful version of a local families index theorem
that relates eventually to deep work in arithmetic intersection theory and Vojta's proof of the Mordell conjecture.

As I understand it, Gilkey's contribution was an invariant theory
argument that tremendously simplified the calculation and allowed a differential form representative for
the $hat{A}$ genus to emerge naturally
in the case of the Dirac operator. And then, I believe there is a $K$-theory argument that deduces the index theorem for a general elliptic operator
from the one for the twisted Dirac operator.

Experts can correct me if I'm wrong, but
from a purely mathematical point of view, essentially all the work on the heat kernel proof was done at this point.
Subsequent interpretations of the proof (more precisely, the supertrace), in terms of supersymmetry, path integrals, loop spaces, etc., were tremendously
influential to many areas of mathematics and physics, but the mathematical core of the index theorem itself appears to have remained largely unchanged for almost forty years. In particular, the terminology I've used myself above, the super- things, didn't occur at all in the original papers of Patodi, Atiyah-Bott-Patodi, or Gilkey.

Added:

Here is just a little bit of geometric-physical intuition regarding the heat kernel in the Gauss-Bonnet case, which I'm sure is completely banal for most people. The density
$$sum_ie^{-tlambda_i}||phi^+_i(x)||^2 dvol(x)-sum_ie^{-tmu_i}||phi^-_i(x)||^2 dvol(x)$$
expresses the heat kernel in terms of orthonormal bases for the even and odd forms. When $trightarrow infty$ all terms involving the positive eigenvalues decay to zero, leaving only contributions from orthonormal harmonic forms. This is one way to to see that the integral of this density, which is independent of $t$, must be the Euler characteristic. On the other hand, as $trightarrow 0$, the operator
$$K^+_t(x,y)dvol(y)=[sum_i e^{-tlambda_i } phi^+_i(x)otimes phi^+_i(y)]dvol(y)$$
literally approaches the identity operator on all even forms (except for the fact that it diverges). That is, the heat kernel interpolates between the identity and the projection to the harmonic forms, in some genuine sense expressing the diffusion of heat from a point distribution to a harmonic steady-state. A similar discussion holds for the odd forms as well. I can't justify this next point even vaguely at the moment, but one should therefore think of $$[K^+_t(x,y)-K^-_t(x,y)]dvol(y)$$ as regularizing the current on $Mtimes M$ given by the diagonal $Msubset Mtimes M$. Thus, the integral of $$[TrK^{+}_t(x,x)-TrK^-_t(x,x)]dvol(x)$$ ends up computing a deformed self-intersection number of the diagonal in $Mtimes M$. From this perspective, it shouldn't be too surprising that the Euler class, representing exactly this self-intersection, shows up.

Added:

I forgot to mention that the Riemann-Roch case is where $$P=bar{partial}+bar{partial}^*$$
going from the even to the odd part of the Dolbeault resolution associated to a holomorphic vector bundle. The limit of the local density is a differential form representing the top degree portion of the Chern character of the bundle multiplied by the Todd class of the tangent bundle. Perhaps it's worth stressing that these special cases all go through the general argument outlined above.

ac.commutative algebra - Field structure for R^n

No if you use the usual additive structure on $mathbb{R}^{n}$ for the field addition; but if you give up commutativity of multiplication, you have the skew-field of hamiltonians, or quaternions, on $mathbb{R}^{4}$, and if you then give up associativity of multiplication, you have the non-associative Cayley algebra, or octonians , on $mathbb{R}^{8}$. The Cayley-Dickson process builds the complex field from the real field, the skew-field of hamiltonians from the complex field, the non-associative Cayley algebra from the hamiltonians, and in general a $2^{n+1}$-dimensional involutive Cayley-Dickson algebra from the $2^{n}$-dimensional involutive Cayley-Dickson algebra. A.A. Albert did much to articulate the state of affairs in the early part of the twentieth century, if memory serves.

mg.metric geometry - Stability of midpoints in CAT(0) spaces

No, even if $X=mathbb R^2$.

Let $A_1$ be (the convex hull of) 4 points with coordinates $(pm 1,pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $sqrt 2$ centered at $(0,0)$. Shift $S_1$ a small distance $varepsilon$ in the horizontal direction, denote the resulting circle by $S_2$. For each vertex of $A_1$, mark its nearest point on $S_2$. The marked points are vertices of a convex quadrangle $A_2$ inscribed in $S_2$ and containing its center $(varepsilon,0)$. Hence $m(A_2)=(varepsilon,0)$ but the Hausdorff distance between $A_1$ and $A_2$ is $approxvarepsilon/sqrt 2$.

soft question - Too old for advanced mathematics?

Kind of an odd question, perhaps, so I apologize in advance if it is inappropriate for this forum. I've never taken a mathematics course since high school, and didn't complete college. However, several years ago I was affected by a serious illness and ended up temporarily disabled. I worked in the music business, and to help pass the time during my convalescence I picked up a book on musical acoustics.

That book reintroduced me to calculus with which I'd had a fleeting encounter with during high school, so to understand what I was reading I figured I needed to brush up, so I picked up a copy of Stewart's "Calculus". Eventually I spent more time working through that book than on the original text. I also got a copy of "Differential Equations" by Edwards and Penny after I had learned enough calculus to understand that. I've also been learning linear algebra - MIT's lectures and problem sets have really helped in this area. I'm fascinated with the mathematics of the Fourier transform, particularly its application to music in the form of the DFT and DSP - I've enjoyed the lectures that Stanford has available on the topic immensely. I just picked up a little book called "Introduction To Bessel Functions" by Frank Bowman that I'm looking forward to reading.

The difficulty is, I'm 30 years old, and I can tell that I'm a lot slower at this than I would have been if I had studied it at age 18. I'm also starting to feel that I'm getting into material that is going to be very difficult to learn without structure or some kind of instruction - like I've picked all the low-hanging fruit and that I'm at the point of diminishing returns. I am fortunate though, that after a lot of time and some great MDs my illness is mostly under control and I now have to decide what to do with "what comes after."

I feel a great deal of regret, though, that I didn't discover that I enjoyed this discipline until it was probably too late to make any difference. I am able, however, to return to college now if I so choose.

The questions I'd like opinions on are these: is returning to school at my age for science or mathematics possible? Is it worth it? I've had a lot of difficulty finding any examples of people who have gotten their first degrees in science or mathematics at my age. Do such people exist? Or is this avenue essentially forever closed beyond a certain point? If anyone is familiar with older first-time students in mathematics or science - how do they fare?

Galois group of a product of irreducible polynomials

If you mean: does knowing the G_i tell you the Galois group of P, then no.

Examples:

$P = (X^2+1)(X^2-2)$ has Galois group $C_2 times C_2$, and both factors have Galois group $C_2$; this works because the splitting fields of the two factors intersect only in $mathbb{Q}$.

But $P = (X^2 + X + 1)(X^2+3)$ has Galois group $C_2$, although both factors again have Galois group $C_2$. Here both factors, though they're coprime, define the same extension $mathbb{Q}(sqrt{-3})$.

I've just seen Robin's answer, so to relate to that: in the first example, the Galois group of P is the whole of $G_1 times G_2$. In the second example, it is the diagonal subgroup of $G_1 times G_2$, which is smaller although still projects surjectively onto each factor.

linear algebra - Hermitian matrices with prescribed number of positive and negative eigenvalues

Let $H$ be a linear subspace of the space of Hermitian $ntimes n$ matrices. Is there a good characterization of those $H$ such that every $Ain H$ has at least $k$ positive and $k$ negative eigenvalues?

For $k=1$ a nice characterization is the following: there is a positive definite matrix $B$ orthogonal to $H$ (w.r.t. the scalar product $(A,B)=mathrm{tr}(AB)$), or equivalently there exists a basis of $mathbb C^n$ such that all matrices in $H$ have zero trace.

Even for $k=2$ I was not able to find any good characterization.

pr.probability - Range of binomial probability, given a certain number of observations?

1.The answer to your specific question can be found in the Wikipedia page:

http://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval

The idea of estimating a distribution parameter is to construct a random variable
whose expectation is the parameter needed to be estimated. In your example, we want to estimate the success probability of a Bernoulli distribution and we construct a binomially distributed random variable by repeated trials. The key point is that the new random variable has the same average (after normalizing by n) but its standard deviation is smaller (by a factor of sqrt(n)) which gives better bounds on the estimated value. The confidence level is just a percetile of the distribution of the random variable used for the estimation (in your example you chose 50%).The interval size is a function of the percentile. The estimated value p* = k/n is inside the interval but the interval is not symmetric in general around this value. In the Wikipedia page, several approximations for large n are given, based on the central limit theorem, which are usually used in real-life estimations.

2.The above solution assumes prior knowledge that the distribution of a single trial is Bernoulli and that the trials are independent. Usually, one needs some prior knowledge to infere a statistical parameter. Specifically in your question, prior knowldge would be that the coin flips are independent, or that at after some flip the success probability p had changed, etc.

at.algebraic topology - is there a general statement about structures on spheres relating to division algebras?

It is classical to take a division algebra over $mathbb{R}$ and defining an H-space structure on the unit spheres by restricting and normalizing.

There are commutative division algebras of dimension 1 and 2 leading to commutative products on $S^0$ and $S^1$ identifying them as Eilenberg-MacLane spaces - Or if we forget some structure as an $E_{infty}$-spaces.

The associative division algebras $mathbb{H}$ defines an associative product on $S^3$, which is also a Lie-group, but forgetting some structure it is an $A_infty$-space.

There division algebra $mathbb{O}$ defines an $A_2$ structure on $S^7$, which is not $A_infty$ (is it $A_3$?).

As is well known it is possible to prove that no other spheres has $A_2$ structure.

Question: Is there a heiraki of structures below $A_2$ yet related such that $S^{15}$ has this structure, but $S^{31}$ does not?

Remark: A heiraki below $A_2$ could be that $A_2=D_infty$ for some definition of structures $D_n$, analagous to $E_1$ being $A_infty$.

Question: Is there an even more general definition of "lower" structures and a statement about all spheres (including possibly non-trivial structures on even-dimensional spheres)?

ag.algebraic geometry - Given a morphism from X to Y, when is the morphism from O_Y to the pushforward of O_X injective

I would like to know under what condition the morphism $mathcal{O}_Ylongrightarrow f_ast mathcal{O}_X$ induced by a morphism $f:Xlongrightarrow Y$ of schemes is injective.

Let me give an example (which I'm not completely sure about though).

I believe, if $X$ and $Y$ are reduced and $f$ is surjective and closed, the morphism $mathcal{O}_Y longrightarrow f_ast mathcal{O}_X$ is injective.

(Thus, proper flat morphisms of varieties have this property.)

Maybe one could forget about schemes and give a condition for locally ringed spaces?

Range of the Fourier transform on L^1

It's not germane to your question, but I can't resist pointing out that it is very hard to exhibit any continuous linear bijection from $L^1$(sensible measure space) onto $C_0$(sensible topological space), and in fact if either space is infinite then I suspect this is never possible, just for reasons of Banach space geometry. Thus, although it doesn't help with what you want to look at, I thought it might be worth mentioning that one can know the answer to "is the FT onto?" must be "no", before looking for an example or using properties of the Fourier transform.

(My caveats are because I don't want to categorically state it can't be done, but in all cases I can think of no such bijection will exist. However, both my general measure theory and my general topology are not what they should be, so I can't remember how to do things precisely in the most general settings.)

Anyway. I claim that there is no continuous linear bijection between $L^1({mathbb R}^d)$ and $C_0(X)$, where $X$ is locally compact Hausdorff (e.g. a metric space). The reason is that we have big powerful results telling us that

(i) every bounded linear operator from $C_0(X)$ to $L^1({mathbb R}^d)$ is weakly compact;

(ii) if the identity map on a Banach space $E$ is weakly compact, then $E$ is reflexive;

(iii) $L^1({mathbb R}^d)$ is not reflexive (ibid).

Unfortunately I can't locate a self-contained proof of the key fact (i). (It can be deduced as a corollary of a rather powerful, fundamental and beautiful result - due to some promising former student of Dieudonné and Schwartz, not sure if he ever went on to do anything important...)

soft question - What should be offered in undergraduate mathematics that's currently not (or isn't usually)?

Computer Science. I know programming has been said already, but computer science isn't programming. (There's the famous Dijkstra quote: “Computer science is no more about computers than astronomy is about telescopes.”)

There is a vast and beautiful field of computer science out there that draws on algebra, category theory, topology, order theory, logic and other areas and that doesn't get much of a mention in mathematics courses (AFAIK). Example subjects are areas like F-(co)algebras for (co)recursive data structures, the Curry-Howard isomorphism and intuitionistic logic and computable analysis.

When I did programming as part of my mathematics course I gave it up. It was merely error analysis for a bunch of numerical methods. I had no idea that concepts I learnt in algebraic topology could help me reason about lists and trees (eg. functors), or that transfinite ordinals aren't just playthings for set theorists and can be immediately applied to termination proofs for programs, or that if my proof didn't use the law of the excluded middle then maybe I could automatically extract a program from it, or that there's a deep connection between computability and continuity, and so on.

soft question - Most helpful math resources on the web

edit by jc: As of May 11, 2010, the work has been completed!

This is a reference that is not yet complete, but it should be very useful when it finally does arrive:

Digital Library of Mathematical Functions (DLMF)
(book and associated website;
will replace Abramowitz & Stegun's Handbook of Mathematical Functions)
NIST / Cambridge University Press
expected 2009/2010
http://dlmf.nist.gov/

This will contain diagrams, tables, properties of, principal values of, and relationships between many important mathematical functions. For example, the trigonometric and other elementary functions are described, with very many formulae relating them.

The Handbook is very good; the Digital Library will be even better.

measure theory - Why is 3 a bad constant in the Vitali covering lemma?

Hi,

recently I had to do with the Hardy-Littlewood maximal function and we used there the Vitali covering lemma with constant 5. Then, given an advice, I proved it with constant k>3. But I cannot find an counterexapmle why 3 is not enough (it is enough in the finite version of the lemma).
Has anyone seen such an example?

characteristic p - If the morphism of root data induced by an isogeny of a reductive group is a Frobenius, is then the isogeny itself a Frobenius?

The answer is "yes", but not in a good way: the descent it arises from is the split form, so this does not encode an interesting $mathbf{F}_ q$-structure. More precisely, $f$ arises from the $q$-Frobenius of a split descent down to $mathbf{F}_ q$ (and so even to the prime field $mathbf{F}_ p$). In particular, this is definitely not the way to encode the information of "interesting" descents of $(G,T)$ to $mathbf{F}_ q$, if that may be your ultimate intention.

To see this, let $(G_0, T_0)$ be the split form over $mathbf{F}_ q$, and $F_0:(G_0, T_0) rightarrow (G_0,T_0)$ the $q$-Frobenius morphism. This induces the self-map of the root datum given by $q$-multiplication on ${rm{X}}(T_0)$. Extending scalars to $k$ has no effect on the root datum for split groups, so $(F_ 0)_ k$ induces the endomorphism of the root datum of $((G_ 0)_ k, (T_ 0)_ k)$ given by $q$-multiplication on the root datum. Pick a $k$-isomorphism between the $k$-split pairs $(G,T)$ and $((G_ 0)_ k, (T_ 0)_ k)$ (as we may, by the
isomorphism Theorem). The resulting isomorphism of root data identifies the maps from $(F_ 0)_ k$ and $f$ since $q$-multiplication is functorial with respect to all homomorphisms between abelian groups. Thus, these maps coincide up to the action of some $t in overline{T}(k) = (overline{T}_ 0)(k)$, where
$overline{T} := T/Z_G$ and
$overline{T}_ 0 := T_ 0/Z_ {G_ 0}$ are the "adjoint tori''. Using an isomorphism
$overline{T}_ 0 simeq mathbf{G}_m^r$, $t$ goes over to some $r$-tuple $(t_i)$ with $t_i in k^{times}$. Since $k$ is separably closed, we can solve $t_i = y_i^{q-1}$ with $y_i in k^{times}$. In other words, $t = f(y)/y$ for some $y in T_0(k)$. So if we modify the identification of $(G_0, T_0)$ as an $mathbf{F}_ q$-descent of $(G,T)$ by composing with the action of $y$ or $y^{-1}$ (depending on direction of maps) then $f = (F_ 0)_ k$ via the new identification of $(G_ 0, T_ 0)$ as a split $mathbf{F} _q$-descent of $(G,T)$.
QED

Remark: A refinement of the argument (using some care if $k$ is not algebraic over
$mathbf{F}_ p$) proves that the split descent to $mathbf{F} _q$ is the only one which can work, but I won't get into that here. (It comes down to the fact that the solutions to $y^{q-1} = 1$ in $k^{times}$ lie in $mathbf{F} _q$.)

Note that the above works over any separably closed field of characteristic $p > 0$, not necessarily algebraically closed, provided the Isomorphism Theorem is valid over such fields. The Existence, Isomorphism, and even Isogeny Theorems are valid for split connected reductive groups over any field. The paper of Steinberg mentioned in the previous answer is really nice, but unfortunately assumes from the outset that the ground field is algebraically closed. Good news, in case you may care, is that one can deduce the results over any field from that case as a "black box'' by using descent theory. See Appendix A.4 of the book "Pseudo-reductive groups''.

gr.group theory - Proving that a group is free

It's a theorem of Stallings and Swan that a group of cohomological dimension one is free.

By a theorem of Serre, torsion-free groups and their finite index subgroups have the same cohomological dimension.

So, a torsion-free group is free if and only if its finite index subgroups are free.

(Here are the references. For Stallings-Swan, see

John R. Stallings, "On torsion-free groups with infinitely many ends", Annals of Mathematics 88 (1968), 312–334.

and

Richard G. Swan, "Groups of cohomological dimension one", Journal of Algebra 12 (1969), 585–610.

Serre's theorem is in Brown's book "Cohomology of Groups.")