3
$\begingroup$

This question is a duplicate of that 2010 MO question.

I am interested in classifying isomorphism classes of $n$-dimensional integral representations of the cyclic group $C_2$ of order $2$. Clearly, any integral representation of $C_2$ is a direct sum of indecomposable integral representations.

The following result is well-known:

Theorem. The group $C_2$ has exactly 3 isomorphism classes of indecomposable integral representations:

(1) trivial;

(2) the sign representation;

(3) the 2-dimensional representation with matrix $\left(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix}\right).$

This result was stated in the answer of Victor Protsak. See also the answer of Todd Leason.

In his comment Victor Protsak gives a reference. He writes: "Curtis and Reiner, Chapter 11. It's a special case of a theorem in Section 74 which classifies integral representations of cyclic groups of prime order. Naturally, this case is much easier and can be done by hand."

Question. How to prove the above theorem "by hand", without reference to the book by Curtis and Reiner?

Motivation: I am working now with algebraic $\mathbb R$-tori. They are classified by integral representations of the Galois group ${\rm Gal}({\mathbb C}/{\mathbb R})$, which is a group of order $2$. In order to understand the well-known classification of indecomposable $\mathbb R$-tori, I need to understand the well-known classification of indecomposable integral representations of ${\rm Gal}({\mathbb C}/{\mathbb R})$.

I asked this seemingly elementary question on Mathematics StackExchange, but got no answers or comments, so I ask it here.

$\endgroup$
4
  • 1
    $\begingroup$ Casselman has a nice write-up of this classification of indecomposable tori … somewhere, but I can't find it right now. $\endgroup$
    – LSpice
    Aug 15, 2020 at 14:01
  • 2
    $\begingroup$ Ah, here we go. Theorem 2, p. 3, of Casselman - Computing with real tori. Let me know if this reference suffices as an answer; I leave it as a comment for now in case you're hoping for a more MO-answer-sized proof. $\endgroup$
    – LSpice
    Aug 15, 2020 at 14:07
  • $\begingroup$ Excellent, thank you! Let us wait... $\endgroup$ Aug 15, 2020 at 14:57
  • 1
    $\begingroup$ @LSpice: If you post this reference to Bill Casselman as an answer, I will be happy to accept it. Many thanks! $\endgroup$ Aug 16, 2020 at 13:43

2 Answers 2

2
$\begingroup$

In Computing with real tori, Casselman has a nice write-up of this theorem from the point of view of not just proving that these are the only indecomposable tori, but, supposing you are given an explicit integral representation of $\operatorname C_2$, explicitly finding/computing its decomposition into these three representations.

In fact, if you (you the general reader, not necessarily @MikhailBorovoi) aren't familiar with Bill Casselman's recent work, it's well worth checking out his page http://www.math.ubc.ca/~cass; he's been very interested for a while in doing actual computations, in the sense of things that can be fed into a computer, relating to algebraic groups. The above is one example; others can be found at http://www.math.ubc.ca/~cass/research/publications.html, including, for example, The computation of structure constants according to Jacques Tits—things that we all know can be done but that most of us (at least I!) would shrink from actually doing, here laid out in a way that demonstrates how to carry it out practically.

(There's also some nice stuff on mathematical graphics!)

$\endgroup$
2
  • $\begingroup$ The first link in your answer does not open. $\endgroup$ Aug 16, 2020 at 13:52
  • $\begingroup$ Huh, sorry; it opens for me. The target is math.ubc.ca/~cass/research/pdf/realtori.pdf . If you can figure out what needs to be done to make it open for you, then please feel free to edit. $\endgroup$
    – LSpice
    Aug 16, 2020 at 13:54
1
$\begingroup$

See Appendix A in M. Borovoi and D. A. Timashev, Galois cohomology and component group of a real reductive group.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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