$G$-invariants of tensor products of $G$-modules Given two $G$-modules $A$ and $B$, are there general criteria which induce
$$
(A \otimes B)^G = A^G \otimes B^G
$$
where $\sigma(a \otimes b) = \sigma a \otimes \sigma b$?
I saw a note stating, that this is not generally true. But I can't find a counter example. (I guess this is invalid, if the tensor product collapses on the left but not on the right) 
So any pointers are welcome. Thanks
 A: Here is a counterexample:
Let $G$ be a finite group and let $\mathbb{Z}[G]$ be its group ring. Then $\mathbb{Z}[G]^{G}\cong \mathbb{Z}$, and hence $\mathbb{Z}[G]^{G}\otimes \mathbb{Z}[G]^{G}\cong \mathbb{Z}$. 
On the other hand $\mathbb{Z}[G]\otimes \mathbb{Z}[G]\cong \mathbb{Z}[G\times G]$ as $G$-modules, and  $\mathbb{Z}[G\times G]^{G}$ is a free abelian group of rank $|G|$.
This will probably only be true in very special cases. I can't really think of any non-trivial examples at the moment. 
One could try looking at the long exact cohomology sequence 
$$ 0\rightarrow A^{G}\otimes B^{G}\rightarrow (A \otimes B)^G \rightarrow C^{G} \rightarrow \mathrm{H}^1(G,A^{G}\otimes B^{G})\rightarrow \ldots $$
associated to the short exact sequence of $G$-modules
$$  0 \rightarrow A^{G}\otimes B^{G} \rightarrow A \otimes B \rightarrow C \rightarrow 0. $$
This says that $ A^{G}\otimes B^{G}= (A \otimes B)^G $ if and only if $C^{G}$ injets into $\mathrm{H}^1(G,A^{G}\otimes B^{G})$. A sufficient condition is: $C^{G}=0$.
A: Another counter example:
Let $V$ be a finite dimensional vector space over $\mathbb{C}$, and let $G = GL(V)$. Then $V^G = 0$. Moreover, $(V^*)^G = 0$.
However, $(V^* \otimes_{\mathbb{C}} V)^G = End(V)^G = End_G(V) \cong \mathbb{C}$. 
Any non-trivial irreducible representation $V$ for a finite group $G$ would provide an example in the same way.
A: Could be useful the perspective of complex representation theory: Let $\rho$ be a complex representation of $G$ (a $G$-module that is a complex vector space and the $G$-action is linear). Then $\rho$ can be decomposed as a direct sum of irreducible representations
$$\rho=n_1\gamma_1\oplus\cdots\oplus n_k\gamma_k,$$with $\gamma_1$ the one-dimensional trivial representation.
The invariant part is $\rho^G=n_1\gamma_1$ is the trivial part of $\rho$, then if we have two representations $\rho$ and $\rho'(=n_1'\gamma\oplus\cdots\oplus n_k'\gamma_k$), then
$$\rho\otimes\rho'=\bigoplus_{i,j}n_in_j'(\gamma_i\otimes\gamma_j)$$
and the invariant part $(\rho\otimes\rho')^G$ contains $\rho^G\otimes\rho'^G$ but is not equal, it depends on which irreducible representations $\gamma_i$ and $\gamma_j$ become trivial when we tensor it.
For example $G=C_2$ the group of order two, $\rho=\rho'=sgn$ the 1-dimensional sign representation, then $\rho^G=0$ but $\rho\otimes\rho'$ is the trivial 1-dimensional representation and then $(\rho\otimes\rho')^G\neq0$.
