nt.number theory - lower bound for torsion of abelian varieties

Now let me address your last question. For the sake of simplicity, let us assume that $K$ is a global field of characteristic $p>2$ and the ring $End(A)$ of all endomorphisms of $A$ (over an algebraic closure of $K$) is the ring $Z$ of integers. Then my old results (Math. Notes: 21 (1978), 415--419 and 22 (1978), 493--498) imply that for all but finitely many primes $ell$ the Galois module $A[ell]$ is absolutely simple and the Galois group $Gal(K(A[ell])/K)$ is noncommutative.

I claim that for all but finitely many primes $ell$ the order of
$Gal(K(A[ell])/K)$ is divisible by $ell$ and therefore $[K(A[ell]):K] > ell$.

Indeed, suppose that for a given $ell$ the order of
$Gal(K(A[ell])/K)$ is not divisible by $ell$. Let us call such $ell$ exceptional. Then the natural representation of $Gal(K(A[ell])/K)$ in $A[ell]$ can be lifted to characteristic zero, i.e.,
$Gal(K(A[ell])/K)$ is isomorphic to a (finite) subgroup of $GL(2dim(A),C)$ where $C$ is the field of complex numbers. Now, by a theorem of Jordan, there exists a positive integer $N$ that depends only on $dim(A)$ anf such that $Gal(K(A[ell])/K)$ contains a normal commutative subgroup, whose index does not exceed $N$. This means that $K(A[ell])/K$ contains a Galois subextension $K_{0,ell}/K$ such that $[K_{0,ell}:K] le N$ and the field extension $K(A[ell])/K_{0,ell}$ is abelian.

Let $S$ be the (finite) set of places of bad reduction for $A$. The field extension $K(A[ell])/K$ is unramified outside $S$; the same is true for the subextension $K_{0,ell}/K$. (In the number field case one should add the divisors of $ell$ but we live in characteristic $p$!). Recall that the Galois extensions of global $K$ with degree $le N$ and ramification only at $S$ constitute a finite set. Let $L/K$ be the compositum of all such extensions. Clearly, $L$ is also a global field while $L/K$ is a finite Galois extension that contains $K_{0,ell}$. In addition, the Galois group $Gal(L(A[ell])/L)$ is commutative. Now considering $A$ as an abelian variety over $L$ and applying previously mentioned results, we obtain that the set of exceptional primes $ell$ is finite.

qa.quantum algebra - Is there a good differential calculus for quantum SU(3)?

For quantum $SU(2)$, Woronowicz gave a well differential calculus. If we denote the generators of quantum $SU(2)$ by $a,b,c,d$, then the ideal of ker($epsilon)$ corresponding to this calculus is
$$
< a+ q^2d - (1+q^2),b^2,c^2,bc,(a-1)b,(d-1)c>.
$$
This calculus can be shown to generalise the classical calculus on $SU(2)$ when $q=1$. Does anyone know of a (good) calculus (and its ideal) for quantum $SU(3)$?

dg.differential geometry - Prescribing a Riemannian metric along a given geodesic

Let $gamma : [0,1] to mathbb R^2$ be a finite $C^2$-curve in the plane which does not intersect itself. Let $p(z)$ be a second-degree polynomial in $z in mathbb R^2$. Can we construct a Riemannian metric $g$ along $gamma$ such that

• $gamma$ is a geodesic of $g$,



• $gamma$ has length 1, and



• $p(z)$ is the 2-jet of $g$ at $gamma(0)$? (i.e. this prescribes $g$ and its first and second derivatives at $gamma(0)$) Edit: As per Sergei's comment, assume that $p$ is chosen so that $gamma$ does in fact solve the geodesic equation.

I think so, and my sketch of an argument follows the proof of existence of Fermi coordinates in reverse. I haven't worked through this in detail yet, though, because I'm more concerned about the next question:

Let $gamma$ and $eta$ both be finite curves $[0,1] to mathbb R^2$ which do not intersect themselves, and such that $gamma(0) = eta(0)$ and $gamma(1) = eta(1)$ with no other intersections (i.e. $gamma cup eta$ is a piecewise, simple, closed $C^2$-curve in the plane). Can we construct a metric $g$ as above? Note that if so, $gamma(0)$ and $gamma(1)$ will be conjugate points along $gamma$.

ag.algebraic geometry - For a line bundle L on a smooth projective variety X, what is meant by Pic^L(X)

Let me address the last part of your question.

Let $X$ be a smooth, projective variety over an arbitrary ground field $k$.

I want to write $Pic^{[L]}(X)$ instead of $Pic^L(X)$ -- i.e., to make explicit that the variety depends only on the Neron-Severi class of $L$ -- for reasons which will become clear shortly.

Suppose first that $L$ is algebraically equivalent to $0$. Then $Pic^{[L]}(X) = Pic^0(X)$, so certainly it is an abelian variety.

Next suppose that $L$ is a $k$-rational line bundle on $X$. Then $Pic^{[L]}(X)$ is not literally an abelian variety, because it is a nonidentity coset of a group rather than a group itself. However, it is canonically isomorphic to the abelian variety $Pic^0(X)$ just by mapping a line bundle $M$ to $M - L$. So it might as well be an abelian variety, really.

Finally, supose that $L$ is not itself $k$-rational but that its Neron-Severi class $L$ is rational -- i.e., $L$ is given by a line bundle over the algebraic closure which is algebraically equivalent to each of its Galois conjugates. Then $Pic^{[L]}(X)$ is a well-defined principal homogenous space of the Picard variety $Pic^0(X)$ but need not have any $k$-rational points. For instance, suppose that $X$ is a curve. Then the Galois action on the Neron-Severi group is trivial, so taking $L/overline{k}$ to be any degree $n$ line bundle, we get $Pic^{[L]}(X) = Pic^n(X) = Alb^n(X)$, a torsor whose $k$-rational points parameterize $k$-rational divisor classes of degree $n$. (Note that here when I write $Pic^0(X)$ I am talking about the Picard variety rather than the degree $0$ part of the Picard group. More careful notation would be $underline{operatorname{Pic}}^0(X)$.)

In particular, if $X$ is a genus one curve, then there is a canonical isomorphism $X cong Pic^1(X)$, so $Pic^1(X)$ can be endowed with the structure of an abelian variety iff $X$ has a $k$-rational point.

Some further material along these lines can be found in Section 4 of

http://math.uga.edu/~pete/wc2.pdf

big list - Counterexamples in Algebra?

Harry Hutchins "Examples of commutative rings" may be of interest.
It is based on his 1978 Chicago Ph.D. thesis under
Kaplansky, and not surprisingly it serves as a useful complement to
Kaplansky's excellent textbook Commutative Rings (most references
to proofs refer to Kaplansky). There is also a 3 page list of
errata, updates,... dated July 1983, which is distributed with the book.

Hutchins, Harry C. 83a:13001 13-02
Examples of commutative rings. (English)
Polygonal Publ. House, Washington, N. J., 1981. vii+167 pp. $13.75. ISBN 0-936428-05-8

The book is divided into two parts: a brief sketch of commutative ring theory
which includes pertinent definitions along with main results without proof
(but with ample references), and Part II, the 180 examples. The examples do
cover a very large range of topics. Although most of them appear elsewhere,
they are enhanced by a fairly complete listing of their properties. Example
67, for instance, is M. Hochster's counterexample to the polynomial
cancellation problem, and it lists a number of properties of the two rings
that were not given in the original paper Proc. Amer. Math. Soc. 34 (1972),
no. 1, 81 - 82; MR 45 #3394. Some of the examples appear more than once,
since many rings exhibit more than one interesting property. (R=Kx, y, z is
used in Examples 6 and 22.) The examples are grouped into areas, but a
drawback is that these have not been labeled and separated off. In addition,
the Index is for Part I and definitions only, and this means that searching
for a specific example with certain properties can be time consuming. The book
can be used as a supplement to one of the standard texts in commutative ring
theory, and it does appear to complement the monograph by I. Kaplansky
Commutative rings, Allyn and Bacon, Boston, Mass., 1970; MR 40 #7234;
second edition, Univ. Chicago Press, Chicago, Ill., 1974; MR 49 #10674.
--Reviewed by Jon L. Johnson

gt.geometric topology - Triangulating surfaces

[Three years later …]

All the published proofs of triangulability of surfaces that I am aware of use the Schoenflies theorem, which is not exactly an easy thing to prove. There is however another line of proof which avoids the Schoenflies theorem and instead uses the Kirby torus trick that underlies Kirby-Siebenmann theory in higher dimensions. There is a 1974 paper by A.J.S.Hamilton that gives much simpler proofs of Moise's theorems on triangulability of 3-manifolds using the torus trick, and the same ideas can be applied even more simply for surfaces. Instead of the Schoenflies theorem one needs a few results about surfaces strictly in the PL (or smooth if one prefers) category. Namely, one needs to know that PL structures are unique up to PL homeomorphism in the following four cases: $S^1times S^1$, $S^1times{mathbb R}$, $[0,1]times{mathbb R}$, and $D^2$. These can be regarded as special cases of the usual classification theorem for compact PL surfaces, extended to include a few noncompact cases.

I haven't seen this proof in the literature, so I've written it up and posted it on the arXiv here, working in the smooth category rather than the PL category.

It's not clear how suitable this proof would be for an undergraduate course. Besides the ingredients mentioned above, a little basic covering space theory is also needed. If one were in the fortunate position of already having covered these things, then this proof might be accessible to undergraduates. On the other hand, it could be of some interest to go through a proof of the often-quoted-but-seldom-proved Schoenflies theorem. (In this connection I might mention a paper by Larry Siebenmann on the Schoenflies theorem in the Russian Math Surveys in 2005, giving history as well as a proof.)

gn.general topology - covers of $Z^infty$

The answer is NO even if we replace $4$ by $3$.
Let me sketch a proof. This is based upon the following lemma.

Lemma. Fix $S>0$ and for an integer $k$ conisder in $mathbb Z^k$ sets $X$ of diameter at most $S$. Denote by $Vol(X)$ the number of points in $X$ and denote by $X1$ the set of points of distance at most $1$ from $X$. Now let $delta(S,k)$ be the supremum over all $X$ of diameter at most $S$ of the ratio:

$$r(S,k)=sup_{Xsubset mathbb Z^k}frac{Vol(X)}{Vol(X1)}.$$

I claim that for a fixed $S$, $lim_{kto infty}r(S,k)=0$.

Let us skip the proof of the lemma and instead deduce the claim. Suppose by contradiction that the answer is positive. Then for every $k$ we will get a solution to the problem in $mathbb Z^k$ with the fixed number of sets ($U^0,...,U^n$) such that each $U^i_j$ is of the diameter at most $S$. Now, chose such $k$ that $r(S,k)<frac{1}{2n}$ and let us deduce the contradiction.

From Lemma it follows that the supremum of asimptotic density of each set $U^i$ in $mathbb Z^k$ is less than $frac{1}{n+1}$. Indeed, since the distance between different components of $U^i$ is $4$, every point of $U^i1$ that does not belong to $U^i$ is on distance one from at most one component of $U^i$. And lemma gives us the inequality (that should be understood as assymptotic in $mathbb Z^k$)
$$Vol(U^i)<frac{1}{2n} Vol(U^i1)le frac{1}{2n}Vol(mathbb Z^k)$$
Hence $mathbb Z^k$ can not be covered by $U^0,...,U^n$.

It is clear where this proof breakes if we conisder $2$-disjoint sets. In this case one point of $U^i1$ can be on distance $1$ to many components of $U^i$ and the above inequality will not hold. But for $3$-disjoint sets this works.

As for the proof of the lemma, I think, it should be rather standard.