Invariants of $2$-torsion group under involution

Question

Let $G=\{1,\tau\}$ be the group with two elements and let $A$ be a free abelian group of finite rank on which $G$ acts (via group homomorphisms). Let $B$ be a $2$-torsion group, also with an action of $G$ (via group homomorphisms). Finally, let $\pi:A\to B$ be a surjective $G$-equivariant group homomorphism. I am interested in the $G$-invariant elements $B^G$ of $B$. It is clear that every element $x$ of $A$ with $\tau(x)=\pm x$ is mapped to $B^G$ under $\pi$. Is every element of $B^G$ of this form, i.e. is it true that $B^G=\{\pi(x): x\in A, \tau(x)=\pm x\}$?

Answer 1

Nope. Let $A = \mathbb{Z}^2$ and let $\tau : A \to A$ be given by the matrix

$$\tau = \left[ \begin{array}{cc} 1 & 0 \\ 2 & - 1 \end{array} \right].$$

Writing $e_1, e_2$ for the standard basis, $\tau$ has eigenvectors $\tau(e_2) = - e_2$ and $\tau(e_1 + e_2) = e_1 + e_2$.

Now let $B = A/2 \cong \mathbb{F}_2^2$ be the $2$-torsion quotient of $A$. On this quotient $\tau$ acts by the identity, so $B^G = B$. This means $\pi(e_1)$ is $G$-invariant but it is not (zero or) the image of an eigenvector. This should be a minimal counterexample.

The idea here was to observe that $b \in B^G$ iff $b = \pi(x)$ where $\tau(x) = x + 2y$ for some $y$, and $\tau^2(x) = x$ requires $x = \tau(x) + 2 \tau(y) = x + 2 (y + \tau(y))$, so (using that $A$ is torsion-free) $\tau(y) = -y$. So $\tau$ above is constructed using $y = e_2$ and $x = e_1$.

Let's also see what is predicted by general theory. We have a short exact sequence $0 \to K \to A \to B \to 0$ of $G$-modules, and taking $G$-invariants produces a long exact sequence in group cohomology beginning $$0 \to K^G \to A^G \to B^G \to H^1(G, K) \to \dots $$ so $\pi : A^G \to B^G$ can fail to be surjective if $H^1(G, K)$ is nonzero. But you are asking not only about $A^G$ but about eigenvectors with eigenvalue $-1$. This corresponds to taking $\text{Hom}(-1, A)$ where $-1$ is the sign representation of $G$ on $\mathbb{Z}$, or equivalently taking $(-A)^G$ where $-A$ is slightly fanciful notation for the tensor $A \otimes (-1)$ of $A$ by the sign representation, and there is again a long exact sequence beginning $$0 \to (-K)^G \to (-A)^G \to (-B)^G \to H^1(G, -K) \to \dots $$ so $(-A)^G \to (-B)^G \cong B^G$ can again fail to be surjective but now if $H^1(G, -K)$ is nonzero. If both maps fail to be surjective then their image in $B^G$ is a union of two proper subgroups, which is never the entire group, and this happens above.