ca.analysis and odes - Isodiametric Inequality

As people have said, obviously in general you don't have A contained in the ball of diameter diamA, more clearly r=diam A/2. What you can have immediately is A contained in the ball of of radius diam A, i.e. r= diamA.

In the proof of Evans - Gariepy, they use Steiner symmetry lemma, meaning that for any set A Lebesgue measurable, you take any line d. You try to make the set to be symmetric w.r.t. the line d but keep the same volume (this can be make by using Fubini's theorem). It's not so hard to show that, by doing that procedure, the volume remains the same while diamA decreases (philosophically it's correct since make somethings more round, more nice will decrease the diameter).

After using Steiner symmetry lemma for the n axis e_1,...,e_n, you will have the new object, name A' that is symmetric w.r.t. e_1,...,e_n and |A|=|A'| while diamA' le diamA. Notice that A' need not to be the sphere, it can have some diamond shape for example. But, the nice thing is A' subset B(0,diamA'/2).

So, we get the result.

sg.symplectic geometry - When is a symplectic manifold equivalent to a cotangent bundle?

In response to your last paragraph, the so-called "twisted" cotangent bundles provide examples where different symplectic forms exhibit very different dynamics with the same Hamiltonian.

Suppose $omega=dalpha$ is the standard symplectic form on a cotangent bundle $pi:T^{*}Xto X$, where $X$ is a closed manifold. Let $sigma$ denote a closed non-exact two-form on $X$, and consider a new family of two-forms $omega_{s}$ for $sin [0,infty)$ defined by $omega_{s}:=omega-spi^{*}sigma$. It's easily checked that $omega_{s}$ is again a symplectic form on $T^{*}X$ for each $sin [0,infty)$ (it's closed as $sigma$ is closed and non-degenerate as $dpi$ vanishes on "vertical" tangent vectors).

Fix a Riemannian metric $g$ on $X$, and let $H:T^{*}Xto X$ denote the standard "kinetic energy" Hamiltonian defined by $H(x,p):=frac{1}{2}|p|^{2}$, and let $xi_{s}$ denote the symplectic gradient of $H$ with respect to $omega_{s}$ (i.e. $i_{xi_{s}}omega_{s}=-dH$). Let $phi_{s}$ denote the flow of $xi_{s}$.

Let $S^{*}X$ denote the unit cosphere bundle of $X$. Since $H$ is autonomous, the flow $phi_{s}$ preserves $S^*{X}$ for each $sin[0,infty)$. The point is that the dynamics of $phi_{s}$ on $S^{*}X$ can vary dramatically depending on $s$.

As a concrete example of this, consider a closed hyperbolic surface $X=mathbb{H}^{2}/Gamma$, where $Gamma$ is a cocompact lattice of $mathrm{PSL}(2,mathbb{R})$. Let $sigma$ denote the area form on $X$. Note that for $s=0$, $phi_{0}$ is just the cogeodesic flow. For $0le s<1$, the dynamics of $phi_{s}$ is Anosov and conjugate (after rescaling) to the cogeodesic flow. All closed orbits are non-contractible. In this case the unit cosphere bundle is a contact type hypersurface in the symplectic manifold $(T^{*}X,omega_{s})$. For $s=1$ we get the horocycle flow. There are no closed orbits at all, and the unit cosphere bundle is not of contact type (in fact, it's not even stable). For $s>1$ all the orbits are closed and contractible. The unit cosphere bundle is again of contact type, but with the opposite orientation.

Perhaps the best place to read about this is Ginzburg's survey article "On closed trajectories of a charge in a magnetic field: An application of symplectic geometry", which is in the book "Contact and symplectic geometry" (CUP,1994). The recent paper "Symplectic topology of Mane's critical values" by Cieliebak, Frauenfelder and Paternain contains lots of examples of this sort of behaviour.

pr.probability - Limit law for the number of local maxima in a square lattice of IID random variables

For $i, j in { 1, ldots, n }$, let $X_{i,j}$ be a real-valued random variable uniformly distributed on the interval $[0,1]$. The $X_{i,j}$ are independent.

Let $A_{i,j}$ be the indicator random variable of the event that $X_{i,j}$ is a local maximum, i. e. it is the largest of the five random variables $X_{i,j}, X_{i,j+1}, X_{i,j-1}, X_{i-1,j}, X_{i+1, j}$. For the sake of not having to think about boundary conditions, interpret all coordinates modulo $n$.

Then it's clear that $E A_{i,j} = 1/5$, by symmetry. It's also not hard to see that:

• $E (A_{i,j} A_{i,j+1}) = 0$, since we can't have local maxima both at $(i,j)$ and at $(i,j+1)$. (The case where $X_{i,j} = X_{i,j+1}$ can be ignored since it occurs with probability zero.




• I believe $E (A_{i,j} A_{i,j+2}) = 2/45$ and $E(A_{i,j} A_{i+1,j+1}) = 1/20$. (In any case, these are clearly constants, and their exact values don't matter.)




• $E(A_{i,j} A_{k,l}) = 1/25$ for all choices of $i, j, k, l$ other than the ones I already listed and those that clearly are related to them by symmetry. That is, $A_{i,j}$ and $A_{k,l}$ are independent unless the rook-neighborhoods of $(i,j)$ and $(k,l)$ overlap.

From this we can compute the mean and variance of $M_n = sum_{i=1}^n sum_{j=1}^n A_{i,j}$, the total number of maxima. Of course $EM = n^2/5$. The variance is a bit harder to compute and I haven't actually written out the computation, but it ought to be asymptotic to $c^2 n^2$ for some positive $c$. (It is the sum of $Theta(n^2)$ covariances each of order 1.)

Let $tilde{M}_n = (M_n-n^2/5)/(cn)$. Then $tilde{M}_n$ has mean $0$ and variance $1$. Is it true that the sequence ${ tilde{M_n} }_{n=1}^infty$ converges in distribution to the standard normal? Intuitively this seems like it should be true -- we're adding up a bunch of small, almost-independent contributions. If it's true, how can this be proven?

This problem came from last year's qualifying exam in probability at Penn; it's been making the rounds around here over the past few days but nobody seems to remember how to do it.

vector bundles - What are some open problems in algebraic geometry?

There's also Fujita's conjecture.

Conjecture: Suppose $X$ is a smooth projective dimensional complex algebraic variety with ample divisor $A$. Then

1. $H^0(X, mathcal{O}_X(K_X + mA))$ is generated by global section when $m > dim X$.


2. $K_X + mA$ is very ample for $m > dim X + 1$

It's also often stated in the complex analytic world.

Also there are many refinements (and generalizations) of this conjecture. For example, the assumption that $X$ is smooth is probably more than you need (something close to rational singularities should be ok). It also might even be true in characteristic $p > 0$.

It's known in relatively low dimensions (up to 5 in case 1. I think?)

computer algebra - Basis for modular forms of half-integral weight

Edit: Here's a rather silly method that should work if SAGE is just giving you cusp forms: $Gamma_0(4)$ has a single normalized cusp form of weight 6, given by $eta(2tau)^{12} = q - 12q^3 + 54q^5 - dots$, so take your basis of cusp forms of weight $k/2 + 6$, and divide each element by this form to get a basis of the space of modular forms of weight $k/2$.

Edit in response to Buzzard: Thanks for pointing out that I should make this argument. Here is a proof that the cusp form has minimal vanishing at all cusps. $Gamma_0(4)$ is conjugate to $Gamma(2)$ by $tau mapsto 2tau$, so it suffices to check that $Delta(tau)$, the square of $eta(tau)^{12}$, vanishes to twice the minimum order at each cusp of $Gamma(2)$. The quotient $Gamma(1)/Gamma(2) cong S_3$ acts transitively on the cusps of $X(2)$ with stabilizers of order 2, so the quotient map to $X(1)$ has ramification degree 2 at each cusp. $Delta(tau)$ is invariant under the weight 12 action of $Gamma(1)$, and $Delta(tau)$ has minimal vanishing at infinity on $X(1)$.

Old answer: If you have a cusp form of weight $k/2$ for $Gamma_0(4)$ (e.g., given to you by SAGE), you can multiply it by the modular function $frac{eta(tau)^8}{eta(4tau)^8} = q^{-1} - 8 + 20q - 62q^3 + 216q^5 - dots$ to get a modular form of the same weight, that is nonvanishing at one of the three cusps and vanishing at the other two. If you want a form that is nonzero at one of the other cusps, you can multiply by $frac{eta(4tau)^8}{eta(tau)^8}$ (has a pole at zero) or by $frac{eta(tau)^{16}eta(4tau)^8}{eta(2tau)^{24}}$ (pole at $1/2$). [Constant term $-8$ added Sept. 23, in response to an email correction from Michael Somos.]

na.numerical analysis - Multivariate Bisection

cross post in StackOverflow

I need an algorithm to perform a 2D bisection method for solving a $2$x$2$ non-linear problem. Example: two equations $f(x,y)=0$ and $g(x,y)=0$ which I want to solve simultaneously. I have very familiar with the 1D bisection ( as well as other numerical methods ). Assume I already know the solution lies between the bounds $x_1 < x < x_2$ and $y_1 < y < y_2$.

In a grid the starting bounds are:

    ^
    |   C       D
y2 -+  o-------o
    |  |       |
    |  |       |
    |  |       |
y1 -+  o-------o
    |   A       B
    o--+------+---->
       x1     x2

and I know the values at $f(A)$, $f(B)$, $f(C)$ and $f(D)$ as well as $g(A)$, $g(B)$, $g(C)$ and $g(D)$. I might even know for which edges $f=0$ and for which $g=0$.

To start the bisection I guess we need to divide the points out along the edges as well as the middle.

    ^
    |   C   F   D
y2 -+  o---o---o
    |  |       |
    |G o   o M o H
    |  |       |
y1 -+  o---o---o
    |   A   E   B
    o--+------+---->
       x1     x2

Now considering the possibilities of combinations such as checking if $f(G)*f(M)<0$ AND $g(G)*g(M)<0$ seems overwhelming. Maybe I am making this a little too complicated, but I think there should be a multidimensional version of the Bisection, just as Newton-Raphson can be easily be multidimed using gradient operators.

Any clues, comments, or links are welcomed.

ag.algebraic geometry - Numerical evidence of Beilinson's conjecture in local fields and function fields

The famous Beilinson's conjecture predicts a relationship between the regulator map in $K$-theory and special value of $L$-function generalizing the Dirichlet's theorem in number theory. Please see this post for details.

In this paper by Tim Dokchitser, Rob de Jeu, Don Zagier, the authors construct families of hyperelliptic curves over $mathbf{Q}$ of arbitrary genus g with (at least) g integral elements in $K_2$. With these elements they compute the regulator numerical and compare the regulator with special value of $L$-function of these curves which give numerical evidence of Beilinson's conjecture for $K_2$ of curves.

Here are my questions:

Is there explicitly formulated conjectures in p-adic fields and function fields as in number fields? Does one can do similar computations for curves over these fields? If not, what are the main obstacles?

books - How to start Game theory?

I asked this same question about a year ago, so I'm very slightly ahead of you. Here's what I know:

As you probably know, there are two major branches for game theory. There's (for lack of a better term) "economics" game theory dealing with real world situations, economics, politics and the like. I know next to nothing about that. However, I do know a decent amount about combinatorial game theory, which is a little bit more ground in mathematics, and deals with two player games such as Go, Chess, Nim, or Tic-Tac-Toe.

The best introductory text is going to be Conway's Winning Ways, any of the volumes 1-4. These are the books to read to get into any other subset of combinatorial games, in my opinion.

My personal specialization thus far is generalizations of Tic-Tac-Toe called achievement games, which you can read about (along with much more) in Tic-Tac-Toe Theory.

However, if you want to go even further in these studies, you are a little bit out of luck. What's very exciting to me about combinatorial game theory is that it's pretty much a brand new field of mathematics, and right now the best techniques we have to study it are educated guessing/brute-forcing and a little bit of discrete mathematics. Although it's disenchanting sometimes, this also means that there is potentially a world of possible links and connections to other branches of math that we don't know about, and is just out there waiting to be discovered.

career - How important are publications for undergrads?

I am (as always) speaking with regard to mathematics, not CS or or some other field:

Undergraduate publications are not viewed as a requirement, nor even necessarily a big plus, in a grad school application. (By cosmic coincidence* I have a stack of grad school applications to look at tomorrow, so if I change my mind on this I'll let you know.)

There is something to the idea that undergraduate papers are more prevalent now than they used to be. I think this is partly because the worldwide mathematical community is more connected and more collaborative now, so it is less critical to be in the right place at the right time in order to do undergraduate research.

The thing about undergraduate papers is that, unsurprisingly, they are most often not very good compared to papers written by more mathematically mature people. (Of course there are some exceptions, my favorite being Furstenberg's one paragraph Monthly article which gives a topological proof of the infinitude of the prime numbers. But even this, while brilliant, certainly does not represent his best work!) There's also the suspicion -- whether true or not -- that the behind the scenes advisor (who may not even deign to appear as a coauthor on undergraduate work) is likely to be the brains behind the operation.

I think it is a good thing to try to do some research as an undergraduate -- I did a summer REU at Indiana University, which was great -- but to realize that it doesn't matter too much whether it gets written up and/or formally published. Surely there should be a stage in one's mathematical training where one doesn't need to feel the pressure to publish!

*: Not really.

geometry - An exterior angle theorem for n-dimensional polytopes?

In the plane, the exterior angle of a vertex is $pi -$ the standard ("interior") angle, which may be negative in some cases. The following is true for non-weird polygons:

The sum of the exterior angles at each vertex is a full turn ($2pi$ radians).

I am informally calling polygons with self-crossing edges or holes as "weird" -- please do let me know what the standard terminology is. I have seen an extension to 3-dimensional polytopes of this form:

The sum of the exterior angles of a polytope is $4pi$ radians.

In this case, the exterior angle of a vertex is $2pi -$ (the sum of the face angles at that vertex). I have not seen a proof, but I think it is true for non-weird polytopes, and a modified version is probably true for polytopes with nonzero genus.

My question is: is there a general n-dimensional version of these properties?

ag.algebraic geometry - When do divisors pull back?

If you want to pull back a Cartier divisor $D$, you can do that provided the image of $f$ is not contained in the support of $D$: just pull back the local equations for $D$.

If this does not happen, on an integral scheme, you can just pass to the associated line bundle $mathcal{O}_X(D)$ and pull back that, obtaining $f^{*} mathcal{O}_X(D)$; of course you lose some information because a line bundle determines a Cartier divisor only up to linear equivalence.

Fulton invented a nice way to avoid this distinction. Define a pseudodivisor on $X$ to be a triple $(Z, L, s)$ where $Z$ is a closed subset of $X$, $L$ a line bundle and $s$ a nowhere vanishing section on $X setminus Z$, hence a trivialization on that open set. Then you can simply define the pullback of this triple as $(f^{-1}(Z), f^{*} L, f^{*} s)$, so you can always pull back pseudo divisors, whatever $f$ is.

The relation with Cartier divisors is the following: to a Cartier divisor $D$ you can associate a pseudodivisor $(|D|, mathcal{O}_X(D), s)$, where $s$ is the section of $mathcal{O}_X(D)$ which gives a local equation for $D$.

This correspondence is not bijective. First, a pseudodivisor $(Z, L, s)$ determines a Cartier divisor if $Z subsetneq X$; note that in this case enlarging $Z$ will not change the associated Cartier divisor, so to obtain a bijective correspondence with Cartier divisors you have to factor out pseudodivisors by an equivalence relation, which I leave to you to formulate. But if $Z = X$, you only obtain a line bundle on $X$, and you have no way to get back a Cartier divisor.

If you want to know more about this, read the second chapter of Fulton's intersection theory.

lie groups - Understanding moment maps and Lie brackets

I'm not an expert, but I don't find the situation very hard to imagine^{(see footnote below)}. It doesn't use anything beyond the fact that "the Lie algebra is the tangent space of the Lie group at the identity". Moreover, for this purpose it is enough to imagine the tangent space as arrows at the identity of the Lie group, pointing in the directions you can move.

Oh, and a side remark that might be helpful to some. Having a map $M to mathfrak g^*$ that I usually thought of as a "moment map" is exactly the same as having a map $mathfrak g to C^infty(M,mathbb R)$ the questioner is talking about, and the latter is easier to visualize.

---

The simplest case is when $G=mathbb R$. This is the standard case of the (time-independent) Hamiltonian flow. The Lie-algebra $mathfrak g$ is one-dimensional, so a linear moment map $f:mathfrak g to C^infty (M,mathbb R)$ is determined by its value on the unit vector $vec e in mathfrak g$. If $H=f(vec e)$, this is precisely the Hamiltonian on your manifold. The flow with respect to the moment map is precisely the same as the flow of this Hamiltonian.

The next simplest case is when $G=mathbb R^n$. The only difference here is that there are several directions in which you could go. Let ${vec e_1 ,ldots vec e_n} in mathfrak g$ be the vectors at the identity of $G$ that represent the possible directions. Now, a moment map $f:mathfrak g to C^infty (M,mathbb R)$ is determined by n Hamiltonians; for each $1leq i leq n$ we have $H_i = f(vec e_i)$.

Now, any path in $G=mathbb R^n$ will correspond to some flow on the manifold $M$. In particular, if you always go "right" (in the direction of $vec e_1$), the flow will be precisely that of the Hamiltonian $H_1$; the existence of other directions won't matter. If you only go in the direction of $vec e_2$, the flow will be that of $H_2$. For any path, you
can approximate it by a piecewise-linear path that's always parallel to a coordinate axes; the flow will be that of $H_i$ whenever you go parallel to the axes of $vec e_i$.

Finally, if G is any Lie group, it's a manifold, so it locally looks like $mathbb R^n$. Everything I said above about this case still holds, with one exception: the coordinate directions may no longer be "independent". (So, going right, then up, then left, then down, might not bring you exactly to the same point you started) Unfortunately, at this point my expertness runs out, so see the other answers for a detailed explanation. However, algebraically, the condition that you need to add is precisely that the moment map f is a Lie-algebra homomorphism; think of this as an error term you need to add to account for the lack of independence when you change the direction of your piecewise-linear path.

Of course, in practice, in calculating flows you don't need to approximate anything with piecewise-linear paths, as there are algebraic equations will give you the Hamiltonian that corresponds to any direction. In the case of $mathbb R ^n$, the map f will simply be linear, in general there is likely an error term.

Oh, and finally, most algebraic equations are probably simpler if you think of your moment maps as maps $M to mathfrak g^*$.

---

Footnote: When I wrote that everything is simple, I hadn't thought of the error terms yet (see above). However, I still think that for most conceptual purposes, it's best to think of your Lie group as $mathbb R^n$ with some error terms added. Somebody correct me if I'm wrong, but when we describe Lie groups using "structure forms", isn't it precisely a way to make this idea precise?

measure theory - Is every smooth function Lebesgue-Lebesgue measurable?

This is motivated by pure curiosity (triggered by this question). A map $f:mathbb R^ntomathbb R^m$ is said to be Lebesgue-Lebesgue measurable if the pre-image of any Lebesgue-measurable subset of $mathbb R^m$ is Lebesgue-measurable in $mathbb R^n$. This class of maps is terribly inconvenient to deal with but it might be useful sometimes. And maybe it is not that bad in the case $m=1$, especially if the answer to the following question is affirmative.

Question: Is every $C^1$ function $f:mathbb R^ntomathbb R$ Lebesgue-Lebesgue measurable? If not, what about $C^infty$ functions?

I could not figure out the answer even for $n=1$. However, there are some immediate observations (please correct me if I am wrong):

• Since the map is already Borel measurable, the desired condition is equivalent to the following: if $Asubsetmathbb R$ has zero measure, then $f^{-1}(A)$ is measurable.


• If $dfne 0$ almost everywhere, then $f$ is Lebesgue-Lebesgue measurable (because locally it is a coordinate projection, up to a $C^1$ diffeomorphism). So the question is essentially about how weird $f$ can be on the set where $df=0$.


• If the answer is affirmative for $C^1$, it is also affirmative for Lipschitz functions (by an approximation theorem).


• The answer is negative for $C^0$, already for $n=1$. An example is a continuous bijection $mathbb Rtomathbb R$ that sends a Cantor-like set $K$ of positive measure to the standard (zero-measure) Cantor set. There is a non-measurable subset of $K$ but its image is measurable since it is a subset of a zero-measure set.

abstract algebra - Variants of Eisenstein irreducibility

Basically all such criteria boil down to some argument involving the Newton polygon as Kevin Buzzard mentions in the comments. While something as general as your statement has trivial counter examples the following generalization holds:

Let $R$ be a unique factorization domain and $f(x) =a_nx^n+cdots +a_0in R[x]$ with $a_0a_nneq 0$. If the Newton polygon of $f$ with
respect to some prime $pin R$ consists of the only line segment from $(0,m)$ to $(n, 0)$
and if $gcd(n,m) = 1$ then $f$ is irreducible in $R[X]$.

I've heard this called the Eisenstein-Dumas criterion of irreducibility (it also proves the example given in the comments). Another generalization of Eisenstein's criterion is the following:

If $p|a_0,a_1,dots,a_k$ but $p^2not | a_0$ then $f(x)$ has an irreducible factor of degree $geq k+1$

(This is how you prove for example, that a polynomial like $x^n+5x^{n-1}+3$ is irreducible, after checking that it has no linear factors.) If not answering your question, at least I hope that this refreshes your memory of the statement you claim above. :)

modular tensor categories - What's the best reference for actual formulas for RT invariants?

Since no one else has jumped in, I'll try again.

Standard references producing a TQFT from a MTC include Turaev's big book and the shorter notes of Bakalov and Kirillov (available online here).

If you want something less algebraic and more combinatorial (which I doubt), I like the book by Kauffman and Lins, and also some paper(s) of Lickorish's from the early 1990's.

I haven't read the original Reshetikhin and Turaev papers in a long time, but I remember thinking at the time that they were perfectly readable.

higher category theory - references for models of stable infinity categories

There seems to be a little confusion in your question about what stable ($infty$,1)-categories are, so I want to address that first.

The definition of a stable infinity category that I'm familiar with is Jacob Lurie's notion (definition 29 DAGI:Stable Infty-Categories). The prototypical examples of stable infinity categories which this definition is designed to capture are the category of spectra (in the topological setting) and the category of chain complexes in an abelian category (in the algebraic setting). Rather this notion is designed to also capture the derived localization of the category of chain complexes, which is a robust version of the derived category.

The advantage of Jacob's succinct definition is that it is expressed categorically. This means that if you have equivalent notions of ($infty$, 1)-category, then they will yield equivalent notions of stable infinity category. So in this sense Julie Bergner's is also comparing stable infinity categories.

The only other model of stable ($infty$,1)-categories that I know which doesn't quite fit this idea but is close is that of stable model categories. These are to stable ($infty$,1)-categories as ordinary model categories are to ordinary ($infty$,1)-categories. That is they are roughly equivalent to a particularly nice class of stable ($infty$,1)-categories.

Your question also asks about $A_infty$-categories as stable ($infty$,1)-categories, and that is the part which is confusing me.

On the one hand, an infinity category can heuristically be defined as a category with topological spaces of morphisms and where composition is associative only up to higher coherence. All the various notions of $infty$-category make this precise in one way or another. Since you mention characteristic zero, I take you are thinking of the algebraic/chain complex version of $A_infty$-category. This can be related to topological $A_infty$-categories via the Dold-Kan correspondence (assuming your complexes are connective).

One the other hand being enriched in chain complexes is like being enriched in topological abelian groups (or in HZ-spectra in the non-connective case). So maybe the concept you are after is that of spectral category, or category enriched in spectra? I know this has been studied, but I'm lean on references.

One last comment. Later in DAGI, Lurie shows that stable infinity categories are automatically enriched in the stable infinity category of spectra. This is totally analogous to the category of chain complexes in an abelian category being enriched in the category of chain complexes of abelian groups. However, just as being enriched in abelian groups is not enough for your category to be abelian, being enriched in spectra is not enough for your infinity category to be stable.

mg.metric geometry - Is the ratio Perimeter/Area for a finite union of unit squares at most 4?

I cannot give a full proof, but I can reduce the problem to another one that I think some people might know the answer to.

Here is the problem:

PROBLEM 1.
Let $U$ be an open subset of the unit square with rectifiable boundary.
Then
$$
Pge 4A,
$$ where $P$ is the length of the boundary of $U$ and $A$ is the area of $U$.

"PROOF:" I can prove this if $Ale pi/4$ and I have a precise idea how to prove it in general.

First assume $Alepi/4$, say $A=pi r^2$ for some $rlefrac12$.
Since the circle minimizes the boundary length for a given area, we get $Pge 2pi r$ which gives the claim.
(Note that this also works if you assume that the diameter of $U$ is $le 1$ and so this gives a second proof of Proposition 7.1 in the mentioned thesis.)

If $A>pi/4$ this proof doesn't work. We have to replace the circle with the corresponding minimizer inside the square.
I do not know what this minimizer looks like, somewhat like a balloon you blow up inside a box. But minimal surface people might know and once you know the shape of that set, you can compute the boundary length to solve the problem as above.
Q.E.D.

Next I show how to solve the original problem once you have Problem 1.
I use induction on the number of squares.
For $n=1$ there is nothing to show.
We do $nto n+1$.
So let a set $F$ in the plane be given which is the union of $n$ unit squares.
Let $A(F)$ be its area and $P(F)$ its boundary length. By induction hyothesis we have $P(F)le 4A(F)$.
Add another unit square $S$, then the boundary length of the new set will be $P(F)+4-P$, where $P$ is the boundary lenght of some open set $U$ inside $S$, more precisely, $U$ is the intersection of the interior of $S$ and the interior of $F$.
Anyway, $U$ is the interior of a polygon, hence has rectifiable boundary.
The area of $Fcup S$ is $A(F)+1-A$, where $A$ is the area of $U$.
Problem 1 now tells us $Pge 4A$.
Together with the induction hypothesis this gives the claim.
Q.E.D.

Representations of finite commutative band semigroups

Firstly, if you're only interested in finite semigroups, I suggest amending the title of the question ;)

Commutative semigroups in which ever element is idempotent are called semilattices, and are a special case of the more general notion of inverse semigroup. They have been much studied, although their infinite-dimensional representation theory seems slightly less well mapped out. In any case, the correspondence you describe between semilattices as semigroups and semilattices as certain kinds of poset is indeed well known.

If you're only interested in finite semilattices, then I think the paper to look at is one by Solomon where he gives explicit formulas for the central idempotents that generate the semigroup ring of a given finite (semi)lattice. Unfortunately I've forgotten the exact reference, but I think similar versions or improvements are discussed in

C. Greene, On the M"obius algebra of a partially ordered set, Adv. in Math. 10 (1973), 177--187.

G. Etienne, On the M"obius algebra of geometric lattices, Eur. J. Combin. 19 (1998), 921--933.

(Off the top of my head, for representations of more general finite semigroups, see recent work of Benjamin Steinberg, which is where I first became aware of some of the older work.)

I would have thought that the representation theory of rectangular bands should actually be much easier than that of general bands, since every rectangular band can be rewritten as $L times R$ for index sets $L, R$ and with multiplication defined by $(l_1,r_1)cdot (l_2,r_2)=(l_1,r_2)$.

lo.logic - Algebrization of second-order logic

Did programme of predicate calculus algebraization succeed? In his essay An autobiography of polyadic algebras Halmos outlined why he is not satisfied with his baby. I suggest that Cylindric algebras are not genuine algebras either; what other algebraic structures have operators parametrized by something (cylindrification)? The situation is strikingly different from either boolean or relation algebra, each having a set of intuitive binary, unary, [and 0-ary] operations.

I suggest that the culprit is positional perspective onto relation attributes. Positional perspective is ubiquitous in math (sequences, function arguments, etc), so is easy to see why it penetrated into the world of relations. Positional perspective makes perfect sense for binary relations, this is why nobody challenged its adequacy for n-ary ones. However, it is easy to see that attribute positions are not essential to the ability to match values of the two different attributes of two different relations. Named attribute perspective is widely used in database theory and practice, which implies yet another algebraization of predicate logic.

cryptography - The Discrete Logarithm problem

I am puzzled with the following discrete logarithm problem:

Given positive integers b, c, m where (b < m) is True it is to find a positive integer e such that

(b**e % m == c) is True

where two stars is exponentiation (e.g. in Ruby, Python or ^ in some other languages) and % is modulo operation. Using general math symbols it looks like:($b^e equiv c (mod m)$).

What is the most effective algorithm (with the lowest big-O complexity) to solve it ?

Example:
Given b=5; c=8; m=13 this algorithm must find e=7 because 5**7%13 = 8

Thank you in advance!

co.combinatorics - Round-Robin Tournaments and Forests

In a round-robin tournament with $n$ teams, each team plays every other team exactly once. Thus, there are $n(n-1)/2$ total games played. How many different standings can result? By a "standing" I mean the ordered sequence $(W_1, ldots, W_n)$ where $W_i$ is the number of wins by the $i$th player. Assume that no game ends in a tie.

I wrote a program to calculate this sequence. For $n$ up to 13, it agrees with the number of forests on $n$ labeled nodes, which is Sequence A001858 in the OEIS. But I can't see the correspondence between tournament standings and forests with labeled nodes. Can anyone explain this?

differential equations - Existence of non-trivial solutions to Dirichlet problem with a potential lying between eigenvalues.

I assume $Omega$ is a bounded open subset of $mathbb{R}^n$ and, say, $Vin L^infty(Omega)$. What you say is correct, but of course you need to assume that the eigenvalues are also consecutive (otherwise e.g. $V$ itself could be another eigenvalue in between [edit: as you actually said]). The comparison principle you are seeking comes from the Courant–Fischer–Weyl variational characterization of the eigenvalues of $-Delta + V,$ by which the $k$-th eigenvalue $lambda_k(-Delta + V\ )$ in the increasing order is expressed as a certain min-max of the Rayleigh quotient, which is monotone wrto $V:$
$$frac{int_Omega left(|nabla u|^2+V(x)u^2 right)dx }{int_Omega u^2 dx }$$
(for a precise statement see e.g. Courant-Hilbert, or Reed-Simon, or Gilbarg-Trudinger, &c).

So if we denote $lambda_k:=lambda_k(-Delta)$ and assume $-lambda_{k+1} < V(x) < -lambda_{k}\ ,$ it follows by the monotonicity
$$lambda_k(-Delta+V)<lambda_k(-Delta - lambda_k)=0 =lambda_{k+1}(-Delta - lambda_{k+1})< lambda_{k+1}(-Delta +V),$$
so that $0$ is not an eigenvalue of $-Delta +V.$

Convergence of a sequence of continuable Dirichlet series

Let's say $f$ is a Dirichlet series which converges on the half-plane $text{Re }s>sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation. Let $g_n$ be a sequence of Dirichlet series, also convergent on $text{Re }s>sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary. We assume that $g_n$ converges to $f$ in the following sense: for every $m>0$ there exists an $N$ for which the series $g_n$ and $f$ match on every term up to the $m$th, for all $n>N$. Note this implies that $g_n(s)$ converges to $f(s)$ for every $text{Re }s>sigma$.

Can it be said that $g_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?

Perhaps that's too much to hope for, and you can't even expect that $g_n(s)$ converges to $f(s)$ even for the point $s=sigma$. I'd certainly be interested in a counterexample which does this!

examples - Does any method of summing divergent series work on the harmonic series?

disclaim: I'm a student major in physics and not in math, with inadequate knowledge of complex analysis, so this answer may have severe mistakes....
It can be understood via Hartogs theorem, at least partially.
Recall that Hartogs extension thm tells us that a complex function with several variables analytic in the connected OS, where O is open and S is compact, can be extended to S. Then the failure of extension to some subset of S indicates the bad behavior of the whole singularity set.
The main idea is to find some f(z,x) with $f(n,x_0)=H_n$, then try to define the harmonic series as $f(infty,x_0)$.
Example1: $Sigma_{n=1}^{infty} frac{x^n}{n}$, the function here is some extension of $f(z,x)=Sigma_{n=1}^{n=z}frac{x^z}{z}$, which can be special value of Lerch function and yet doesn't matter here, (in the following we may use the $z=infty$ or $x=infty$ charts implicitly, so you should apply $z to 1/z$ and something like that) and O is some neighborhood of $(z=infty,x=1)$. Yet $f(infty,x)=-text{Log}(1-x)$ has a brunch point at $x=1$ and a brunch cut running to $x=infty$, ie, S is noncompact.
Example2: $Sigma_{n=1}^{infty} frac{n^x}{n}$, this time the value at $z=infty$ is zeta function with an isolated pole, yet $f(z,1-x)=zeta(x)-zeta(z,x)$, the Riemann zeta is analytic for $xneq1$, and Hurwitz zeta is usually defined for $z>0$ and $xneq 1$. Roughly the picture is that f is singular at $x=0$, which is removable, and at a family of x-planes located at {z=negative integers} acumulated around z=infty, thus S is noncompact.
Example3: for the $Sigma_{n=1}^{infty} frac{n^x+n^{-x}}{2n}$ regularization appearing in http://math.stackexchange.com/questions/20005/is-it-possible-to-use-regularization-methods-on-the-harmonic-series, the reason is that the singularities are cancelled exactly in pairs and $z=infty, x=0$ is removable for Hartogs thm.
Posible relation with renormalization: the trick here is to choose proper "conter-terms" cancelling the poles exactly - this is the regularization sheme, just like dimensional regularization, but this leaves constant factors unfixed, then the condition $f(n,x_0)=H_n$ comes to rescue - this is alike the renormalization scheme: we use renormalization conditions to connect the regularized results with true values (experimental values). Yet I think it's differnt from other types of resummation methods since substracting poles will change the value of convergent series.

gr.group theory - Does subgroup structure of a finite group characterize isomorphism type?

The answer is no in general. I.e, there are finite non-isomorphic groups G and H such that there exists a bijection between their elements which also induces a bijection between their subgroups.

For this, I used two non-isomorphic groups which not only have the same subgroup lattice (which certainly is necessary), but also have the same conjugacy classes. There are two such groups of size 605, both a semidirect product $(C_{11}times C_{11}) rtimes C_5$ (see this site for details on the construction). In the small group library of GAP, these are the groups with id [ 605, 5 ] and. [ 605, 6 ]. These are provably non-isomorphic (you can construct the groups as described in the reference I gave, and then use GAPs IdSmallGroup command to verify that the groups described there are the same as the ones I am working with here). With a short computer program, one can now construct a suitable bijection.

First, let us take the two groups:

gap> G:=SmallGroup(605, 5);    
<pc group of size 605 with 3 generators>
gap> H:=SmallGroup(605, 6);
<pc group of size 605 with 3 generators>

The elements of these groups are of order 1, 5 or 11, and there are 1, 484 and 120 of each. We will sort them in a "nice" way (that is, we try to match each subgroup of order 5 to another one, element by element) and obtain a bijection from this. First, a helper function to give us all elements in "nice" order:

ElementsInNiceOrder := function (K)
    local elts, cc;
    elts := [ One(K) ];
    cc := ConjugacyClassSubgroups(K, Group(K.1));
    Append(elts, Concatenation(List(cc, g -> Filtered(g,h->Order(h)=5))));
    Append(elts, Filtered(Group(K.2, K.3), g -> Order(g)=11));
    return elts;
end;;

Now we can take the elements in the nice order and define the bijection $f$:

gap> Gelts := ElementsInNiceOrder(G);;
gap> Helts := ElementsInNiceOrder(H);;
gap> f := g -> Helts[Position(Gelts, g)];;

Finally, we compute the sets of all subgroups of $G$ resp. $H$, and verify that $f$ induces a bijection between them:

gap> Gsubs := Union(ConjugacyClassesSubgroups(G));;          
gap> Hsubs := Union(ConjugacyClassesSubgroups(H));;
gap> Set(Gsubs, g -> Group(List(g, f))) = Hsubs;
true

Thus we have established the claim with help of a computer algebra system. From this, one could now obtain a pen & paper proof for the claim, if one desires so. I have not done this in full detail, but here are some hints.

Say $G$ is generated by three generators $g_1,g_2,g_3$, where $g_1$ generates the $C_5$ factor and $g_2,g_3$ generate the characteristic subgroup $C_{11}times C_{11}$. We choose a similar generating set $h_1,h_2,h_3$ for $H$. We now define $f$ in two steps: First, for $0leq n,m <11$ it shall map $g_2^n g_3^m$ to $h_2^n h_3^m$.

This covers all elements of order 1 or 11, so in step two we specify how to map the remaining elements, which all have order 5. These are split into four conjugacy classes: $g_1^G$, $(g_1^2)^G$, $(g_1^3)^G$ and $(g_1^4)^G$. We fix any bijection between $g_1^G$ and $h_1^H$ and extend that to a bijection on all elements of order 5 by the rule $f((g_1^g)^n)=f(g_1^g)^n$. With some effort, one can now verify that this is a well-defined bijection between $G$ and $H$ with the desired properties. You will need to determine the subgroup lattice in each case; linear algebra helps a bit, as well as the fact that all subgroups have order 1, 5, 11, 55, 121 (unique) or 605. I'll leave the details to the reader, as I myself am happy enough with the computer result.

terminology - How do you pronounce "Hartshorne"?

He prefers it be pronounced as in Hart's Horn. I asked him a few years ago, our brief common ground being assisting Marvin Jay Greenberg with revisions for the fourth edition of his book on Euclidean and non-Euclidean geometry. That is not to say that I have ever heard anyone else say it that way. But then few people get my name right.

fa.functional analysis - Good books on theory of distributions

Grubb's recent Distributions And Operators is supposed to be quite good.

There's also the recommended reference work, Strichartz, R. (1994), A Guide to Distribution Theory and Fourier Transforms

The comprehensive treatise on the subject-although quite old now-is Gel'fand, I.M.; Shilov, G.E. (1966–1968), Generalized functions, 1–5,.

A very good,though quite advanced,source that's now available in Dover is Trèves, François (1967), Topological Vector Spaces, Distributions and Kernels That book is one of the classic texts on functional analysis and if you're an analyst or aspire to be,there's no reason not to have it now. But as I said,it's quite challenging.

That should be enough to get you started.And of course,if you read French,you really should go back and read Schwartz's original treatise.

dg.differential geometry - Representations of SU(2) and tensors on SU(2)

A cop-out answer is to read Chapters 22-3 (maybe 21 also) of Lam's Topics In Contemporary Mathematical Physics. These chapters are viewable in Google Books at

http://books.google.com/books?id=bXq_M3_t1voC&dq=Topics+in+Contemporary+Mathematical+Physics

The treatment is accessible to undergraduates. I have included notes and errata below.

---

Chapter 21
p. 223: NB. Regarding equation (21.19), note that the Young diagrams ${6,5,4,1}$, ${6,4,4,2}$, and ${6,4,4,1,1}$ aren't included on the RHS because they have more than $n = 3$ rows.

p. 228. For Corollary 21.1, see also Theorem 19.19 in Fulton and Harris. At the bottom of the page, other legitimate Young diagrams with two rows should be included.

p. 229. At the top of the page, legitimate Young diagrams of the form (e.g.) ${r_1, r_2,1}$ should be included. Regarding the first full paragraph, see section 19.5 of Fulton and Harris for Weyl's ``associated" diagrams.

p. 230: The reference to equation (20.71) should be to (20.72); just below, note that $dim([mu_1]) = tbinom{mu_1 + n - 1}{n - 1}$ equals the expressions given for $n=3$. The ``Completely" after equation (21.50) should be decapitalized. It may be worth highlighting (or elaborating on) the reference to spin.

Chapter 22
p. 234: In the first line of equation (22.3), the term $S_{2j}(e_1^{(2j)})$ should be $S_{2j}(e_1^{2j})$. In the text immediately following, some kind of reference should be made along the lines of the sentence beginning ``Note that in the above equation..." in the paragraph after equation (22.9).

p. 235: NB. $u' equiv left(
begin{smallmatrix}
u^1 & u^2
end{smallmatrix}
right) left(
begin{smallmatrix}
a & b \
-overline{b} & a
end{smallmatrix}
right)$, cf. (22.22).

p. 237: In equations (22.17) and (22.19), factorial symbols need to accompany all the terms under the square root. The reference to equation (22.25) near the end of the page should be to (22.15).

p. 238: In equations (22.24-5), factorial symbols need to be added for the terms in the square roots.

p. 239: quqntum" should readquantum".

p. 240: NB. $a = e^{-iphi/2}$ and $b=0$ implies equation (22.30).

p. 241: There should be a mention to the effect that there exists $pi$ such that the eighth equality of equation (22.42) holds.

p. 242: Factorial symbols need to go in the square root in equation (22.47).

p. 245: In the penultimate sentence of the middle paragraph, a reference could profitably be made to (22.35).

p. 246: I skipped these exercises, but the theorem numbers appear to be low by 1.

Chapter 23
p. 249: Regarding the first sentence of the first full paragraph, note that $P^1$ is invariant under $SO(3)$, and the invariance of $P^k$ follows. Equation (23.13) should be $H^k equiv ker(nabla^2)|_{P^k}$.

p. 250: NB. At the bottom of the page, draw a correspondence $2theta rightsquigarrow phi$. Note that
begin{equation}
D(R_3(2theta)) cdot begin{cases}
x pm iy \
z
end{cases} = begin{cases}
(x cos 2theta - y sin 2theta) pm i(x sin 2theta + y cos 2theta) \
z
end{cases} = begin{cases}
(x pm iy)e^{pm 2itheta} \
z
end{cases}.
end{equation}

p. 251: The unnumbered equation and its preceding text is not quite right: see (the results of) exercise 23.2.

p. 252: Regarding the comment for p. 251 and exercise 23.2, note that $frac{x pm iy}{r} = e^{pm iphi}sin theta$ and $frac{z}{r} = cos theta$. It follows that $Y_2^0 = sqrt{frac{5}{16pi}}left( 2 left(frac{z}{r}right)^2 - left(frac{x+iy}{r}right)left(frac{x-iy}{r}right) right), quad Y_2^1 = -sqrt{frac{15}{8pi}} left(frac{z}{r}right)left(frac{x+iy}{r}right), quad Y_2^2 = -sqrt{frac{15}{32pi}} left(frac{x+iy}{r}right)^2;$
$Y_3^0 = sqrt{frac{7}{16pi}}left( 2 left(frac{z}{r}right)^3 - 3left(frac{z}{r}right)left(frac{x+iy}{r}right)left(frac{x-iy}{r}right) right), quad Y_3^1 = -sqrt{frac{21}{64pi}} left(frac{x+iy}{r}right) left( 4 left(frac{z}{r}right)^2 - left(frac{x+iy}{r}right)left(frac{x-iy}{r}right) right),$
$Y_3^2 = sqrt{frac{105}{32pi}} left(frac{z}{r}right)left(frac{x+iy}{r}right)^2, quad Y_3^3 = -sqrt{frac{35}{64pi}} left(frac{x+iy}{r}right)^3.$

p. 253: Before equation (23.44) reference equations (18.5-6) and discussion preceding them; also see (23.52-3). For equation (23.46), include $J^3|El0rangle = 0$. The reference to (22.42) should be to (22.41).

nt.number theory - Are there pairs of consecutive integers with the same sum of factors?

Background/Motivation

I was planning to explain Ruth-Aaron pairs to my son, but it took me a few moments to remember the definition. Along the way, I thought of the mis-definition, a pair of consecutive numbers with the same sum of divisors. Well, that's actually two definitions, depending on whether you are looking only at proper divisors. Suppose all divisors. I quickly found (14,15) which both have a divisor sum (sigma function) of 24. Some more work provided (206,207) and then a search on OEIS gave sequence A002961.

What about proper divisors only? (2,3) comes quickly, but then nothing for a while. Noting that the parity of this value ($sigma(n) -n$) is the same as that of $n$ unless $n$ is a square or twice a square, any solution pair must include one number of that form. With that much information in hand, I posted this problem at the reference desk on Wikipedia. User PrimeHunter determined that there were no solutions up to $10^{12}$, but there were no general responses.

Aside from the parity issue, I haven't found other individual constraints that would filter the candidates--the number of adjacent values identical modulo $p$ for other small primes is at least as great as would be expected by chance, and there are a fair number of pairs that are arithmetically close.

Other than (2,3), are there pairs of consecutive integers such that $sigma(n)-n = sigma(n+1)-(n+1)$?

ac.commutative algebra - Is a valuation domain PID when its maximal ideal is principal?

I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x in K^{times}$, at least one of $x,x^{-1}$ lies in $R$.

In this case, I believe the answer is no. Let $R$ be any valuation domain whose value group $K^{times}/R^{times}$ is isomorphic, as a totally ordered abelian group, to $mathbb{Z} times mathbb{Z}$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal.

For some information on valuation rings, see e.g. Section 17 of

http://math.uga.edu/~pete/integral.pdf

nt.number theory - Hasse principle for a group

Given a group $g$, a possibly nonabelian $g$-module $G$, and a family of subgroups $(h_i)_{i in I}$ of $g$, Ono defines a pointed Shafarevich-Tate set, say $(S,0)$. He says the Hasse Principle holds for $G$ (with respect to the data of the $g$-action and the set of subgroups ${h_i}$) if the $S = {0}$.

Namely, for each subgroup $h_i$ of $g$ there is a restriction map in Galois cohomology

$r_i: H^1(g,G) rightarrow H^1(h_i,G)$.

(The restriction map is defined on one-cocycles merely by pulling back by the inclusion map $h_i hookrightarrow g$.) Restriction carries the distinguished (trivial) class to the trivial class.

Then $S$ is defined as the intersection of the kernels of all the $r_i$'s. Evidently the
trivial class $0$ lies in $S$, so $(S,0)$ is a pointed set.

All of the above was just a detailed review of the beginning of Ono's paper. Now let me explain why this generalizes the notion of the Shafarevich-Tate group of an abelian variety A over $mathbb{Q}$ [or take a more general global field, if you like].

The Shafarevich-Tate group $Sha(A,mathbb{Q})$ is the set of all principal homogeneous spaces (henceforth phs) $X$ under $A$ which have $mathbb{Q}_p$-points for every prime $p$ and also $mathbb{R}$-points. Because the automorphism group of a phs under a group $A$ is just $A$ itself, by [what I call] the first principle of Galois cohomology, the pointed set of all phs under A is isomorphic to the Galois cohomology set

$H^1(mathbb{Q},A) = H^1(mathfrak{g}_{mathbb{Q}},A(overline{mathbb{Q}}))$,

where $mathfrak{g}_{mathbb{Q}} = Aut(overline{mathbb{Q}}/mathbb{Q})$ is the absolute Galois group of $mathbb{Q}$. [Since $A$ is commutative, the $H^1$ is itself a commutative group, whereas for nonabelian $A$ it would in general be only a pointed set.]

Thus here we have $G = A(overline{mathbb{Q}})$ and $g = mathfrak{g}_{mathbb{Q}}$. What are the $h_i$'s? For each prime $p$, $h_p$ is the Galois group of $mathbb{Q}_p$,
viewed as a subgroup of $mathbb{Q}$ (i.e., as a decomposition group at $p$) via choosing an embedding of the algebraic closure of $mathbb{Q}$ into the algebraic closure of $Q_p$; also we define $h_{infty}$ to be the restriction to the subgroup generated by a complex conjugation, i.e., a group isomorphic to $Aut(mathbb{C}/mathbb{R})$. A cohomology class lies in the kernel of $h_p$ iff the corresponding phs acquires a point after base extension to $Q_p$ (and similarly for $h_{infty}$.

Thus the Shafarevich-Tate pointed set of $A$ is indeed a special case of Ono's construction.

That's the motivation I can give you. As to exactly why Ono's particular choice of Shafarevich-Tate set for an arbitrary group $G$ -- namely take $g = G$ with the conjugation action, and let $(h_i)_{i in I}$ be the family of cyclic subgroups of $G$ -- is interesting and natural...I can't help you there, and I'd like to know myself.

Why are you interested in this paper?

cv.complex variables - Example of continuous function that is analytic on the interior but cannot be analytically continued?

I found some neat stuff in Remmert's Classical topics in complex function theory.

Fabry's gap theorem gives a way to construct many examples including some already mentioned. Stated for the unit disk, it says:

If $m_1,m_2,ldots$ is a sequence of positive integers such that $displaystyle{lim_{ntoinfty}}frac{m_n}{n}=infty$ and if $displaystyle{f(z)=sum_{n=1}^infty a_nz^{m_n}}$ has radius of convergence 1, then the unit disk is the domain of holomorphy of $f$.

For example, if $p_n$ is the $n^{th}$ prime, then $$f(z)=sum_{n=1}^infty frac{z^{p_n}}{n^2}$$ converges uniformly on the closed disk and is therefore continuous. It is not analytically extendable to any larger set because it satisfies the hypotheses of Fabry's theorem.

---

An interesting result that yields many such functions in a nonconstructive way is a theorem of Fatou-Hurwitz-Pólya:

If $displaystyle{f(z)=sum_{n=0}^infty a_n z^n}$ has radius of convergence 1, then the set of functions $$f_epsilon(z)=sum_{n=0}^infty epsilon_na_nz^n$$ for $epsilon_nin{pm1}$ whose domain of holomorphy is the unit disk has cardinality $2^{aleph_0}$.

Hausdorff showed further that if $displaystyle{lim_{ntoinfty} |a_n|^{1/n}}$ exists (and equals 1) then the set of such functions whose domain of holomorphy is not the unit disk is at most countable. This applies in particular to the function $displaystyle{f(z)=sum_{n=1}^infty frac{z^n}{n^2}}$, which therefore yields examples by changing the signs of the coefficients in all but countably many ways.

---

One more, this time an explicit example from Remmert: The series $$f(z)=1+2z+sum_{n=1}^inftyfrac{z^{2^n}}{2^{n^2}}$$ is one-to-one and has real derivatives of all orders on the closed disk, and has the open disk as domain of holomorphy.

Reference: Remmert's Classical topics in complex function theory, pages 252-258. (Fatou-Hurwitz-Pólya is stated on a page without preview.)

exposition - Is it best to run or walk in the rain?

According to the Norwegian meterological institute, the answer is that it is best to run. According to Mythbusters (quoted in the comments to that article), the answer is that it is best to walk.

My guess would be that this is something that can be properly modelled mathematically and solved. I suspect that I'm not alone in making a start on this, but as it's usually during a rainstorm that I think of it, I don't get far before deciding that the best strategy is to call a taxi!

I would also guess that someone has already done this and figured out the answer. If I'm right about that, I would like to read about it and learn what the real answer is. Thus my question is most likely: can someone provide a reference for this?

(If I'm wrong that it's already been solved, then some pointers as to how to solve it or even an explanation of what the main issues are, would be most interesting.)

As well as simply satisfying my curiosity, this is also one of those questions that non-mathematicians often ask (or at least, that non-mathematicians can easily understand the motivation for the question) so if there's a reasonable answer then it would be a good way to show off some mathematics.

---

Update This question has a long and curious history. There's a discussion about it at the meta site: http://tea.mathoverflow.net/discussion/90/question-about-walking-in-the-rain

Overview of automorphic representations for $SL(2)/{mathbf{Q}}$?

In short: what does Labesse-Langlands say?

Slightly more precise: what are the cuspidal automorphic representations of $SL_2(mathbf{A}_{mathbf{Q}})$, together with multiplicities? Let's say that I have a complete list of the cuspidal automorphic representations of $GL_2/mathbf{Q}$ and I want to try and deduce what is happening for $SL_2$. I am looking for "concrete examples of the phenomena that occur".

Now let me show my ignorance more fully. My understanding from trying to read Labesse-Langlands is the following. The local story looks something like this: if $pi$ is a smooth irreducible admissible representation of $GL(2,mathbf{Q}{}_p)$ then its restriction to $SL(2,mathbf{Q}{}_p)$ is either irreducible, or splits as a direct sum of 2 non-isomorphic representations, or, occasionally, as a direct sum of 4. One interesting case is the unramified principal series with Satake parameter $X^2+c$ for any $c$; this splits into two pieces (if I've understood correctly) (and furthermore these are the only unramified principal series which do not remain irreducible under restriction). Does precisely one of these pieces have an $SL(2,mathbf{Z}{}_p)$-fixed vector? And does the other one have a fixed vector for the other hyperspecial max compact (more precisely, for a hyperspecial in the other conj class)? Have I got this right?

Packets: Local $L$-packets are precisely the J-H factors for $SL(2,mathbf{Q}{}_p)$ showing up in an irreducible $GL(2,mathbf{Q}{}_p)$-representation. So they have size 1, 2, or 4.

Now globally. a global $L$-packet is a restricted product of local $L$-packets (all but finitely many of the components had better have an invariant vector under our fixed hyperspecial max compact coming from a global integral model). Note that global automorphic $L$-packets might be infinite (because $a_p$ can be 0 for infinitely many $p$ in the modular form case). My understanding is that it is generally the case that one element of a global $L$-packet is automorphic if and only if all of them are, and in this case, again, generally, each one shows up in the automorphic forms with the same multiplicity. What is this multiplicity? Does it depend?

Finally, my understanding is that the above principle (multiplicities all being equal) fails precisely when $pi$ is induced from a grossencharacter on a quadratic extension of $mathbf{Q}$. In this case there seems to be error terms in [LL]. Can someone explain an explicit example where they can say precisely which elements of the packet are automorphic, and what the multiplicities which which the automorphic representations occur in the space of cusp forms?

I find it very tough reading papers of Langlands. My instinct usually would be to press on and try and work out some examples myself (which is no doubt what I'll do anyway), but I thought I'd ask here first to see what happens (I know from experience that there's a non-zero chance that someone will point me to a website containing 10 lectures on Labesse-Langlands...)

Edit: I guess that there's no reason why I shouldn't replace "cuspidal" by "lies in the discrete series" with the above (in the sense that the questions then still seem to make sense, I still understand everything (in some sense) for $GL_2$ and I still don't know the answers for $SL_2$)

nt.number theory - Can every finite graph be represented by one prescribed sequence of natural numbers?

The answer is "yes": there is such a family of $F$ functions. In fact, a single computable function, acting on a single integer argument, suffices. We may do this by storing essentially complete information about the graph, and about the process of "constructing" the graph $G$ (that is, the process of computing suitable integers $n_j$ representing vertices $v_j in V(G)$), in the integers $n_j$ themselves.

Edit: My original answer contained an error in which the correct adjacency relations were not properly produced. I correct this below, and have taken the opportunity to make other minor revisions.

We may specify a graph on $n$ vertices using a single integer by a number of different methods, such as using binary representation to use an $n^2$ bit integer to give the adjacency matrix of the graph. (The leading 1 represents an entry in the diagonal, which may serve to define the size of the graph without denoting any edges if we consider only simple graphs.) Denote this binary representation of the graph $G$ by $g in mathbb N$. We compute the integers $n_j$ by carrying $g$ as the exponents of different primes, and using other prime factors to "induce" adjacency between the integers representing different integers.

Let $p_j$ denote the $j^{textrm{th}}$ prime, with $p_1 = 2$, $p_2 = 3$, etc. We define
$$ N_j = p_j^g $$
which will act as a "label" of sorts for our vertices; each vertex $v_j in V(G)$ (for an arbitrary ordering of the vertices) will be represented by an integer which is divisible by the prime $p_j$, but not by any other prime $p_k$ for $1 le k le n$. At the same time, this labelling carries with it an entire description of the graph, as well as the vertex ordering (in this case, given by the ordering of the rows/columns of the adjacency matrix of $G$). The smallest prime factor of the integer $n_j$ corresponding to the vertex $v_j$ will indicate which vertex it is in the order, and carry a complete description of the graph $G$.

We may induce adjacency among the vertices by giving them appropriate common prime factors. A simple way to do this is to associate a prime to each edge, and give any two vertices belonging to a common edge the corresponding prime factor. (Vertices which do not share an edge in common will have no common prime factors, and thus be coprime.) We order the possible edges by considering the lexicographic ordering on all ordered pairs $(v_j, v_k)$ such that $v_j < v_k$: thus the possible edge $v_j v_k$ (for $j < k$) will be edge number $varepsilon(j,k) = binom{k-1}{2} + j$ in the enumeration. More generally, we may define
$$ varepsilon(j,k) = binom{max {j,k} - 1}{2} + min {j,k} .$$
We may represent the adjacency of two vertices $v_j, v_k$ by giving their corresponding integers $n_j, n_k$ a common prime factor, namely the prime $p_{n+varepsilon(j,k)}$. The exponents of the primes $p_{n+1}$ through $p_{n+binom{n}{2}}$ in the integers $n_j$ then form the incidence matrix of edges to vertices in $G$.

Thus, we may let $F(x)$ be the function which computes the following:

1. Determine the smallest prime $p_{j-1} mid x$, and the corresponding index $j$.
   
2. Determine the exponent, $g$, of the largest power of $p_{j-1}$ which divides $x$.
   
3. Extract from $g$ complete information about the graph $G$, including its size $n$.
   
4. Determine the next largest prime $p_j > p_{j-1}$.
   
5. Compute $N_j = p_j^g$.
   
6. Compute $M_j = p_{n+j}$.
   
7. For each $1 le k le n$, $k ne j$:
   
   • Compute $varepsilon(j,k)$.
     
   • If $v_j$ is adjacent to $v_k$ in $G$, set $M_j leftarrow M_j p_{n+varepsilon(j,k)}$.
     
8. Return $y = N_j M_j$.
   

To produce an appropriate $F$-sequence which represents $G$, it then suffices to compute $n_1$, which is
$$ n_1 = p_1^g prod_{k ne j} {p_{n+varepsilon(j,k)}}^{A_{j,k}} , $$
where $A$ is the adjacency matrix of $G$; this suffices to iteratively produce the other integers $n_k$ using $F$.

The above can be readily extended to accommodate hypergraphs (allow more than two vertices per 'edge') and other generalizations; one only needs to choose a suitable representation, and store it in an integer in some way which can be extracted. These are standard tricks in computability theory; we are just adapting Godel numbering for a special purpose in this case.

If you want an "interesting" way of representing graphs by models in integer sequences, which does not just amount to packing and unpacking of data structures in the integers, you're going to have to impose some non-trivial bounds --- on the integer sizes, on the computational efficiency of the procedure, etc. --- and then, you will almost certainly have to be satisfied with obtaining only some special class of graphs. But that special class may still be an interesting one, if your computational constraints are well-chosen.

at.algebraic topology - K-theory as a generalized cohomology theory

1 is doubly wrong. First, you need to distinguished generalized cohomology theories and reduced generalized cohomology theories. If you want to work with the latter, you should replace "a point" in 1 by "$S^0$", and then the corrected version of 3 no longer holds. But even this new version 1' is false; a generalized cohomology theory is not determined by its coefficients, unless they are concentrated in a single degree (example: complex K-theory vs. integer cohomology made even periodic).

set theory - Definable collections of non measurable sets of reals

(Edit.) With a closer reading of your question, I see that you asked for a very specific notion of definability.

If you allow the family to have size larger than continuum, there is a trivial Yes answer. Namely, let phi(x) be the assertion "x is a non-measurable set of reals". In any model of ZFC, this formula defines a family of non-measurable sets of reals, and it is not difficult to show in ZFC that there are at least continuum many such sets (for example, as in the comment of Qiachu Yuan). Thus, ZFC proves that {x | phi(x)} is a family of non-measurable sets of size at least continuum.

But if you insist that the family have size exactly the continuum, as your question clearly states, then this trivial answer doesn't work. Indeed, one can't even take the class of all Vitali sets in this case, since there are 2^continuum many sets of reals that contain exactly one point from each equivalence class for rational translations.

Qiachu Yuan's suggestion about translations of a single Vitali set does have size continuum, but there is little reason to expect the Vitali set to be definable in the way that you have requested, and so it does not provide the desired definable family.

In my earlier posted answer, I considered the possibility that you might have meant some other notion of definability, or whether parameters are allowed in the definition, and so on. And I find some of these other versions of the question to be quite interesting and subtle.

I pointed out that it is surely consistent with ZFC that there is the desired definable family of non-measurable sets, since in fact any set at all can be made definable in a forcing extension that adds no reals and no sets of reals. So you can take any family of non-measurable sets that you like and go to a forcing extension where this family is definable.

Perhaps a stronger notion of definibility would be to use the notion of projective definitions, where one wants to define the sets within the structure of the reals, using quantification only over reals and natural numbers (rather than over the entire set-theoretic universe). Thus, we want a projective formula phi(x,z), such that A_z={x | phi(x,z)} is always non-measurable for any z and all A_z are different. Such a formula would be a strong example of the phenomenon you seek.

The first answer to this way of asking the question is that it is consistent with ZFC that there is such a projective family. The reason is that I have mentioned in a number of questions and answers on this site, under the Axiom of Constructibility V=L, there is a projectively definable well-ordering of the reals. Thus, under V=L, one can projectively define a Vitali set, and then take the family of its translations. There is no need for a parameter in this definition, since a particular Vitali set can be projectively defined without parameters from the projectively definable well-ordering of the reals.

The second answer to this version of the question, however, is that under certain set-theoretic assumptions such as Projective Determinacy, every projective set of reals is Lebesgue measurable. In this case, there can be no such projectively defined family of non-measurable sets. The assumption of PD is consistent with ZFC from large cardinals, but perhaps one needs a much weaker hypothesis meerely to get every projective set measurable.

In summary, if one wants a projectively definable family of non-measurable sets, then it is independent of ZFC, if large cardinals are consistent. (Perhaps the need for large cardinals can be reduced.)

nt.number theory - Modular forms reference

Have a look at Section 6.6 of Diamond and Shurman, A First Course in Modular Forms:

As an aside, the theorem states a bit more than you have said: for instance, when the field of Fourier coefficients is $mathbb{Q}$, you are just asserting the existence of an elliptic curve $E_{/mathbb{Q}}$ with $operatorname{End}_{mathbb{Q}}(E) otimes_{mathbb{Z}} mathbb{Q} = mathbb{Q}$: every elliptic curve over $mathbb{Q}$ has this property. You want to require an equality of L-series between the abelian variety and the modular form.

gr.group theory - alternative construction of the quotient group

I have tried a method with a combinatorial group theory class, which is first to look at the universal quotient property relative to equivalence relations on a set. You explore this idea, and show that the set of equivalence classes is the answer you want and you derive the `first iso theorem' in that context. Exploring it to death if you like, with trivial do-able examples and constructions that students know. (e.g. different pairs (a,b) of integers giving the same rational numbers etc.) A function defines an equivalence relation and the equivalence relation defines a function and they are related. NB. no algebra at all in sight. This is strictly constructions on sets and functions.

The idea of the universal property is now centre stage i.e. where it should be. You can now ask if something similar is going to happen for groups and homomorphisms. The point is that it is not dividing out by a (normal) subgroup that is at stake here but by an equivalence relation (and that will not work of course). Equivalence relations seem to be more basic perhaps.

We all know the problem and the answer but the student has then something nearer to their previous experience. The equivalence relation does not work since although the quotient does seem to have an 'obvious' multiplication, that does not work and simple examples show it not to be well-defined (NB. 'Well definition' is a simple concept but students can (and do) find it very hard to grasp. This way, it is met as 'ill definition' first. The notion of well definition is perhaps not understood because it is usually presented as being a solution to a problem that the student has not ever met themselves.)

You now examine what goes wrong and get to the idea that you need a congruence not simply an equivalence relation. (Here I have used a categorical approach without the use of the term category, and really looked as a congruence as being an equivalence relation 'internal' to the category of groups... and no need to mention categories unless you feel like it. You can make the transition from equivalence relation to congruence in any way that seems appropriate for the background of the students.) This is the hard bit, not cosets, and it is hard, but sort of avoided in the usual treatment.

Now normal subgroups can be shown to be a reformulation of congruences and you can flip from one to the other and back again with no loss of info.

I like this treatment as the cosets only appear at the last moment, and then are the group theory analogue of equivalence classes. (If we have not got those across before we do cosets then there is no hope!)

This treatment also concentrates on the algebraic details that really need understanding. I do not mean group theory details as they are more specialised. We have equivalence classes, well definition and a universal property, three big ideas.

Normal subgroups come out as being natural.

Did this approach work? Not all the students could handle the idea of cosets, but they did seem happier with equivalence classes, well definition etc, and also seemed to like the universal property idea, which they met in various other contexts (product groups, free product, free group, etc.) in that course.

ca.analysis and odes - $fcirc f=g$ revisited

This may be related to solving f(f(x))=g(x). Let
$C(mathbb{R})$ be the linear space of all continuous functions from
reals to reals, and let $mathcal{S}$ $:=$ { $gin C(mathbb{R})$ $;$ $exists$ $fin C(mathbb{R})$ s.t. $fcirc f=g$ } . Is there some infinite dimensional (or, at least,
bidimensional) linear subspace of $C(mathbb{R})$ contained in $mathcal{S}$
?

P.S. As a remark, there is a [maybe] interesting connection between
How to solve f(f(x)) = cos(x) ? and Borsuk pairs of Banach spaces .
Namely, let $E$ be the closed subspace of $C[-1,1]$ consisting of
all even functions, and let $K$ be the closed unit ball of $E$.
Then the continuous mapping $Psi:$ $K$ $rightarrow$ $E$ expressed
by $Psi(f)$ $:=$ $fcirc f$ $+$ $left(leftVert frightVert _{infty}-1right)cdotcos$

is odd on $partial K$, and has no zeros in $K$.

Generating Classical Groups over Finite Local Rings

I am interested in classical groups (in particular $SL_n$, $Sp_{2n}$, $SO_n^{+}$) over finite
rings of the form $$R_k=mathbb{F}_q[t]/(t^k)$$ for some prime power $q$ (where $q$ is odd in the
orthogonal case) and $k in mathbb{N}$. Over a finite field, the maximal subgroups of the
classical groups are known, and so it is for example known that any two semisimple elements of
orders $q^n+1$ and $q^n-1$ generate the group $Sp_{2n}(mathbb{F}_q)$ (and similar results for
the other classical groups). I am wondering whether
it is true that any two semisimple elements of orders $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate
$Sp_{2n}(R_k)$. Is anyting like this known? Are there lists of maximal subgroups for classical
groups over finite rings? I would also be interested in similar results over $mathbb{Z}_p/(p^k)$.

EDIT: The answer to the question as asked is usually vacuously yes. Instead, asm meant actually to ask about the group generated by two tori (not just two semisimple elements) of the specified orders, specifically they said (in a now deleted answer):

"I misphrased my question. What I meant to ask is whether any two maximal tori of order $q^{n(k-1)}(q^n+1)$ and $q^{n(k-1)}(q^n-1)$ generate $Sp_{2n}(R_k)$. Of course these tori are far from cyclic. I guess my head was still in the cyclic case $Sp_{2n}(mathbb{F}_q)$."

nt.number theory - What's the "best" proof of quadratic reciprocity?

The question asked for the nicest proof for a first undergraduate course. Has anyone who offered a proposal used their favorite choice in a course? (Obviously the suggestions referring to $K$-theory or Hilbert symbols weren't suggested in that spirit.)
I've taught an undergrad number theory class several times and initially I gave the Gauss sum proof. But I realized afterwards that to the students this truly comes out of nowhere (it seems too magical), so I hunted around for other proofs, preferably some which build on more basic ideas that I could present earlier in the course. The Eisenstein (sine-function) proof doesn't fit that requirement, and Zolotarev seems too far-out if the students have not had group theory (which most have not). So what else is available?

There is a proof due to V. Lebesgue (not H. Lebesgue!) that is based on counting points on hyperspheres mod $p$. It can be found in Ireland-Rosen's book. For an odd prime $p$ and positive integer $n$, let

$$N_n(p) = #{(x_1,dots,x_n) in ({mathbf Z}/(p))^n : x_1^2 + cdots + x_n^2 = 1}.$$

This is the number of mod $p$ points on the sphere in $n$-space mod $p$. Earlier in the course I have the students discover numerically that every number mod $p$ is a sum of two squares. That is,

$$#{(x,y) in ({mathbf Z}/(p))^2 : x^2 + y^2 = a}$$

is positive for every $a$ in ${mathbf Z}/(p)$. This could be shown by the pigeonhole principle, since $x^2$ and $a - y^2$ each take $(p+1)/2$ values mod $p$ and thus have an overlap. But more precisely, if you look at examples, you quickly discover that this 2-variable count is independent of $a$ when $a$ is nonzero (from a more adv. point of view, the independence is because the norm map on unit groups $(({mathbf Z}/(p))[t]/(t^2+1))^times to ({mathbf Z}/(p))^times$ is a homomorphism so its fibers have the same size, but that is a crazy explanation in an elem. number theory course). Enough data suggest what that uniform value is for any nonzero $a bmod p$, and then we prove that in class. With this 2-variable count we return to the hypersphere count and get a simple recursive formula connecting $N_n(p)$ to $N_{n-2}(p)$. If you let $n = q$ be an odd prime, the recursive formula involves $p^{(q-1)/2}$ plus $N_{q-2}(p)$ times a multiple of $q$, so $N_q(p) bmod q$ involves $(frac{p}{q})$. [Note: Although the application will use $N_q(p)$, you must think about $N_n(p)$ for general odd $n$ first since the recursion from $n$ back to $n-2$ makes no sense in general when the number of variables is only an odd prime: $n-2$ usually isn't prime when $n$ is.]

At the same time, the set being counted by $N_n(p)$ is invariant under cyclic shifts of the coordinates. On the very first problem set I have the students discover numerically that the number of cyclic shifts of an $n$-tuple is always a divisor of $n$. So when we let $n = q$ be an odd prime, $N_n(p) = N_q(p)$ is the number of constant $q$-tuples on the unit sphere mod $p$ plus a multiple of $q$. A constant $q$-tuple is basically counting whether or not $q bmod p$ is a square. So $N_q(p) bmod q$ is related to $(frac{q}{p})$.

In the two approaches to counting $N_q(p) bmod q$, one involves $(frac{p}{q})$ and the other involves $(frac{q}{p})$. This implies the QR relation mod $q$, which is actual equality since $1$ is not $-1 bmod q$.

One nice thing about this approach is that it can also be used to prove the supplementary law for $(frac{2}{p})$, by counting

$$#{(x,y) in ({mathbf Z}/(p))^2 : x^2 + y^2 equiv 1 bmod p}$$

in two ways. First there is the exact formula for the count (not just mod $p$) which I mentioned before. Second, most solutions in this count come in packets of size 8 (permute coordinates and change signs to get 8 solutions out of one solution). The exceptions which don't fall into packets of size 8 (when $x$ is $pm y$ mod $p$ or $x$ or $y$ is $0$ mod $p$) depend on whether or not 2 mod $p$ is a square (does $2x^2 equiv 1 bmod p$ have a solution?), and comparing these two formulas mod 8 implies the usual rule for $(frac{2}{p})$.

Since I am able to get the students to work on ideas that are used in the proof much earlier on during the semester (one doesn't need quadratic residues to numerically look at solutions of $x^2 + y^2 equiv a bmod p$, for instance), this proof nicely ties together things they have seen throughout the course. So that's why this is my vote for the best proof to give in an undergrad course.