Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want some reference to learn basic things about Hodge-Tate decomosition, and what back ground I need.

share|cite|improve this question
As I know Hodge-Tate decomposition can be done for all $p$-divisible groups. It should be well-explained in Tate's paper $p$−divisible groups. 1967 Proc. Conf. Local Fields (Driebergen, 1966) pp. 158–183 Springer, Berlin. Some notes are also available on the net click here – Edvin Goey Sep 6 '11 at 14:39
You might want to give more details about your background and motivation.I think of a Hodge--Tate decomposition as something attached to the $p$-adic etale cohomology of a variety over a $p$-adic field, or perhaps a $p$-divisible group (as in Edvin Goey's comment), or more generally to a $p$-adic representation of the Galois group of a $p$-adic local field. So I'm not sure what you mean by a Hodge--Tate decomposition of an algebraic group; this may well be my ignorance, but in any case, more details would help me figure out what you actually want to know. Regards, – Matt E Sep 6 '11 at 15:35
I am working on Galois representations (more precisely Galois representations with value in an algebraic group over $Q_p$). So if you have something about HT decomposition of representations, then it will help me. Is Edvin's link relevant for this situation ? – user10676 Sep 6 '11 at 16:23
Yes, Edvin's link is very relevant: it is to Tate's original paper on Hodge--Tate decompositions for Tate modules of $p$-divisible groups. On the other hand, the theory has developed a lot since then, and so it is a bit of a distance from Tate's original article to what you are asking about. I can try to write some kind of answer when I have more time. Regards, – Matt E Sep 6 '11 at 17:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.