User anon - MathOverflow

Answer by anon for Permission to use Online Notes

2012-01-05T18:39:36Z

Rule of thumb: it's OK to print out one copy of online notes for your own private use without requesting permission (that's why they are posted). So if you and your students will each be printing your own copies, there should be no need to request permission. If you are going to print a stack of copies to hand out, or place them in the library, you should probably request permission. Some notes say what can be done with them.

Added: to answer your specific question, no, it is not customary. I have several online notes which googling shows are often used in courses. I never get requests for permission, except sometimes when the instructor is planning to print copies and hand them out (or sell them at cost) to the class, or when the notes are to be placed in a library.

Answer by anon for What are non-abelian $L$-functions?

2011-10-23T21:42:09Z

"I have heard people discussing the utility of L-functions, claiming that since they are essentially cohomological entities, they are "abelian" and therefore lack force." This sounds like twaddle to me. Artin L-series are non-abelian; the L-series of representations of the Weil group are non-abelian; automorphic L-series are nonabelian (unless they are on GL1); the L-series of motives are usually nonabelian. Perhaps you could clarify your question.

Answer by anon for isomorphism of fibre functors

2011-10-02T02:48:15Z

The answer is yes. As already noted, it is true if the category is generated by a finite set of objects. Let $G$ be the group of tensor automorphisms of one of the fibre functors. Then $G$ is an affine group scheme and the isomorphisms from one fibre functor to the second form a torsor for $G$. The group $G$ is an inverse limit of algebraic groups in which the transition maps are surjective. Correspondingly, the torsor is an inverse limit of trivial torsors in which the transition maps are surjective. Therefore the torsor itself is trivial (i.e., it has $k$-point).

Answer by anon for Is $ord(xy)$ independent of $ord(x)$ and $ord(y)$ in a finite group?

2011-09-23T21:55:06Z

Let $a$ and $b$ be elements of a group $G$. If $a$ has order $m$ and $b$ has order $n$, what can we say about the order of $ab$? The next theorem shows that we can say nothing at all.

THEOREM: For any integers $m,n,r>1$, there exists a finite group $G$ with elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ has order $r$.

PROOF: We shall show that, for a suitable prime power $q$, there exist elements $a$ and $b$ of $SL_{2}(F_{q})$ such that $a$, $b$, and $ab$ have orders $2m$, $2n$, and $2r$ respectively. As $-I$ is the unique element of order $2$ in $SL_{2}(F_{q})$, the images of $a$, $b$, $ab$ in $SL_{2}(F_{q})/{\pm I}$ will then have orders $m$, $n$, and $r$ as required.

Let $p$ be a prime number not dividing $2mnr$. Then $p$ is a unit in the finite ring $\mathbb{Z}/2mnr\mathbb{Z}$, and so some power of it, $q$ say, is $1$ in the ring. This means that $2mnr$ divides $q-1$. As the group $F_{q}^{\times}$ has order $q-1$ and is cyclic, there exist elements $u$, $v$, and $w$ of $F_{q}^{\times}$ having orders $2m$, $2n$, and $2r$ respectively.

Let $$ a=\left( \begin{array}{cc} u & 1\\ 0 & u^{-1} \end{array} \right)$$ and $$b=\left( \begin{array}{cc}% v & 0\\ t & v^{-1}% \end{array} \right)$$ (elements of $SL_{2}(F_{q})$), where $t$ has been chosen so that $$ uv+t+u^{-1}v^{-1}=w+w^{-1}. $$

The characteristic polynomial of $a$ is $(X-u)(X-u^{-1})$, and so $a$ is similar to $diag(u,u^{-1})$. Therefore $a$ has order $2m$. Similarly $b$ has order $2n$. The matrix $$ ab=\left( \begin{array}{cc} uv+t & v^{-1}\\ u^{-1}t & u^{-1}v^{-1}% \end{array} \right) , $$ has characteristic polynomial $$ X^{2}-(uv+t+u^{-1}v^{-1})X+1=(X-w)(X-w^{-1})\text{,} $$ and so $ab$ is similar to $diag(w,w^{-1})$. Therefore $ab$ has order $2r$.

I don't know who found this beautiful proof. Apparently the original proof of G.A. Miller is very complicated; see MO24940.

Comment by anon

2012-04-22T00:00:40Z

For proofs, see: Grauert and Remmert, Math. Ann. (1958), 245--318; SGA 4, XI.4.3; and SGA 1, XII. If you are willing to use resolution of singularities, the proof is not too hard (see SGA 1).

Comment by anon

2012-03-25T23:42:26Z

It's not a textbook, but the theorem is proved in the setting of schemes in an expose of Artin in SGA 3. Artin also gives a very sketchy proof of the theorem in his first article in Arithmetic Geometry 1986 (Cornell/Silverman) --- I think I would look there.

Comment by anon

2012-03-23T11:11:12Z

"This is usually proved using the Leray spectral sequence" I hope not (Example 1). The hypotheses imply that $f_*$ takes an injective resolution of $F$ to an acyclic resolution of $f_*F$, which can be used to compute its cohomology. This gives the result.

Comment by anon

2012-03-18T02:28:55Z

Look at Hom from G to the twist (as a functor). This is a G-torsor.

Comment by anon

2012-03-10T19:17:35Z

Well first you need to do it over Q. If H and G are semisimple and G is split over Q, you can take the split form of H. If G is not split, it's a twist of the split form, so you need the cocycle to come from one on H. Looks dubious to me.

Comment by anon

2012-03-07T05:45:43Z

A semisimple Lie group is a covering of a semisimple algebraic group .... Alternatively, use that its Lie algebra is a product of simple Lie algebras.

Comment by anon

2012-02-18T18:52:13Z

Even simpler, an abelian Lie algebra is a Cartan subalgebra of itself.

Comment by anon

2012-02-15T08:04:15Z

My guess is the answer is no. Specifically, I'd guess that there exist really different fibre functors that become isomorphic when you forget their monoidal structures. For example, two fibre functors send an object to vector spaces of the same dimension, so they become equal on objects when you replace the category of vector spaces with its skeleton. Perhaps this can be pushed further to show that sometimes (often? always?) two fibre functors will become isomorphic when you forget their monoidal structures.

Comment by anon

2012-01-23T08:44:49Z

The group of isometries of D as a Riemannian manifold, the group of holomorphic isometries of D, and the group of holomorphic automorphisms of D all have the same identity component. This is discussed a bit in the appendix to the original paper of Baily-Borel.

Comment by anon

2012-01-19T18:53:30Z

It shouldn't "restart the review procedure" --- the editor will just send the new version of the article to his current referee.

Comment by anon

2012-01-13T00:01:46Z

"Langlands did quite a good job of suggesting that the Jugendtraum was some sort of wrong turning." Where did he do that?

Comment by anon

2012-01-12T05:04:20Z

I agree with Anton. However, not every journal will be able to make use of your code. For example, if you write your article in AMS-TeX, then even the AMS will retype it, increasing their costs and your proof reading.

Comment by anon

2012-01-05T09:16:05Z

Over a field of characteristic $p$, there is an obvious action of $\mathbb{G}_m$ on $\alpha_p$, and hence an action of $\mu_p$ on $\alpha_p$. The semi-direct product is a noncommutative connected group scheme of order $p^2$.

Comment by anon

2012-01-03T03:17:49Z

"I would like to exclude Galois groups". Actually, all profinite groups are Galois groups, so you may be in trouble. More seriously, I agree with KConrad: for most us of a profinite group is a projective limit of finite groups, so it's better to start with that as the definition.

Comment by anon

2011-12-20T10:03:43Z

Have you tried looking at the exposition of the theorem in: Berthelot, P. Le théorème de dualité plate pour les surfaces (d'après J. S. Milne). (French) [The flat duality theorem for surfaces (according to J. S. Milne)] Algebraic surfaces (Orsay, 1976--78), pp. 203--237, Lecture Notes in Math., 868, Springer, Berlin-New York, 1981. MR0638601?