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Let $G$ be a group which acts linearly on a vector space of dimension $n$ over a field $k$. Denote by $\rho$ this representation and consider the associated adjoint representation $Ad\rho$ which is just the representation of $G$ on $M_n(k)$ where $G$ acts by conjugation via $\rho$.

Now, working with group cohomology, we have the cup product $\cup : H^1(G,Ad \rho) \times H^1(G,Ad \rho) \to H^2(G,Ad\rho\otimes_{\mathbb{Z}}Ad\rho)$.

If I am not mistaken, this cup product is skew symmetric, hence if $u$ denotes the class of a 1-cocycle in $H^1(G,A\rho)$, then $2(u \cup u )= 0 \in H^2(G,Ad\rho\otimes_{\mathbb{Z}}Ad\rho)$.

Can we find an "explicit" function $f : G \to M_n(k)\otimes M_n(k)$ such that for all $\sigma,\tau \in G$, $2(u \cup u) (\sigma,\tau) = u(\sigma)\otimes \rho(\sigma)u(\tau)\rho(\sigma)^{-1} = \sigma.f(\tau) - f(\sigma \tau) + f(\sigma)$ ?

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  • $\begingroup$ You should check that it's skew-symmetric and possibly the answer to your question would pop up. Have you tried? $\endgroup$
    – YCor
    May 4, 2015 at 19:36
  • $\begingroup$ I did. The only proof of skew-symmetry I have seen appeared in several places (and I don't know if one can prove it with an other argument) and involve dimension shifting, i.e. we embed $Ad\rho$ into its Coinduction, call it $A$ and call $A*$ the corresponding quotient. Then $H^n(Ad\rho) \simeq H^{n+1}(A*)$ (at least when n>1, for n=0 it is surjective) via the connexion homomorphism and you use formulae involving this map to show the skew-symmetry. $\endgroup$
    – JWM
    May 4, 2015 at 20:22
  • $\begingroup$ I'm not sure that the product is skew-symmetric, because the pairing $Ad(\rho) \otimes Ad(\rho) \to (Ad(\rho) \otimes Ad(\rho))$ is not symmetric. $\endgroup$ May 4, 2015 at 22:05
  • $\begingroup$ Yes but there is an isomorphism $A\otimesB = B \otimes A$ $\endgroup$
    – JWM
    May 5, 2015 at 6:46
  • $\begingroup$ @JWM In general there is a cup product $H^p(G;V) \otimes H^q(G;V') \to H^{p+q}(G;V \otimes V')$. Skew-commutativity is expressed by saying that if $\tau$ is the twist isomorphism $V \otimes V' \to V' \otimes V$, then $\tau_*(\alpha \cup \beta) = (-1)^{pq} \beta \cup \alpha$. Even if $V = V'$ we cannot eliminate the $\tau_*$ and get skew-symmetry because $\tau$ is not the identity. We need a pairing $m: V \otimes V \to W$ such that $m \circ \tau = m$; then $m_*(\alpha \cup \beta) = (-1)^{pq} m_*(\beta \cup \alpha)$. $\endgroup$ May 5, 2015 at 15:10

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