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I’ve just started writing my revision notes for Riemann surfaces, and it’s suddenly dawned on me that it gives a hazy explanation of why the sheaf cohomology might give the same answer as the kind of homology theory you get by messing about with paths.

If you take a function defined on a little ball in your space, and then analytically continue it along a path, you get a unique analytic continuation. However, if you take a path, a loop say, which corresponds to nontrivial homology (like going around zero when you’ve got hold of a branch of the logarithm), you end up collecting something you didn’t have when you started. In particular, you’re not extending to a well-defined global section. I like to believe that maybe we also have some kind of converse… namely, if you have a local section failing to be global, then a measure of that failure is the same as the number of different ‘loops’ you can analytically continue the function along (so maybe I want non-homologous paths to always give rise to different local sections in some sense).

Maybe once I’ve finished revising the course properly (and probably after exams when there might be time to think about things like sheaves and cohomology), I’ll have some more intelligent things to say (after all, now I think about it, that the etale space provides a universal cover, at least in the sense of analysis, is pretty suggestive… although this would suggest analogues closer to that of the fundamental group…. so something has to get Abelian somewhere…) but I thought this was worth recording. Perhaps the wonderful world of Riemann surfaces is the best way of leaving paths behind and accepting local function rings as the gatekeepers to the geometry of a space… if indeed they are.

Let $X_1,X_2,....$ be some random variables converging to $X$ in probability, and which are uniformly integrable.

Inituitively, the convergence in probability says that they converge nicely most of the time’ and the uniform integrability says that where they don’t converge nicely, they don’t converge too nastily. This is the intuitive reason why we can deduce that these two conditions suffice for convergence in $L^1$, and I attach what I hope is an intuitive proof (though it differs wildly from that in the official notes).

Let $\epsilon >0$.

Firstly, fix $\delta >0$ such that for every $n$, $\int_B |X_n - X| \leq \int_B |X_n| + \int_B |X| < \frac{\epsilon}{2}$ whenever $\mathbb{P}(B) < \delta$ (using uniform integrability).

Secondly, fix $N$ such that whenever $n\geq N$, $\mathbb{P}(|X_n - X| \geq \frac{\epsilon}{2}) < \delta$ (possible by convergence in probability.

We deduce that for all $n \geq N$, $\mathbb{E}|X_n - X| = \int_{\{ |X_n - X|<\frac{\epsilon}{2} \}} |X_n - X| + \int_{\{ |X_n - X| \geq \frac{\epsilon}{2} \}} |X_n - X| < \epsilon.$

So we are done, for exactly the reasons we expected. In both official sets of course notes I’ve been using, the proof takes an entirely different and what feels to me much more obscure route of picking an a.s. convergent subsequence, using Fatou’s lemma and another lemma about uniform integrability which seem to be just missing the point, but if there is a flaw in my above proof I’d be glad to know about it.

EDIT: Thanks to Spencer for pointing out that the flaw was assuming that X is uniformly integrable. Fortunately, this can be proved relatively cleanly, and essentially the proof I want is documented well in Martin Orr’s notes, also posted below. Thanks for the feedback guys.

Just a brief note in which I aim to decode the ideas of the proof that the Galois group of an integer polynomial reduced modulo p is a subgroup of the Galois group of the polynomial over the rationals.

Here is our idea. Let $f(x) = (x-a_1)...(x-a_n)$ be a polynomial with distinct roots in a splitting field $L/\mathbb{Q}$. Then instead of viewing the Galois group of $f$ over $\mathbb{Q}$ as something mysterious about $L$, we concretely identify it as a subgroup $G$ of the symmetric group on n letters, corresponding to its action on the roots. Note that this is a perfectly sensible thing to do, since any element of the Galois group which fixes all the roots of $f$ acts as the identity on $L$, by the definition of a splitting field.

It helps us to visualise this algebraically, so let $F(X,T_1,...,T_n) = \prod_{\sigma \in S_n} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n))$, a generic resolvent’ which contains all the possible permutations and is therefore stable under any permutation. In particular, it is stable under the subgroup $G$ and hence has rational coefficients. Furthermore, if we factorise it $F=F_1 F_2... F_k$ into irreducibles (over $\mathbb{Q}$) these correspond to cosets of $G$.

Let us see why. On the one hand, if $(X - (a_1 T_1 + ... + a_n T_n))| F_1$ then since $G$ stabilises $F_1$, $(X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n))| F_1$ for all $\sigma \in G$. On the other hand, $\prod_{\sigma \in G} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n)$ is a rational polynomial, by Galois theory, so since $F_1$ is irreducible, we have in fact

$F_1 = \prod_{\sigma \in G} (X - (a_{\sigma(1)} T_1 + ... + a_{\sigma(n)} T_n)$.

Similarly, the other irreducible factors represent G-cosets by exactly the same arguments. Therefore we can literally read off the Galois group by looking at the factorisation of a single irreducible factor $F_1$. What a pleasant way to study the Galois group of a polynomial!

We can use a couple of algebraic tricks to push this idea further over the field $\mathbb{Q}$ in the case where $f$ is a monic integer polynomial. Firstly, by Gauss’ Lemma we can take $F_1,F_2,...,F_k$ to be integer polynomials (pass from base ring $\mathbb{Q}(T_1,...,T_n)$ to $\mathbb{Z}[T_1,...,T_n]$). Next, now we have integer coefficients, we can reduce the whole setup modulo a prime $latex p$, and we are now studying the Galois group of $f$ over the field $\mathbb{F}_p$, but all our above arguments about irreducible factors correspond to our reading off subgroups still work.

However, note that since $F_1$ is a $\mathbb{Z}$-polynomial, when we reduce the whole game modulo p, it becomes a perfectly respectable $\mathbb{F}_p$-polynomial, and therefore we deduce that the irreducible factor of $F$ reduced modulo $p$ which is divisible in a splitting field over $\mathbb{F}_p$ by $(X - (a_1 T_1 + ... + a_n T_n))$ is an irreducible factor of the reduction of $F_1$, and therefore has factors corresponding to a subset of the factors of $F_1$. Thus, reading off the Galois group from the irreducible factors now, we deduce that

$\text{Gal}_{\mathbb{F}_p}(f ) \hookrightarrow \text{Gal}_{\mathbb{Q}}(f ).$

I think this reading off the Galois group by looking (in a splitting field) at the linear factors of the irreducibles can make one feel a little suspicious, but such suspicions are completely unfounded: by identifying the Galois group with the relevent subset of $S_n$ (having fixed an ordering of the roots in a splitting field), it is completely rigorous and elegantly transparent.

This is a brief post to record a cool trick from a talk by Vladimir Dokchitser today at the BSD conference.

In computing the parity of the analytic rank of an elliptic curve he used the following trick. Observe that if f(x) = f(-x) for f an analytic function then f is even, so has an even order of vanishing at x=0. Conversely, if f(x)=-f(-x), it has an odd order of vanishing at x=0. So weirdly, by computing whether a function is odd or even, one can seem to determine the parity of its order of vanishing.

The Lefschetz fixed point theorem tells us that if an endomorphism on a space has Lefschetz number (the alternating sum of the traces of its induced maps on homology) not equal to zero, then it must have a fixed point. Groups famously have lots of topological automorphisms with no fixed points (left-multiply by any nontrivial element g). One can also prove that such a function is homotopic to the identity whenever the group is path connected (just take any path from 1 to g and appropriate translations thereof performed simultaneously).

Hence, since homotopic maps induce the same maps on homology, the Lefschetz number of the identity map must be zero, and the Lefschetz number of the identity map is precisely the Euler characteristic!

This is really cool. In particular, let’s think about 2-manifolds. The only compact triangulable 2-manifolds of Euler characteristic 0 are the torus and the Klein bottle. We know the torus has a natural group structure. What Lefschetz has told us is in fact, no other orientable surfaces have a group structure! In particular, we recover the reason elliptic curves have street cred. An algebraic curve (over the complex numbers) has a topological group structure iff it is an elliptic curve.

This also raises the question of whether the Klein bottle also has a nice group structure. Given it’s sort of a twisted torus, maybe we can build one somehow…