gr.group theory - Almost but not quite a homomorphism

Related to Henry Wilton's comments: the following might not quite be what you're looking for, but seems interesting given that quasi-morphisms have been mentioned. I'm doing this from memory so if there's a gap, someone please let me know!

Let $E$ be a Hilbert space, $B(E)$ the algebra of all bounded linear operators on $E$.
(Even the case $E={mathbb R}^n$ is of interest.) Fix a small $epsilon>0$. Then there exists $delta>0$ with the following property:

Let $G$ be an abelian group, and let
$f:G to B(E)$ be a bounded function (i.e. $sup_{xin G} | f(x) | < infty$) which satisfies
$$ sup_{x,y}| f(x)f(y) - f(xy) | leq delta. $$
Then there is some representation $rho: G to B(E)$ such that $sup_x | f(x)- rho(x)| leq epsilon$.

So, less formally, bounded "almost representations" of abelian groups are "near to" genuine representations.

I imagine this could be proved by an averaging argument: the way I learned of this result is as a special case of a more general one, in which the word "abelian" is replaced by the word "amenable", and the word "Hilbert" is replaced by "nice reflexive Banach". That in turn is a special case of a general result on almost multiplicative maps between Banach algebras satisfying certain conditions (due to B. E. Johnson).

Anyway, sorry this has wandered off track. The point was to say that there are contexts where things which are close to being group homomorphisms $Hto K$, might under a small perturbation be genuine homomorphisms when restricted to a specified abelian subgroup of $H$. However, in general this can't be done so as to work simultaneously for all abelian subgroups of $H$.

nt.number theory - Intuition for the last step in Serre's proof of the three-squares theorem

The Lemma in the aswer by Bjorn Poonen can be sharpened to give an exact relationship
between $mathrm{den}(x)$ and $mathrm{den}(x')$.

Lemma 1.
$~$Let $f=f_2+f_1+f_0inmathbb{Z}[X]=mathbb{Z}[X_1,ldots,X_n]$ ($ngeq 1$),
with each $f_i$
homogeneous of degree $i$.
Let $y,vinmathbb{Z}^n$, where $v$ is primitive and $f_2(v)neq0$,
and define $F=AT^2+BT+C:=f(y+Tv)inmathbb{Z}[T]$.
Suppose that the two zeros $t$ and $t'$ of $F$ are rational,
so that the two rational points $x=y+tv$ and $x'=y+t'v$ are zeros of $f$.
If $x=a/b$ and $x'=a'/b'$ are the reduced representations (with $b,b'>0$),
and $xneq y$ (which is certainly true if $xnotinmathbb{Z}^n$),
then
begin{equation*}
b' ,=, mathrm{sgn}(A)cdot
frac{f_2(x-y)}
{mathrm{gcd}(A,B,C),mathrm{gcd}(a-by)^2}cdot b~; tag{1}
end{equation*}
exchanging $x$ and $x'$ gives the analogous identity (provided $yneq x'$).

Remark.
$~$For any $u=(u_1,ldots,u_n)inmathbb{Z}^n$ we write $mathrm{gcd}(u):=mathrm{gcd}(u_1,ldots,u_n)$.

Proof. $~$Since $tv=x-y=(a-by)/b=(c/b)v$,
where $c=pm,mathrm{gcd}(a-by)$ and $mathrm{gcd}(c,b)=1$,
we have $t=c/b$, and similarly $t'=c'/b'$ with $c'=pm,mathrm{gcd}(a'-b'y)$
and $mathrm{gcd}(c',b')=1$. The leading coefficient of $F$ is $A=f_2(v)=(b/c)^2f_2(x-y)$
($cneq 0$ because $a-by=b(x-y)neq 0$).
By Gauss lemma $F=d(bT-c)(b'T-c')$ with $d=mathrm{sgn}(A)cdotmathrm{gcd}(A,B,C)$.
Then $dbb'=A=(b/c)^2f_2(x-y)$ gives us $b'$ expressed
as in the lemma.$~$
Done.

The Lemma in Bjorn Poonen's post is an immediate consequence.

Lemma 1 is about the geometric background of Davenport-Cassels lemma:
it relates the reduced representations of two rational zeros of $f$
that lie on an integral line $L=y+mathbb{Q}v$
whose direction vector $v$ is an anisotropic vector of the quadratic form $f_2$.
(An integral line is an affine line in $mathbb{Q}^n$
that contains an integral point and hence infinitely many integral points.)
It is not required that one or the other of the two zeros is non-integral,
the lemma says something interesting even when both zeros are integral.
Also, the two zeros may coincide, in which case the line $L$
is tangent to the quadric ${f=0}$.

If we actually write out the other identity mentioned in the lemma
and then compare the two identities,
we obtain the identity (supposing $yneq x,x'$)
begin{equation*}
f_2(x-y)f_2(x'-y)
,=, bigl(mathrm{gcd}(A,B,C),mathrm{gcd}(a-by),mathrm{gcd}(a'-b'y)bigr)^2~.tag{2}
end{equation*}
It is not in the least surprising that $f_2(x-y)f_2(x'-y)$ is a square:
if $q$ is any quadratic form on a vector space $V$ over some field $K$,
and $uin V$ and $lambda,muin K$,
then it is trivial that $q(lambda u)q(mu u)=bigl(lambdamu q(u)bigr)^2$.
It may seem slightly surprising that $f_2(x-y)f_2(x'-y)$ is a square of an integer,
but this is
a consequence of $A=dbb'$: in the situation of the lemma
the trivial identity satisfied by a general quadratic form reads
begin{equation*}
f_2(x-y)f_2(x'-y) ,=, Bigl(frac{c}{b},frac{c'}{b'}f_2(v)Bigr)^2~,
end{equation*}
and substituting $f_2(v)=A=dbb'$ yields
begin{equation*}
f_2(x-y)f_2(x'-y) ,=, (dcc')^2 ,=, C^2 ,=, f(y)^2
end{equation*}
(this time without the restriction $yneq x,x'$).
Let $L_{mathbb{Z}}$ denote the set of all integral points on the line $L$:
$L_{mathbb{Z}}:=Lcapmathbb{Z}^n=y+mathbb{Z}v$.
For any $zin L_{mathbb{Z}}$
the identity (2) still holds when $y$ is replaced by $z$,
but we must be careful and write it as
begin{equation*}
f_2(x-z)f_2(x'-z)
,=, bigl(mathrm{gcd}(A,B_z,C_z),mathrm{gcd}(a-bz),mathrm{gcd}(a'-b'z)bigr)^2~,
end{equation*}
because the coefficients $B_z$ and $C_z$ of $F_z=f(z+Tv)$ depend on $z$.
However,
note that the coefficient $A_z=A=f_2(v)$,
as well as the greatest common divisor of the coefficients of $F_z$
(the content of $F_z$),
$mathrm{gcd}(A,B_z,C_z)=left|Aright|/bb'=left|dright|$, do not depend on $z$.
For $zin L_{mathbb{Z}}$ we define $c(z),c'(z)inmathbb{Z}$
by $x-z=bigl(c(z)/bbigr)v$ and $x'-z=bigl(c'(z)/b'bigr)v$.
Since $mathrm{gcd}(a-bz)=left|c(z)right|$ and $mathrm{gcd}(a'-b'z)=left|c'(z)right|$,
we have
begin{equation*}
f_2(x-z)f_2(x'-z) ,=, bigl(d,c(z),c'(z)bigr)^2 ,=, C_z^2 ,=, f(z)^2~,
qquadquad zin L_{mathbb{Z}},.
end{equation*}
Remark. $~$Idiot me!
This is just a very special case of the general power-of-a-point theorem,
which does not rely on specific factorization properties of integers
and is almost trivial to prove:

Let $K$ be a field,
let $fin K[X] = K[X_1,ldots,X_n]$ be of degree $m$ (where $m, ngeq 1$),
and denote by $f_m$ the homogeneous component of $f$ of degree $m$.
Let $L$ be an affine line in $K^n$ with a direction vector $v$, where $f_m(v)neq 0$.
Let $yin L$, and define $F_y := f(y+Tv)in K[T]$, a polynomial of degree $m$.
Suppose that $F_y$ has $m$ zeros (counting multiplicities) $t_1$, $ldots$, $t_m$ in $K$.
Then the points $x_i=y+t_ivin L$, $1leq ileq m$, are zeros of $f$,
the multiset of the $x_i$'s does not depend on the choice of $yin L$,
and
begin{equation*}
f_m(y-x_1)f_m(y-x_2)cdots f_m(y-x_m) ,=, f(y)^m~.
end{equation*}

The independence is easy: if $z=y+sv$,
then $F_z$ has the zeros $t_i-s$, whence $z+(t_i-s)v = y+t_iv = x_i$.
The leading coefficient of $F_y$ is $f_m(v)$ and its constant term is $f(y)$.
From $F_y=f_m(v)(T-t_1)cdots(T-t_m)$ we get $f(y)=(-1)^m t_1cdots t_m f_m(v)$,
whence $f_m(y-x_1)cdots f_m(y-x_m) = bigl((-t_1)cdots(-t_m)f_m(v)bigr)^m = f(y)^m$.

We digress.
Let's return to the situation in Lemma 1.

We regard the point $y$ as fixed, serving as an origin of $L_{mathbb{Z}}$.
Let us determine $c(z)$ for a general point $z=y+kvin L_{mathbb{Z}}$, $kinmathbb{Z}$:
from
begin{equation*}
frac{c(y+kv)}{b},v ,=, x-(y+kv) ,=, (x-y)-kv ,=, frac{c(y)-kb}{b},v
end{equation*}
we see that
begin{equation*}
c(y+kv) ,=, c(y) - kb~. tag{3}
end{equation*}
For any $zin L_{mathbb{Z}}$ we have
begin{equation*}
f_2(x-z) ,=, frac{c(z)^2}{b^2}f_2(v)
,=, frac{c(z)^2}{b^2},dbb'
,=, frac{db'}{b},c(z)^2
,=, frac{e_0}{b_0},c(z)^2~, tag{4}
end{equation*}
where $b_0=b/mathrm{gcd}(b,db')$ and $e_0=db'/mathrm{gcd}(b,db')$.
Since $mathrm{gcd}bigl(b,c(z)bigr) = 1$, and hence $mathrm{gcd}bigl(b_0,c(z)bigr) = 1$, it follows that
begin{equation*}
mathrm{den}bigl(f_2(x-z)bigr) ,=, b_0 qquadquad text{for every $zin L_{mathbb{Z}}$},.
end{equation*}
Combining (3) and (4) we obtain
begin{equation*}
f_2bigl(x-(y+k)vbigr) ,=, frac{e_0}{b_0}bigl(c(y)-kbbigr)^2~, qquadquad kinmathbb{Z},.
end{equation*}
In the special case $x=x'$, when the line $L$ is a tangent of the quadric ${f=0}$,
we have $b=b'$, whence
begin{equation*}
f_2(x-z) ,=, d,c(z)^2~, qquadquad zin L_{mathbb{Z}},,
end{equation*}
thus $f_2(x-z)$ is an integer for every integral point $z$ in $L$.

The discussion above has demonstrated that the identity (1) and its brethren have uses
unrelated to Davenport-Cassels lemma.
Now we return to applications of (1) to Davenport-Cassels lemma and (a little way) beyond it.
The assumptions $xinmathbb{Q}^nsetminusmathbb{Z}^n$ and $0<left|f_2(x-y)right|<1$
imply the premises $f_2(v)neq 0$ and $xneq y$ of Lemma 1
and then yield the instantaneous result $b'<b$.
But besides the $f_2(x-y)$ in the numerator on the right hand side of (1)
there are also the factors $mathrm{gcd}(A,B,C)$ and $mathrm{gcd}(a-by)^2$ in the denominator
whose product can be greater than $1$ and can help make $b'$ smaller than $b$
even when $left|f_2(x-y)right|geq 1$.
This leads to the idea of walking with Aubry 'on the far side':
when there is no integral point $y$ satisfying $left|f_2(x-y)right|geq 1$
we may still find an integral point $y$
so that we can make a step along the line $L$
from a zero $x=a/b$ of $f$ to a zero $x'=a'/b'$ of $f$ with $b'<b$.
The folowing two examples attest that this idea actually works.

But first, a definition.
A quadratic form $q$ on $mathbb{Q}^n$ is said to be Euclidean
if for every $xinmathbb{Q}^nsetminusmathbb{Z}^n$
there exists $yinmathbb{Z}^n$ such that $0 < left|q(x-y)right| < 1$.

For the first example let $q(X_1,X_2,X_3)=X_1^2+X_2^2+5X_3^2$
and $f(X)=q(X) - m$, where $m$ is a positive integer.
The positive-definite quadratic form $q$ is not Euclidean on $mathbb{Q}^3$,
since for any $xinmathbb{Z}^3+bigl(frac{1}{2},frac{1}{2},frac{1}{2}bigr)$
and any $yinmathbb{Z}^3$ we have $q(x-y)geq 7/4$.
Consider the set $F$ of all points $x$ in the cube
$bigl{(x_1,x_2,x_3)inmathbb{R}^3 bigm| 0leq x_1,x_2,x_3leq frac{1}{2}bigr}$
at which $q(x)geq 1$;
the set $F$ ('the far side') is shown as the darker shaded part of the cube
in the following figure:

The function $q(1-x_1,1-x_2,1-x_3)$ of a point $(x_1,x_2,x_3)$ in the set $F$
attains its largest value $8-2sqrt{5}$ at the point $P=bigl(0,0,1/sqrt{5}bigr)$.
$quad$Suppose that $xinmathbb{Q}^3setminusmathbb{Z}^3$ is a zero of $f$,
and let $y:=mathrm{round}(x)$.
If $q(x)<1$ (we certainly have $q(x)>0$) then fine,
we make a step to the next zero of $f$ with smaller denominator
this side of the Euclidean horizon.
Otherwise $q(x-y)geq 1$, we are on the far side, and must tread more carefully.
Let $x=a/b=(a_1,a_2,a_3)/b$ be the reduced representation.
We claim that since $q(x)=m$ is an integer, the denominator $b$ is odd.
Suppose that $b$ is even;
then at least one of $a_1$, $a_2$, $a_3$ is odd,
and so in $q(x)=q(a)/b^2$ the numerator $q(a)$ is congruent to $1$, $2$, or $3$ modulo $4$,
while the denominator is divisible by $4$, contradiction.
Note that $b$ being odd implies $0 leq left|x_i-y_iright| < frac{1}{2}$, $i=1,2,3$.
Now we choose an integral point $z$, close to the integral point $y$.
If $a_1-by_1$ is even, then we set $z_1:=y_1$.
If $a_1-by_1$ is odd, we let $z_1$ be a second closest integer to $x_1$
(there are two possible choices for $z_1$ iff $x_1$ is an integer,
and we may choose either of them);
then $z_1=y_1pm1$, $a_1-bz_1$ is even, and $left|x_1-z_1right|=1-left|x_1-y_1right|$.
In either case we have $left|x_1-z_1right|leq1 - left|x_1-y_1right|$.
The coordinates $z_2$ and $z_3$ are chosen analogously.
We step to the next zero $x'=a'/b'$ of $f$ along the line laid through the points $x$ and $z$.
Since $bigl(left|x_1-y_1right|,left|x_2-y_2right|,left|x_3-y_3right|bigr)in F$,
it follows that
$$q(x-z)leq qbigl(1-left|x_1-y_1right|,1-left|x_2-y_2right|,1-left|x_3-y_3right|bigr)
leq 8-2sqrt{5} < 4~.$$
By the choice of the point $z$ all three coordinates of $a-bz$ are even,
thus $mathrm{gcd}(a-bz)geq 2$,
and Lemma 1 tells us that $b'<b$.
$quad$Done.

For the second example we consider $q(X_1,X_2,X_3)=X_1^2+X_2^2+2X_3^2$
and $f(X)=q(X)-m$ with $m$ a positive integer.
The quadratic form $q$ is barely non-Euclidean:
for every $xinmathbb{Q}^3setminusmathbb{Z}^3$ there is $y=mathrm{round}(x)inmathbb{Z}^3$
such that $q(x-y)leq 1$.
The problem is that there exist points $x$ for which $q(x-y)=1$ is the best we can do:
if $xin M := mathbb{Z}^3+bigl(frac{1}{2},frac{1}{2},frac{1}{2}bigr)$,
then $q(x-y)geq 1$ for any $yinmathbb{Z}^3$.
Note that $q(x)$ is an odd integer for every $xin M$,
thus there do exist integers $m$, all of them odd,
so that $f$ has rational zeros,
but woe, it also has a rational zero,
with all three coordinates precisely halfway between consecutive integers,
at which we get stuck, because there is no Euclidean step from it to another zero.
On the other hand, if $m$ is even we never get stuck,
there is always a Euclidean step from a non-integral zero;
that is, though $f$ is not Euclidean, it is 'conditionally' Euclidean
on the set of its zeros.
$quad$Now suppose that $m$ is odd and that $f$ has rational zeros,
and that we walked ourselves into a point $x=a/2in M$.
In this case the trick we have used in the preceding example does not work,
because $mathrm{gcd}(a-2y)^2leq f_2(x-y)$ for any $yinmathbb{Z}^3$;
we must seek help from the other factor $mathrm{gcd}(A,B,C)$
in the denominator on the right hand side of (1).
Note that we can round the point $x$ to any of the eight integral points
$x+frac{1}{2}(delta_1,delta_2,delta_3)$, where $delta_1,delta_2,delta_3in{-1,1}$;
let $y$ be one of these eight points.
We have $v=(delta_1,delta_2,delta_3)$,
and $F(T) = f(y+Tv) = q(v)T^2 + 2langle y,vrangle T + q(y) - m$,
where $langletext{-},text{-}rangle$
is the bilinear form associated with the quadratic form $q$,
$langle X,Yrangle = X_1Y_1+X_2Y_2+2X_3Y_3$.
The leading coeffient $A=q(v)=4$ is even,
the next coefficient $B=2langle y,vrangle$ is also even,
thus it remains to make $C=q(y)-m$ even.
But this is easy: choose $delta_1$ and $delta_2$ so
that one of $y_1$, $y_2$ is even and the other one is odd.
Such a choice makes $mathrm{gcd}(A,B,C)geq 2$,
and since $q(x-y)=1$, we can step to a zero $x'=a'/b'$ of $f$ with $b'<2$,
that is, to an integral zero (whence $mathrm{gcd}(A,B,C)=2$).
$quad$Done.

at.algebraic topology - Mumford Conjecture

One application that I know of the Mumford conjecture is Teleman's proof of Givental's conjecture in this paper. Givental's conjecture states that when the quantum cohomology of a smooth projective variety (or compact symplectic manifold) is (generically) semisimple (meaning that the algebra is semisimple for generic values of the deformation parameter), then the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the quantum cohomology. Since quantum cohomology consists of genus 0 information, another way to say this is that the higher genus Gromov-Witten invariants are uniquely and explicitly determined by the genus 0 Gromov-Witten invariants, when the quantum cohomology is semisimple.

The proof, extremely roughly, uses the Mumford conjecture in the following way: semisimplicity allows you to go back and forth between low genus and high genus without loss of information. The Mumford conjecture tells you what things look like in very high genus, so if you want to know what something in some arbitrary genus looks like, you kick it up to very high genus, identify it by the Mumford conjecture, and then kick it back down to its original genus.

---Begin large parenthetical remark---

You might then ask, which smooth projective varieties (or compact symplectic manifolds) have semisimple quantum cohomology? Smooth projective toric varieties form a class of examples. Arend Bayer showed that semisimple quantum cohomology is preserved under blowing up points. There are other examples...

There is an interesting conjecture of Dubrovin which states that a variety has semisimple quantum cohomology if and only if its derived category has a full exceptional collection. (This conjecture is based heavily on mirror symmetry philosophy...) I don't know the status of this conjecture. But I think it is true, for example, that having a full exceptional collection is preserved under blowing up points. (Proved by Orlov? Bondal?)

---End large parenthetical remark---

It is also apparently possible to view the Mumford conjecture as a special case of the cobordism hypothesis. Take a look at Jacob Lurie's paper on TFTs, particularly section 2.5.

pr.probability - Can you explain the description of the Lovasz Local Lemma by Moser+Tardos?

Yes, one can prove the refinement stated in your question by a slight adaptation of the proof of the classical case.

More precisely, the proof is by recursion over the number of events in $mathbf{A}$. Like in the classical case, assuming that the statement holds for every collection of $n$ events, it is enough to show that $P(A|bar{B})le x(A)$ for every collection $mathbf{A}=${A}$cup${$A_i;1le ile n$}`, where $B$ denotes the union of the events $A_i$ for $1le ile n$ and $bar B$ denotes the complement of $B$.

The first step is to decompose $B$ into $B=Ccup F$ where $C$ is the union of the events in $Gamma(A)$ and $F$ the union of the events not in $Gamma(A)$. Hence, $P(A|bar{B})=P(A|bar{C},bar{F})=N/D$ where $N=P(A,bar{C}|bar{F})$ and $D=P(bar{C}|bar{F})$.

As regards the numerator, $Nle P(A|bar{F})=P(A)$, where the equality stems from the fact that $A$ is independent of the $A_i$ not in $Gamma(A)$, which define $F$.

Some additional work is required to deal with the denumerator. Assume wlog that $Gamma(A)=${$A_i;1le ile q$}and write $bar C$ as $bar C=bar A_1cap bar A_2capcdotscap bar A_q$. Then, by Bayes formula, $P(bar C|bar{F})$ is the product over $1le ile q$ of $P(bar A_i|bar C_i,bar{F})$, where each $C_i$ is the union of $A_j$ for $jle q$, $jne i$. Now, each of these probabilities is conditional on an event which involves at most $q-1le n$ events in $mathbf A$ hence $P(bar A_i|bar C_i,bar{F})ge1-x(A_i)$ by the recursion hypothesis applied to the collection{$A_i;1le ile n$}`.

Finally, $Nle P(A)le x(A)prod_{ile q}(1-x(A_i))$ and $Dgeprod_{ile q}(1-x(A_i))$ hence $P(A|bar{B})le x(A)$ for every collection $mathbf A$ of size $n+1$ as above, which proves the theorem.

universal algebra - Computing colimits in a Lawvere theory

Just saw this question and thought I would mention the ``explicit construction". You seem to be happy with reflexive coequalisers, as they are sifted colimits can just be computed pointwise in Prod(L,Set). From coproducts and reflexive coequalisers in a category one can get arbitrary coequalisers:
To give coequalisers in a category A is to give a left adjoint to $Delta:A to Graph(A)$, to give reflexive coequalisers is to give a left adjoint to $Delta:A to RGraph(A)$. But if you have coproducts then then the forgetful functor $RGraph(A) to Graph(A)$ has a left adjoint, where you freely add identities to a graph $d,c:A to B$ (two arrows meant), giving a new graph $d_{1},c_{1}:A+B to B$ with the evident description. Therefore composing this left adjoint $Graph(A) to RGraph(A)$ by the reflexive coequaliser functor $RGraph(A) to A$ gives arbitrary coequalisers. Therefore reflexive coequalisers and coproducts gives coequalisers and coproducts and thus all colimits. The problem then is to understand coproducts. This is probably most easily seen on the monads side which again you seem happy with.
Any algebra for a monad is a coequaliser of frees, via its canonical presentation:
$(T^{2}A,mu_{TA}) to (TA,mu_{A}) to (A,a)$ (two arrows meant from $(T^{2}A,mu_{A})$ to $(TA,mu_{A})$ . This explicit presentation of an algebra contains all we need to construct an explicit presentation for coproducts. Well we must have that to give $(A,a)+(B,b) to (C,c)$ is to give a pair $(A,a) to (C,c)$ and $(B,b) to (C,c)$ which, by the above presentation is to give maps $(TA,\mu_{A}) to (C,c)$ and $(TB,\mu_{B}) to (C,c)$ each coequalising those morphisms of the respective presentation. But to give $(TA,\mu_{A}) to (C,c)$ and $(TB,\mu_{B}) to (C,c)$ is to give an arrow from their sum, $(T(A+B),mu_{A+B}) to (C,c)$, the free algebra functor preserving coproducts. Now those maps of the respective presentations induce a pair $(T^{2}(A+B),mu_{T(A+B)}) to (T(A+B),mu_{A+B})$ and to say that the maps $(A,a) to (C,c)$ and $(B,b) to (C,c)$ coequalise the previous pairs amounts to saying that the induced map $(T(A+B),mu_{A+B}) to (C,c)$ coequalises this pair; thus the coproduct of $(A,a)$ and $(B,b)$ must be the coequaliser of the maps $(T^{2}(A+B),mu_{T(A+B)}) to (T(A+B),mu_{A+B})$ constructed (good exercise to work this out). This explains why the coproduct of a pair of groups and so on is a quotient of the free group on the disjoint union of the underlying sets. This construction may be interpreting just the same in the Lawvere theory setting.

co.combinatorics - Orthogonal matrices with small entries

Here's an idea which I think might be expandable to a solution once some details are filled in. (I am rather tired at the moment, though, so apologies if there is a cretinous error in what follows.)

We'll do the case $n=4m-1$ where $m$ is an integer; the case $n=4m-3$ is similar.

Let $C$ be a $2mtimes 2m$ matrix which has the required form. Let $A$ be the $ntimes n$ matrix with $C$ in the top left corner, $1$ on the remaning $2m-1$ diagonal entries, and zero elsewhere. Let $B$ be the $ntimes n$ matrix with $C$ in the bottom right corner, $1$ on the remaining $2m-1$ diagonal entries, and zero elsewhere.

$A=left[begin{matrix} C & 0 \\ 0 & I_{2m-1} end{matrix} right]quad,quad B= left[begin{matrix} I_{2m-1} & 0 \\ 0 & C end{matrix} right]$

Both $A$ and $B$ will be real orthogonal since $C$ is.
Consider the matrix $AB$, which being the product of real orthogonal matrices will also be orthogonal. I claim that the entries will all be $O(sqrt{n})$ as required.

In more detail:

-- If both $i$ and $j$ are $leq 2m-1$, then $(AB)_{ij}=A_{ij}=C_{ij}$ which is small by our choice of $C$; by symmetry, we can dispose of the case where both $i$ and $j$ are $geq 2m+1$ in a similar way.

-- If $ileq 2m-1$ and $jgeq 2m+1$, then on considering $sum_r A_{ir}B_{rj}$ we see that the only nonzero contribution comes when $rleq 2m$ and $rgeq 2m$, i.e. when $r=2m$ and so $(AB)_{ij}=A_{i,2m}B_{2m,j}$ is small.

-- If $i=2m$ or $j=2m$ then a similar analysis shows that $(AB)_{ij}$ can't be bigger than the entries of $C$ (at least up to some constant independent of $m$).

-- If $igeq 2m+1$ and $jleq 2m-1$ then $(AB)_{ij}=0$.

That should handle the case $n=4m-1$. The case $n=4m-3$ can be done in a similar fashion, but this time we will have extra factors of $3$ floating around since we have $3times n$ and $ntimes 3$ regions to consider, rather than just $1times n$ and $ntimes 1$ regions.

j-invariant of a supersingular elliptic curve

In characteristic $p$, every map $E_1 to E_2$ factors as a power of the Frobenius $varphi_r colon E_1 to E_1^{(p^r)}$ followed by a separable morphism $E_1^{(p^r)} to E_2$, and we find $r$ by looking at the inseparable degree of our map (if the map is separable, then $r=0$, as Pete pointed out).

Now, in the case of interest, if $E$ is supersingular, $widehat{varphi}$ is inseparable (as this is equivalent to multiplication by $p$ being purely inseparable). But then $widehat{varphi} colon E^{(p)} to E$ factors as $E^{(p)} to E^{(p^2)} to E$ by comparing degrees, where the first map is the Frobenius and the second is an isomorphism.

It then follows that $j(E) = j(E^{(p^2)}) = j(E)^{p^2}$ so $j(E) in mathbb{F}_{p^2}$.

books - Introductions to Disease- and Price-Modeling

I can't comment on the economics side of things, but for epidemiology a good reference is:

Brauer, Driessche, and Wu, "Mathematical Epidemiology"

(http://books.google.com/books?id=gcP5l1a22rQC&dq)

It gives a series of articles in the field that cover everything from basic SIR models up, though it does not attempt to unify the presentation--it's just a series of articles.

J.D. Murray's "Mathematical Biology," which is in two volumes, is also widely used and liked, and has several sections on population and disease modeling.

gt.geometric topology - Are compact submanifolds of "S X (0,1)" with one boundary component handlebodies, where S is a closed surface?

There is a class of three-manifolds which are called Handlebodies with Wormholes. M is a handlebody with wormholes if it is homeomorphic to the result of removing the regular neighborhood of a system of properly embedded arcs from a handlebody, that includes the example that Richard gave above. There is a Topology paper by Menasco and Thompson
"Compressing Handlebodies with holes" that studies this class of manifolds.

If you also allowed the removal of the regular neighborhoods of graphs that did not touch the boundary of $S$, that class of manifolds would include every compact three-manifold with boundary that can be embedded in $Stimes [0,1]$.

More simply, suppose you decided to remove the neighborhood of one properly embedded arc
in $Stimes [0,1]$ , whose boundary consisted of a point in $Stimes {0}$ and one in $Stimes {1}$, I proved that the complement is a handlebody if and only if the arc
is isotopic to a vertical arc ${pt}times [0,1]$. That appeared in,

Minimal surfaces and Heegaard splittings of the three-torus. Pacific J. Math. 124 (1986), no. 1, 119--130

I then developed an unknotting lemma for systems of arcs

An unknotting lemma for systems of arcs in $Ftimes I$. Pacific J. Math. 139 (1989), no. 1, 59--66.

There is related work of Gordon, that was done about the same time, but the referee lost my paper so that there was a two year lag publication.

On primitive sets of loops in the boundary of a handlebody.
Topology Appl. 27 (1987), no. 3, 285--299.

appeared.

lo.logic - Is all ordinary mathematics contained in high school mathematics?

By high school mathematics I mean Elementary Function Arithmetic (EFA), where one is allowed +,×, x^y, and a weak form of induction for formulas with bounded quantifiers. This is much weaker than primitive recursive arithmetic, which is in turn much weaker than Peano arithmetic, which is in turn much weaker than ZFC that we normally work in.

However there seem to be very few theorems (about integers) that are known to require anything more than this incredibly weak system to prove them. The few theorems that I know need more than this include:
*Consistency results for various stronger systems (following Godel). This includes results such as the Paris Harrington theorem and Goodstein sequences that are cleverly disguised forms of consistency results.
*Some results in Ramsey theory, saying that anything possible will happen in a sufficiently large set. Typical examples: Gowers proved a very large lower bound for Szemeredi's lemma showing that it cannot be proved in elementary function arithmetic, and the Robertson-Seymour graph minor theorem is known to require such large functions that it is unprovable in Peano arithmetic.

I can think of no results at all (about integers) outside these areas (mathematical logic, variations of Ramsey theory) that are known to require anything more than EFA to prove.
A good rule of thumb is that anything involving unbounded towers of exponentials is probably not provable in EFA, and conversely if there is no function this large then one might suspect the result is provable in EFA.

So my question is : does anyone know of natural results in "ordinary" mathematics (number theory, algebraic geometry, Lie groups, operator algebras, differential geometry, combinatorics, etc...) in which functions larger than a finite tower of exponentials occur in a serious way? In practice this is probably more or less equivalent to asking for theorems about integers unprovable in EFA.

Related links:
http://en.wikipedia.org/wiki/Grand_conjecture about Friedman asking a similar question.

By the way, encoding deep results as Diophantine equations and so on is cheating. And please do not make remarks suggesting that Fermat's last theorem needs inaccessible cardinals unless you understand Wiles's proof.

nt.number theory - What's the relationship between Gauss sums and the normal distribution?

It's true (as the answer below and some of the commenters note) that it's easy to interpret this question in a way that makes it seem trivial and uninteresting. I'm quite sure, however, that pursuing typographical similarity between $e^{x^2}$ and $zeta^{m^2}$ leads to interesting mathematics, and so here's a more serious attempt at propoganda for some of Ivan Cherednik's work.

Pages 6,7,8 and 9 of Cherednik's paper "Double affine Hecke algebras and difference Fourier transforms" explain how to ``interpolate'' between integral formulas relating the Gaussian to the Gamma function and (a certain generalization of) Gauss sums.

More explicitly, he shows that the formula (for many people, it's really just the definition of the Gamma function)

$$int_{-infty}^{infty} e^{-x^2} x^{2k} dx=Gamma left( k+frac{1}{2} right)$$

(for $k in mathbb{C}$ with real part $>-1/2$) and the Gauss-Selberg sum

$$sum_{j=0}^{N-2k} zeta^{(k-j)^2/4} frac{1-zeta^{j+k}}{1-zeta^k} prod_{l=1}^j frac{1-zeta^{l+2k-1}}{1-zeta^l}=prod_{j=1}^k (1-zeta^j)^{-1} sum_{m=0}^{2N-1} zeta^{m^2/4}$$

(where $N$ is a positive integer, $zeta=e^{2pi i/N}$ is a prim. $N$th root of $1$, and $k$ is a positive integer at most $N/2$) can both be obtained as limiting cases of the same $q$-series identity. The common generalization of the Gaussian and the function $k mapsto zeta^{k^2}$ is the function $x mapsto q^{x^2}$, and the measures weighting the integral and sum get replaced by Macdonald's measure---essentially the same one that shows up in the constant term conjecture for $A_1$, and that produces the Macdonald polynomials and kick-started the DAHA. The Fourier transform is deformed along with everything else to produce the "Cherednik-Fourier" transform.

I don't know how much of the roots of unity story generalizes to higher rank root systems.

Note: In the Gauss-Selberg sum, replacing $k$ by the integer part of $N/2$ and manipulating a little (as in the nice exposition by David Speyer linked to in the question above) gives the usual formula for the Gauss sum.

Is every Adams ring morphism a lambda-ring morphism?

A lambda-ring $R$ is called "special" if it satisfies the $lambda^ileft(xyright)=...$ and $lambda^ileft(lambda^jleft(xright)right)=...$ relations, or, equivalently, if the map $lambda_T:RtoLambdaleft(Rright)$ given by $lambda_Tleft(xright)=sumlimits_{i=0}^{infty}lambda^ileft(xright)T^i$ (where the $sum$ sign means addition in $Rleft[left[Tright]right]$, not addition in $Lambdaleft(Rright)$) is a morphism of lambda-rings. If you are wondering what the hell I am talking about, most likely you belong to the school of algebraists that denote only special lambda-rings as lambda-rings at all.

Anyway, let $A$ and $B$ be two special lambda-rings, and for every $i>0$, let $Psi_A^i$ and $Psi_B^i$ be the $i$-th Adams operations on $A$ and $B$, respectively. Let $f:Ato B$ be a ring homomorphism such that $fcircPsi_A^i=Psi_B^icirc f$ for every $i>0$. Does this yield that $f$ is a lambda-ring homomorphism, i. e. that $fcirclambda_A^i=lambda_B^icirc f$ for every $i>0$ ?

Note that this is clear if both $A$ and $B$ are torsion-free as additive groups (i. e., none of the elements $1$, $2$, $3$, ... is a zero-divisor in any of the rings $A$ and $B$), but Hazewinkel, in his text Witt vectors, part 1 (Lemma 16.35), claims the same result for the general case. I am writing a list of errata for his text, and I would like to know whether this should be included - well, and I'd like to know the answer anyway, as I am writing some notes on lambda-rings as well.

For the sake of completeness, here is a definition of Adams operations: These are the maps $Psi^i:Rto R$ for every integer $i>0$ (where $R$ is a special lambda-ring) defined by the equation

$sumlimits_{i=1}^{infty} Psi^ileft(xright)T^i = -Tfrac{d}{dT}logleft(lambda_{-T}left(xright)right)$ in the ring $Rleft[left[Tright]right]$ for every $xin R$.

Here, even if the term $logleft(lambda_{-T}left(xright)right)$ may not make sense (since some of the fractions $frac{1}{1}$, $frac{1}{2}$, $frac{1}{3}$, ... may not exist in $R$), the logarithmic derivative $frac{d}{dT}logleft(lambda_{-T}left(xright)right)$ is defined formally by

$displaystyle frac{d}{dT}logleft(lambda_{-T}left(xright)right)=frac{frac{d}{dT}lambda_{-T}left(xright)}{lambda_{-T}left(xright)}$.

lo.logic - Where's the notion of interpretation (model) originally introduced?

Updated answer:

As best as I have been able to figure out, the pre-Tarskian notions of "semantics" in mathematical logic grew out of the "algebra of logic" introduced by Boole ("An investigation into the laws of thought," 1854, and some earlier papers) and elaborated by Charles Peirce, Schröder, and others.

It's difficult for me to follow all the arguments in the old papers, but roughly the idea behind Boole's logic was to study logical equations such as "$x + (1 - x) = 1$" or "$x times (1 - x) = 0.$" Here, + means exclusive or, multiplication is "and," and subtracting $x$ means taking a conjunction with not-$x$. The number 1 should be interpreted as an always-true proposition, 0 as an always-false proposition, and $x$ as a propositional variable (or as Boole might say, a proposition with "indeterminate truth value").

Boole was interested in this analogy between logic an algebra, and here maybe we see the beginnings of the notion of interpretation in logic: we can check the validity of these formulas by considering whether they are true for all possible propositions $x$. (At least, I think this is what Boole meant -- you should track down the Dover reprinting of The Laws of Thought if you want to do some more historical investigation.)

Peirce considered the possibility of different "domains of individuals," which could be finite, infinite, or even uncountable. I think it was Peirce who first generalized Boole's calculus of logical propositions to the "calculus of relatives," where a "relative" is the interpretation of some $n$-ary predicate in a domain of individuals. To track down the beginnings of this, I would try Peirce's 1870 "Description of a notation for the logic of relatives," which unfortunately I cannot access right now from where I am.

Peirce, Schröder, and even Löweinheim in his 1915 "On possibilities in the calculus of relatives" continued to use algebraic notation along the lines of Boole, with many $0$'s and $1$'s. Even the "domain of individuals'' was denoted by $1^1$!

One thing in particular that is confusing about reading Löweinheim's paper is that, while he is clearly aware that the domain of individuals $1^1$ could be one of any number of possible collections of things (some finite, others infinite), he seems to insist on talking about "the domain of individuals $1^1$" and referring to every possible domain by the same name $1^1$! Obviously, this is confusing if you want to think about comparing two different such domains, and maybe one of Tarski's key contributions here was simply to introduce a notation ''$mathfrak{A} models varphi$'' which explicitly names the universe $mathfrak{A}$ and suggests comparison with other universes $mathfrak{B}, mathfrak{C}, ldots$.

My original answer (missing the key point):

My knee-jerk answer to this question was going to be, "Tarski!" But you seem to already be aware of Tarski's work, so maybe you're looking for something different?

In particular: Alfred Tarski's 1933 article "The concept of truth in formalized languages" (in Polish, unfortunately) seems to be generally regarded as the first place where the concept of "logical satisfaction" (in the modern sense) was first defined.

There was already an "application" of semantic methods in logic by 1940: Gödel's proof of that Con(ZFC) implies Con(ZFC + GCH + AC). (It might be fun to try to find an even earlier application of semantic methods to prove a syntactic result.) Certainly by the 1960's the field of model theory was coming into its own with the work of A. Robinson, Vaught, Morley, and others.

soft question - How to write math well?

One trick that my advisor, Ronnie Lee, advocated was to use a descriptive term before using the symbolic name for the object. Thus write, "the function $f$, the element $x$, the group $G$, or the subgroup $H$. Most importantly, don't expect that your reader has internalized the notation that you are using. If you introduced a symbol $Theta_{i,j,k}(x,y,z)$ on page 2 and you don't use it again until page 5, then remind them that the subscripts of the cocycle $Theta$ indicate one thing while the arguments $x,y,z$ indicate another.

Another trick that is suggested by literature --- and can be deadly in technical writing --- is to try and find synonyms for the objects in question. A group might be a group for a while, or later it may be giving an action. In the latter case, the set of symmetries $G$ that act on the space $X$ is given by $ldots$. Context is important.

Vary cadence. Long sentences that contain many ideas should have shorter declarative sentences interspersed. Read your papers out loud. Do they sound repetitive?

My last piece of advice is one I have been wanting to say for a long time. Don't write your results up. Write your results down. You figure out what I mean by that.

gn.general topology - Functions separting points in Hausdorff spaces

This is really a comment on t3suji's answer, but it's too long to be a comment as such.

t3suji's answer is the canonical one in the following precise sense. Let $e: X to Y$ be a morphism in any category. It's an elementary exercise to show that the following conditions on $e$ are equivalent:

1. $e$ is an epimorphism




2. the square
   $$
   begin{array}{ccc}
   X &stackrel{e}{to} &Y \
   edownarrow & &downarrow 1_Y \
   Y &stackrel{1_Y}{to} &Y
   end{array}
   $$
   is a pushout




3. for some morphism $f: Y to Z$, the square
   $$
   begin{array}{ccc}
   X &stackrel{e}{to} &Y \
   edownarrow & &downarrow f \
   Y &stackrel{f}{to} &Z
   end{array}
   $$
   is a pushout.

I'll only use the equivalence 1 $iff$ 3 here. The other implications are just scene-setting.

Suppose we want to show that a particular morphism $e$ is not epi. Assuming that there are enough pushouts around, we can argue as follows. Form the pushout square
$$
begin{array}{ccc}
X &stackrel{e}{to} &Y \
edownarrow & &downarrow f \
Y &stackrel{g}{to} &Z.
end{array}
$$
If $f neq g$ then the implication 1 $Rightarrow$ 3 tells us that $e$ is not epi. Moreover, this strategy is bound to work, in the sense that if $f = g$ then the implication 3 $Rightarrow$ 1 tells us that $e$ is epi after all.

It only remains to see that this is indeed what t3suji did. In his/her situation, $e$ was the inclusion $X to Y$. He/she then took the coequalizer of the two obvious maps $X to Y + Y$ (where $+$ means coproduct, i.e. disjoint union). For elementary and totally general reasons, this is the same thing as taking the pushout just mentioned. The morphisms that t3suji called $iota_1$ and $iota_2$, I called $f$ and $g$. Finally, although t3suji's pushout is in the category of all topological spaces, he/she then verified that the space $Z$ is indeed Hausdorff, from which it follows that it's also a pushout in Hausdorff spaces.

So now you know, in principle, how to answer any question of the form "prove that such-and-such a morphism isn't epi".

lie groups - how many injective homomorphism between two lie algebra sl2 and sp6 up to conjugate by Sp6?

As a follow-up to Jim's answer (which came in as I was typing an inferior answer), let me add that the 7 possible embeddings are given in the $C_3$ entry of Table VI in the paper: Classification of semisimple subalgebras of simple Lie algebras by Lorente and Gruber. It's of course based on Dynkin, but they work out the details up to rank 6.

---

Added

The defining vectors for the 7 embeddings are given by: (1,0,0), (1,1,0), (1,1,1), (2,2,0), (3,1,0), (3,1,1) and (5,3,1). Recall that the embedding with defining vector $(a,b,c)$ is one for which the Cartan generator $H$ of the $mathfrak{sl}(2)$ subalgebra is given by $H = a H_1 + b H_2 + c H_3$, where $(H_i)$ is an orthonormal basis of a Cartan subalgebra of $C_3$ containing $H$.

ag.algebraic geometry - Analysis analogue of Orlov's theorem?

Mukai's theorem states that if $X$ is an abelian variety, and $check{X}$ is the dual abelian variety, then the Fourier-Mukai transform corresponding to the Poincare line bundle on $X times check{X}$ is a derived equivalence. This is analogous to Pontryagin duality for locally compact abelian topological groups.

Orlov's theorem states (I think) that if $X$ and $Y$ are smooth projective varieties, then every derived equivalence $D^b(X) to D^b(Y)$ is induced by a Fourier-Mukai transform.

Is there an analysis analogue of Orlov's theorem? If abelian varieties are analogous to locally compact abelian groups, then are varieties analogous to locally compact topological spaces? What is the analogue of projective variety? What is the analogue of smooth? Is it true that any isomorphism of algebras of functions (suitably defined) on [[whatever the analogue of a smooth projective variety is]] is induced by some kind of integral transform?

(I came up with these questions while writing up my answer to this question about the Fourier-Mukai transform.)

gr.group theory - Spectral properties of Cayley graphs

I know at least one special case where your second question makes sense. If $G$ is a compact group, it has a category $text{Rep}(G)$ of finite-dimensional unitary representations which break up into direct sums of irreducible representations. Fix a representation $V$ such that every irreducible representation appears in $V^{otimes n}$ for some $n$. One can construct a graph $Gamma(V)$ whose vertices are the irreducible representations of $G$ and where the number of edges from $A$ to $B$ is the multiplicity by which $B$ appears in $A otimes V$. By the assumption, $Gamma(V)$ is connected, and its combinatorial properties encode information about the behavior of the tensor powers of $V$, hence behavior about $G$.

When $G$ is finite, this graph has the property that its eigenvalues are precisely the character values $chi_V(g)$ as $g$ runs through all conjugacy classes. But the great thing is that this statement still makes sense even when $G$ is infinite in a sense which is made precise in this blog post.

Finally, if $G$ is abelian, all of the finite-dimensional irreducible representations are one-dimensional. They can be identified with the Pontryagin dual $G^{vee}$, which is discrete, and $Gamma(V)$ becomes precisely the Cayley graph of $G^{vee}$ with respect to the generators that make up $V$! So this is one sense in which the Cayley graph of an infinite group gives you algebraic data, but about its dual group.

soft question - How to select a journal?

Many shall find this criterion worthless, but personally I can't help taking the production quality of the journal into account (even if it is not of primary importance).

Quality paper and ink improve the reading comfort, and some journals (some of Elsevier particularly) have pages with pale ink, or thin paper that let the verso appear. On the other end of the scope are journals like Acta mathematica of Publications mathématiques de l'IHES with nice paper and fonts.

Another aspect of this criterion is the processing quality. To compare two very different experiences, in my first article incredible mistakes have been added after the proofs (expressions like $n/2$ replaced by $n^2$) while the AMS journals do an amazing job, showing you exactly what they changed in your paper (corrected spellings e.g.) and sometimes asking for confirmation. This part you cannot judge before being published in the journal, or discussing with colleagues.

at.algebraic topology - Combinatorics of the Stasheff polytopes

After a change of variables, the answer is sequence A033282 in the Encyclopedia of Integer Sequences: $T(n,k)$ is the number of diagonal dissections of a convex $n$-gon into $k+1$ regions. The page gives the wonderful formula,
$$T(n,k) = frac{1}{k+1}binom{n-3}{k}binom{n+k-1}{k},$$
for relevant values of $k$.

I find it easier to also keep track of the dual Stasheff polytope, which can be realized as a simplicial complex based on dissections of a convex polygon. The polytope $K_n^*$ has a vertex for each diagonal of an $(n+1)$-gon, and it has a face for every collection of disjoint diagonals. So, in terms of your original parameters, $K_n$ has $T(n+1,n-2-k)$ faces of dimension $k$.

Also: One reason that I like the dual Stasheff polytope is that it has an amazing infinite generalization called the Hatcher-Thurston arc complex. You again take collections of disjoint arcs that connect marked points, but in the generalization you can take any surface with or without boundary, as long as it has at least one marked point total and at least one on each boundary component. (And I suppose in the disk case it needs at least three marked points.) Each isotopy class of arcs is a vertex, and each disjoint collection is a face. It is a combinatorial model of Teichmüller spaces or moduli spaces of curves (with the requisite marked points).

---

Gil Kalai in the comments asks for a few more details of the Hatcher-Thurston arc complex, and he gives a reference to one of the original papers, "On triangulations of surfaces, Topology Appl. 40 (1991), 189–194," by Allen Hatcher. Briefly: Suppose that $Sigma$ is a fixed surface with some marked points. There should be enough marked points so that there exists at least one generalized triangulation of $Sigma$ whose vertex set is the marked points. A question that could be taken as motivation is the following: Can you find a complete set of moves on triangulations, moves on moves, moves on moves on moves, etc.? Whether the set of moves is complete is not entirely a rigorous question, but there is an interesting answer. The moves and higher moves just come from erasing edges of the triangulation. The main theorem is that the resulting simplicial complex is contractible. The mapping class group acts on the complex, and it acts freely on the high-dimensional simplices.

More precisely, the arc complex has a vertex $v$ for every isotopy class of an arc between two of the marked points, among properly embedded arcs that miss all of the marked points in the interior. If a collection of arcs can be made disjoint after isotopy, then the corresponding vertices subtend a simplex. The disk case is an exception in which the arc complex is not quite contractible, but rather a sphere.

nt.number theory - Computing (on a computer) higher ramification groups and/or conductors of representations.

I am supervising an undergraduate for a project in which he's going to talk about the relationship between Galois representations and modular forms. We decided we'd figure out a few examples of weight 1 modular forms and Galois representations and see them matching up. But I realised when working through some examples that computing the conductor of the Galois representation was giving me problems sometimes at small primes.

Here's an explicit question. Set $f=x^4 + 2x^2 - 1$ and let $K$ be the splitting field of $f$ over $mathbf{Q}$. It's Galois over $mathbf{Q}$ with group $D_8$. Let $rho$ be the irreducible 2-dimensional representation of $D_8$. What is the conductor of $rho$? Note that I don't particularly want to know the answer to this particular question, I want to know how to work these things out in general. In fact I think I could perhaps figure out the conductor of $rho$ by doing calculations on the modular forms side, but I don't want to do that (somehow the point of the project is seeing that calculations done in 2 different ways match up, rather than using known modularity results to do the calculations).

Using pari or magma I see that $K$ is unramified outside 2, and the ideal (2) is an 8th power in the integers of $K$. To compute the conductor of $rho$ the naive approach is to figure out the higher ramification groups at 2 and then just use the usual formula. But the only computer algebra package I know which will compute higher ramification groups is magma, and if I create the splitting field of $f$ over $mathbf{Q}_2$ (computed using pari's "polcompositum" command)

Qx<x>:=PolynomialRing(Rationals());
g:=x^8 + 20*x^6 + 146*x^4 + 460*x^2 + 1681;
L := LocalField(pAdicField(2, 50),g);
DecompositionGroup(L);

then I get an instant memory overflow (magma wants 2.5 gigs to do this, apparently), and furthermore the other calculations I would have to do if I were to be following up this idea would be things like

RamificationGroup(L, 3);

which apparently need 11 gigs of ram to run. Ouch. Note also that if I pull the precision of the $p$-adic field down from 50 then magma complains that the precision isn't large enough to do some arithmetic in $L$ that it wants to do.

I think then my question must be: are there any computer algebra resources that will compute higher ramification groups for local fields without needing exorbitant amounts of memory? Or is it a genuinely an "11-gigs" calculation that I want to do?? And perhaps another question is: is there another way of computing the conductor of a (non-abelian finite image) Galois representation without having to compute these higher ramification groups (and without computing any modular forms either)?

at.algebraic topology - Killing the torsion in homotopy

I don't have an answer to this question, but for the analogous question for homology it looks like it can't be done. By the universal coefficient theorem, a construction like this for homology would give a construction for cohomology as well. To get a counterexample in cohomology, take an Eilenberg-MacLane space $K({mathbb Z},n)$ with $n$ even. This has rational cohomology a polynomial ring ${mathbb Q}[x]$ with $x$ of degree $n$. It follows from this that if you factor the torsion out of the integral cohomology ring you get a polynomial ring ${mathbb Z}[x]$ with $x$ of degree $n$, as one can see by looking at a map from ${mathbb C}P^infty$ to $K({mathbb Z},n)$ that induces an isomorphism on $H^n(--;{mathbb Z})$, using the fact that the integral cohomology of ${mathbb C}P^infty$ is a polynomial ring. However, there is no space whose integral cohomology ring is a polynomial ring on a generator of degree $n$ if $n > 4$, as one sees by looking at Steenrod squares and at Steenrod powers for the prime $p=3$. (This is Corollary 4L.10 in my book.)

In the context that Baez was talking about, rationalizing the homotopy groups is equivalent to rationalizing the homology groups, so it seems to be worth knowing that one can't kill torsion in homology, at least.

dg.differential geometry - Integration and Stokes' theorem for vector bundle-valued differential forms?

After looking at this question for a few days in the context of the Riemann curvature tensor, holonomy for a given affine connection, and the (false) conjecture that the parallel transport around the boundary curve could equal the integral of the Riemann tensor within the span of the closed curve, I've concluded that the Stokes theorem cannot be applied to this conjecture except when the connection is flat.

The reason for the failure of the conjectured relation between curvature and parallel transport is that the Stokes theorem's integrals are themselves not really well defined. But it is not that simple. I'll explain...

When even constructing a simple Riemann integral from the fundamentals, one has to add vectors at different points inside the region. Even if you have a connection, you have to decide which paths to use to connect the points of the region. You can do a kind of a "raster scan" of the image of a rectangular region of $mathbb{R}^2$, parallel transporting the vectors back to the left of the scan to add them on the left hand side, and then you can parallel transport all of these X-scans down the Y-axis by transporting them down to the bottom left of the rectangle. But then what do you have? It's clearly not geometrically meaningful. And then you have the same problem with the boundary integral of a vector function.

A second conclusion which I came to is that if you do apply the Stokes theorem to this scenario, you get a mathematically correct identity, which has a practical value as the first iteration of the Picard iteration procedure to compute the parallel transport around the curve. This, clearly, is not extremely useful. But in my opinion, the Stokes theorem is applicable to this situation. It just doesn't give anything geometrically meaningful for a non-flat geometry, and it has limited value for relating curvature to parallel transport and holonomy. On the other hand, it does get the right answer in the limit of a shrinking region to a point, which gives the correct answer for the "Cartan characterization of curvature".

This issue is related to questions 16850 and 50051.

ct.category theory - Definition of homotopy limits

Here's a definition for homotopy limits that isn't quite right, but seems salvageable. Does anyone know how to fix it?

Suppose the category $C$ is some reasonable setting for homotopy theory, say it's enriched over some kind of category of spaces (e.g. chain complexes, simplicial sets, ...).

Def: Let F: D^op $to$ C be a diagram (functor). An object X together with a map η from X to the diagram is a LIMIT for the diagram iff the induced natural transformation of functors Hom_C(-, X) $to lim$ Hom_C(-,F) is an isomorphism. A pair (X,η) is a homotopy limit for the diagram F iff the induced transformation of functors Hom_C(-,X) $to lim$ Hom_C(-, F) is a weak equivalence.

This definition doesn't quite cut it since, in most of the motivating examples I know, though the homotopy limit object X does come equipped with a morphism to each object in the diagram, these do not commute with the morphisms in the diagram---they only commute up to homotopy. So a homotopy limit won't even come with a map to the diagram, so it doesn't come with an induced natural transformation. How then can I characterize the object X by a similar universal property as the (strict) limit?

reference request - Proofs without words

$$arctan frac{1}{3} + arctan frac{1}{2} = arctan 1$$

It's easy to generalize this to

$$ arctan frac{1}{n} + arctan frac{n-1}{n+1} = arctan 1, text{ for } n in mathbb{N}$$

which can further be generalized to

$$ arctan frac{a}{b} + arctan frac{b-a}{b+a} = arctan 1, text{ for } a,b in mathbb{N}, a leq b $$

Edit: A similar result relating Fibonacci numbers to arctangents can be found here and here.

gr.group theory - Stable w-length

Here are some weak observations that don't quite answer any of your questions. Let $g$ be a positive integer, and consider the free group $F_{2g}$ generated by $a_k$ and $b_k$ for $k = 1$ to $g$. Consider the word:

$$w_g = [a_1,b_1][a_2,b_2][a_3,b_3] ldots [a_g,b_g].$$

Suppose that $lambda_g = sl(w_g,w_g)$. I claim that for any $x$ in the commutator of $F_2$ with $cl(x) = g$, the stable commutator length $scl(x)$ of $x$ is $le g cdot lambda_g$. Suppose otherwise. First of all, note that for large $n$ we can write $w^n_g$ as the product of (roughly) $n cdot lambda_g cdot g$ commutators. Since the commutator length of $x$ is $g$, there exists a map from $F_{2g}$ to $F_{2}$ such that the image of $w_g$ is $x$. On the other hand, we see that the image of $w^n_g$ is $x^n$, and thus the commutator length of $x^n$ is (asymptotically) at most by $n cdot lambda_g cdot g$, and thus $scl(x) le lambda_g cdot g$.

Example: $cl([x,y]^3) = 2$ and $scl([x,y]^3) = 3/2$, and thus $lambda_3 ge 3/4$. In general, the fact that $scl([x,y]) = 1/2$ implies that that $lambda_g$ tends to one as $g$ increases.

I think one can promote this example to a word in $F_2$. Consider the characteristic homomorphism $phi_n:F_2 rightarrow mathbf{Z} oplus mathbf{Z}
rightarrow mathbf{Z}/nmathbf{Z} oplus mathbf{Z}/nmathbf{Z}$. Suppose that $n$ is odd, and write $2g = n^2 + 1$. The kernel of $F_2$ is free of rank $2g$. Pick generators for $ker(phi_n)$ once and for all, and call them $a_k$ and $b_k$ for $k = 1$ to $g$. We may think of $a_k$ and $b_k$ as elements in $F_2$, but also as formal words. Since $ker(phi_n) = F_{2g}$ is characteristic, the formal words $a_k$ and $b_k$ always yield elements of $F_{2g}$ (alternatively, the images of $a_k$ and $b_k$ in $mathbf{Z} oplus mathbf{Z}$
are divisible by $n$, and this will be so for any substitution of elements of $F_2$ for the generators). Let

$$w_g = [a_1,b_1][a_2,b_2] ldots [a_g,b_g].$$

The argument proceeds as above. If $sl(w_g,w_g) = mu_g$, then we can write $w^n_g$ (for large $n$) as the product of $n cdot mu_g cdot g$ commutators, each of which is the commutator of a pair of elements of $F_{2g}$ (by the characteristic property of the words $a_k$ and $b_k$ described above). Hence, choosing an appropriate map from $F_{2g}$ to $F_2$, we may deduce that for any $x in F_2$ with $cl(x) = g$ that $scl(x) le g cdot mu_g$. Thus we have found words $w_g$ in $F_2$ such that $sl(w_g,w_g)$ tends to $1$ as $g$ goes to infinity. Of course, this says nothing about whether $sl(w_g,w_g)$ actually equals $1$ for any $g$.

Finally, a random other example. If $w = [a,b^2]$, then

$$w^3 = [ab^2a^{-1},b^{-2} ab^2a^{-2}][b^{-2} a b^2,b^4] = [b^{-2} a b^2 a^{-2},(aba^{-1})^2]^{-1}[b^{-2} a b^2,(b^2)^2],$$

so $sl(w,w) le 2/3$.

I wrote this on a very old computer that was too slow for previewing LaTeX, but hopefully this can still be read.

co.combinatorics - Sum of 'the first k' binomial coefficients for fixed n

Jean Gallier gives this bound (Proposition 4.16 in Ch.4 of "Discrete Math" preprint)

$$f(n,k) < 2^{n-1} frac{{n choose k+1}}{n choose n/2}$$

where $f(N,k)=sum_{i=0}^k {Nchoose i}$, and $kle n/2-1$ for even $n$

It seems to be worse than Michael's bound except for large values of k

Here's a plot of f(50,k) (blue circles), Michael Lugo's bound (brown diamonds) and Gallier's (magenta squares)

n = 50;
bisum[k_] := Total[Table[Binomial[n, x], {x, 0, k}]];
bibound[k_] := Binomial[n, k + 1]/Binomial[n, n/2] 2^(n - 1);
lugobound[k_] := Binomial[n, k] (n - (k - 1))/(n - (2 k - 1));
ListPlot[Transpose[{bisum[#], bibound[#], lugobound[#]} & /@ 
   Range[0, n/2 - 1]], PlotRange -> All, PlotMarkers -> Automatic]

Edit
The proof, Proposition 3.8.2 from Lovasz "Discrete Math".

Lovasz gives another bound (Theorem 5.3.2) in terms of exponential which seems fairly close to previous one

$$f(n,k)le 2^{n-1} exp (frac{(n-2k-2)^2}{4(1+k-n)}$$
Lovasz bound is the top one.

n = 50;
gallier[k_] := Binomial[n, k + 1]/Binomial[n, n/2] 2^(n - 1);
lovasz[k_] := 2^(n - 1) Exp[(n - 2 k - 2)^2/(4 (1 + k - n))];
ListPlot[Transpose[{gallier[#], lovasz[#]} & /@ Range[0, n/2 - 1]], 
 PlotRange -> All, PlotMarkers -> Automatic]

fa.functional analysis - Continuous automorphism groups of normed vector spaces?

The following answer gives a partial description of the isometry groups of finite-dimensional normed spaces.

I assume that an isometry is a bijection preserving the distance function. By the Mazur-Ulam theorem it then follows that an isometry is a linear transformation composed with a translation. Thus we may assume without loss of generality that an isometry fixes the origin, so the isometry group is a subgroup of $GL(n)$.

Then the isometry group of any (real) finite-dimensional normed space is conjugate in $GL(n)$ to a closed subgroup of $O(n)$ that contain $-id$. This is seen as follows.

Consider the John ellipsoid $E$ of the unit ball $B$ of some $n$-dimensional normed space. This is the ellipsoid of largest volume contained in $B$ and, crucially, it is the unique such ellipsoid.

After some choice of basis we may assume that $E$ is the Euclidean ball. An isometry maps $B$ onto $B$, so it must map the John ellipsoid to the John ellipsoid. It follows that the isometry group is a subgroup of $O(n)$ containing $-id$. This subgroup is clearly closed, hence compact.

The converse is surely false. The following is an attempt at constructing a norm from such a subgroup. Fix a Euclidean unit vector $v$. Then its $Gv$ is a compact set of Euclidean unit vectors, symmetric with respect to the origin. Its convex hull $Gv$ is still compact and symmetric, so gives a unit ball $B_0$ of some norm on the linear span of $Gv$. If this linear span is not all of $mathbb{R}^n$, then the unit ball has to be made full-dimensional in a sufficiently rough way, so as not to add any more isometries.

However, as pointed out by Leonid Kovalev in the comments, there are closed subgroups of $O(n)$, such as $U(n)$, where this construction gives a norm with a strictly larger isometry group (in the case of $U(n)$, the Euclidean norm).

As pointed out by Bill Johnson in a comment to his answer, it was shown by Gordon and Loewy that any $finite$ subgroup of $O(n)$ that contains $-id$ is the isometry group of some norm on $mathbb{R}^n$. It's still my guess that the only way you can get infinite isometry groups (in the finite-dimensional case) is by having Euclidean subspaces, and for the norm to be so symmetric that it shares all the symmetries of this subspace.

pr.probability - Is there a way to analytically compute the recurrence time of a finite Markov process?

This is a response to a comment.

The coupon collector's problem is elementary. I don't have a particular scholarly reference in mind, but rather the technique of the proofs. There are a few proofs of the $n H_n$ expected time to collect all coupons. One possibility is that you can compute the expected time to collect the $k$th new coupon,
$n/(n-k+1)$. That uses a lot of symmetry you don't have for a general Markov process. Here, you have transition probabilities and times on (current location, subset visited so far).

Analogous to what I did here, you can use inclusion-exclusion. The expected time to cover everything (with discrete time) is the sum of the probability that you haven't covered everything by time $t-1$, which you can express as

$$sum_t sum_{Ssubset V} -1^{|S|+1}Prob({X_i}_{ilt t}cap S = emptyset) $$

where $V$ = ${1,...,n}$. You can switch the order of summation to get about $2^n$ analytically solvable problems about avoiding particular subsets.

$$sum_{Ssubset V} -1^{|S|+1} A(S)$$

where $A(S)$ = expected time before you first enter $S$.

The same holds for continuous time.

noncommutative algebra - Q-Divisor and Determinant Map on a Maximal Order

Given a smooth projective surface $X$, let $A$ be a sheaf of maximal orders in a division ring.
Let us for simplicity assume $A$ ramifies in one curve $C$ with ramification index $e$. Let $A^*$ be the dual sheaf.

How can I see that the determinant map is a map from $A^*$ to $O_X((1-frac{1}{e})C)$?
And how to understand the invertible sheaf $O_X((1-frac{1}{e})C)$? How to handle the rational coefficients?

Since $A$ only ramifies in C, we have that $A$ is Azumaya on $U:=Xbackslash C$. So on
$U$ we have $A^*cong A$. There the determinant map induces a map $A^{*} rightarrow O_U$ since $A$ is just a matrix algbera etale locally. So I see that we have a map $A^* rightarrow O_X(rC)$. But how to find $r=1-frac{1}{e}$? How can i determine $A^*$ on $C$?

The question arose reading Theorem 7.1.4. on page 157 of this script: http://www.math.lsa.umich.edu/courses/711/ordersms-num.pdf

matrices - What's known about the 3rd coefficient in the BMV conjecture?

The most appealing statement of the Bessis-Moussa-Villani conjecture is as follows:

Conjecture: For all Hermitian positive semidefinite $ntimes n$ matrices $A$ and $B$,
and all positive integer $m$, the polynomial function
$$t in mathbb{R}mapsto g(t) equiv tr[(A + t B)^m] =
sumlimits_{
k=0}^m
a_kt^k$$
has only nonnegative coefficients $a_k, k=1,cdots,m$.

Most recent and past research concerns on the quantities $m$ and $n$. What about trying to prove the conjecture in the following way: first show that $a_mge0$, $a_0ge0$, $a_{m-1}ge0$, $a_1ge0$, $a_{m-2}ge0$, $a_2ge0$. And then go on to show $a_3ge 0$ (and so is $a_{m-3}ge 0$)...

But it seems difficult to show $a_3ge 0$. Can anyone share some idea on this particular coefficient?

UPDATED
What is the largest term in $S_{2m,m}(AB)$ ? More precisely, consider the word in two positive definite letters $A^{k_1}BA^{k_2}Bcdots A^{k_m}B$ , where $(k1,k2,cdots,k_m)$ is a pair of nonnegative integer solution of $k_1+k_2+cdots+k_m=m$ . Is it true that $tr(A^{k_1}BA^{k_2}Bcdots A^{k_m}B)le tr(A^mB^m)$ ?

nt.number theory - Integer points of an elliptic curve

There are precisely two available "serious" implementations of the standard algorithm for computing integral points on an elliptic curve: a non-free one in Magma (http://magma.maths.usyd.edu.au/magma/) and a free one in Sage (http://sagemath.org). The one in Sage was done by Cremona and two German masters students a few years ago, and when refereeing the Sage code, I compared the answers with Magma, and uncovered and reported numerous bugs in Magma, which were subsequently fixed. Here's how to use Sage to find all integral (or S-integral!) points on a curve over Q:

sage: E = EllipticCurve([1,2,3,4,5])
sage: E.integral_points()
[(1 : 2 : 1)]
sage: E.S_integral_points([2])
[(-103/64 : -233/512 : 1), (1 : 2 : 1)]

and here is how to use Magma:

> E := EllipticCurve([1,2,3,4,5]);
> IntegralPoints(E);
[ (1 : 2 : 1) ]
> SIntegralPoints(E, [2]);                                                     
[ (1 : 2 : 1), (-103/64 : -233/512 : 1) ]

Note that in both cases by default the points are only returned up to sign. In Sage you get both signs like this:

sage: E.integral_points(both_signs=True)
[(1 : -6 : 1), (1 : 2 : 1)]

Finally, you can use Magma for free online here: http://magma.maths.usyd.edu.au/calc/
and you can use Sage free here: http://demo.sagenb.org/. With Sage, you can also just download it for free and install it on your computer. With Magma, you have to pay between $100 and a few thousand dollars, depending on who you are, and deal with copy protection.

NOTE: Technically a system called SIMATH (http://tnt.math.se.tmu.ac.jp/simath/) had an implementation of computing integral points. But it was killed by our friends at Siemens Corp.

ca.analysis and odes - Convergence of a general Bertrand series

Let $ S= sum 1/n log^1n log^2n log^3n ..log^{TL(n)}n $.

Is it convergent when $n$ runs on integers say above 2 ?

$log^i n$ denotes the i'th iterate of $log$ (in base 2 ) of $n$, $log^2n$ means $loglogn$ .

$T(n)$ is the tower of $n$ (stack of $n$ 2's) that is $T(1)=2$ , $T(n+1)=2^{T(n)}$.

$TL(n)$ is the towerian log:
$ TL(n) = Sup ( k : T(k) <= n < T(k+1) ) $.

MOTIVATION : Generalizing the following that are called Bertrand series (I think):

$sum 1/n$ is the harmonic serie , $sum 1/nlogn$ , $sum 1/nlognlog^2n $ and $sum 1/nlognlog^2nlog^3n $ are all known to be divergent.

Here the product of iterated logs is pushed as far as possible and its size depends on the parameter $n$.

na.numerical analysis - Which method to apply to this problem?

If you really have equations like

$ 5xa + 8yb + 5zc + 3xd + 2ye + 2zf + 6xg + 7yh leq w $

and all of the letters above are variables, then in the most general case you have an instance of Multivariate Quadratic Equations, which is $NP$-complete. (Even without the "if...then" rules.) The hardness is really independent of the domain of the variables. As long as each variable domain takes on at least two distinct possible values, you are in the land of $NP$-hard. Note you can always force a variable $x$ to take on exactly two of the possible values $v_1$, $v_2$ by imposing that

$(x-v_1)(x-v_2) = 0$.

As mentioned in other comments, a tractable special case would be if $x$, $y$, $z$ are fixed and $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ vary over the rationals. In this case your problem is an instance of Linear Programming, known to be solvable in polynomial time (and efficiently in practice; note these two properties do not always coincide!).

Another potentially tractable case (in practice, not in theory) is when $x$, $y$, $z$ are fixed and some of the $a$, $b$, $c$, $d$, $e$, $f$, $g$, $h$ can only be $0$ or $1$. (Typically this condition can be translated to: some variables in the range $a$ through $g$ can only take on one of two possible values such as $a=0.9$ or $a=1$, through a linear transformation.) This case is called Integer Linear Programming. Although it is also $NP$-complete, there is software available that can sometimes solve instances of these fairly efficiently. Moreover, depending on the "if...then" rules you have, you may be able to cleverly translate them into integer linear programming constraints. (Note an "if...then" rule has one of two possible outcomes for a variable, and similarly an integer-valued variable will have one of two possible outcomes.)

Here's a simple example of what I mean, though I don't think this example will help you directly. Suppose $x$ is a variable that's either 0 or 1 and $d$ is a coefficient. You want to translate: "if ($d geq 35$) then $x=0$ else $x=1$". Look at the inequalities $d geq (d-35)x + 35$, $d leq (34.999-d)x+d$. When $x=0$, we have $d geq 35$, $d leq d$. When $x=1$, we have $d geq d$, $d leq 34.999$.

To get any more specific about things, I would need to know more properties of the problems you are trying to solve. Hopefully by searching for the names of the above problems you can find more relevant references. Good luck!

differential equations - Mathematical modeling - how to calculate the displacement at any point in a membrane

I'm trying to calculate the displacement of a wall at any point due to a point source of vibration. The vibration is considered to be directly perpendicular to the surface of the wall, for calculation purposes.

Input: random force applied to some point (call it white noise)

Output: Displacement at any other part of the wall (located (x,y) from the original point)

I assume that the displacement will at a simple level be related to some K(x,y). The relationship is probably K/(x^2 + y^2), I just need to figure out what goes into K

As far as I can say, the stiffness of the material will increase the speed and distance with which displacement waves travel, and decrease the total displacement.

Also, things like standing waves with the borders of the wall, vibrational characteristics of the material, and others come into play. I'm not really sure how to model all these things.

Can anyone give me some more input into what I should be considering, or maybe some actual equations that can help me relate the material properties of the wall to the displacement at a specific location?

Thanks.

What is the relation between characters of a group and its lie algebra?

The primary reason for studying Lie algebras is the following fundamental fact: the representation theory of a Lie algebra is the same as the representation theory of the corresponding connected, simply connected Lie group.

Of course, the representation theory of a Lie group in general is very complicated. First of all, it might not be connected. Then the Lie algebra cannot tell the difference between the whole group and the connected component of the identity. This connected component is normal, and the quotient is discrete. So to understand the representation theory of the whole group requires, at the very least, knowing the representation theory of that discrete quotient. Even in the compact case, this would require knowing the representation theory of finite groups. For a given finite group, the characters know everything, but of course the classification problem in general is completely intractable.

The other thing that can go wrong is that even a connected group need not be simply connected. Any connected Lie group is a group quotient of a connected, simply connected group with the same Lie algebra, where the kernel is a discrete central subgroup of the connected, simply connected guy. So the representation theory of the quotient is the same as the representations of the simply connected group for which this central discrete acts trivially.

There is a complete classification of connected compact groups. You start with the steps above: classifying the disconnected groups is intractable, but a connected one is a quotient of a connected simply connected one. This simply connected group is compact iff the corresponding Lie algebra is semisimple; otherwise it has some abelian parts. In general, any compact group is a quotient by a central finite subgroup of a direct product: torus times (connected simply connected) semisimple. Torus actions are easy, and the representation theory of semisimples is classified as well. Whether your representation descends to the quotient I'm not entirely sure I know how to read off of the character. When the group is semisimple (no torus part), I do: finite-dimensional representations are determined by their highest weights (which can be read from the character), which all lie in the "weight" lattice; quotients of the simply-connected semisimple correspond exactly to lattices between the weight lattice and the "root" lattice, and you can just check that your character/weights are in the sublattice.

All of this should be explained well in your favorite Lie theory textbook.

nt.number theory - hard diophantine equation: $x^3 + y^5 = z^7$

I may be misunderstanding the question, but I do not believe that it has any integer solutions. At the very least, none are known to exist at the moment. Any solutions would be counterexamples to the Fermat-Catalan conjecture with {m,n,k} = {3,5,7} (since 1/3 + 1/5 + 1/7 = 71/105 < 1). The most I can tell you is that, for coprime {x,y,z}, there are finitely many solutions to your equation. I think your (x,y) = 1 means that they're coprime, anyway, so it follows that z must be coprime. Therefore, any solution at all would disprove the related Beal's conjecture.

linear algebra - Proving "almost all matrices over C are diagonalizable".

This is an elementary question, but a little subtle so I hope it is suitable for MO.

Let $T$ be an $n times n$ square matrix over $mathbb{C}$.

The characteristic polynomial $T - lambda I$ splits into linear factors like $T - lambda_iI$, and we have the Jordan canonical form:

$$ J = begin{bmatrix} J_1 \ & J_2 \ & & ddots \ & & & J_n end{bmatrix}$$

where each block $J_i$ corresponds to the eigenvalue $lambda_i$ and is of the form

$$ J_i = begin{bmatrix} lambda_i & 1 \ & lambda_i & ddots \ & & ddots & 1 \ & & & lambda_i end{bmatrix}$$

and each $J_i$ has the property that $J_i - lambda_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J_i - lambda_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. Thus, Jordan canonical form gives the closest possible to a diagonal matrix. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization.

So far, so good. What I want to prove is the assertion that "Almost all square matrices over $mathbb{C}$ is diagonalizable". The measure on the space of matrices is obvious, since it can be identified with $mathbb{C}^{n^2}$. How to prove, perhaps using the above Jordan canonical form explanation, that almost all matrices are like this?

I am able to reason out the algebra part as above, but is finding difficulty in the analytic part. All I am able to manage is the following. The characteristic equation is of the form

$$(x - lambda_1)(x - lambda_2) cdots (x - lambda_n)$$

and in the space generated by the $lambda_i$'s, the measure of the set in which it can happen that $lambda_i = lambda_j$ when $i neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$.

But here I have cheated, I used only the characteristic equation instead of using the full matrix. How do I prove it rigorously?

ct.category theory - Generation of object-internal structure in a strict 2-category

It's not entirely clear to me what you're asking, but I'll have a go at answering it anyway.

An internal category in a category with pullbacks consists of objects $A_0$ and $A_1$, maps $s,tcolon A_1to A_0$, $icolon A_0to A_1$, and $ccolon A_1times_{A_0}A_1to A_1$ satisfying the usual axioms. An equivalent way to describe this is that for each object $X$ you have a category whose objects are $Hom(X,A_0)$ and whose morphisms are $Hom(X,A_1)$, which varies with $X$ in a natural way. Of course the latter makes sense even if the category has no pullbacks. This is all just like the case of groups, and it requires only an ambient category, not even a 2-category.

Now if I have a strict 2-category $C$ which has strict finite 2-limits, in particular its underlying 1-category $C^1$ has pullbacks. Moreover, any object $A$ gives rise to a canonical internal category in $C^1$ with $A_0 = A$ and $A_1 = A^{mathbf{2}}$, the cotensor with the "walking arrow" $mathbf{2} = (cdot to cdot)$. This construction defines a strict 2-functor from $C$ to the 2-category $Cat(C^1)$ of internal categories, functors, and natural transformations in the 1-category $C^1$. Moreover, I believe that this 2-functor is strictly 2-fully-faithful, i.e. an isomorphism on hom-categories. For this reason, Street has called a strict 2-category with strict 2-pullbacks and strict cotensors with $mathbf{2}$ (which is all you really need for this argument) a "representable 2-category": all the 2-dimensional structure of its objects can be "represented" as internal categories in some 1-category. Cf. for instance "Fibrations and Yoneda's lemma in a 2-category."

If $C$ is a non-strict 2-category, then it doesn't have an underlying 1-category, so it's not as clear how to write this down. But I think that morally, it should still be true, if one takes the care to phrase it correctly. The question of $n$-categories for $n>2$ is quite a different matter, though; I don't know if anyone's thought about it.