rt.representation theory - How many ways are there to globalize Harish Chandra modules?

Suppose $G$ a reductive Lie group with finitely many connected components, and suppose in addition that the connected component $G^0$ of the identity can be expressed as a finite cover of a linear Lie group. Denote by $mathfrak{g}$ the complexified Lie algebra, and denote by $K$ a maximal compact in the complexification of $G$.

Denote by $mathbf{HC}(mathfrak{g},K)$ the category of admissible $(mathfrak{g},K)$-modules or (Harish Chandra modules), and $(mathfrak{g},K)$-module homomorphisms. Denote by $mathbf{Rep}(G)$ the category of admissible representations of finite length (on complete locally convex Hausdorff topological vector spaces), with continuous linear $G$-maps.

The Harish Chandra functor $mathcal{M}colonmathbf{Rep}(G)tomathbf{HC}(mathfrak{g},K)$ assigns to any admissible representation $V$ the Harish Chandra module of $K$-finite vectors of $V$. This is a faithful, exact functor. Let us call an exact functor $mathcal{G}colonmathbf{HC}(mathfrak{g},K)tomathbf{Rep}(G)$ along with a comparison isomorphism $eta_{mathcal{G}}colonmathcal{M}circmathcal{G}simeqmathrm{id}$ a globalization functor.

Our first observation is that globalization functors exist.

Theorem. [Casselman-Wallach] The restriction of $mathcal{M}$ to the full subcategory of smooth admissible Fréchet spaces is an equivalence. Moreover, for any Harish Chandra module $M$, the essentially unique smooth admissible representation $(pi,V)$ such that $Mcongmathcal{M}(pi,V)$ has the property $pi(mathcal{S}(G))M=V$, where $mathcal{S}(G)$ is the Schwartz algebra of $G$.

If we do not restrict $mathcal{M}$ to smooth admissible Fréchet spaces, then we have a minimal globalization and a maximal one.

Theorem. [Kashiwara-Schmid] $mathcal{M}$ admits both a left adjoint $mathcal{G}_0$ and right adjoint $mathcal{G}_{infty}$, and the counit and unit give these functors the structure of globalization functors.

Construction. Here, briefly, are descriptions of the minimal and maximal globalizations. The minimal globalization is

$$mathcal{G}_0=textit{Dist}_c(G)otimes_{U(mathfrak{g})}-$$

where $textit{Dist}_c(G)$ denotes the space of compactly supported distributions on $G$, and the maximal one is

$$mathcal{G}_{infty}=mathrm{Hom}_{U(mathfrak{g})}((-)^{vee},C^{infty}(G))$$

where $M^{vee}$ is the dual Harish Chandra module of $M$ (i.e., the $K$-finite vectors of the algebraic dual of $M$).

For any Harish Chandra module $M$, the minimal globalization $mathcal{G}_0(M)$ is a dual Fréchet nuclear space, and the maximal globalization $mathcal{G}_{infty}(M)$ is a Fréchet nuclear space.

Example. If $Psubset G$ is a parabolic subgroup, then the space $L^2(G/P)$ of $L^2$-functions on the homogeneous space $G/P$ is an admissible representation, and $M=mathcal{M}(L^2(G/P))$ is a particularly interesting Harish Chandra module. In this case, one may identify $mathcal{G}_0(M)$ with the real analytic functions on $G/P$, and one may identify $mathcal{G}_{infty}(M)$ with the hyperfunctions on $G/P$.

[I think other globalizations with different properties are known or expected; I don't yet know much about these, however.]

Consider the category $mathbf{Glob}(G)$ of globalization functors for $G$; morphisms $mathcal{G}'tomathcal{G}$ are natural transformations that are required to be compatible with the comparison isomorphisms $eta_{mathcal{G}'}$ and $eta_{mathcal{G}}$. Since $mathcal{M}$ is faithful, this category is actually a poset, and it has both an inf and a sup, namely $mathcal{G}_0$ and $mathcal{G}_{infty}$. This is the poset of globalizations for $G$.

I'd like to know more about the structure of the poset $mathbf{Glob}(G)$ — really, anything at all, but let me ask the following concrete question.

Question. Does every finite collection of elements of $mathbf{Glob}(G)$ admit both an inf and a sup?

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[Added later]

Emerton (below) mentions a geometric picture that appears to be very well adapted to the study of our poset $mathbf{Glob}(G)$. Let me at introduce the main ideas of the objects of interest, and what I learned about our poset. [What I'm going to say was essentially outlined by Kashiwara in 1987.] For this, we probably need to assume that $G$ is connected.

Notation. Let $X$ be the flag manifold of the complexification of $G$. Let $lambdainmathfrak{h}^{vee}$ be a dominant element of the dual space of the universal Cartan; for simplicity, let's assume that it is regular. Now one can define the twisted equivariant bounded derived categories $D^b_G(X)_{-lambda}$ and $D^b_K(X)_{-lambda}$ of constructible sheaves on $X$. Now let $mathbf{Glob}(G,lambda)$ denote the poset of globalizations for admissible representations with infinitesimal character $chi_{lambda}$, so the objects are exact functors $mathcal{G}colonmathbf{HC}(mathfrak{g},K)_{chi_{lambda}}tomathbf{Rep}(G)_{chi_{lambda}}$ equipped with natural isomorphisms $eta_{mathcal{G}}:mathcal{M}circmathcal{G}simeqmathrm{id}$.

Matsuki correspondence. [Mirkovic-Uzawa-Vilonen] There is a canonical equivalence $Phicolon D^b_G(X)_{-lambda}simeq D^b_K(X)_{-lambda}$. The perverse t-structure on the latter can be lifted along this correspondence to obtain a t-structure on $D^b_G(X)_{-lambda}$ as well. The Matsuki correspondence then restricts to an equivalence $Phicolon P_G(X)_{-lambda}simeq P_K(X)_{-lambda}$ between the corresponding hearts.

Beilinson-Bernstein construction. There is a canonical equivalence $alphacolon P_K(X)_{-lambda}simeqmathbf{HC}(mathfrak{g},K)_{chi_{lambda}}$, given by Riemann-Hilbert, followed by taking cohomology. [If $lambda$ is not regular, then this isn't quite an equivalence.]

Now we deduce a geometric description of an object of $mathbf{Glob}(G,lambda)$ as an exact functor $mathcal{H}colon P_G(X)_{-lambda}tomathbf{Rep}(G)_{chi_{lambda}}$ equipped with a natural isomorphism $mathcal{M}circmathcal{H}simeqalphacircPhi$, or equivalently, as a suitably t-exact functor $mathcal{H}colon D^b_G(X)_{-lambda}to D^bmathbf{Rep}(G)_{chi_{lambda}}$ equipped with a functorial identification between the (complex of) Harish Chandra module(s) of $K$-finite vectors of $mathcal{H}(F)$ and $mathrm{RHom}(mathbf{D}Phi F,mathcal{O}_X(lambda))$ for any $Fin D^b_G(X)_{-lambda}$. In particular, as Emerton observes, the maximal and minimal globalizations can be expressed as

$$mathcal{H}_{infty}(F)=mathrm{RHom}(mathbf{D}F,mathcal{O}_X(lambda))$$

and

$$mathcal{H}_0(F)=Fotimes^Lmathcal{O}_X(lambda)$$

Note that Verdier duality gives rise to an anti-involution $mathcal{H}mapsto(mathcal{H}circmathbf{D})^{vee}$ of the poset $mathbf{Glob}(G,lambda)$; in particular, it exchanges $mathcal{H}_{infty}$ and $mathcal{H}_0$.

I now expect that one can show the following (though I don't claim to have thought about this point carefully enough to call it a proposition).

Conjecture. All globalization functors are representable. That is, every element of $mathbf{Glob}(G,lambda)$ is of the form $mathrm{RHom}(mathbf{D}(-),E)$ for some object $Ein D^b_G(X)_{-lambda}$.

Question. Can one characterize those objects $Ein D^b_G(X)_{-lambda}$ such that the functor $mathrm{RHom}(mathbf{D}(-),E)$ is a globalization functor? Given a map between any two of these, under what circumstances do they induce a morphism of globalization functors (as defined above)?

In particular, note that if my expectation holds, then one should be able to find a copy of the poset $mathbf{Glob}(G,lambda)$ embedded in $D^b_G(X)_{-lambda}$.

rt.representation theory - Classifying strata for the adjoint representation of GL from first principles

I'm going to give a partial answer here for two reasons: (1) I am lazy and (2) this is starting to feel a little homeworky to me. Obviously, no one would assign this material as homework, but part of reading a math paper is taking the time to work out lots of simple examples and see how the definitions work. I feel like you are pushing the boundaries of how much of this work it is reasonable to ask other people to do. Not a major criticism, certainly not a vote to close the question, but my input.

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On to the math. I've scanned the first 3 pages of Hesselink's paper. He make the following definitions. G acts on V, v is a point of V and $star$ a chosen base point of V fixed by G. In your setting, G is $GL_n$, V is the $n times n$ matrices where G acts by conjugation, and $star$ is zero. Hesselink writes Y(G) for what is essentially $mathrm{Hom}(mathbb{C}^*, G)$. More precisely, Hesselink tensors with $mathbb{Q}$, so that he can talk about maps like $t mapsto left( begin{smallmatrix} t^{1/3} & 0 \\ 0 & t^{-2/7} end{smallmatrix} right)$. I'll ignore this detail.

For $lambda in Y(G)$, Hesselink defines a rational number $m(lambda)$. We talked about this in your previous question. In this setting, where V is an $N$-dimensional vector space, Hesselink gives an explicit formula for m on the bottom of page 142/top of page 143: Diagonalize the action of $lambda$ as $t mapsto mathrm{diag}(t^{m_1}, cdots, t^{m_N})$ and write $v = sum v_i e_i$.. Then $m(lambda) = min(m_i : v_i neq 0)$ if this number is nonnegative, and is $- infty$ if this minimum is negative.

Let's see what this definition means in your setting. We can conjugate any $lambda$ into diagonal form as $t mapsto mathrm{diag}(t^{c_1}, cdots, t^{c_n})$. I've replaced $m_i$ by $c_i$ to point out that these $c$'s are not the $m$'s of the previous paragraph. In our notation, the $N$ of the previous paragraph is $n^2$. The vector space $V$ has dimension $n^2$ with basis $e_{ij}$. The action of $lambda(t)$ on $e_{ij}$ is by $t^{c_i - c_j}$. (Exercise!).

So $m(lambda) > 0$ if and only if $c_i leq c_j$ implies $v_{ij} =0$.

We may as well order our basis such that $c_1 geq c_2 geq cdots geq c_n$.
If $c_1 > c_2 > cdots >c_n$ then we see that $m(lambda) > 0$ if and only if $v$ is a strictly upper triangular matrix. When there are some equalities among the $c$'s, you want $v$ to be strictly block upper triangular. For such a $v$, $m(lambda) = min(c_i - c_j : v_{ij} neq 0)$. In particular, notice that there exists a $lambda$ such that $m(lambda) > 0$ if and only if $v$ is nilpotent.

Hesselink defines $Lambda(v)$ to be the locus in ${ lambda : m(lambda) = 1 }$ where $q(lambda)$ is minimized, where $q$ is the inner product from your previous question. What you want to show is that $Lambda(v)$ determines the Jordan normal form of $v$.

I must admit that I haven't thought out how to prove this. But I hope this makes things explicit enough that you can attack it.

gr.group theory - Conjugacy classes insersecting subgroups of finite groups

If we let $k(G)$ denote the number of conjugacy classes of the finite group $G$,
then for any subgroup $H$ of $G$ (normal or not), a Theorem of P.X. Gallagher
states that $[G:H]^{-1}k(H) leq k(G) leq [G:H]k(H)$ (this has probably been discovered
and rediscovered many times). I find that the easiest way to see it is using irreducible
complex characters. Of course, $k(G)$ is also the number of complex irreducible characters of $G$,
and likewise for $H$. For each irreducible character $chi$ of $G$, there is an irreducible
character $mu$ of $H$ such that $mu$ occurs with non-zero multiplicity in the restriction of $chi$ to $H$. By Frobenius reciprocity, $chi$ is an irreducible constituent of the character of $G$
induced from the character $mu$ of $H$. On the other hand, using Frobenius reciprocity again, each irreducible constituent of $mu$ induced to $G$ must have degree at least $mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of $mu$ induced to $G$
Since this is true for each irreducible character of $H$, and since each irreducible character of $G$ must appear with non-zero multiplicity in at least one such character, we have
$k(G) leq [G:H]k(H).$ Going in the other direction, if $chi$ is an irreducible character of $G$
and $mu$ is an irreducible constituent of te restriction of $chi$ to $H$, then we have
$chi(1) leq [G:H]mu(1)$, since $chi$ occurs as a constituent of $mu$ induced to $G$.
Thus $mu(1) geq frac{chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible
constituents of the restriction of $chi$ to $H$. Since each irreducible character of $H$ occurs
as a constituent of some such irreducible character of $G$ (consider, for example, the restriction
of the regular character), we have $k(H) leq [G:H] k(G).$

C++ Sieve of Atkin overlooks a few prime numbers

Hello All!

Recently I've been working on a C++ prime generator that uses the Sieve of Atkin ( http://en.wikipedia.org/wiki/Sieve_of_atkin ) to generate its primes. My objective is to be able to generate any 32-bit number. I'll use it mostly for project euler problems. mostly it's just a summer project.

The program uses a bitboard to store primality: that is, a series of ones and zeros where for example the 11th bit would be a 1, the 12th a 0, and the 13th a 1, etc. For efficient memory usage, this is actually and array of chars, each char containing 8 bits. I use flags and bitwise-operators to set and retrieve bits. The gyst of the algorithm is simple: do a first pass using some equations I don't pretend to understand to define if a number is considered "prime" or not. This will for the most part get the correct answers, but a couple nonprime numbers will be marked as prime. Therefore, when iterating through the list, you set all multiples of the prime you just found to "not prime". This has the handy advantage of requiring less processor time the larger a prime gets.

I've got it 90% complete, with one catch:
some of the primes are missing.

Through inspecting the bitboard, I have ascertained that these primes are omitted during the first pass, which basically toggles a number for every solution it has for a number of equations (see wikipedia entry). I've gone over this chunk of code time and time again. I even tried increasing the bounds to what is shown in the wikipedia articles, which is less efficient but I figured might hit a few numbers that I have somehow omitted. Nothing has worked. These numbers simply evaluate to not prime. Most of my test has been on all primes under 128. Of this range, these are the primes that are omitted:

23 and 59.

I have no doubt that on a higher range, more would be missing (just don't want to count through all of them). I don't know why these are missing, but they are. Is there anything special about these two primes? I've double and triple checked, finding and fixing mistakes, but it is still probably something stupid that I am missing.

anyways, here is my code:

#include <iostream>
#include <limits.h>
#include <math.h>

using namespace std;

const unsigned short DWORD_BITS = 8;

unsigned char flag(const unsigned char);
void printBinary(unsigned char);


class PrimeGen
{
    public:
        unsigned char* sieve;
        unsigned sievelen;
        unsigned limit;
        unsigned bookmark;


        PrimeGen(const unsigned);

        void firstPass();
        unsigned next();

        bool getBit(const unsigned);
        void onBit(const unsigned);
        void offBit(const unsigned);
        void switchBit(const unsigned);

        void printBoard();
};


PrimeGen::PrimeGen(const unsigned max_num)
{
    limit = max_num;
    sievelen = limit / DWORD_BITS + 1;
    bookmark = 0;

    sieve = (unsigned char*) malloc(sievelen);
    for (unsigned i = 0; i < sievelen; i++) {sieve[i] = 0;}

    firstPass();
}


inline bool PrimeGen::getBit(const unsigned index)
{
    return sieve[index/DWORD_BITS] & flag(index%DWORD_BITS);
}


inline void PrimeGen::onBit(const unsigned index)
{
    sieve[index/DWORD_BITS] |= flag(index%DWORD_BITS);
}


inline void PrimeGen::offBit(const unsigned index)
{
    sieve[index/DWORD_BITS] &= ~flag(index%DWORD_BITS);
}


inline void PrimeGen::switchBit(const unsigned index)
{
    sieve[index/DWORD_BITS] ^= flag(index%DWORD_BITS);
}


void PrimeGen::firstPass()
{
    unsigned nmod,n,x,y,xroof, yroof;

    //n = 4x^2 + y^2
    xroof = (unsigned) sqrt(((double)(limit - 1)) / 4);
    for(x = 1; x <= xroof; x++){
        yroof = (unsigned) sqrt((double)(limit - 4 * x * x));
        for(y = 1; y <= yroof; y++){
            n = (4 * x * x) + (y * y);
            nmod = n % 12;
            if (nmod == 1 || nmod == 5){
                switchBit(n);
            }
        }
    }

    xroof = (unsigned) sqrt(((double)(limit - 1)) / 3);
    for(x = 1; x <= xroof; x++){
        yroof = (unsigned) sqrt((double)(limit - 3 * x * x));
        for(y = 1; y <= yroof; y++){
            n = (3 * x * x) + (y * y);
            nmod = n % 12;
            if (nmod == 7){
                switchBit(n);
            }
        }
    }

    xroof = (unsigned) sqrt(((double)(limit + 1)) / 3);
    for(x = 1; x <= xroof; x++){
        yroof = (unsigned) sqrt((double)(3 * x * x - 1));
        for(y = 1; y <= yroof; y++){
            n = (3 * x * x) - (y * y);
            nmod = n % 12;
            if (nmod == 11){
                switchBit(n);
            }
        }
    }
}


unsigned PrimeGen::next()
{
    while (bookmark <= limit)
    {
        bookmark++;

        if (getBit(bookmark))
        {
            unsigned out = bookmark;

            for(unsigned num = bookmark * 2; num <= limit; num += bookmark)
            {
                offBit(num);
            }

            return out;
        }
    }

    return 0;
}


inline void PrimeGen::printBoard()
{
    for(unsigned i = 0; i < sievelen; i++)
    {
        if (i % 4 == 0)
            cout << endl;

        printBinary(sieve[i]);
        cout << " ";
    }
}


inline unsigned char flag(const unsigned char bit_index)
{
    return ((unsigned char) 128) >> bit_index;
}


inline void printBinary(unsigned char byte)
{
    unsigned int i = 1 << (sizeof(byte) * 8 - 1);

    while (i > 0) {
        if (byte & i)
            cout << "1";
        else
            cout << "0";
        i >>= 1;
    }
}

I did my best to clean it up and make it readable. I'm not a professional programmer, so please be merciful.

Here is the output I get, when I initialize a PrimeGen object named pgen, print its initial bitboard with pgen.printBoard() (please note that 23 and 59 are missing before next() iteration), and then iterate through next() and print all of the returned primes:

00000101 00010100 01010000 01000101
00000100 01010001 00000100 00000100
00010001 01000001 00010000 01000000
01000101 00010100 01000000 00000001

5
7
11
13
17
19
29
31
37
41
43
47
53
61
67
71
73
79
83
89
97
101
103
107
109
113
127

DONE

Process returned 0 (0x0)   execution time : 0.064 s
Press any key to continue.

gr.group theory - Topologies on an infinite symmetric group

Let $X$ be an infinite set, and let $G$ be the symmetric group on $X$. I want to understand $G$ by putting a topology on it, without imposing any more structure on $X$. What 'interesting' possibilities are there and what is known about them?

In particular, I have heard of the pointwise convergence topology (an open neighbourhood of g is a set of permutations that agree on some specified finite set of points), and found some papers on this, but are there any other topologies that have been studied?

What if I take the coarsest topology compatible with the group operations such that either a) the stabiliser of any subset is closed, b) the stabiliser of any partition is closed, or c) both are closed? I think these will be coarser than the pointwise convergence topology, because in the pointwise convergence topology the stabiliser of any first-order structure is closed. Is there a useful characterisation of the open subgroups and/or closed subgroups?

peano arithmetic - Naturally definable sets of natural numbers

In general, I consider this question interesting. But on the other hand - well, you don't specify what literals you accept in your formal System.

If you have $le$ for example, it is likely that you define $a < b$ by $a le b wedge aneq b$, so you implicitly get an $=$ into your formula, and hence, you also cant allow $x < n$ for fixed $n$. Same for $le$. So actually, I would consider it more reasonable to disallow explicit numbers in general. In fact, anything that depends on an explicitly given (and therefore exchangable) number doesnt seem "natural" to me.

(Also, being even doesnt really seem natural to me - it is divisibility by 2, why not divisibility by 3? But as soon as you have a $+$-Function, you cannot suppress this being natural.)

Lets assume we are in a classical setting. Then it is sufficient to have $wedge$, $vee$, $lnot$, $forall$ to express anything we want, and the relation symbols $=$, <, and the function symbols $S$, $+$, $cdot$ (normally, we could spare $S$ and add $0$ and $1$ if we have $+$, but since we dont want explicit numbers anyway, lets just add the successor-function and disallow $0$ and $1$ - so the "naturality" is a consequence rather than an enforcement).

Of course, we can express $mathbb{N}$ and $emptyset$ by $x=x$ and $x < x$, and conjunctions and disjunctions of it. We can express the set of even numbers by $lnotforall y lnot x=y+y$, and the set of all primes by $forall a,b,c,d . x=acdot d rightarrow x=bcdot d rightarrow x neq c cdot d$, and we can express finite intersections and unions.

And - we still can define predicates depending on explicit numbers, since $forall y (x < y vee x=y) leftrightarrow x=0$. So lets - additionally - disallow <. Then the atoms we have left are $=$-expressions between terms consisting of variables, $S$, $+$ and $cdot$. Trivially, every of these terms is equal to some polynomial with natural coefficients. On the other hand, we can express every polynomial with natural coefficients of at least degree 1 in every variable by these natural numbers. And thus, our atomic relationships between variables are equivalent to the relationships between polynomials with natural coefficients of at least degree 1, which can always be expressed as sets of roots of a polynomial with integer coefficients of at least degree 1. So what we get as atoms are relations describing sets of natural roots of polynomials with integer coefficients.

Anyway, enough of this, well, we can still get $0$ by saying $x+x=x$ and $1$ by saying $x*x=x wedge x+x neq x$.

Disallowing $+$ and $cdot$ would seem strange to me, as then you would only have the successor-function, which would only allow atomic relations equivalent to $x=y+n$ for fixed $n$ - and still, here you would be able to get $0$ by saying $forall y.x neq S(y)$.

Maybe you could also disallow negations instead, then you would get something similar to algebraic varieties. Or maybe you should take a look at Presburger Arithmetic.

I myself would just define a set of natural numbers as "natural" if it is decidable.

ho.history overview - What to do with antique math books?

My grandfather had a PhD in math. When he died, he left a lot of math textbooks, which I took. These include things like Van der Waerden's 2-volume algebra set from the 1970s,
"Studies in Global Geometry and Analysis" by Shiing-Shen Chern, a series called "Mathematics: it's content, methods, and meaning," and many more.

I'm keeping about 20 of them, but there are 103 which I don't want to keep, but which I don't know what to do with. I obviously don't want to throw them away, and I don't really know what will happen to them if I donate them to the giant used-books depository in downtown Baltimore (called "the book thing," where people drop off and pick up used books for free). I'd like to donate them to some math collector or math library. But maybe there are just too many used antique math books floating around.

RECAP: I have 103 antique used math books which I cannot keep. Do you have a suggestion for what to do with them?

Thanks,
David

Common Computations in Group Cohomology

The automorphisms of this extension are basically the same as group cohomology $H^1(B; A)$, so I will focus on that first.

So we want to show that this must be zero given that B and A are finite abelian and of coprime order. This is the same as the twisted cohomology of the classifying space BB. Now we can look at the bundle $p:EB to BB$. This is a covering space with fiber the discrete space B.

Now because the fiber is discrete we have a wrong way transfer map in twisted cohomology:

$$p_{!}: H^1(EB; A) to H^1(BB; A)$$

where the first group is twisted cohomology in the pulled-back local coefficient system. As with all transfers we have that

$$p_{!} circ p^*: H^1(B; A) to H^1(B; A)$$

is multiplication by the order of the fiber, i.e. $|B|$. Since the orders of A and B are coprime this is an isomorphism. But since $EG simeq pt$ is contractible, this map factors through the zero group and hence $H^1(B; A) = 0$.

Any proof that the cohomology of a group is torsion for the order of the group (there are more concrete ones then the above) will yield the same result that $H^1(B; A) = 0$. There are many ways to prove this (as the comments point out), the above is just my favorite.

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So what is the difference between the $H^1(B;A)$ and the isomorphisms of the extension G? Well as you pointed out the isomorphisms of G (which restrict to the identity on A and the quotient B) are the same as the bar resolution coycycles Z^1(B;A). So we have an exact sequence,

$$A= C^0(B;A) to Z^1(B; A) to H^1(B;A) to 0$$

but as we saw, the term $H^1(B;A) = 0$. So we must compute the boundaries. You have one such potential homomorphism for each element of $A$, although different elements might give rise to the same automorphism of $G$. They are of the form:

$$b mapsto a - b cdot a$$

where $a in A$ is fixed. If the action of B on A is trivial, then these vanish, but in general they can be non-zero. My favorite example is the quaterion group which we view as

$$mathbb{Z}/4 to Q_8 to mathbb{Z}/2$$

with the $mathbb{Z}/4$ the group $( 1, i, -1, -i )$. An element $x in A = mathbb{Z}/4$ induces the homomorphism

$$y mapsto x - y cdot x$$

which sends the non-trivial element of $mathbb{Z}/2$ to $2x$. In particular it is non-trivial for a generator of A. This corresponds to the isomoprhism of $Q_8$ which sends $i$ to $i$ and $j$ to $-j$.

ca.analysis and odes - Does the exponential function have a square root?

For any nonpositive $x$, define the sequence $(x_0, x_1, ...)$ by $x_0 = x$, $x_{i+1} = e^{x_i}$. If we let $y$ be another nonpositive real number, and define $(y_0, ...)$ similarly, then if we define $h$ on the union of these two sequences by $h(x_i) = y_i$, $h(y_i) = x_{i+1}$, we get $h(h(x_i)) = x_{i+1}$, $h(h(y_i)) = y_{i+1}$. So, to define $h$ everywhere, we just pair off the nonpositive reals any old way and define $h$ as above for each pair (every number can be written as an $x_i$ for a unique $i ge 0, x le 0$, so this defines $h$ on the entire real line). Thus, there are actually uncountably many square roots of the exponential function, almost all of which are horribly discontinuous.

Edit: in fact, any such $h$ must be of this form - if $h(x_0) = y_i$, then we must have $h(y_i) = x_1, h(x_1) = y_{i+1}, ..., h(y_{i+j}) = x_{j+1}, h(x_j) = y_{i+j}$, if $y = x$ then we have a problem when we plug in $j = i$, so we must have $y ne x$, and by a similar argument if $h(y_0) = z_k$ for some $k$, then $x_1 = h(y_i) = z_{i+k}$ so $x = z, i+k = 1$, so either $h(x_0) = y_0, h(y_0) = x_1$ or vice-versa.

To find such a pairing producing a continuous $h$, you can take any homeomorphism $f : (-infty, -1] rightarrow (-1, 0]$, such as $f(x) = e^{x+1}-1$, and have your pairs be of the form $(x, f(x))$.

cv.complex variables - What does the incidence algebra of the lattices in C tell us about modular forms?

I have two different and probably unrelated questions that can both be superficially described by the title, so I hope you'll forgive me if I ask them together. They both fall under the category of trying to internalize some of the basic definitions of the theory.

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It's commonly said that one way to think about the definition of a modular form $f(z)$ of weight $k$ is that the $k$-fold differential $f(z) (dz)^k$ is invariant under $Gamma(N)$, i.e. it defines a $k$-fold differential on $Y(N)$. According to Milne, these two definitions are only equivalent for meromorphic modular and differentials respectively, and the analogous relationship between modular forms and holomorphic $k$-fold differentials is more complicated.

Question 1: What is the nature of the conceptual relationship between modular forms and holomorphic differentials? In other words, to what extent is the construction of modular forms a special case of a more general construction for an arbitrary Riemann surface, and to what extent does it depend on special properties of $Y(N)$ (and what are those properties)?

Now that I've read more carefully, this question is more or less resolved by Lemma 4.11 in Milne. The point seems to be that the two definitions of the order of a pole coming from modular forms and from differentials disagree at the elliptic points and cusps because fixed points of the group action count with different multiplicity, or something. So now I'm only interested in the second question. (The original title of this question was, somewhat tongue-in-cheek, "what is a modular form?")

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Another way to think about modular forms is that they are particularly well-behaved functions on the set of lattices in $mathbb{C}$. Now, the set of lattices in $mathbb{C}$ forms a locally finite poset under inclusion, and Gian-Carlo Rota has taught me to think about incidence algebras whenever I see functions on a locally finite poset. This perspective seems relevant to the combinatorial definition of the Hecke operators so I want to know if it can be developed more thoroughly.

Question 2: What is the relationship, if any, between modular forms and the incidence algebra of the poset of lattices in $mathbb{C}$ under inclusion? In particular, does Mobius inversion have any significance?

(I'm not really sure how to tag this.)

soft question - Magic trick based on deep mathematics

Here is a trick much in the spirit of the original number-adding example; moreover I'm sure Richard will appreciate the type of "deep mathematics" involved.

On a rectangular board of a given size $mtimes n$, Alice places (in absence of the magician) the numbers $1$ to $mn$ (written on cards) in such a way that rows and columns are increasing but otherwise at random (in math term she chooses a random rectangular standard Young tableau). She also chooses one of the numbers say $k$ and records its place on the board. Now the she removes the number $1$ at the top left and fills the empty square by a "jeu de taquin" sequence of moves (each time the empty square is filled from the right or from below, choosing the smaller candidate to keep rows and columns increasing, and until no candidates are left). This is repeated for the number $2$ (now at the top left) and so forth until $k-1$ is gone and $k$ is at the top left. Now enters the magician, looks at the board briefly, and then points out the original position of $k$ that Alice had recorded. For maximum surprise $k$ should be chosen away from the extremities of the range, and certainly not $1$ or $mn$ whose original positions are obvious.

All the magician needs to do is mentally determine the path the next slide (removing $k$) would take, and apply a central symmetry with respect to the center of the rectangle to the final square of that path.

In fact, the magician could in principle locate the original squares of all remaining numbers (but probably not mentally), simply by continuing to apply jeu de taquin slides. The fact that the tableau shown to the magician determines the original positions of all remaining numbers can be understood from the relatively well known properties of invertibility and confluence of jeu de taquin: one could slide back all remaining numbers to the bottom right corner, choosing the slides in an arbitrary order. However that would be virtually impossible to do mentally. The fact that the described simple method works is based on the less known fact that the Schútzenberger dual of any rectangular tableau can be obtained by negating the entries and applying central symmetry (see the final page of my contribution to the Foata Festschrift).

lo.logic - Deficiency of necessary conditions

First, let me mention that one must be careful when asserting that an implication fails. Taken literally, the assertion "D(x) does not imply P(x)" is logically equivalent to the assertion that D(x) is true and P(x) is false. This meaning of material implication that is used in mathematics is not the same as the natural language interpretation of if-then. For example, if the professor says to a student "It is not true that if you pass the final, then you pass the class", most people would not want the students to deduce logically that he or she will pass the final, but fail the class. But this does follow logically from the mathematical usage of material implication. So your definition of "defect" may not be what you intend.

To be sure, mathematicians are often sloppy about this. One often hears people say that such-and-such condition does not imply another condition. What they mean is that it does not necessarily imply the other condition. For example, suppose I have a function f, and someone says "its not true that if f is continuous, then f is differentiable." This statement is logically equivalent to the assertion that f is indeed continuous and not differentiable. What they meant to say, of course, was that "not every continuous function is differentiable".

In your case, you assert two implication failures: one if the definition of defect and another in the definition of minimal. When you clarify exactly what you mean more precisely, you will be led to the conclusion that the only sensible (minimal) defect is simply the assertion D(x), asserting that "either P(x) holds, or Q(x) fails". This statement does not imply P(x), except for those values of x for which Q(x) already implies P(x), and also if D(x) ∧ Q(x), then P(x) follows immediately. If D'(x) is any other statement such that D'(x) ∧ Q(x) implies P(x), then D'(x) implies that either Q(x) fails or P(x) holds, and so D'(x) implies D(x).

rt.representation theory - Is there a "correct" general setting for the principle: "tensoring any object with a projective object yields another projective"?

Apparently this principle was first formulated for left modules over the group algebra $A=kG$ of a finite group, where $k$ is a field of characteristic $p>0$ dividing $|G|$. (See Exercise 2 on p. 426 of Curtis & Reiner, Representation Theory of Finite Groups and Associative Algebras, 1962.) Here the Hopf algebra structure of A yields a natural left module structure on the tensor product of two left modules over k.

By the mid-1970s similar tensor product behavior was observed in other special cases for left A-modules and their tensor products, where A is a Hopf algebra over a commutative ring k: (1) the (finite dimensional) restricted enveloping algebra of a restricted Lie algebra $mathfrak{g}$ over a field of prime characteristic; (2) more generally the hyperalgebra of a higher Frobenius kernel when $mathfrak{g}$ is the Lie algebra of a reductive algebraic group; (3) the universal enveloping algebra of a Kac-Moody algebra in characteristic 0; (4) the full hyperalgebra of a reductive algebraic group in prime characteristic (with "projective" replaced by "injective" as in J.C. Jantzen's book Representations of Algebraic Groups, I.3). Relevant references:

B. Pareigis, Kohomologie von p-Lie Algebren, Math. Z. 104 (1968); Lemma 2.5

J.E. Humphreys, Projective modules for SL(2,q), J. Algebra 25 (1973); Thms. 1, 2 (and note
added in proof referring to Pareigis)

J.E. Humphreys, Ordinary and modular representations of Chevalley groups, Springer
Lect. Notes in Math. 528 (1976); Appendix T (following Sweedler's suggestion)

H. Garland and J. Lepowsky, Lie algebra homology and the Macdonald-Kac formulas, Invent. Math. 34 (1976); 1.7 and Remark

J.E. Humphreys, On the hyperalgebra of a semisimple algebraic group, in Contributions to Algebra, Academic Press, 1977; 3.1

The arguments here typically involve special cases of a general theorem suggested by Sweedler (and closely related to the "tensor identity" discussed in a recent MO post 37709 ): Let $A$ be a Hopf algebra (with antipode) over a commutative ring $k$, with Hopf subalgebra $B$ (possibly k). Given an $A$-module $M$ and a $B$-module $N$, there is a natural $A$-module isomorphism:
$$(A otimes_B N) otimes_k M cong A otimes_B (N otimes_k M)$$ On the left side, A acts via comultiplication, while on the right it acts on the first factor.

Is this the optimal generality, and if so is there a textbook reference?

riemann surfaces - Teichmuller Theory introduction

In addition to the ones already mentioned:

J. Harer's lecture notes on the cohomology of moduli spaces (doesn't have all the proofs, but describes the main ideas related to the cell decomposition of the moduli spaces; Springer LNM something, I believe; unfortunately I'm away for the holidays and can't access Mathscinet to find a precise reference).

K. Strebel, Quadratic differentials (careful exposition of the complex analytic results used to construct the cell decomposition mentioned above; not much about moduli spaces or Teichm"uller theory though; Springer Erbebnisse).

L. Ahlfors, Lectures on quasi-conformal mappings (construction of Teichmuller spaces).

L. Ahlfors' and L. Bers's papers in Analytic functions, Princeton, 1960.

matrices - Splitting matrix of rank one

I am having trouble understanding your English. But, if I understand you correctly, the following is a counter-example:

Let $k$ be a field and let $R$ be the ring $k[a,b,c,d]/(ab-cd)$. Then $R$ is normal and $left( begin{smallmatrix} a & c \\ d & b end{smallmatrix} right)$ has rank 1. However, we can not write this matrix as $left( begin{smallmatrix} w \\ x end{smallmatrix} right) left( begin{smallmatrix} y & z end{smallmatrix} right)$ for any $w$, $x$, $y$, $z in R$.

I think your condition should almost imply that the ring is a UFD. If I have any non-unique factorization $ab=cd$, I can use it to build a counter-example like this one.

UPDATE Here are two more examples: $R=k[a,b,c]/(ac-b^2)$ and $left( begin{smallmatrix} a & b \\ b & c end{smallmatrix} right)$.

$R=mathbb{Z}[sqrt{-5}]$ and $left( begin{smallmatrix} 2 & 1+sqrt{-5} \\ 1-sqrt{-5} & 3 end{smallmatrix} right)$.

These examples rule out most attempts I could think of to find a class of rings larger than UFDs for which the result holds.

ds.dynamical systems - Moshe Rosenfeld's Salmon Problem

As an amusement at the start of this talk, Moshe Rosenfeld poses the following question.

Suppose that there are n salmon which
begin at distinct points on a unit
circle, each facing either clockwise
or counterclockwise. On a signal, each salmon moves around the circle in its chosen direction at a constant speed (the same for all salmon). When two salmon meet, they both instantly reverse directions. If any salmon ever returns to its starting point, it dies. (If two salmon meet at one of their starting points, there is a death and no change of direction; as Rosenfeld says, "Death comes first.")

1. Is it true that all the salmon will eventually die?


2. (assuming the answer to part 1 is yes) Give an algorithm to find the last survivor.

I spent a certain amount of time on buses and planes tinkering with this. It's quite easy to show that every configuration is preperiodic, as a start. I have some ideas about how one might finish. Eventually I decided just to look for more information on the problem, with no real success.

One of the themes of his talk is how some problems become popular and some gather dust on the shelf. Is the latter what happened to this problem?

His second question is a bit mysterious. The problem setup itself is algorithmic in nature, so what does it mean? Is there anything besides "elegance" that would distinguish the kind of answer we should have in mind from a stupid answer like "just watch the salmon"? ("Running time" could be an answer, but it seems likely that just letting the salmon swim wouldn't take all that long.)

I am really asking three subquestions on this topic.

1. Did this question ever get solved or taken up seriously? If so, where?


2. Is there a natural, nonvacuous interpretation of the second part of
   the question?


3. What is the solution? (This is actually the subquestion I am least
   interested in, but it felt wrong not
   to ask it.)

(Please feel free to re-tag, still getting used to things here.)

pr.probability - Constructing Bernoulli random variables with prescribed correlation

Here's a generalization of gowers's construction that is practical for Bernoulli RVs with $pne1/2$. You want to generate $n$ Bernoulli RVs, each taking on value 0 or 1, each with mean $p<1/2$. (For $p>1/2$, do as described below for $1-p$, then complement the results.) Let $d=sqrt{2}text{ erfc}^{-1}left(2pright)$. (This is just the inverse survival function for the standard normal distribution.) Take unit vectors $v_1,...v_n$ as before. Generate a random $n$-vector $z$ whose components are IID standard normal RVs. Let $B_i=1$ iff $zcdot v_i>d$. $zcdot v_i$ is standard normal, so obviously gives the desired mean of $p$.

What about correlations? As in gowers's construction, these depend uniquely on the angle between vectors $v_i$ and $v_j$. Let $c_{ij}$ be the coincidence frequency between $B_i$ and $B_j$, i.e., the frequency with which both are 1, which is related to the correlation. If $theta_{ij}=cos^{-1}left(v_icdot v_jright)$, then

$$c_{ij}=int_d^infty Phileft(frac{ucostheta_{ij}-d}{sintheta_{ij}}right)phileft(uright)du$$

where $Phi(z)$ and $phi(z)$ are the standard normal CDF and PDF, respectively. $c$ decreases monotonically from $p$ at $theta=0$ to 0 at $theta=pi$. In a practical problem you'd probably want the inverse: you'd know $p$ and $c$ and want to get $theta$. I doubt that can be done other than numerically, but $c$ is a single function of two bounded variables $p$ and $theta$, so you can tabulate it numerically once and invert the interpolated function if you're going to be doing a lot of this.

Now you know what all dot products $v_icdot v_j=cos{theta_{ij}}$ need to be, it is simple to construct vectors at these angles. Let $v_1=left(1,0,...,0right)$. Then $v_2=left(costheta_{12},sintheta_{12},0,...,0right)$. For $v_3$, solve

$$pmatrix{v_{11}&v_{12}cr v_{21}&v_{22}}pmatrix{v_{31}cr v_{32}}=pmatrix{1&0cr costheta_{12}&sintheta_{12}}pmatrix{v_{31}cr v_{32}}=pmatrix{costheta_{13}crcostheta_{23}}$$

... then let $v_{33}=sqrt{1-v_{31}^2-v_{32}^2}$. Continue to generate the rest of the $v_i$. Since the matrix at every stage is lower triangular, the solution is unique as long as the diagonal is positive. The construction fails only if the norm of the first $i-1$ components of $v_i$ is $ge1$. I'm going to speculate that that occurs only if you give it a set of impossible coincidence frequencies (for instance, $c_{12}=c_{13}=p$, $c_{23}=0$), but I haven't attempted to show that.

Edit: Nope, I was too optimistic. For instance, if you have three mutually exclusive Bernoulli RVs with $ple1/3$, which is clearly possible, this construction fails. Alas.

rt.representation theory - Signed and unsigned Hecke algebra canonical basis

You probably know all of this already, but here goes...

Write $C'_w = T_w + sum_{x < w} p_{x,w} T_x$ where $p_{x,w} in umathbb{Z}[u]$. Now, the other basis can be defined by applying the involutive automorphism $b: mathcal{H}_n to mathcal{H}_n$, given by $b(T_w)=T_w$ and $b(u)=-u^{-1}$.

Define $C_w := b(C'_w)$.

Since, $b$ commutes with the bar involution, this basis is bar invariant as well.

Explicitly, $C_w = T_w + sum_{x < w} (-1)^{ell(w)+ell(x)} bar p_{x,w} T_x$.

So $C_w = bar{P}^{-1} P C'_w$ which seems hard to compute in general.

ca.analysis and odes - The characteristic (indicator) function of a set is not in the Sobolev space H1

The answer posted by Tom, as written is actually not true. A function in $H^1$ will not in general be differential almost everywhere; it depends on the dimension. In one dimension however it is indeed true that $H^1$ functions are differentiable almost everywhere (they are in fact absolutely continuous). There are two ways of seeing it is not in $H^1$. The simple answer is that if you differentiate the characteristic function of say $[0,infty)$ then you will get the dirac measure. However let me just answer your question first:

Answer 1:
Take any smooth compactly supported $phi:mathbb{R} to mathbb{R}$. By definition of weak derivative we have
$int phi g^{prime} dx = - int phi^{prime} g dx$ where I've set $g=1_{[0,infty)}$. This would have to be true for all such $phi$ if the weak derivative existed. Now take $phi^{epsilon}$ to be supported in a neighborhood $(-epsilon,epsilon)$ of $0$. We are making the crucial assumption that $g^{prime}$ is an integrable and hence it follows that $int phi^{epsilon} g^{prime} to 0$ as $epsilon to 0$. However, $phi^{epsilon}$ is smooth and so
$int partial_xphi^{epsilon}(x)g(x)dx = phi^{epsilon}(0)$ since $phi$ was assumed to have compact support in $(-epsilon,epsilon)$. Now just fix $phi^{epsilon}(0)=1$ and we have that $phi^{epsilon}(0) to 0$ by the first integral equality. This is a clear contradiction.

Notice that in fact that this really shows that $g' dx = delta(x)$.

Answer 2:
Take $1_{[0,1]}$ instead so that it is an $L^2([0,1])$ function. This is in fact the fourier transform of a "sinc" function, $sin(k)/k$ up to some normalization constants. If we consider the $H^1$ norm in frequency space we would need $int_0^{infty} |k|^2frac{sin(k)^2}{|k|^2} < infty$ which is clearly false. This requires being at ease with the fourier transform so if you're not, answer 1 is probably best.

It is true in $mathbb{R}^n$ that if $u in W^{1,p}$ for $p > n$ then $u$ is a.e differentiable and equals a.e its weak gradient (see Evans chapter 5). This is to correct what Tom had said although perhaps we was thinking about the $n=1$ case in which case $2 > 1$.

Hope this helps!
Dorian

nt.number theory - (nontrivial) isotrivial family of elliptic curves

Hint: use quadratic twists.

Edit: So as not to drag things out, I hope it's okay if I just give you a standard example. Let

$E_0: y^2 = x^3 + Ax + B$

be your favorite elliptic curve over $mathbb{Q}$ (i.e., any will do). Consider the
elliptic curve
$E: t y^2 = x^3 + Ax + B$
over the rational function field $mathbb{Q}(t) = mathbb{Q}(mathbb{P}^1)$. Spreading this out as a scheme over $mathbb{P}^1_{/mathbb{Q}}$, we see that there are two singular fibers, at $t = 0$ and $t = infty$. Discarding these we get an elliptic curve over $mathbb{A}^1 setminus {0}$ which is isotrivial -- the $j$-invariant over every fiber is $j(E_0)$ -- but nontrivial: the isomorphism classes of the fibers are in bijection with $H^1(mathbb{Q},mathbb{Z}/2mathbb{Z}) cong mathbb{Q}^{times}/mathbb{Q}^{times 2}$.

It's a good bet that you'll find this example somewhere in the chapter on elliptic surfaces in Silverman's Advanced Topics in the Arithmetic of Elliptic Curves.

Can it be done over any base scheme? Unless I misunderstand, of course not, e.g. not over the spectrum of a field.

homotopy theory - BU with tensor product H-space structure

I'll write $U(X)=[X,BU_otimes]$. So $U(X)subset K(X)=[X,Ztimes BU]$ is the multiplicatively closed subset corresponding to "virtual bundles of rank 1".
Likewise, I'll write $I(X)=[X,BU]subset K(X)$ for the ideal of "virtual bundles of rank 0".
There's a bijection $amapsto 1+a$ from $I(X)$ to $U(X)$.

Here's the claim: if $X$ is compact (let's say it's a finite CW-complex of dimension $n$), then $I(X)$ is a nilpotent ideal (in fact, $I(X)^{n+1}=0$). This is a generic fact about multiplicative cohomology theories, and it can be proved in a number of ways; you can think about it in terms of the multiplicative properties of the Atiyah-Hirzebruch spectral sequence, for instance.

For such $X$, it is clear that elements of $1+ain U(X)$ are invertible, given by the series $(1+a)^{-1}=1+a+a^2+cdots$ which terminates since $ain I(X)$ and $I(X)$ is nilpotent.

For infinite dimensional $X$, you need to argue a little harder. If $X=lim_{to} X_i$ where the $X_i$ are finite CW-complexes, then there is a surjection $K(X)to lim_{leftarrow} K(X_i)$, which restricts to surjections for $I(X)$ and $U(X)$.
The kernel is a $lim^1$-term. If the $lim^1$-term vanishes, then it's clear that elements of $U(X)$ are invertible, since their images in the $U(X_i)$ are invertible. It's enough to check that $lim^1$-vanishes in the case that $X=BU_otimes$, which is a standard calculation.

reference request - Classification of finite groups of isometries

There is a vast literature on the classification of finite linear groups over various fields. Over the complex or real fields, all finite linear groups are conjugate to subgroups of the respective unitary or orthogonal group, so as remarked in one of the comments above, studying finite groups of isometries in this context is the same as studying all the finite subgroups of ${rm GL}(n,mathbb{C})$ or ${rm GL}(n,mathbb{R}).$ As Richard Borcherds remarked, this soon becomes a complicated
problem. But strategies have evolved since the birth of representation theory to tackle the problem (for general fields) difficult as it is, in a systematic way. I'll discuss the real and complex cases. Generally speaking, we want to concentrate attention on linear groups which can't be described
in some "obvious" way in terms of linear groups in smaller dimensions. The first reduction, then,
is to concentrate on irreducible groups, those which leave no proper non-zero subspace invariant.
Maschke's Theorem tells us that no information is lost in the reduction. Another question,
for real representations, is what changes if we extend scalars to the complex field, where life
is generally easier. An irreducible real linear group may become reducible when the scalars are
extended to the complex numbers (this only happens when its character has squared-norm $2$ or $4$).
In each case, the real finite linear group is isomorphic to a finite complex linear group
in half the original dimension. So now I only speak of finite complex linear groups.
As remarked in someone's earlier comment, the next natural reduction is to the case of primitive
linear groups, those which (up to equivalence) be induced from linear groups of smaller dimension.
There are strong restrictions on normal subgroups of finite primitive linear groups. In particular,
the structure of primitive solvable finite linear groups is very tight, and is well-understood.
Having reduced to the primitive case (back to the general finite group), the next question
is whether the underlying module is a tensor product of two non-trivial modules of smaller
dimension. At this point, it may be necessary to take (still finite) central extensions of the group you started with. If there is a non-trivial tensor factorization, then we are reduced to questions
in smaller dimension. If there is no such factorization (even allowing for central extensions), then the structure of the residual groups is very restricted indeed. The given representation may
be "tensor induced" from a representation (of smaller dimension) of a proper subgroup. Tensor induction was introduced by Serre. If it can't be tensor induced from a lower dimensional
representation (again, even allowing for central extensions), then the only possibility that
remains is subgroup of a central extension of the automorphism group of a finite simple group
(containing all inner automorphisms). Many mathematicians, for example, Guralnick, Tiep, Zalesski,
have calculated (relatively) low dimensional complex representations of (central extensions of)
finite simple groups in recent years. My answer is therefore: yes, it is a difficult question,
but one which can be addressed systematically in any given case, and for which much hard-won
theory is available in the mathematical literature. Addendum: Just as it becomes impractical to list
all groups of a given finite order relatively soon, and we have to content ourselves with understanding the "building blocks", that is, the finite simple groups, so it is with finite
linear groups. There are three types of building blocks for finite complex linear groups:
a) 1-dimensional cyclic linear groups.
b) Finite complex linear groups $G$ of dimension $p^{n}$, for some prime $p$ and integer $n > 0$,
which have an irreducible normal $p$-subgroup $E$ (extraspecial of order $p^{2n+1}$ and
exponent $p$ when $p$ is odd; either extraspecial or the central product of an extraspecial
group of order $p^{2n+1}$ with a cyclic group of order $4$ when $p = 2.$). In this case,
$G/EZ(G)$ is isomorphic to an irreducible subgroup of the finite symplectic group ${rm Sp}(2n,p)$.
c) Finite complex linear groups $G$ of degree $m$ which have an irreducible quasisimple
subgroup $S$ ( this means that $S = S^{prime}$ and $S/Z(S)$ is a non-Abelian simple group).
Then $G/SZ(G)$ is a subgroup of the outer automorphism group of $S/Z(S)$.
The third type of building block naturally does not occur for solvable linear groups.
In both cases b) and c), the respective subgroups $E$ and $S$ are minimal subject to being normal,
but not central.

abstract algebra - when are epimorphisms of algebraic objects surjective?

Update: the following exchange appeared on the categories mailing list several years ago: http://article.gmane.org/gmane.science.mathematics.categories/3094. Walter Tholen's response strongly suggests that the answer to Martin's question is that the converse does not hold, although I don't have access to the four-author article he cites as reference. It's probably worth a look though, and if I learn anything more I'll post another update.

Second update: My surmise was correct. Walter Tholen kindly emailed to me the relevant two pages (pp. 88-89) of the four-author paper

• 
  E.W. Kiss, L. Marki, P. Prohle, W. Tholen: Studia Sci. Math. Hungaricum 18
  (1983) 79-141.
  

where the following example is given on page 89: in the category of semigroups with zero such that all 4-fold products are zero, all epimorphisms are surjective but not all monos are regular [a specific nonregular mono is described]. (I can forward this email if you write to me at topological dot musings at gmail dot com.)

Assuming amalgamated products exist (as they do in categories of algebras of a Lawvere theory), a mono i: A >--> B is regular if it is the equalizer of the pair of canonical maps from B to the amalgamated product B *_A B (i.e., the coprojections of the pushout of i with itself, aka the cokernel pair of i). The equalizer of the cokernel pair defines a closure operator on the lattice of subalgebras Sub(B), called the dominion operator Dom_B. So to prove a subalgebra is not regular is to show that it is not Dom-closed. The key technical result needed to prove the claim above is Isbell's Zig-Zag theorem (given in his paper Epimorphisms and Dominions in the 1965 La Jolla conference proceedings on categorical algebra), as recalled here, which gives a precise and useful criterion for an element to belong to the dominion (= Dom-closure) of a subalgebra.

Hope this helps. I am voting up your question, Martin, since it's rather nontrivial!

soft question - A single paper everyone should read?

I am surprised to see that so many people suggest meta-mathematical articles, which try to explain how one should do good mathematics in one or the other form. Personally, I usually find it a waste of time to read these, and there a few statements to which I agree so wholeheartedly as the one of Borel:

"I feel that what mathematics needs least are pundits who issue prescriptions or guidelines for presumably less enlightened mortals."

The mere idea that you can learn how to do mathematics (or in fact anything useful) from reading a HowTo seems extremely weird to me. I would rather read any classical math article, and there are plenty of them. The subject does not really matter, you can learn good mathematical thinking from each of them, and in my opinion much easier than from any of the above guideline articles. Just to be constructive, take for example (in alphabetical order)

• Atiyah&Bott, The Yang-Mills equations over Riemann surfaces.


• Borel, Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts.


• Furstenberg, A Poisson formula for semi-simple Lie groups.


• Gromov,Groups of polynomial growth and expanding maps.


• Tate, Fourier analysis in number fields and Hecke's zeta-functions.

I am not suggesting that any mathematician should read all of them, but any one of them will do. In fact, the actual content of these papers does not matter so much. It is rather, that they give an insight how a new idea is born. So, if you want to give birth to new ideas yourself, look at them, not at some guideline.

real algebra - Artin Schreier Theorem for Rings

This has been in my mind for quite some time. Looking at Artin Schreier Theorem for fields:

If L is a field and K its algebraic closure and if 1< [K:L] < infinity then L=K[i] and L is a real closed field (Thus L has characteristic 0. Here i is just the square root of -1).

I was wondering if a "generalized" Artin Schreier exist or if someone could refer to me to some paper that attempts this. There is a concept of real closedness and "algebraic closedness" of reduced commutative rings, but I doubt that the statement would hold.

So one has the following conjecture:

If L is a reduced commutative ring and K is its total integral closure (this is an equivalent notion of algebraic closure if K and L were fields) and if 1<[K:L]< infinity (here I mean that K is a finite L-module that is not the same as L) then L is real (thus its characteristic is 0.. and one can add that L is real closed in the sense of reduced commutative rings).

Can one easily show this, even at least prove that L has characteristic 0?

linear algebra - What optimization criteris should be used for this problem?

The real world version:

I have a united value (e.i. 12in, 120V 1.414 kg*m/s) where the units are specified as the rational exponents of the 5 base units; m, s, kg, C and K. Additionally, I have a set of non-base units and I want to find the "simplest" combination of those units that matches my value.

The abstract version:

With a little manipulation this can be converted to a very under defined linear combination problem. With a little more work it can be restricted to integer solutions, making it (I think) a linear programming problem (a subject I know almost nothing about).

The part I actualy care about:

What additional constraints/criteria can be applied to this problem to make it solvable and what are the implications of the options? Specifically, I'm looking for a solution that is well documented, simple to evaluate and produces approximately the same results as people would expect.

ct.category theory - Monoidal operations on categories where the maps on Aut, End are injective

Your conditions don't seem to obtain very often, unfortunately.

Let's begin with the one-object case (where the one object is the unit, as it must be). This is the same thing as a monoid object in monoids, or equivalently, a commutative monoid $A$, by Eckmann-Hilton. Your requirement (1) appears to be that the multiplication $Atimes Ato A$ is injective; requirement (2) is the same, applied to the invertible elements. This says that every element of $A$ can be uniquely factored, so that $A$ is trivial. So in the one-object case, (1) happens exactly once, and (2) is equivalent to the nonexistence of nontrivial invertible elements.

Using this, we deduce that the unit in a monoidal category satisfying either (1) or (2) is either very rigid in the sense that it admits no nontrivial endomorphisms or, respectively, rigid in the sense that it admits no nontrivial automorphisms.

If the category $mathcal{C}$ has an initial object $varnothing$ that is preserved by the monoidal structure (so that $Xotimesvarnothing=varnothingotimes X=varnothing$ for all $Xinmathcal{C}$), then every object has to (1) very rigid or (2) rigid. This kills off the bulk of the "algebraic" examples (like modules over a ring, etc.)

Also, note that your conditions are symmetric, i.e., (1) or (2) holds for $mathcal{C}$ iff it holds for $mathcal{C}^{mathrm{op}}$. So Reid's example showing that not all cartesian categories satisfy (2) also works to show that not all cocartesian categories satisfy (2). (We contemplate the coproduct in $mathrm{Set}^{mathrm{op}}$...)

I don't want to be overly negative; so here's a situation in which I think your conditions do hold: suppose $mathcal{C}$ a category with all finite coproducts. Suppose the morphism $Xto Xsqcup Y$ is monic for any $X,Yinmathcal{C}$. Then $mathcal{C}$ (equipped with the coproduct) satisfies your condition. (So this works for finite sets with disjoint union, for instance.) Dually, if $mathcal{C}$ is a category with all finite products such that the morphism $Xtimes Yto X$ is epic for any $X,Yinmathcal{C}$, then $mathcal{C}$ (equipped with the product) satisfies your condition. (So this works for the product on nonempty finite sets, for instance.)

Difference between connected vs strongly connected vs complete graphs

What is the difference between

connected

strongly-connected and

complete?

My understanding is:

connected: you can get to every vertex from every other vertex.

strongly connected: every vertex has an edge connecting it to every other vertex.

complete: same as strongly connected.

Is this correct?

ac.commutative algebra - Classification of finite commutative rings

Yes, a finite ring $R$ is a finite direct sum of local finite rings. As a first step, for each prime $p$ there is a subring $R_p$ of $R$ corresponding to the elements annihilated by the powers of $p$. $require{enclose} enclose{horizontalstrike}{R_p style{font-family:inherit;}{text{is then an}}hspace{-7mm}}$
$enclose{horizontalstrike}{style{font-family:inherit;}{text{algebra over}} mathbb{Z}/p.}$ $R_p$ then resembles an algebra over $mathbb{Z}/p$ and it could be one, but it can also have a more complicated structure as an abelian $p$-group (see below). This step generalizes to maximal ideals: For each maximal ideal $m$, $R_m$ is the subring of elements annihilated by $m^n$ for some $n$, and $R$ is the direct sum of these subrings, which are local rings.

It is not difficult to write down a rough partial classification of of local finite rings. If $R$ is local with maximal ideal $m$, it is resembles an algebra over the finite field $F = R/m$; the associated graded ring is such an algebra. If you choose a basis $x_1,ldots,x_n$ for $m/m^2$, then $R$ or its associated graded is a quotient of the polynomial ring $F[vec{x}]$ in which only finitely many monomials are non-zero. You can make a diagram of these non-zero monomials; they can be any order ideal in the $n$-dimensional orthant. Or, in basis-independent form, $R$ has a length, which is the largest nonvanishing power of $m$, and each $m^k/m^{k+1}$ is some quotient of the $k$th symmetric power of the generating vector space $V = m/m^2$.

After that, the non-zero monomials may be linearly dependent (and never mind that $R$ might be more complicated than its associated graded). Informally, there will be an endless stream of partial results and there will never be a complete classification when the length of the local ring is 3 or more. To see this, suppose that $m^4 = 0$, and suppose that $m^3$ is only one dimension shy of $S^3(V)$. Then the ring is defined by an arbitrary symmetric trilinear form in $V$. These make a "wild" sequence of algebraic varieties, in the same sense that people say that the representation theories of certain rings are wild. For instance, I think (not quite sure) that it is NP-hard to determine when two such trilinear forms are equivalent. NP-hardness is not by itself rigorously equivalent to no classification, but informally the classification is an intractable mess.

If the nonvanishing monomials in $R$ are linearly independent, then it is a toric local ring. Toric local rings are certainly a tractable class of finite rings.

The situation is similar to non-commutative $p$-groups, which are also wild and will never be classified. In both cases, certain classes have a nice structure. It is also interesting to make estimates for how many there are.

Note: Corrected per comment.

matrices - Freeness of the Canonical $SU(n)$ Action

I have another question about $SU(n)$, again I hope it's not too basic. For $n=2$, the action of $SU(2)$ on $C^2$ is free since it's equal to the group of rotations. In general, the group of rotations is properly contained in $SU(n)$, does this mean that its action on $C^n$ is no longer free.

nt.number theory - Hyperspecial subgroup of a product of semisimple algebraic groups

To allow all characteristics, the semisimplicity requirement on the Lie algebras should be replaced with the requirement that the $G_i$ are semisimple as $F$-groups. (In positive characteristic the Lie algebras of connected semisimple groups can often fail to be semisimple.) In fact, the simply connected and semisimplicity hypotheses are stronger than necessary; connected reductive is sufficient. Taking into account the definition of "hyperspecial maximal compact subgroup" (of the group of $F$-rational points), the question in this modified form immediately reduces to a more "algebraic" assertion having nothing to do with local fields, as follows.

Consider a reductive group scheme $mathcal{G}$ (with connected fibers, following the convention of SGA3) over a normal noetherian scheme $S$, and let $G$ be its fiber over the scheme $eta$ of generic points of $S$. Assume $G$ decomposes as a direct product $G = G_1 times G_2$ with $G_i$ necessarily connected reductive over $eta$. Then does there exist a pair of reductive closed $S$-subgroup schemes (with connected fibers) $mathcal{G}_i$ in $G_i$ such that $mathcal{G}_ 1 times mathcal{G}_2 = mathcal{G}$ compatibly with the given identification on generic fibers? Below is a proof that the answer is ``yes''.

First observe that such $mathcal{G}_i$ are necessarily unique if they exist, being the Zariski closures of the $G_i$ in $mathcal{G}$ (since $S$ is reduced). In view of this uniqueness, by descent theory it follows that to prove the existence we may work 'etale-locally on the base. (This step tends to ruin connectedness hypotheses, since $S' times_S S'$ is generally cannot be arranged to be connected, even if $S$ is connected and $S' rightarrow S$ is a connected 'etale cover.) Hence, we now assume that $mathcal{G}$ is $S$-split in the sense that it admits a (fiberwise) maximal $S$-torus $mathcal{T}$ that is $S$-split. We may also work separately over each connected component of $S$, so we can now assume $S$ is connected (as we will make no further changes to $S$ in the argument).

The maximal $eta$-torus $T := mathcal{T}_ {eta}$ in $G$ uniquely decomposes as $T = T_1 times T_2$ for necessarily $eta$-split maximal $eta$-tori $T_i$ in $G_i$. By the classification of split pairs (i.e., fiberwise connected reductive group equipped with a split maximal torus) over any base scheme in terms of root data, to give a direct product decomposition of $(mathcal{G}, mathcal{T})$ is the same as to decompose its root datum into a direct product (i.e., direct product of the $X$ and $X^{vee}$ parts, and corresponding disjoint union for the $R$ and $R^{vee}$ parts). By connectedness, the normal $S$ is irreducible (i.e., $eta$ is a single point), so the root data of $(mathcal{G}, mathcal{T})$ and its generic fiber $(G,T)$ are canonically identified. QED

proof theory - Proving inequalities over algebraic structures

The desired degree of generality seems to vary greatly among the various parts of the question and the subsequent comment. Here's a simple answer to one aspect of the question. There is no algorithm for computably enumerating all true inequalities between polynomials with natural number coefficients. (Here "true" means identically true, for all natural number values of the variables.) In particular, there is no deductive system that can prove exactly those true inequalities.

The reason is that any statement of the form "$P(x_1,dots,x_n)$ has no integer solutions," with $P$ a polynomial over the integers, can be rewritten as an inequality between polynomials with natural number coefficients, as follows: Replace each of the integer variables $x_i$ with the difference $y_i-z_i$ of two natural number variables (so that we can talk about solutions in $mathbb N$ instead of $mathbb Z$); then express the absence of zeros of the polynomial as $P^2 geq 1$; and finally, transpose any terms with negative coefficients to the right side.

Now suppose, toward a contradiction, that we could enumerate the true inequalities over the semiring $mathbb N$. So, we could enumerate all true statements of the form "$P(x_1,dots,x_n)$ has no integer solutions," considered above. Then we could decide whether any given Diophantine equation $P=0$ has a solution. Just run both the enumeration (to see whether it turns up a proof that there's no solution) and a systematic search for a solution. Eventually, you'll get the answer. But such a decision algorithm is impossible, by the solution of Hilbert's 10th problem.

reference request - Proofs without words

I'm quite surprised no-one pointed out this one yet:

Theorem. The trefoil knot is knotted.

Proof.

$square$

Some comments: a 3-colouring of a knot diagram D is a choice of one of three colours for each arc D, such that at each crossing one sees either all three colours or one single colour. Every diagram admits at least three colourings, i.e. the constant ones. We'll call nontrivial every 3-colouring in which at least two colours (and therefore all three) actually show up. It's easy to see (one theorem, more pictures!) that Reidemeister moves preserve the property of having a nontrivial 3-colouring, and that the unknot doesn't have any nontrivial colouring.

The picture shows a (nontrivial) 3-colouring of the trefoil.

EDIT: I've made explicit what "nontrivial" meant ― see comments below. Since I'm here, let me also point out that the number of 3-colourings is independent of the diagram, and is itself a knot invariant. It also happens to be a power of 3, and is related to the fundamental group of the knot complement (see Justin Robert's Knot knotes if you're interested).

ag.algebraic geometry - Tautological bundle on G$(n,k)$ and Chern classes

Depending on what you need to do with the Chern roots, it may be cleaner to just ask for the Chern classes of $S$.

In this case, let $Q$ be the quotient bundle, i.e., there is a trivial bundle ${bf C}^k$ which contains $S$ as a subbundle, and $Q = {bf C}^k / S$. The $i$th Chern class of $Q$ is the cohomology class of $sigma_i$, the special Schubert class of codimension $i$ (see Proposition 3.5.5 of Manivel's book Symmetric Functions, Schubert Polynomials, and Degeneracy Loci). Using the relation $c(S)c(Q) = 1$, and knowledge of the cohomology ring of $G(n,k)$ should be enough to perform any usual calculations.

For computing with these Schubert classes, one only needs to learn the Pieri rule (and perhaps the Littlewood-Richardson rule depending on the circumstance), both of which can be found in Manivel's book (chapter 1) or see Wikipedia.

ct.category theory - Category of categories as a foundation of mathematics

My personal opinion is that one should consider the 2-category of categories, rather than the 1-category of categories. I think the axioms one wants for such an "ET2CC" will be something like:

• Firstly, some exactness axioms amounting to its being a "2-pretopos" in the sense I described here: http://ncatlab.org/michaelshulman/show/2-categorical+logic . This gives you an "internal logic" like that of an ordinary (pre)topos.


• Secondly, the existence of certain exponentials (this is optional).


• Thirdly, the existence of a "classifying discrete opfibration" $elto set$ in the sense introduced by Mark Weber ("Yoneda structures from 2-toposes") which serves as "the category of sets," and internally satisfies some suitable axioms.


• Finally, a "well-pointedness" axiom saying that the terminal object is a generator, as is the case one level down with in ETCS. This is what says you have a 2-category of categories, rather than (for instance) a 2-category of stacks.

Once you have all this, you can use finite 2-categorical limits and the "internal logic" to construct all the usual concrete categories out of the object "set". For instance, "set" has finite products internally, which means that the morphisms $set to 1$ and $set to set times set$ have right adjoints in our 2-category Cat (i.e. "set" is a "cartesian object" in Cat). The composite $set to settimes set to set$ of the diagonal with the "binary products" morphism is the "functor" which, intuitively, takes a set $A$ to the set $Atimes A$. Now the 2-categorical limit called an "inserter" applied to this composite and the identity of "set" can be considered "the category of sets $A$ equipped with a function $Atimes Ato A$," i.e. the category of magmas.

Now we have a forgetful functor $magma to set$, and also a functor $magma to set$ which takes a magma to the triple product $Atimes A times A$, and there are two 2-cells relating these constructed from two different composites of the inserter 2-cell defining the category of magmas. The "equifier" (another 2-categorical limit) of these 2-cells it makes sense to call "the category of semigroups" (sets with an associative binary operation). Proceeding in this way we can construct the categories of monoids, groups, abelian groups, and eventually rings.

A more direct way to describe the category of rings with a universal property is as follows. Since $set$ is a cartesian object, each hom-category $Cat(X,set)$ has finite products, so we can define the category $ring(Cat(X,set))$ of rings internal to it. Then the category $ring$ is equipped with a forgetful functor $ring to set$ which has the structure of a ring in $Cat(ring,set)$, and which is universal in the sense that we have a natural equivalence $ring(Cat(X,set)) simeq Cat(X,ring)$. The above construction then just shows that such a representing object exists whenever Cat has suitable finitary structure.

One can hope for a similar elementary theory of the 3-category of 2-categories, and so on up the ladder, but it's not as clear to me yet what the appropriate exactness properties will be.

lo.logic - Categorical foundations without set theory

On the subject of categorical versus set-theoretic foundations there
is too much complicated discussion about structure that misses the
essential point about whether "collections" are necessary.

It doesn't matter exactly what your personal list of mathematical
requirements may be -- rings, the category of them, fibrations,
2-categories or whatever -- developing the appropriate foundational
system for it is just a matter of "programming", once you understand
the general setting.

The crucial issue is whether you are taken in by the Great Set-Theoretic
Swindle that mathematics depends on collections (completed infinities).
(I am sorry that it is necessary to use strong language here in order to
flag the fact that I reject a widely held but mistaken opinion.)

Set theory as a purported foundation for mathematics does not and cannot
turn collections into objects. It just axiomatises some of the intuitions
about how we would like to handle collections, based on the relationship
called "inhabits" (eg "Paul inhabits London", "3 inhabits N"). This
binary relation, written $epsilon$, is formalised using first order
predicate calculus, usually with just one sort, the universe of sets.
The familiar axioms of (whichever) set theory are formulae in first order
predicate calculus together with $epsilon$.

(There are better and more modern ways of capturing the intuitions about
collections, based on the whole of the 20th century's experience of algebra
and other subjects, for example using pretoposes and arithmetic universes,
but they would be a technical distraction from the main foundational issue.)

Lawvere's "Elementary Theory of the Category of Sets" axiomatises some
of the intuitions about the category of sets, using the same methodology.
Now there are two sorts (the members of one are called "objects" or "sets"
and of the other "morphisms" or "functions"). The axioms of a category
or of an elementary topos are formulae in first order predicate calculus
together with domain, codomain, identity and composition.

Set theorists claim that this use of category theory for foundations
depends on prior use of set theory, on the grounds that you need to start
with "the collection of objects" and "the collection of morphisms".
Curiously, they think that their own approach is immune to the same
criticism.

I would like to make it clear that I do NOT share this view of Lawvere's.

Prior to 1870 completed infinities were considered to be nonsense.

When you learned arithmetic at primary school, you learned some rules that
said that, when you had certain symbols on the page in front of you,
such as "5+7", you could add certain other symbols, in this case "=12".
If you followed the rules correctly, the teacher gave you a gold star,
but if you broke them you were told off.

Maybe you learned another set of rules about how you could add lines and
circles to a geometrical figure ("Euclidean geometry"). Or another one
involving "integration by parts". And so on. NEVER was there a "completed
infinity".

Whilst the mainstream of pure mathematics allowed itself to be seduced
by completed infinities in set theory, symbolic logic continued and
continues to formulate systems of rules that permit certain additions
to be made to arrays of characters written on a page. There are many
different systems -- the point of my opening paragraph is that you can
design your own system to meet your own mathematical requirements --
but a certain degree of uniformity has been achieved in the way that they
are presented.

• We need an inexhaustible supply of VARIABLES for which we may substitute.




• There are FUNCTION SYMBOLS that form terms from variables and other terms.




• There are BASE TYPES such as 0 and N, and CONSTRUCTORS for forming new
  types, such as $times$, $+$, $/$, $to$, ....




• There are TRUTH VALUES ($bot$ and $top$), RELATION SYMBOLS ($=$)
  and CONNECTIVES and QUANTIFIERS for forming new predicates.




• Each variable has a type, formation of terms and predicates must respect
  certain typing rules, and each formation, equality or assertion of a
  predicate is made in the CONTEXT of certain type-assignments and
  assumptions.




• There are RULES for asserting equations, predicates, etc.

We can, for example, formulate ZERMELO TYPE THEORY in this style. It has
type-constructors called powerset and {x:X|p(x)} and a relation-symbol
called $epsilon$. Obviously I am not going to write out all of the details
here, but it is not difficult to make this agree with what ordinary
mathematicians call "set theory" and is adequate for most of their
requirements

Alternatively, one can formulate the theory of an elementary topos is this
style, or any other categorical structure that you require. Then a "ring"
is a type together with some morphisms for which certain equations are
provable.

If you want to talk about "the category of sets" or "the category of rings"
WITHIN your tpe theory then this can be done by adding types known as
"universes", terms that give names to objects in the internal category
of sets and a dependent type that provides a way of externalising
the internal sets.

So, although the methodology is the one that is practised by type theorists,
it can equally well be used for category theory and the traditional purposes
of pure mathematics. (In fact, it is better to formalise a type theory
such as my "Zermelo type theory" and then use a uniform construction to
turn it into a category such as a topos. This is easier because the
associativity of composition is awkward to handle in a recursive setting.
However, this is a technical footnote.)

A lot of these ideas are covered in my book "Practical Foundations of
Mathematics" (CUP 1999), http://www.PaulTaylor.EU/Practical-Foundations
Since writing the book I have written things in a more type-theoretic
than categorical style, but they are equivalent. My programme called
"Abstract Stone Duality", http://www.PaulTaylor.EU/ASD is an example of the
methodology above, but far more radical than the context of this question
in its rejection of set theory, ie I see toposes as being just as bad.

fa.functional analysis - Hilbert spaces are induced by a bilinear form. How about n-linear forms?

As long as the form is positive definite and the unit ball is convex, you get a perfectly good Banach space using any symmetric $n$-linear form on a real vector space $V$. The degree $n$ is necessarily even. It is equivalent to defining the norm as the $n$th root of a homogeneous degree $n$ polynomial. $ell^p$ is an example for any even integer $p$. There are many other examples. I found a paper, Banach spaces with polynomial norms, by Bruce Reznick, that studies these norms. He obtains various results; the most appealing one to me at a glance is that these Banach spaces are all reflexive.

Off-hand I can't think of any simple way to recover positive definiteness starting with odd polynomials. The cube of the norm on $ell^3$ is a polynomial in the absolute values of the coordinates rather than the coordinates themselves.

Addendum: To address Darsh's comment, what you would look at in the complex case is self-conjugate polynomials of degree $(n,n)$. Equivalently, as with all complex Banach norms, the realification is a real Banach norm which is invariant under complex scalar rotation.