Let $(R, \mathfrak m)$ be a local ring (not necessarily an integral domain) and $T$ be a free $R$-module of finite rank $n\geq 2$. Let $\rho: G \to \mathrm{Aut}_{R\text{-linear}}(T)$ be a represenation of a group G. Is it true that if the residual representation $\overline \rho: = \rho \textrm{ mod } \mathfrak m $ has no nonzero $G$-invariant, then the representation $\rho$ has no nonzero $G$-invariant? It seems that the answer is no in general. In fact, the short exact sequence

$0 \to \mathfrak mT \to T \to T/\mathfrak m T \to 0$

gives the exact sequence

$0 \to (\mathfrak mT)^G \to (T)^G \to (T/\mathfrak m T)^G.$

This gives

$(\mathfrak mT)^G \simeq (T)^G$

as $\overline \rho = T/\mathfrak m T$ has no nonzero $G$-invariant. So my question is equivalent to asking for an example of a representation $T$ over a local ring such that $\mathfrak mT$ has a nonzero $G$-invariant.

  • $\begingroup$ Note that under the condition $(T/\mathfrak{m}T)^G = \{0\}$ you not only get $(\mathfrak{m}T)^G \cong T^G$ but in fact $(\mathfrak{m}T)^G = T^G$, that is, $\mathfrak{m} T^G = T^G$. In particular, by Nakayama lemma, $T^G$ is either trivial or not finitely generated. $\endgroup$ – Matthias Klupsch Mar 26 '14 at 8:31
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    $\begingroup$ I don't see how the inclusion $\mathfrak mT^G \subset (\mathfrak mT)^G$ is an equality (if at all). $\endgroup$ – Jyoti Prakash Saha Mar 26 '14 at 22:22
  • $\begingroup$ Yeah, you are right, this is not clear at all, I had not put enough thought into it. $\endgroup$ – Matthias Klupsch Mar 27 '14 at 9:46

Note that $\mathfrak m^n T/\mathfrak m^{n+1}$ is naturally isomorphic to $(\mathfrak m^n/\mathfrak m^{n+1})\otimes_k T/\mathfrak m$ as a $G$-representation, with $G$-acting through the right-hand factor. In particular, if $T/\mathfrak m$ has trivial $G$-invariants, so does $\mathfrak m^n/\mathfrak m^{n+1}$.

From this, an easy devissage shows that $T/\mathfrak m^{n+1}$ has trivial $G$-invariants for every $n$, and hence so does the $\mathfrak m$-adic completion of $T$. If $R$ is Noetherian (so that $T$ embeds into its $\mathfrak m$-adic completion) or complete, we then see that $T$ itself has trivial $G$-invariants.

Any counterexample thus has to be non-Noetherian and not complete. I don't have enough feeling for those contexts to say for sure whether a counterexample actually exists without thinking more about it.

  • $\begingroup$ Thank you for this nice answer. $\endgroup$ – Jyoti Prakash Saha Jul 2 '14 at 8:49

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