5
$\begingroup$

Let $p$ be a prime, let $\mathbb{Z}_p$ be the ring of $p$-adic integers, and let $G$ be a cyclic group of order $p$. It is rather well-known that finite rank $\mathbb{Z}_p$-free representations of $G$ are just direct sums of three indecomposable representations (trivial rank 1 representation, the group ring of rank $p$, and the augmentation ideal of rank $p-1$). Because $G$ is cyclic, the Tate cohomology of such any such representation is 2-periodic, and can be viewed as a finite rank $\mathbb{Z}/2\mathbb{Z}$-graded $\mathbb{F}_p$-vector space.

There is a map that takes any representation $V$ to the "super-dimension" $\dim \hat{H}^0(G, V) - \dim \hat{H}^1(G,V)$ of its Tate cohomology. This is additive on all short exact sequences, and is multiplicative for tensor products. There is another map that takes any representation to the "total dimension" $\dim \hat{H}^0(G, V) + \dim \hat{H}^1(G,V)$ of its Tate cohomology. This map is additive on split short exact sequences, and also multiplicative on tensor products, but I've only seen a proof by direct calculation on indecomposables (e.g., Corollary 2.2 in Borcherds's Modular moonshine III).

We can rephrase this in terms of the the Green ring $\operatorname{Rep}_{\mathbb{Z}_p}(G)$, which is given by the usual Grothendieck ring construction, but restricting the exact sequences relation to split exact sequences. The previous paragraph essentially says that both "super-dimension" and "total dimension" give homomorphisms from the Green ring to the integers, and in fact "super-dimension" factors through a homomorphism from the Grothendieck ring.

My question is: Can these results be extended to the case $G$ is a cyclic group of prime-power order?

Naturally, Tate cohomology is no longer a vector space over a field, so dimension is a bit hairy to generalize, but "number of elementary factors" still seems to be a reasonable characteristic zero lift of "trace of identity". I am reasonably confident the super-dimension case can be done functorially, just because Euler characteristic is so natural. On the other hand, it is no longer so convenient to check multiplicativity of total dimension by hand, because (by a result in Berman's 1961 dissertation) any cyclic group of order $p^n$ for $n \geq 3$ has infinitely many indecomposable representations over $\mathbb{Z}_p$.

$\endgroup$

0

Your Answer

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