9
$\begingroup$

I asked this question at math stackexchange but did not get any answer and I was suggested to post the question here.

I am studying basic theory of elliptic curves. I encountered Galois cohomology. But two introductory textbooks I read used only $H^0$ and $H^1$. I am curious why higher cohomologies did not appear. I even do not know the definition of higher Galois cohomologies.

  1. Is there any interesting application of higher cohomologies ($H^2, H^3, H^4$ etc) in the study of elliptic curves?

  2. If so, what are these? Can you describe the idea?

Thank you.

$\endgroup$
11
$\begingroup$

When thinking of cohomology as describing a defect to a functor being exact, it has to be expected that the first few $H^i$ appear more often. But there are of course higher coholomology groups and they appear naturally.

However over local and global fields, they can be described in terms of $H^0$ and $H^1$ usually. First over a non-archimedean local field $k$, $H^2(k, E[m])$ is dual to $H^0(k,E[m])=E(k)[m]$ and $H^i(k,E[m])=0$ for $i>2$ when $m$ is coprime to the characteristic, and $H^2(k,E)$ vanishes. That is local Tate duality.

For a global field $k$, the kernel of $H^2(k, E[m])$ to all its local versions is dual to the kernel of $H^1(k,E[m])$ to all its local versions. Again $H^i(k,E[m])$ vanishes for $i>2$ and odd $m$ coprime to the characteristic. This is part of the long exact sequence of Poitou-Tate.

etc. There is much more to be told. But in fact most of the above is not specific to elliptic curves and a book like "Cohomology of number fields" will describe this in all details.

$\endgroup$
  • 3
    $\begingroup$ I think you need to be slightly more careful with the real places. Namely, the cohomology $H^i(\mathbb{R}, E[m])$ can be non-zero for $i > 2$; though of course it is periodic in $i$ with period $2$. Also, for number fields with real embeddings one can have a non-trivial $H^3$. $\endgroup$ – Daniel Loughran Nov 8 '15 at 22:04
  • $\begingroup$ That is why I took $m$ odd. Oh, but I forgot that in the local case. I correct that now. Thanks. $\endgroup$ – Chris Wuthrich Nov 9 '15 at 10:58
8
$\begingroup$

$H^2$ appears quite prominently in the duality theory of elliptic curves. For a local field such as a finite extension of $\mathbb Q_p$, one has $H^2(\text{Gal}(\bar K/K),\bar K^*)=\mathbb Q/\mathbb Z$, and for a global field such as a number field, one similarly has $H^2(\text{Gal}(\bar K/K),\mathsf{A}_{\bar K}^*)=\mathbb Q/\mathbb Z$, where $\mathsf{A}_{\bar K}^*$ is the group of ideles. This makes these groups ideal targets for pairings via cup product from, e.g., two $H^1$ groups. Tate's local pairing on $E(K)\times H^1(\text{Gal}(\bar K/K),E(\bar K))$ and the Cassels-Tate pairing on the Shafarevich-Tate group can be formulated in these terms; see Tate's address to the International Congress. An interesting side note is that Tate's definition of the global pairing relies at one point on the fact that $H^3(\text{Gal}(\bar K/K),\bar K^*)=0$ in order to modify a 2-cochain and turn it into a cocycle. So there's a situation where an $H^3$ appears.

You ask "But two introductory textbooks I read used only H0 and H1. I am curious why higher cohomologies did not appear." I'm guessing one of those textbooks is mine. The reason is that the material in the text only needed $H^0$ and $H^1$, and I wanted to keep the exposition as accessible as possible, so I did not see the need to develop the general theory. In principle, the whole book could have been written without any cohomology. But cohomology is very important in modern mathematics, so my idea was to introduce the $H^0$ and $H^1$ very concretely and show the reader how useful they are, which I hoped would help motivate the more general (but more complicated) theory of higher $H^i$'s when the reader eventually ran across them. I will note that in the first paragraph of Appendix B of The Arithmetic of Elliptic Curves, I did provide 4 references for people who wanted to see more of the theory of group cohomology, including the definition and construction of the higher $H^i$'s.

$\endgroup$
  • 1
    $\begingroup$ Does one not actually have $H^2(\text{Gal}(\bar K/K),\mathsf{A}_{\bar K}^*)=\prod_{v \text{ real}} \mathbb Z/2\mathbb Z \prod_{ v\nmid \infty} \mathbb Q/\mathbb Z$, where products are over places $v$ of $K$? $\endgroup$ – Daniel Loughran Nov 9 '15 at 15:42
  • 1
    $\begingroup$ @DanielLoughran Right, I always forget those darn real places! But in the interest of consistency, I'd write them as $\frac12\mathbb Z/\mathbb Z$, so they look like subgroups of $\mathbb Q/\mathbb Z$. $\endgroup$ – Joe Silverman Nov 9 '15 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.