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Suppose $G$ is a finite group, and $V$ is a $\mathbb C G$-module. We can equip $V\otimes V$ with the structure of a $\mathbb C G$-module via $$g(v\otimes w) = gv\otimes gw.$$ I am trying to understand how this defines a $\mathbb C G$-module structure on the symmetric power $Sym^2(V)$ and the exterior power $\Lambda^2(V)$ as subrepresentations of $V\otimes V$ (Fulton & Harris, p.4).

I was trying to show that the action of any $g\in G$ on $(v\otimes w - w\otimes v)$ and on $(v\otimes v)$ always gives $0$, in which case, the action of $G$ induces an action on $Sym^2(V)$ and $\Lambda^2(G)$, respectively. However, I don't see why this should be true. I am also not sure why representations of these quotient of $V\otimes V$ would be "sub"-representations.

algibberishian
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I was trying to show that the action of any $g\in G$ on $(v\otimes w - w\otimes v)$ and on $(v\otimes v)$ always gives $0$, in which case, the action of $G$ induces an action on $Sym^2(V)$ and $\Lambda^2(G)$, respectively.

This isn't the statement you want to show. The symmetric and exterior squares are defined as quotients by certain subspaces; what you want to show is that the action of $G$ preserves these subspaces, so you want to show that the action of $g \in G$ on a vector of the form $v \otimes w - w \otimes v$ or $v \otimes v$ is another vector of the same form, which is true: you get $gv \otimes gw - gw \otimes gv$ and $gv \otimes gv$ respectively.

Qiaochu Yuan
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  • Oh makes sense of course, because we want it to be zero but again in the quotient space. Thanks! – algibberishian Sep 17 '22 at 18:47
  • Sorry if this is silly, but wouldn't that only prove $\bigwedge^n(V)$ and $\text{Sym}^n(V)$ are quotient representations? How does it also prove they are subrepresentations of the $n$-th tensor product? – user760 Jul 11 '23 at 18:53
  • Shouldn't we need to show $gP_n=P_n gP_n$, for every $g\in G$? Here $P_n$ is the projection from $V^{\otimes n}$ onto the subspace identified with $\bigwedge^n (V)$ or $\text{Sym}^n(V)$? I'm also a beginner. Apology if I'm wrong. – user760 Jul 11 '23 at 19:20