dg.differential geometry - If the second derivative of a function on $mathbb R^n$ is everywhere nondegenerate, does it follow that the first derivative is an injection?

The counter example given in the comments by Brian Conrad (and Dylan Thurston) is very nice. However, it oscillates wildly at $infty$, and I believe that if you assume some nice properties at $infty$ you will get a possitive answer.

Translating $f$ by a linear function does not change the assumptions on $f$ and so the claim is equivalent to the fact that all such $f$ has no or a unique critical points.

If we add assumptions such that e.g. the Conley index (or homotopy index) is well-defined and a sphere then this modified claim would follow. Indeed, if two or more critical points existed they would by assumptions be non-degenerate with same Morse index and a small pertubation would yield a Conley index which is a vedge of two or more spheres - a contradiction.

rt.representation theory - Representations of reductive groups over finite rings

Lusztig's results mentioned above have been generalised to groups over arbitrary finite local rings, see Unramified representations of reductive groups over finite rings, Represent. Theory 13 (2009), 636-656.

There is a recent paper extending the above construction to certain ramified maximal tori, cf. Extended Deligne-Lusztig varieties for general and special linear groups, arXiv:0911.4593v1. It is shown here that the above "unramified" construction omits certain interesting representations. Some of these missing representations are connected with ramified maximal tori, and it is therefore desirable to have generalised Deligne-Lusztig varieties attached also to such tori.

In addition to the cohomological approaches to constructing representations, there are some partial purely algebraic constructions. There is a paper by G. Hill (Regular Elements and Regular Characters of $text{GL}_n(mathcal{O})$, which gives a construction of many so-called regular representations. These include most of the interesting representations, such as the strongly cuspidal ones.
There is also an approach, due to U. Onn, which defines a new type of induction functor (called infinitesimal induction) which complements the classical parabolic induction. This construction considers general automorphism groups of finite modules over DVRs with finite residue field. So far, this approach has led to a classification of all the representations in the rank-2 case (which includes $text{GL}_2(mathcal{O}/mathfrak{p}^r)$).

Since any finite commutative ring is the direct product of finite local rings, it is enough to study representations of groups over the latter. This is however a very hard problem, and one cannot expect an explicit list of all the representations in general. As in the representation theory of most infinite groups, the most fruitful approach seems to be to define a nice category of representations which is both possible to control, and at the same time is rich enough to include most (or all) the representations one is interested in. A candidate for such a category for $text{GL}_n(mathcal{O})$ is the one consisting of regular representations, mentioned above. One can ask interesting questions about the regular representations, such as which of them are given by the cohomological constructions, or which of them are types for supercuspidal representations of $text{GL}_n(F)$, where $F$ is a local field with ring of integers $mathcal{O}$.

ag.algebraic geometry - Local structure of the surface of bitangents to a quartic

Let $S subset mathbb{P}^3$ be a (possibly singular) quartic. I need some information about the local structure of the surface $Bit(S)$ of bitangents to $S$. I have done the computations, but they are a bit heavy, and I'd be happier if I could just cite a reference. In particular I need a local description of $Bit(S)$ around a line joining two isolated nodes.

The book Abel-Jacobi isogenies for certain types of Fano threefolds by Welters studies the surface of bitangents in some detail, but it soon restricts to the case of smooth $S$.

rt.representation theory - The equivalence of cateogry of equivariant sheaves on principal bundle and category of sheaves on base space.

Let $pi:Pmapsto B$ is a $G$-principal bundle, which means $G$ acts on $P$ freely and $pi$ is a locally trivial fibration. Here is a well-known theorem:

THeorem: The inverse image functor $pi^*$ gives an equivalence from $Sh_G(P)$ to $Sh(B)$, and the inverse functor is given by $pi_* ^G$.

Let me explain some notations. the object in $Sh_G(P)$ is a pair $(mathcal{F},alpha)$, where $alpha:p^*mathcal{F}simeq a^* mathcal{F}$, and satisfy cocycle condition. Here $p:Gtimes Pmapsto P$ is the projection, and $a:Gtimes Pmapsto P$ is the action.
The functor $pi_* ^G$ is given by assigning open subset V to $mathcal{F}(pi^{-1}(U))^G$.
We need to show that
(1). $pi^*$ is fully-faithful, i.e., for any two sheaves $mathcal{F}$ and $mathcal{G}$, $Hom(mathcal{F},mathcal{G})simeq Hom_G(pi^*(mathcal{F}),pi^*(mathcal{G}))$.

(2). $pi^*$ is essentially surjective, i.e., for any $G$-equivariant sheaf $mathcal{H}$, there exists an isomorphism $mathcal{H}simeq pi^* pi_*^G(mathcal{H})$.

(1) follows from $pi_*^Gcircpi^*simeq Id$, which is relatively easy.
For (2), I have checked when $B$ is a point. For general case, I can reduce it to the following isomorphism: $pi_*(mathcal{H})_b ^G simeq$

$ Gamma(pi^{-1}(b), mathcal{H}|_{pi^{-1}(b)})^G$, where $bin B$.

I know for nice group $G$, for example when $G$ is compact, this is really an isomorphism, because we have base change. For general group, I have no idea at all.

Who can help to finish this proof? Or give some other idea?

soft question - What out-of-print books would you like to see re-printed?

I'm REAL excited about this question,but I don't have the time right now to think about it enough to post a list. I was actually going to compile one for Dover this summer-a long one. But I'll think about it and try and post a few at this thread.Here's a few to get started:

Elements of Homotopy Theory by George Whitehead:A classic by the master and it would be a fantastic resource for classical homotopy theory from a geometrical standpoint that can serve as a foundation for the modern,high tech treatment via model categories.Why it's out of print baffles me.

Analysis And Solution of Partial Differential Equations by Robert L.Street:There are so few good undergraduate textbooks on this subject and a nice inexpensive reissue of this book would go a long way towards assisting this situation.Wonderful discussion and lots of nice examples.

Notes on Differential Geometry by Noel J.Hicks: An absolute classic and it needs to be brought back for a new generation of graduate students-after being proofread carefully,of course.Graduate students learning differential geometry will wonder why people have been hiding it from them.

The Foundations of Geometry by K.Borsuk and Smilew:A lost classic on axiomatic treatment of the classical plane geometries from a modern standpoint.Another book that baffles me why it's out of print.

There-that'll get you guys started. I actually hope to post the full list at my blog this summer. I'll let you guys know when it's up for the world to see.

ho.history overview - Who invented the gamma function?

The first person who gave a representation of the so called gamma function was Daniel Bernoulli in a letter to Goldbach from 1729-10-06. The letter can be seen here.

The formula reads in modern notation as given by Gronau in the article cited in the answer:

$ x! = lim_{nrightarrow infty}left(n+1+frac{x}{2}right)^{x-1} prod_{i=1}^nfrac{i+1}{i+x} $

Gronau also observes that "Numerical experiments show that the formula of Bernoulli converges much faster to its limit than that of Euler ...", "that of Euler" refers here to a formula Euler has given in a letter to Goldbach dated 1729-10-13.

Gronau writes: "Euler who, at that time, stayed together with D. Bernoulli in St. Petersburg gave a similar representation of this interpolating function. But then, Euler did much more. He gave further representations by integrals, and formulated interesting theorems on the properties of this function."

Though this justifies the name 'Euler gamma function' Euler's representation was historically only second to Daniel Bernoulli's.

The correspondence between Goldbach, Daniel Bernoulli and Euler which undoubtedly gave birth to the gamma function is well documented in Paul Heinrich Fuss's „Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIeme siècle ..“, St. Pétersbourg, 1843.

ag.algebraic geometry - Affine morphisms in different settings coincide?

Emerton and Zoran's answers completely answer the question as stated, but there's another way to think about affine morphisms that is worth mentioning.

Given any quasi-compact and quasi-separated morphism of schemes $f:Xto Y$, $newcommand{O}{mathcal O}f_*O_X$ is a quasi-coherent sheaf of $O_Y$-algebras. This functor has an adjoint, called relative $Spec_Y$, or relative Spec. Given a quasi-coherent sheaf of $O_Y$-algebras $newcommand{A}{mathcal A}A$, we get a scheme over $Y$, $phi^A:Spec_Y Ato Y$, with the property that $phi^A_*(O_{Spec_Y A})=A$ and $Hom_Y(X,Spec_Y A)cong Hom_{O_Ytext{-alg}}(A,f_*O_X)$ for any $f:Xto Y$. A morphism $f:Xto Y$ is affine if and only if $Xcong Spec_Y(A)$ (as a $Y$-scheme) for some $A$ (which must be $f_*O_X$). See EGA II §1 for this development of affine morphisms.

I find this way of thinking about affine morphisms is useful for two reasons. First, if I'm working with a bunch of schemes affine over $Y$, it's often easier for me to think about a bunch of $O_Y$-algebras with algebra morphisms between them. Second, the adjunction in the previous paragraph tells you that any (quasi-compact quasi-separated) morphism $f:Xto Y$ has a canonical factorization through an affine morphism $Xto Spec_Y(f_*O_X)to Y$, called the Stein factorization (the first morphism is Stein, meaning that the structure sheaf pushes forward to the structure sheaf). This factorization is often extremely handy; for example, if a morphism is quasi-affine, the Stein factorization is a witness of its quasi-affineness.

lo.logic - What is the general opinion on the Generalized Continuum Hypothesis?

I'm community wikiing this, since although I don't want it to be a discussion thread, I don't think that there is really a right answer to this.

From what I've seen, model theorists and logicians are mostly opposed to GCH, while on the other end of the spectrum, some functional analysis depends on GCH, so it is much better tolerated among functional analysts. In fact, I considered myself very much +GCH for a while, but Joel and Francois noted some interesting stuff about forcing axioms, (the more powerful ones directly contradict CH).

What is the general opinion on GCH in the mathematical community (replace GCH with CH where necessary)? Does it happen to be that CH/GCH doesn't often come up in algebra?

Please don't post just post "I agree with +-CH". I'd like your assessment of the mathematical community's opinion. Maybe your experiences with mathematicians you know, etc. Even your own experiences or opinion can work. I am just not interested in having 30 or 40 one line answers. Essentially, I'm not looking for a poll.

Edit: GCH=Generalized Continuum Hypothesis
CH= Continuum Hypothesis

CH says that $aleph_1=mathfrak{c}$. That is, the successor cardinal of $aleph_0$ is the continuum. The generalized form (GCH) says that for any infinite cardinal $kappa$, we have $kappa^+=2^kappa$, that is, there are no cardinals strictly between $kappa$ and $2^kappa$.

Edit 2 (Harry): Changed the wording about FA. If it still isn't true, and you can improve it, feel free to edit the post yourself and change it.

ag.algebraic geometry - Holomorphic and antiholomorphic forms of projective space

Greg's otherwise excellent answer gives the impression that computing Chern classes on projective space requires a computer algebra system. I'm writing to repell this impression. The cohomology ring of $mathbb{P}^{n-1}$ is $mathbb{Z}[h]/h^n$ where $h$ is Poincare dual to the class of a hyperplane. We have the short exact sequence

$$0 to S to mathbb{C}^n to Q to 0$$,

where $S$ is the tautological line bundle, whose fiber over a point of $mathbb{P}^{n-1}$ is the corresponding line in $mathbb{C}^n$. The line bundle $S$ is also called $mathcal{O}(-1)$, and has Chern class $1-h$. So $$c(Q) = 1/(1-h) = 1+h+h^2 + cdots h^{n-1}.$$

As Greg explained, the tangent bundle is $mathrm{Hom}(S, Q) = S^{star} otimes Q$. The formula for the Chern class of a general tensor product is painful to use in practice, but we can circumvent that here by tensoring the above exact sequence by $S^{star}$.

$$0 to mathbb{C} to (S^{star})^{oplus n} to T_{mathbb{P}^{n-1}} to 0$$,

So the Chern class of the tangent bundle to $mathbb{P}^{n-1}$ is
$$(1+h)^n = 1 + n h + binom{n}{2} h + cdots + n h^{n-1}.$$

Yes, I deliberately ended that sum one term early. Remember that $h^n$ is $0$.

As Greg says, if this is to be expressible as a direct sum¹ of line bundles, then this should be a product of $n-1$ linear forms, $c(T) = prod_{i=1}^{n-1} (1+ a_i h)$. Since this is an equality of polynomials of degree $n-1$, we can forget that we are working modulo $h^n$ and just check whether the honest polynomial $f(x) := 1 + n x + binom{n}{2} x + cdots + n x^{n-1}$ factors in this way.

The answer is it does not, except when $n=2$ (the case of $mathbb{P}^1$.)
The unique factorization of $f(x)$ is
$$prod_{omega^n=1, omega neq 1} (1+x (1-omega)).$$

Thus does raise an interesting question. When $n$ is even, $(1+2x)$ divides $f(x)$. This suggests that $mathcal{O}(2)$ might be a subquotient of $T_{mathbb{P}^{n-1}}$. I can't figure out whether or not this happens.

---

The case of Grassmannians is going to be worse for three reasons. The two minor ones are that (1) we may honestly have to use the formula for the chern class of a tensor product. and (2) the polynomials in question will be multivariate polynomials. The big problem will be that $H^{star}(G(k,n))$ has relations in degree lower than $dim G(k,n)+1$, so we can't pull the trick of forgetting that $h^n=0$ and working with honest polynomials.

I suspect that your question was more "give a nice description of the tangent bundle to the Grassmannian" than "can that tangent bundle be expressed as a direct sum of line bundles?". If you seriously care about the latter, I'll give it more thought.

¹ Direct sum is the natural thing to ask for in the categories of smooth, or of topological, complex vector bundles. If you like the algebraic or holomorphic categories, as I do, it is more natural to ask for the weaker property that there is a filtration of the vector bundle, all of whose quotients are line bundles.

gn.general topology - Are the C(S^n, S^n)'s homeomorphic ?

Let m, n > 1. Is it true that C(S^m, S^m), and C(S^n, S^n) are homeomorphic ?
[both endowed with the sup metric (or equivalently the compact-open topology)]

Generally, C(S^n, S^n), with n >= 1, is a countably infinite (disjoint) union of
path-connected (due to Hopf) components C_{n,k}, k in Z.

I think each of these components may be viewed as an infinite dimensional
Frechet manifold. Unfortunately, they are not contractible. However, the question
is [somewhat vaguely] motivated by the
Henderson's Theorem.

Also, I have some related questions :
- Fixing n, are the C_{n,k}'s homeomorphic to each other (at least for k <> 0) ?
- Are there some m, n, k, l with m <> n s.t. C_{m,k} ~ C_{n,l} ?

What I was able to do in this direction until now is to show the existence of a
proper, one-to-one, degree-preserving map from C(S^m, S^m) into C(S^n, S^n).
Even for m >> n, but far to be surjective.

at.algebraic topology - Connectivity after Geometric Realization?

Yes it is a surjection on $pi_0$, because each component of $|Y|$ has at least one component of $Y_0$.

Beyond that there are no restrictions. For instance, you can get any homotopy type for $|X|$ and $|Y|$ and any homotopy type for the map between them with $X_0$ and $Y_0$ just one point, as long as you ask that $pi_0(|X|)$ and $pi_0(|Y|)$ are trivial.

(I'm taking "simplicial space" to mean a simplicial object in the category of topological spaces, say the compact Hausdorff ones.)

gn.general topology - A question about indecomposable continua.

As pointed out by Jeff, the notion you define may not really be what you are after, since indecomposable continua are not 'indecomposable' in your sense. However, we can ask:

Is there a nontrivial connected metric space $X$ such that $X$ cannot be written as the union of two proper connected subsets?

The answer, as Jeff suggested, is no.

Indeed, let $X$ be a nontrivial connected metric space. If $X$ does not have any cut-points, then clearly we can write
$$X = (xsetminus{x_0}) cup (Xsetminus{x_1})$$
for some $x_0neq x_1$, and are done.

If $X$ does have a cut-point $x_0$, let $A$ and $B$ be open subsets of $X$ such that
$$Acap B = {x_0}; quad Asetminus{x_0},Bsetminus{x_0}neqemptyset quadtext{and}quad Acup B = X.$$

We claim that $A$ and $B$ are connected. Indeed, if $Uni x_0$ is relatively open and closed in $A$, then $Ucup B$ is open and closed in $X$, so we must have $U=A$ (since $X$ is connected).

Regarding your question on the number of proper connected subsets, we can still ask the following question:

If $X$ is any nontrivial connected metric space, what can be said about the cardinality of the set $S$ of proper connected subsets of $X$?

It seems plausible that the set $S$ has at least the cardinality of the continuum, but I wasn't able to find a reference (and haven't thought very deeply about it). Certainly the set $S$ must be infinite.

nt.number theory - Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $

EDIT: Hendrik Lenstra emailed me a proof of Conjecture 2. I'll append it below. So Jagy's question is now solved.

---

OK so I think that Jagy wants to make the following conjecture:

CONJECTURE 1: an integer $C$ is not representable by the form F(x,y,z)=2x^2+xy+3y^2+z^3-z if, and only if, $C$ is odd and $27C^2-4=23D^2$ with $D$ an integer.

[EDIT/clarification: Jagy only asks one direction of the iff in his question, and this answer below gives a complete answer to the question Jagy asks. I came back to this question recently though [I am writing this para a year after I wrote the original answer] and tried to fill in the details of the argument in the other direction (proving that if C was not an odd integer solution to $27C^2-4=23D^2$ then $C$ was represented by the form) and I failed. So the "hole" I flag in the answer below still really is a hole, and this post still remains an answer to Jagy's question, but not a complete proof of Conjecture 1, which should still be regarded as open.]

I have a proof strategy for this. I am too lazy to fill in some of the details though, so maybe a bit of it doesn't work, but it should be OK. However, I am also reliant on a much easier-looking conjecture (which I've tested numerically so should be fine, but I can't see why it's true):

CONJECTURE 2: if $C$ is odd and $27C^2-4=23D^2$, then there's no prime p
dividing D of the form $2x^2+xy+3y^2$.

So I am claiming Conj 2 implies the "only if" version of Conj 1. I don't know how to prove Conj 2
but it looks very accessible [edit: I do now; see below]. Note that the Pell equation is related to units
in $mathbf{Q}(sqrt{69})$ and the $2x^2+xy+3y^2$ is related to factorization
in $mathbf{Q}(sqrt{-23})$. I've seen other results relating the arithmetic
of $mathbf{Q}(sqrt{D})$ and $mathbf{Q}(sqrt{-3D})$.

---

Ok, so assuming Conjecture 2, let me sketch a proof of the "only if" part of Conjecture 1.

The Pell equation is intimately related to the recurrence relation

$$t_{n+2}=25t_{n+1}-t_n$$

with various initial conditions. For example the positive $C$s which
are solutions to $27C^2-4=23D^2$ are all generated by this recurrence
starting at $C_1=C_2=1$, and the $D$s are all generated by the same
recurrence with $D_1=-1$ and $D_2=1$. Note that $C_n$ is even iff $n$
is a multiple of 3, and (by solving the recurrence explicitly) one
checks easily that $C_{3n}=(3C_{n+1})^3-(3C_{n+1})$, so we've represented
the even solutions to the Pell equation as values of $F$ (with $x=y=0$).

Let's then consider the odd solutions to the Pell equation. Say $C$
is one of these. We want to prove that there is no solution in
integers $x,y,z$ to

$$2x^2+xy+3y^2=z^3-z+C.$$

Let's do it by contradiction. Consider the polynomial $Z^3-Z+C$. First
I claim it's irreducible. This is because it is monic, of degree 3,
and has no integer root, because $C$ is odd. Next I claim that
the splitting field contains $mathbf{Q}(sqrt{-23})$. This is
because of our Pell assumption and the fact that the discriminant
of $Z^3-Z+C$ is $4-27C^2$. Next I claim that the splitting
field of $Z^3-Z+C$ is in fact the Hilbert class field of
$mathbf{Q}(sqrt{-23})$. I only know an ugly way of seeing this:
if $theta$ is a root of $Z^3-Z+1=0$ then I know recurrence relations
$e_n$, $f_n$ and $g_n$ (all defined using the relation above but with
different initial conditions) with $e_ntheta^2+f_ntheta+g_n$ a root of
$Z^3-Z+C_{3n+1}$, and other relations giving roots of $Z^3-Z+C_{3n+2}$
and $Z^3-Z-C_{3n+1}$ and $Z^3-Z-C_{3n+2}$. Most unenlightening but it
does the job because it embeds $mathbf{Q}(theta)$ into the splitting
field, and the Galois closure of $mathbf{Q}(theta)$ is the Hilbert
class field of $mathbf{Q}(sqrt{-23})$.

Right, now for the contradiction, assuming Conjecture 2. Let's assume
that $C$ is a solution to the Pell, and $z^3-z+C$ can be written $2x^2+xy+3y^2$.
Now $C$ is odd so $z^3-z+C$ isn't zero, and hence it's positive,
so it's the norm of a non-principal ideal~$I$ in the integers $R$ of
$mathbf{Q}(sqrt{-23})$. This ideal $I$ is a product of prime ideals,
and $I$ isn't principal, so one of the prime ideals had better also not
be principal. Say this prime ideal has norm $p$. We conclude that $p$
divides $z^3-z+C$ and $p$ is of the form $2x^2+xy+3y^2$. Note in
particular that this implies $pnot=23$. Also $pnot=3$, because $C$
is odd and (because of general Pell stuff) hence prime to 3.

CASE 1: $p$ is coprime to $D^2$ (with $27C^2-4=23D^2$). In this
case the polynomial $Z^3-Z+C$ has non-zero discriminant mod $p$
(because $pnot=23$) and furthermore has a root $Z=z$ mod $p$.
Hence mod $p$ the polynomial either splits as the product of a linear
and a quadratic, or the product of three linears. This tells us
something about the factorization of $p$ in the splitting field
of $Z^3-Z+C$: either $p$ remains inert in $mathbf{Q}(sqrt{-23})$,
or it splits into 6 primes in the splitting field and hence splits
into two principal primes in $mathbf{Q}(sqrt{-23})$ (because the
principal primes are the ones that split completely in the Hilbert
class field). In either case $p$ can't be of the form $2x^2+xy+3y^2$,
so this case is done.

CASE 2: This is simply Conjecture 2.

In both cases we have our contradiction, and
so we have proved, so far, assuming Conjecture 2, that a solution $C$ to $27C^2-4=23D^2$
is representable as $2x^2+xy+3y^2+z^3-z$ iff it's even.

Note that Conjecture 2 can be verified by computer for explicit values
of $C$, giving unconditional results---for example I checked in just
a few seconds that any odd $C$ with $|C|<10^{72}$ and satisfying the
Pell equation was not representable by the form, and that result
does not rely on anything. At least that's something concrete for Jagy.

---

OK so what about the other way: say $27C^2-4$ is not 23 times a square.
How to go about representing $C$ by our form? Well, here I am going to
be much vaguer because there are issues I am simply too tired to deal
with (and note that this is not the question that Jagy asked anyway).
Here's the idea. Look at the proof of Theorem 2 in Jagy's pdf Mordell.pdf.
Here Mordell gives a general algorithm to represent certain integers
by (quadratic in two variables) + (cubic in one variable). If you
apply it not to the form we're interested in, but to the following
equation:

$$x^2+xy+6y^2=z^3-z+C$$

then, I didn't check all the details, but I convinced myself that they
could easily be checked if I had another hour or two, but I think that
the techniques show that whatever the value of $C$ is, this equation
has a solution. The idea is to fix $C$, let $theta$ be a root of
the cubic on the right (which we can assume is irreducible, as if it
were reducible then we get a solution with $x=y=0$), to rewrite the right
hand side as $N_{F/mathbf{Q}}(z-theta)$, with $F=mathbf{Q}(theta)$
and now to try and write $z-theta$ as $G^2+GH+2H^2$ with
$G,Hinmathbf{Z}[theta]$. Mordell does this explicitly (in a slightly
different case) in the pdf. The arguments come out the same though,
and we end up having to check that a certain cubic in four variables
has a solution modulo~23 with a certain property. I'll skip the painful
details. The cubic depends on $C$ mod 23, and so a computer calculation
can deal with all 23 cases.

Once this is done properly we have a solution to $x^2+xy+6y^2=z^3-z+C$,
so we have written $z^3-z+C$ as the norm of a principal ideal in
the integers of $mathbf{Q}(sqrt{-23})$. What we need to do now is
to write it as the norm of a non-principal ideal, and of course we'll
be able to do this if we can find some prime $p$ dividing $z^3-z+C$
which splits in $mathbf{Q}(sqrt{-23})$ into two non-principal
primes, because then we replace one of the prime divisors above $p$
in our ideal by the other one. What we need then is to show that
if the discriminant of $z^3-z+C$ is not $-23$ times a square,
then there is some prime $p$ of the form $2x^2+xy+3y^2$ dividing
some number of the form $z^3-z+C$ which is the norm of a principal
ideal. This should follow from the Cebotarev density theorem, because
Mordell's methods construct a huge number of solutions to $x^2+xy+6y^2=z^3-z+C$
which are "only constrained modulo 23", and so one should presumably
be able to find a prime which splits in $mathbf{Q}(sqrt{-23})$,
splits completely in the splitting field of $z^3-z+C$ and doesn't
split completely in the splitting field of $z^3-z+1$. I have run out
of energy to deal with this point however, so again there is a hole here.
This issue seems analytic to me, and I am not much of an analytic guy.
[edit: I came back to this question a year later and couldn't do it,
so this should not be regarded as a proof of the "if" part of Conj 1]

---

EDIT: OK so here, verbatim, is an email from Lenstra in which he establishes
Conjecture 2.

(EDIT: dollar signs added - GM)

Fact. Let $theta$ be a zero of $X^3-X+1$, let $eta$ in ${bf Z}[theta]$ be
a zero of $X^3-X+C$ with $C$ in $bf Z$ odd, and let $p$ be a prime
number that is inert in ${bf Z}[theta]$. Then $p$ does not divide
index$({bf Z}[theta]:{bf Z}[eta])$.

Proof. By hypothesis, ${bf Z}[theta]/p{bf Z}[theta]$ is a field of size $p^3$.
Let $e$ be the image of $eta$ in that field. Since $X^3-X+C$ is
irreducible in ${bf Z}[X]$ (even mod 2), it is the characteristic
polynomial of $eta$ over $bf Z$. Hence its reduction mod $p$ is the
characteristic polynomial of $e$ over ${bf Z}/p{bf Z}$. If now $e$ is in
${bf Z}/p{bf Z}$, then that characteristic polynomial also equals $(X-e)^3$,
so that in ${bf Z}/p{bf Z}$ we have $3e = 0$ and $3e^2 = -1$, a contradiction.
Hence $e$ is not in ${bf Z}/p{bf Z}$, so $({bf Z}/p{bf Z})[e] = {bf Z}[theta]/p{bf Z}[theta]$,
which is the same as saying ${bf Z}[theta] = {bf Z}[eta] + p{bf Z}[theta]$. Then
$p$ acts surjectively on the finite abelian group ${bf Z}[theta]/{bf Z}[eta]$,
so the order of that group is not divisible by $p$. End of proof.

mathematics education - To what extent can algorithms in undergraduate linear algebra be made continuous/polynomial/etc.?

I feel like many of the algorithms that I learned — indeed, that I have taught — in undergraduate linear algebra classes depend sensitively on whether certain numbers are $0$. For example, many a linear algebra homework exercise consists of a matrix and a request that the student calculate a basis for the kernel or image. The standard approach consists of row-reducing the matrix and reading off the answer. The algorithm to row-reduce a matrix has many steps of the form "if $a neq 0$, do something that involves a division by $a$, and if $a = 0$, do something else". Such a step is unfortunate from many points of view. In particular, it is not even continuous in $a$, so if you only have partial data about the value of $a$ (say, a truncated decimal expansion), then you cannot hope to apply this algorithm.

For the purpose of calculating kernel and image bases, perhaps this is not too surprising. Indeed, the dimensions of the kernel and image of a basis do not depend polynomially on the coefficients — they don't even depend continuously, only semicontinuously — and so there is really no hope in writing an algorithm that computes the coefficients of a basis for either and that is algebraic in the matrix.

On the other hand, even the usual algorithm we teach to compute inverses to invertible matrices again uses row reduction. The final answer is algebraic in the matrix coefficients, and there are algorithms that run algebraically (something to do with minors). So my (slightly ambiguous and open-ended) questions are:

What problems in undergraduate linear algebra can be solved by algorithms that are totally polynomial/algebraic/continuous in the matrix coefficients? In settings where there is not complete information about a matrix — e.g. when the coefficients are given by truncated decimal expansions — what types of algorithms do people actually use to "approximate" the answers to problems that cannot be solved continuously? And how do people who do such problems measure/define how good their "approximations" are?

My motivation for asking this question was the (closed) question matrix that annihilates matrix, in which the asker asks for an algorithm that, when fed an $ntimes m$ matrix $A$, computes an $mtimes n$ matrix $B$ so that $ker B = operatorname{im} A$. This is another example of a problem that is easy to do by row reduction, but it's not clear to me if there are solutions that run "more" continuously than that. (There won't be a completely continuous algorithm. Set $m=n$ and feed in an invertible matrix $A$. Then $B = 0$. But the invertibles are dense among all matrices.)

ct.category theory - What's a groupoid? What's a good example of a groupoid?

Let me expand a bit on what Dave said.

The Yoneda lemma tells us that given an object X of a category C, the (covariant, contravariant, whatever) functor h_X : C -> Set, which sends an object Y to the set Hom(Y,X), can be thought of as the "same" as the object X. There are many situations in which we are interested in a functor F : C -> Set, and we might like to know whether F is isomorphic to h_M for some object M, because that reduces the study of F to the study of a single object M. In such a case we say that F is represented by M. The letter M here, suggestively, stands for "moduli".

Example: Given a group G, the functor BG' : Top -> Set is the functor which sends a topological space X to the set of isomorphism classes of principal G-bundles over X. (You can also do the analogous thing for schemes.)

Example: The functor M_g' : Sch -> Set is the functor which sends a scheme X to the set of isomorphism classes of flat families of genus g curves over X.

In both of the above examples, there is no object M for which h_M is isomorphic to the functor. So this is perhaps not so nice. But, without getting into too many details, there is a natural "fix", namely we can instead consider the functor BG : Top -> Groupoid (resp. M_g : Sch -> Groupoid) which sends a topological space (resp. a scheme) to the groupoid of G-bundles (resp. flat families of genus g curves). This groupoid has objects G-bundles and morphisms isomorphisms of G-bundles (resp. the obvious analogous thing). The original set-valued functor is just the composition of this functor with the functor Groupoid to Set which takes a groupoid and returns the set of isomorphism classes of objects in the groupoid.

Anyway, despite the fact that the set-valued functors are not so "geometric", since they are not represented by a "geometric" object (topological space and scheme, respectively), the groupoid-valued functors are more "geometric". In the case of M_g, the "geometric" structure we get is that of a "Deligne-Mumford stack", which essentially means that we can for practical purposes pretend that it is represented by a scheme with only some slightly "weird" properties. In the case of BG (the topological one) you can take a "geometric realization" and recover the classifying space BG that we know and love.

Another very important reason for studying groupoids and another very important class of groupoids comes from, as others have already mentioned, group actions. When a group acts on a manifold or a variety, the naive quotient may be badly behaved, for example it may no longer be a manifold (e.g. it might not be smooth, or it might not be Hausdorff) or respectively a variety (or it may not even be clear how to take the quotient at all!), which makes it harder to study geometric properties of the alleged "quotient". However, the groupoid viewpoint allows us to get a better handle on the quotient and its geometry. More precisely, if G is a group acting on a space (manifold, scheme, variety, whatever) X, then the "correct" quotient is actually the functor X/G : C -> Groupoid (where C is the category of manifolds, schemes, whatever) which sends an object Y to the groupoid of pairs (G-bundles E over Y, G-equivariant morphism from the total space of E to X). The functor BG is a special case of this; it's pt/G.

There's some further discussion on this sort of stuff at the nLab:

http://ncatlab.org/nlab/show/moduli+space

http://ncatlab.org/nlab/show/classifying+space

gr.group theory - when is Aut(G) abelian

From MathReviews:

---

MR0367059 (51 #3301)
Jonah, D.; Konvisser, M.
Some non-abelian $p$-groups with abelian automorphism groups.
Arch. Math. (Basel) 26 (1975), 131--133.

This paper exhibits, for each prime $p$, $p+1$ nonisomorphic groups of order $p^8$ with elementary abelian automorphism group of order $p^{16}$. All of these groups have elementary abelian and isomorphic commutator subgroups and commutator quotient groups, and they are nilpotent of class two. All their automorphisms are central. With the methods of the reviewer and Liebeck one could also construct other such groups, but the orders would be much larger.

---

FYI, I found this via a google search.

The first to construct such a group (of order $64 = 2^6$) was G.A. Miller* in 1913. If you know something about this early American group theorist (he studied groups of order 2, then groups of order 3, then...and he was good at it, and wrote hundreds of papers!), this is not so surprising. I found a nice treatment of "Miller groups" in Section 8 of

http://arxiv.org/PS_cache/math/pdf/0602/0602282v3.pdf

(*): The wikipedia page seems a little harsh. As the present example shows, he was a very clever guy.

gn.general topology - Topologies making a class of functions continuous

Let $X:={f: mathbb{C}to mathbb{C}}$ be a class of total functions on $mathbb{C}$ closed under composition, addition, multiplication, and scalar multiplication. Does there exist a topology on $mathbb{C}$ making these functions and only these functions continuous?

If it's not true in general (it probably isn't), are there any interesting known cases where it is true?

Note: I emphasize total functions because we want them to be everywhere defined. This avoids functions with bad singularities.

Edit: Obviously, continuous functions in the standard topology fit this bill, but this is tautological and not in the spirit of the problem.

Edit 2: Apparently the way I asked this question made it seem like I was looking for an answer to the "general case" which seems pretty untrue although I haven't actually worked it out. Rather, the real question was interesting cases where it is true.

ag.algebraic geometry - Line Bundles on Torus Quotient

Suppose you have a scheme $X$ that is acted on by a torus $T$. Then the action induces a grading on the functions on $X$ by the character lattice of $T$. So for a fixed character $lambda$, we can consider $mathcal{O}_{X,lambda}$, the $lambda$ graded part. Assuming the quotient $X/T$ exists, these graded parts should descend to quasicoherent sheaves on the quotient.

My question is, when are these sheaves line bundles?

In the basic examples I know, they are always line bundles. For example, if you take $X= mathbb{A}^{n+1} - 0$ and $T = mathbb{C}^*$, then on $X$ you get the ordinary grading by homogeneous degree. When you descend to the quotient, you get the line bundles $mathcal{O}(k)$ on $mathbb{P}^n$. You can also take $G$ a complex semi-simple group, $B$ a Borel subgroup, $U$ the maximal unipotent. Then $G/U rightarrow G/B$ is a torus quotient, and the graded pieces descend to line bundles. I think, a similar story is true for all homogeneous spaces, but I'm having a little trouble phrasing it in terms of torus quotients.

In fact, in these situations, these are all the line bundles.

So more generally, my question is, what properties can you require of the general $X$ so it behaves like the two examples above?

topos theory - Is there a finitely complete category with terminal object but NO subobject classifier?

I am pretty sure I proved that the category of groups has no subobject classifier at some point. I will try and edit this post with the proof when I have more time, but I think this is an example for you to think about.

EDIT: Ya, this isn't too bad. If there was a subobject classifier $Omega$ in Groups, then by looking at the characteristic map of the injection of $0 rightarrow A$ for each map, you will see by writing out the diagrams that A will have to inject into $Omega$. But then $Omega$ is bigger than every cardinal since there is a group of every cardinality. That doesn't fly.

EDIT of EDIT: I guess I should flesh this out in greater detail.

The first thing to note is that the category of groups has a zero object (it's terminal object is also initial). I will write 0 for this.

Say Groups had a subobject classifier $0 stackrel{true}longrightarrow Omega$. Note that the map true must be the unique map out of 0. Also note that the unique map $0 stackrel{!}rightarrow A$ is a monomorphism (i.e. injection) for every A. Thus

                                       0----->0
                                       |      |
                                     ! |      | true
                                       |      | 
                                       /     /
                                       A ---->Ω 

is a pullback square, where the lower map, $chi$, is the characteristic map of !. I claim that $chi$ is a monomorphism. This is because the ker($chi$) maps to both A and 0 to make the diagram commute, so the inclusion of ker($chi$) into A factors through 0 by the definition of a pullback. In other words, the kernel is trivial, so $chi$ is an injection. Thus every group A admits an injection to $Omega$ which is bad for set theoretic reasons.

soft question - Justifying a theory by a seemingly unrelated example

As for category theory, I don't think that there is a motivating example which has not already a category theoretic flavor. The leading theme is to unify and then generalize constructions resp. arguments, which come up in all areas of mathematics. Historically, natural transformations were introduced for the foundations of homology theory of topological spaces. But to start with an easy example, you may observe that for abelian groups $A,B,C$ there is a canonical isomorphism $(A oplus B) oplus C cong A oplus (B oplus C)$, which reminds you of other associativity results such as $(X cup Y) cup Z cong X cup (Y cup Z)$ for sets (here $cup$ means disjoint union). Within category theory, you can see what's the real content of this: direct sum and disjoint unions are examples of coproducts, and coproducts are always associative. Even more striking, Yoneda's Lemma, which lies at the heart of foundations of category theory, tells you that the case of sets already settles the general case!

But category theory is more than just a language, it also provides general constructions: Assume you want to approximate a theory with another theory. This may be formalized by finding an adjunction between two categories. Freyds/Special Adjoint Functor Theorem tell you when this is possible. Although in some situations you can't write down the adjunction, the only thing you need is to know that it exists. For example, what is the categorical coproduct of an infinite family of compact hausdorff spaces? Can you write it down without using Stone-Cech?

There are also somewhat global motivations: Some Theories behave like other theories, and thus you may develope a theory for a large class of categories at once: monodial categories, topological categories, algebraic categories, locally presentable categories, etc. Of course, the same is true for other notions of category theory (functors, natural transformations, types of morphisms, etc.).

But if one has not heard of category theory before, the first motiviation should be to think in categories (in the colloquial sense). For example the set, which underlies a group, really differs from the group. In almost every book and lecture, this is absorbed by abuse of notation. The existence of bases in vector spaces is no reason at all to restrict linear algebra to vector spaces of the form $K^{(B)}$. Similarily, vector bundles should not be defined as bundles which are locally isomorphic to some $mathbb{R}^n times X$, which most topologists still ignore! Rather, it is first of all a vector space object in the category of bundles over $X$.

Let's conclude with an example which both introduces functors and algebraic geometry: Assume you have a system of polynomial equations $f_1(x)=...=f_n(x)=0$ in $m$ variables defined over $mathbb{Z}$ and you want to study the solutions in arbitrary rings at once, using a single mathematical object, e.g. having in mind some local-global results of algebraic number theory. So for every ring $R$, we put $F(R) = {x in R^m : f_1(x)=...=f_n(x)=0}$. Observe that for every ring homomorphism $R to S$ there is a set map $F(R) to F(S)$ and that this is compatible with composition of homomorphisms. This exactly means that $F$ is a functor from the category of rings to the category of sets. Algebraic geometry studies functors which locally look like the functor above.

st.statistics - Multinomial transformation for matrices

Suppose we have a vector of probabilities $mathbf{p}=(p_1,...,p_n)$, where $p_i>0$ for $i=1,...n$ and $sum p_i=1$. Define new vector $mathbf{r}=(r_1,...,r_{n-1})$ in a following way:

$r_i=log(p_i/p_n)$

This defines the transformation $T:(0,1)^ntomathbb{R}^{n-1}$, $mathbf{r}=Tmathbf{p}$. This transformation can be called multinomial transformation (or to be more precise inverse multinomial transformation), since similar formula is used in http://en.wikipedia.org/wiki/Multinomial_logit>multinomial logit model.

This transformation is useful for modelling, since resulting $r_i$ can be any real number, and there is an easy way to transform $r_i$ back to probabilities:

$p_n=dfrac{1}{1+sum exp(r_i)},$

$p_i=exp(r_i)p_n$.

My question is whether there exists a similar transformation for matrices. Suppose we have two probability vectors $mathbf{p}=(p_1,...,p_n)$, $mathbf{q}=(q_1,...q_m)$ and $ntimes m$ matrix $P=(p_{ij})$, satisfying

$sum p_i=1$, $sum q_i=1$

$sum_{j=1}^m p_{ij}=p_i$, for each $i=1,...,n$, (1)

$sum_{i=1}^n p_{ij}=q_j$, for each $j=1,...,m$, (2)

(what we actualy have is a bivariate discrete probability distribution with given marginal distributions).

Now what I am looking for is a transformation which transforms $p_{ij}$ to unbounded real numbers, but such that the inverse would satisfy constraints (1) and (2). In effect I am looking for the bijection from subset of $(0,1)^{nm}$ to $R^{k}$, where $k$ should be $(n-1)(m-1)$.

I suspect that maybe copulas can be involved here, or some properties of stochastic matrices. If somebody could give me any pointers I would be very grateful.

ca.analysis and odes - Asymptotic series for roots of polynomials

If you have a good initial guess (such as your case), then you don't need either undetermined coefficients or Newton's method (although they both work), you can in fact get a procedure which will give you as many coefficients as you want. Harald's first step is excellent, so we'll start from that. Given $F(epsilon,w) = epsilon^2w+epsilon w^2+w^3-1 = 0$, we can derive a differential equation for $w(epsilon)$. With the help of Maple's gfun[algeqtodiffeq], giving
$$ 9+ left( 6{epsilon}^{5}+7{epsilon}^{2} right) w ( epsilon ) + left( 27-6{epsilon}^{6}-7{epsilon}^{
3} right) {frac {d}{depsilon}}w(epsilon) + left( -3{epsilon}^{7}-14{epsilon}^{4}-27epsilon right) {frac {d^{2}}{d{epsilon}^{2}}}w ( epsilon ) $$
with initial conditions $w (0) =1,w^{'''}(0) =-4/9 $ [the DE is singular at $epsilon=0$ so the initial conditions are non-standard). Nevertheless, from there one can continue and use gfun[diffeqtorec] to get a recurrence for the coefficients of the series. The result is
$$ left( 6-3n-3{n}^{2} right) u(n) + left( -98-77n-14{n}^{2} right) u (n+3) + left( -
648-27{n}^{2}-270n right) u (n+6 ) $$
with starting conditions $u(0) =1, u(1) =-1/3,u (2) =-2/9,u (3) ={frac {7}{81}},u(4) =0,u(5) ={frac {14}{729}}$. These are sufficient to allow you to 'unwind' the recurrence as much as you want.

Bjorn's answer gives the underlying 'analytic' answer that justifies that the above gives a convergent result. Actually, the procedure above works in other cases as well, but then the result is only valid in a sector, and computing the size of that sector can be fiendishly difficult.

soft question - What are examples of theorems get extensions based on simple observation?

Perhaps Euler's polyhedral formula:

V(vertices) + F(faces) - E(edges) = 2

provides an example of what you mean?

Euler did not give a proper proof but shortly thereafter this result inspired huge advances that had dramatic effects on the evolution of geometry, topology, convexity, and what today is called graph theory.

One measure of how rich this topic is can be seen from the many types and styles of proofs that can be found for this result collected below by David Eppstein:

http://www.ics.uci.edu/~eppstein/junkyard/euler/

set theory - Explicitly constructing an infinite set with particular size

I would like to preface by saying that I have no significant experience working with set theory, so I'm probably making an intuitive mistake. I have figured out where the mistake probably is, but I can't figure out why it IS a mistake. I figured that this was the best outlet to ask my question.

I was reading about the Continuum Hypothesis on Wikipedia recently, particularly about the fact that it's undecidable in ZFC -- which means that it is undecidable whether or not there is an infinite set whose size is strictly between that of the natural numbers and that of the real numbers. Now, intuitively, a proof that it cannot be disproved would immediately give way to the fact that no counterexample could be constructed (for such a counterexample would disprove it, which is impossible), and thus it must therefore be true. But that's not where I'm going with this.

Cantor proved that the rational numbers are countable -- that there exists a counting method such that, given any integer, you could determine the unique rational number which corresponds to it, and given any rational number, you could determine the unique integer which corresponds to it. The proof is fairly cool, but that's not where I'm going, either. Essentially, this demonstrated that $aleph_0^2 = aleph_0$. Then, he went on to prove that all algebraic numbers were countable, which proved the stronger statement that for any finite n, $aleph_0^n = aleph_0$. But yet, the cardinality of the real numbers is still strictly greater.

He explicitly determined the cardinality of the real numbers as $2^{aleph_0}$, or, strictly speaking, for any $alpha > 1, mathfrak c = alpha^{aleph_0}$, because, no matter which base you're in, the number of real numbers doesn't change. This means that the cardinality of the real numbers is strictly exponential, whereas the cardinality of the countable numbers is strictly polynomial.

This is where my confusion arises. If I construct a set whose size after an infinite number of steps is bounded by any polynomial, it is countable, whereas a set whose size is greater than every polynomial would not be countable.

The Adleman–Pomerance–Rumely_primality_test has running time, for a given n, of $n^{Olog(log(n))}$, which is of super-polynomial running time -- there exists no polynomial that is strictly greater than that function. However, it is also sub-exponential -- there exists no exponential function that is strictly less than it, either. Therefore, it exists between the polynomials and exponentials.

Using this, I can construct a set of numbers whose size after n steps is equal to $f(n) = n^{Olog(log(n))}$ by appending approximately f'(n) unique values to the end of the set. I have now explicitly constructed an infinite set whose size is $aleph_0^{log(log(aleph_0))}$, have I not? And, as I said before, it is, eventually, larger than any set whose size grows polynomially. But it is also smaller than every set whose size grows exponentially.

Of course, it also turns out that this set is countable -- I can give you the numbers in the set, if you so wish. But that's only part of the problem.

Using the same method, I can also construct a set whose size grows exponentially, if I only change my function to $f(n) = 2^n$. I then append f'(n) unique values to the end of my set and, voila, as I take a countably infinite number of steps, namely $aleph_0$, the size of my set becomes $2^{aleph_0}$, the cardinality of the continuum -- but, as before, I can tell the n^th item in my set, and if you give me any item in my set, I can tell you exactly where it lies.

Thus is my question: What did I do wrong? Either I have shown that $2^{aleph_0} = aleph_0$, which is exceedingly unlikely, or my assumption that my new set's size is equal to $2^{aleph_0}$ is incorrect, and I cannot understand why.

Any help would be appreciated, thanks!

--Gabriel Benamy

ac.commutative algebra - elementary classification of artinian rings

this may be too elementary for mathoverflow, but I'll give it a try.

rings are commutative here. it is well-known that every $0$-dimensional noetherian ring is artinian. the standard proof uses a filtration argument; then it's left to show that every finitely generated vector space is artinian (dimension!) and that extensions of artinian by artinian modules are artinian (tage the images and the preimages of the chain, finally both are stable). by a sheaf argument, it's easy to reduce to: every noetherian ring with exactly one prime ideal is artinian.

is there a proof which is somehow more direct? perhaps a clever manipulation of chains of ideals? I don't expect it, but it would be great for the students in my tutorial, which had to solve this as an exercise without knowing anything about artinian or noetherian rings going beyond the definitions.

at.algebraic topology - Meaning of orientation/orientability over rings other than the integers

I just wanted to mention that while orientability for cohomology with arbitrary coefficients is governed solely by cohomology with coefficients in ℤ, there are other cohomology theories for which is is not true. For example, if you have an action of $pi_1(X)$ on an abelian group M, then you can talk about (co)homology with twisted coefficients in M. For any vector bundle there is a coefficient module such that the bundle is orientable with respect to these twisted coefficients (or, to paraphrase Matthew Ando, "every bundle is orientable if you're twisted enough").

Also, one can ask whether a vector bundle is orientable with respect to topological K-theory, real or complex, or many other generalized cohomology theories, which capture interesting information about the manifold.

So while ℤ/m-coefficients may not be the most interesting coefficient systems to study orientability in, they're part of a larger systematic family of questions (and they don't take much extra work if you're already doing ℤ and ℤ/2-coefficients).

Finally, for something like real coefficients you might think of orientations that differ by a real scalar geometrically, e.g. according to some volume form.

ca.analysis and odes - What's a nice argument that shows the volume of the unit ball in $mathbb R^n$ approaches 0?

Edit: As Matthias pointed out, the following argument only works for the ball with radius 1/2.

To measure volume, we need to agree on a unit of volume [1]. The traditional way of doing this is to set the volume of the unit cube to one.

Now, think about the $n$-ball inscribed in the unit $n$-cube. As we increase $n$, the ball's diameter stays constant, but what happens to its volume? When $n = 1$, the ball takes up the whole unit cube, so its volume is one. When $n = 2$, the ball no longer takes up the whole unit cube, so its volume is less than one. When $n = 3$, the ball takes up even less of the unit cube, so its volume is even smaller.

There's an easy way to see that when you go from $mathbb{R}^n$ to $mathbb{R}^{n + 1}$, the fraction of the unit cube occupied by the inscribed ball goes down. Start with an $n$-ball inscribed in the unit $n$-cube, and extrude both objects into the $(n + 1)$st dimenion. Now you have an $(n + 1)$-cylinder inscribed in an $(n + 1)$-cube. The fraction of the $(n + 1)$-cube occupied by the $(n + 1)$-cylinder is clearly the same as the fraction of the $n$-cube occupied by the $n$-ball. It's easy to see, however, that the $(n + 1)$-ball inscribed in the $(n + 1)$-cube fits inside the inscribed $(n + 1)$-cylinder with room to spare.

This argument only shows that the volume of the unit-diameter $n$-ball decreases as $n$ grows; it doesn't show that the volume goes to zero. I'm hopeful, however, that a more sophisticated version of the same argument might do the trick!

Edit: A more sophisticated version of the same argument does do the trick, and Matthias posted it while I was writing my post! Hooray!

---

[1] To be more sophisticated about it: the differential n-form in $mathbb{R}^n$ is only unique up to multiplication by a constant, so we need to settle on a constant.

mathematics education - Interesting applications of max-flow and linear programming

You can prove the Birkhoff-von Neumann theorem directly with linear programming. Depending on your taste it is a quite elegant way to prove that result. There are basically two ways - one to use the conditions for a vertex of a polytope given by constraints to show that a doubly stochastic matrix which is a vertex of the Birkhoff polytope must have a row or column with only one nonzero entry, then induce. This does not use the full "fundamental theorem of linear programming".

The other approach is to observe that at a vertex there is a full dimensional set of linear objectives for which the vertex is optimal, formulate the dual program and then show that the 2n unconstrained dual variables lie on an n dimensional space; complementary slackness then shows that the primal variable has only n nonzero elements, double stochasticity then guarantees there must be one in each row, one in each column, and each must be unity - therefore a permutation matrix. Obviously this approach really does exploit the linear program structure, if that is what you want to teach.

I came up with this myself so don't know of an actual reference, but it should not be that novel.

You can also prove Birkhoff-von Neumann are a max flow/min cut theorem (which is pretty well known) but I do not find that as elegant. However if you are emphasizing max flow/min cut as opposed to the linear programming structure, then you might want to do that one.

examples - What is the first interesting theorem in (insert subject here)?

In most students' introduction to rigorous proof-based mathematics, many of the initial exercises and theorems are just a test of a student's understanding of how to work with the axioms and unpack various definitions, i.e. they don't really say anything interesting about mathematics "in the wild." In various subjects, what would you consider to be the first theorem (say, in the usual presentation in a standard undergraduate textbook) with actual content?

Some possible examples are below. Feel free to either add them or disagree, but as usual, keep your answers to one suggestion per post.

• Number theory: the existence of primitive roots.


• Set theory: the Cantor-Bernstein-Schroeder theorem.


• Group theory: the Sylow theorems.


• Real analysis: the Heine-Borel theorem.


• Topology: Urysohn's lemma.

Edit: I seem to have accidentally created the tag "soft-questions." Can we delete tags?

Edit #2: In a comment, ilya asked "You want the first result after all the basic tools have been introduced?" That's more or less my question. I guess part of what I'm looking for is the first result that justifies the introduction of all the basic tools in the first place.

group actions - Is there an invariant theory explanation of the orbit structure of GL₂ acting on second-diagonal symmetric matrices by g∙X = gXJg^tJ ?

Statement of the Specific Result

Let $J$ denote the matrices with ones on the "second diagonal", meaning the diagonal between the (1,n) and (n,1) entry, and zeros elsewhere. So in the case $n=2$, which is all we'll be concerned with here:

$$
J=begin{pmatrix}
0&1
newline
1&0
end{pmatrix}
$$

Let the transpose of $g$ be denoted by ${}^tg$.

Let $F$ be a field (not necessarily alg. closed: for example, a number field), and consider the action of $G:=mathrm{GL}_2(F)$ on the vector space $mathcal{L}$ of matrices with entries in $F$ which are symmetric about the second diagonal, by $gcdot X= g X J{}^tgJ$ for $ginmathrm{GL}_2(F)$. In explicit coordinates,

$$g=begin{pmatrix}a&b
newline
c&dend{pmatrix}
$$

$$X = begin{pmatrix}gamma&beta
newline
alpha&gammaend{pmatrix}$$

$$J{}^tgJ=begin{pmatrix}d&b
newline
c&aend{pmatrix}
$$

so that this procedure of "taking the transpose of $g$ and conjugating by $J$" amounts to taking the transpose of $g$ along the second diagonal. (See below for Context). In several papers, I find the following stated:

1. $G$ acts on $mathcal{L}$ with an open orbit.
   
2. Given a representative $X_0inmathcal{L}$ of the open orbit, the stabilizer $G_0(X_0)$ is in general reductive, and in this specific example, a one-dimensional torus.
   
3. An point $X$ is called generic if the stabilizer $G_0(X)$ is of type $G_0(X_0)$, and an orbit is called generic if one, equiv. all, its points are generic points. A complete set of representatives of the generic orbits (with exactly one representative from each orbit) is given by the matrices
   $$
   begin{pmatrix}
   0&beta\
   alpha&0
   end{pmatrix}
   $$
   
   with
   $$alpha,betain {F^*}^2backslash F^*$$
   (so with the diagonal element $gamma=0$ and the off-diagonal elements ranging over nonzero square classes of $F^*$ independently).
   

My questions about this result

What I would like to know...

1. 
   is there any tidy way of "characterizing" generic points or orbits, as referred to in items 1 and 2?
   
2. Is there any conceptual or "coordinate-free" way of characterizing the representative set given in item 3?
   

I have a feeling that "standard, classical" invariant theory, especially that of the symplectic group, may give an answer to this. I am not sufficiently familiar with the invariant-theory literature to find this, so if you could point me to a specific reference that I could read and which would allow me to answer these questions, that would be great. Although I have a (partial) confirmation of these facts by brute-force matrix calculations, these are not really ideal to use for my purposes, nor is it clear that they could be carried out by anyone in higher dimensions than 2!

The context, and more on why I expect invariant theory to play a role

The context of this problem is that the symplectic group $mathrm{Sp}_4(F)$ has a standard ("Siegel") parabolic $P$ with Levi factor $M$ isomorphic to $G$, which embeds into $mathrm{Sp}_4$ by $mathrm{diag}(g,J{}^t g^{-1}J)$ (according to one of the two or so common matrix models of $mathrm{Sp}_4$). The nilpotent radical of $P$ can be identified with $mathcal{L}$, and the conjugation action of $M$ is then identified with the action above.

If one considers the analogous situation with $mathrm{SO}_{5}$ (say), and the parabolic $Q$ with Levi factor $mathrm{GL_1}times mathrm{SO}_{3}$, the nilpotent radical of $Q$ can be identified with "row vectors of length 3", and the genericity condition can clearly be expressed as a row-vector representative having non-zero length. Then it is easy to divide "generic" vectors into different $M(F)$-conjugacy classes by the square-class of their (non-zero) lengths. So this very simple invariant-theory interpretation gives me the feeling there is something conceptual going on in the situation I have described, which I am unfortunately missing at the moment.

Thanks for reading and I will greatly appreciate any help!

ca.analysis and odes - How to prove that rational functions satisfy a Lipschitz condition in the *chordal metric*?

By compactness, you only need to prove the Lipschitz property locally. First prove that Möbius transforms are Lipschitz (easy – they are compositions of translations, multiplications by constants, and inversions). Then, by composing with suitable Möbius transforms, you only need to show that a rational function which maps 0 to 0 is Lipshitz on a neighbourhood of 0. This is trivial, I hope you agree.

computability theory - Ackermann function in the Primitive recursive arithmetic

You can express the totality of any computable function in PRA, using Kleene's T predicate, which is primitive recursive. So if you pick any index $e$ for the Ackermann function, the formula $(forall n)(exists t) T(underline{e}, n, t)$ is already in the language of PRA.

However, you cannot prove the totality of the Ackermann function in PRA. One way to see this is to note that PRA is a subtheory of $text{I-}Sigma^0_1$, modulo an interpretation of the language of PRA into $text{I-}Sigma^0_1$. The provably total functions of $text{I-}Sigma^0_1$ are well-known to be exactly the primitive recursive functions.

There is a lot of proof theory literature on provably total functions, which are also called provably recursive functions. But I don't know how much of it focuses specifically on primitive recursive arithmetic. One place to look might be Hájek and Pudlák, Metamthematics of First-Order Arithmetic.

ag.algebraic geometry - Langlands Dual Groups

As Peter mentioned, reductive groups are determined by their root data, and the Langlands dual is given by switching weights with coweights, and roots with coroots.

There is a "construction" of a group from a root datum in SGA III Exp 25 (in vol 3). It starts by reducing to the simply connected semisimple case (meaning there are ways of going from this case to the general case). The weights give you a torus T, and the positive/negative roots give you unipotent groups U+ and U-. You form a scheme Omega = U- x T x U+, and create G by gluing a few disjoint copies of Omega together, and writing down a composition law. This yields a split group over an arbitrary base scheme.

gr.group theory - Poincaré Theorem on presentation from a fundamental polyhedra

Poincaré Theorem on Kleinian groups (groups acting discontinously on Euclidean or hyperbolic spaces or on spheres) provides a method to obtain a presentation of a Kleinian group from a fundamental polyhedra.

I know the proof in Maskit book (Kleinian groups) but I would like to know other proofs.
I also know other proofs for Fuchsian groups (dimension 2) which does not generalize to higher dimension (e.g. Beardon's book, The geometry of discrete groups).

I have two motivations:
1) Maskit proof also proves Poincaré Polyhedra Theorem, which states the necessary and sufficient conditions for a polyhedra to be fundamental polyhedra of some Kleinian group.
I have the filling that a direct proof of the "presentation theorem" should be possible and simpler than proving the "Polyhedra Theorem".

2) Does Poincaré Theorem generalizes to direct product of hyperbolic spaces?

at.algebraic topology - triviality of fibre bundles

are there some general method in judging if a fible bundle is trivial?

At least,for vector bundles,there is a well-developed theory,that is Charicteristic Classes.the triviality of vector bundles is equivalent to the vanishment of its characteristic classes.

for principal bundles,the triviality is equivalent to the exsience of a cross section.
(any good perspective on this assertion?besides,how to tell if there is a cross section)

for general fiber bundle (E,F,B,G),(here,E is total space,F the fiber,B the base space,G the structure group),we can construct its associate principal bundle (E',G,B,G),i.e.to replace the fiber F with the topological group G.there is a theorem that a fiber bundle is trivial iff its associate principal bundle is trivial.

hence the problem is reduced to find a cross section of principal bundle.

I want to know if there is some other methods that are more usable?
Thank you!

algebraic groups - Comparing Iwahori Decompositions

I believe this is true in any building (which IG is). Equivalently, this is one of the usual properties of BN-pairs. The way you can think about it is that I is generated by $Icap w_2 I w_2^{-1}$ and $Icap w_1^{-1}Uw_1$ (since no positive root space can be sent to a negative one by both $w_2$ and $w_1^{-1}$).

convexity - Maximal Ellipsoid

John's Theorem can be stated as "To every compact, convex body, there is a unique inscribed ellipsoid, whose volume is maximal among all inscribed ellipsoids." It goes on to classify this maximal ellipsoid. By using this theorem, one can prove that the ellipsoid of maximal volume which is contained in a square is a circle.

This strikes me as a problem which was probably studied well before Fritz John, and yet I have been unable to prove the statement about squares and circles in an elegant, but low-brow manner. Any thoughts?

game theory - Fairest way to choose gifts

Here's an idea. For any partition $(A,B)$ of $[2n]$, where $|A|=|B|=n$, we can ask each child if they prefer $A$ or $B$. If one prefers $A$ and the other prefers $B$, then we are done. Otherwise, they have the same preference function over all such partitions.

Lemma. There exists partitions $(A,B)$ and $(A',B')$ such that

1. both children prefer $A$ over $B$,




2. both children prefer $B'$ over $A'$, and




3. $(A',B')$ is obtained from $(A,B)$ by performing a single swap.

Proof. Perform swaps until $(A,B)$ becomes $(B,A)$. At some point, each child must switch preferences.

Given the assumption that the gifts are all roughly the same value, it seems fair to offer such an $(A,B)$ as a choice and to flip a coin to decide who gets $A$.

inequalities - subadditive implies concave

Here is a sketch of how you could do this.

Start with a concave function f (ADDED: also, f satisfies all the conditions stated in the question). Consider now the interval $(1,1.1)$. We will try to modify f by increasing its value on this interval while preserving the values at the endpoints. Let's set aside smoothness for now.

What properties should the modified f have? Subadditivity is threateneded only in cases where the $x + y$ happens to land inside $(1,1.1)$. Because of the concavity up to 1, we can see that it suffices to ensure that the new variant $f_1$ of f satisfy:

$$f_1(1 + epsilon) le f(1) + f(epsilon)$$

Basically, we can move the subadditivity condition to the boundary because of the known concavity property of f.

Thus, $f_1$ can be chosen arbitrarily subject to taking the correct boundary values and to being bounded from above by the expression indicated. Assuming strict concavity, there is some free space between the actual current function $f$ and the upper bound for $f_1$ given by the equation. We can therefore choose a $f_1$ that works. Maintaining the condition of being increasing is not problematic -- even if it were, we could just add a huge coefficient linear function to $f_1$ and make it increasing. Smoothness is the relatively harder part, but it could be fixed using bump functions.

EDIT: There's another constraint, that again is achievable: subadditivity when both $x$ and $x + y$ are in the interval $(1,1.1)$.

I also think that functions such as $2x + sqrt{x}e^{-(x-1)^2}$ or variants thereof may work directly, but I don't know of any easy analytical proof of it.

ra.rings and algebras - How many rings exist when ring is subspace of finite dimentional vector space?

Suppose we have ring $R[M]$ over monoid $M$ in real number $R$.The number of generators for the monoid, is finite. Now suppose that every ring element $r$ has decomposition in finite linear basis of n-elements. So ring is subspace of n-dimensional vector space.

How many rings, depending on n has this property? Could You give me some references ( books, preprints, articles) when such theorem is stated or proved?

---

Motivation: Here monoid ring and some structure within it - how is it called? in a comment to the question, Scott Carnahan wrote:

The existence of the decomposition of
elements of R[M] implies the ring is a
subspace of a 4-dimensional real
vector space. There are many such
rings, but only finitely many monoid
rings of dimension less than 5.

Clarification of the problem:

I have finitely presented noncommutative monoid with unity and two generators $g_1,g_2$: M = F/Rel where $Rel = {g_1^2 = e , g_2^2 = g_2 }$, $e$ is unit element and $F$ is free monoid over two generators. Because of relations $Rel$ every elemet in monoid has form for example $g = g_1g_2g_1g_2...g_1g_2$ ( alternating finite sequence with subscripts 1212... or 2121...). Different monoid elements contains different number of multiplications. It is very simple although infinite multiplicative structure.

Then I consider monoid ring over reals $R[M]$. Every element in $R[M]$ has form:

(1) $t = r_1g_1 + r_2g_2 r_3g_1g_2+r_4g_2g_1+ r_5g_1g_2g_1 + ...+r_p g_ig_kg_i...g_s+ ...$ and so on. $r_i in R$ and $g_i in M$.

Note that in general monomial element $g_ig_k...g_s$ every subscript has value in ${1,2}$ and no two following each other subscripts are the same ( they alternate like in sequence like $1212..$ or $2121..$. Of course this is standard ring definition.

In structure, I would like to describe You here, I have strange additional property: there is element $g_3$ in ring $R[M]$ ( but it is not monoid element!) which allows following decomposition:

For every $r in R[M]$ we have

$r= r_0 e + r_1 g_1 +r_2 g_2 +r_3 g_3$

Look: there are only four terms in decomposition, even if You decompose general ring element in the form of (1). However after such decomposition I may only multiply such elements and not add them. In some way it looks like vector space. From the other side such decomposition may be treated as another monoid. So in fact decomposition as above, I trying to treat as some kind of "parametrization" of ring elements. Indeed it is element of some vector space, but there are further requirements on coefficients $r_i$ ( which I do not need to describe here, they in fact are part of monoid definition).

As far as I know this is not standard ring property - maybe I am wrong. If I think about for example polynomial ring (that in simple case is real ring over multiplicative monoid generated by one generator $x$) such decomposition is not possible.

ag.algebraic geometry - Is there a way to check if a relative Hilbert Scheme is reduced?

More specifically, suppose I have a rational curve on a complete intersection, and I know that the relative Hilbert Scheme is not smooth at the point corresponding to my rational curve. Is there any algorithm that will eventually tell me whether the Hilbert Scheme is reduced or not there?

Just to make it harder, this curve happens to pass through the singular locus of the complete intersection.

set theory - Determining set membership from ordering relationships among disjoint sets

Suppose we have a set $P$ (an infinite set), and we have a partition of $P$ into (finitely many) disjoint subsets $P_i$, so that $P = cup_i P_i$, and $P_i cap P_j = emptyset$ for $i neq j$.

Suppose now that we have a set of partial orderings of $P$, such that for each of the orderings, we know that they strictly order some of the partitions. I'm not sure if that is proper math-speak, but what I mean is, for example, for the first partial ordering given, it guarantees $a < b$ for all $a in P_1$ and for all $b in P_2$ (I will write this as $P_1 < P_2$). There may, for example, be another partial ordering given that guarantees $P_2 < P_1$.

My question is, given a finite set of $p_i in P$ with a finite number of partitions $P_j$, as well as a finite set of partial orderings, is it possible to infer set membership of the $p_i$'s into the partitions ($P_j$'s)? If so, then what is required of the $p_i$ and the partial orderings? How many and which kinds of orderings need to be supplied?

Since I don't know the proper math terminology, I'm making this a wiki.

big list - What are your favorite puzzles/toys for introducing new mathematical concepts to students?

Replying to Dan Brumleve's recommendation of the card game "Set" as an answer because my comment was exceeding the maximum allowed size.

The card game "Set" is most analogous to playing tic-tac-toe on a 4-dimensional lattice of size $3 times 3times 3times 3$, allowing for lines to wrap around in that 4-space.

As Douglas Zare described it, the "game asks you to recognize lines in affine 4-space over the field with 3 elements." There are $3^4=81$ cards, with each card showing a design that can be described by 4 attributes, with each attribute containing 3 elements: 3 colors, 3 shapes, 3 levels of shading (outlined, striped, solid), and 3 cardinalities (1,2, or 3).

Twelve cards are initially dealt with players competing to find a grouping of 3 cards such that their attributes are collinear in the 4-d space: either all the same or all different. Thus each line in affine space can also be described by the vector $vin ${$-1, 0,+1$}$^4$ (but excluding ${0}^4$) and a representative member of that grouping. Each card can be also be seen as describing a permutation on the group of the 81 cards.

One quick question that comes out of this is

• Are 12 cards sufficient to guarantee that the cards dealt contains such a collinear grouping?

The answer to that is no. If the people playing concur that the 12 cards initally dealt does not contain 3-points in the 4-d lattice that are collinear, then 3 more cards are dealt, etc.

• What is the minimum number of cards that must be selected to guarantee that that it contains a collinear group of three? The answer to that (21) must be one more than the answer to

What is the maximum number of cards that can be played which do not contain a collinear group of three? (ans=20)

The card game misuses the mathematical term "set" in its name and in its directions for playing the game, since it asks the players to yell out "set!" when they find such a collinear grouping in the 4-d lattice. It should be rightfully called "line", or perhaps most correctly "4-dimensional affine space line over the field with 3 elements". But yelling that out each time would certainly slow down playing the game. :)

http://www.springerlink.com/content/l816l24678517v44/ has an article from the Mathematical Intelligencer about this game by Benjamin Lent Davis, Diane Maclagan and Ravi Vakil.