The Dimension of the Space of Tensors Over the Group Algebra

Now we can return to the space of tensor products over the group algebra and take a more solid pass at calculating its dimension. Key to this approach will be the isomorphism $V\otimes_GW\cong(V\otimes W)^G$ .

First off, we want to calculate the character of $V\otimes W$ . If $V$ — as a left $G$ -module — has character $\chi$ and $W$ has character $\psi$ , then we know that the inner tensor product has character

$\displaystyle\chi\otimes\psi(g)=\chi(g)\psi(g)$

Next, we recall that the submodule of invariants $(V\otimes W)^G$ can be written as

$\displaystyle(V\otimes W)^G\cong V^\mathrm{triv}\otimes\hom_G(V^\mathrm{triv},V\otimes W)$

Now, we know that $\dim(V^\mathrm{triv})=1$ , and thus the dimension of our space of invariants is the dimension of the $\hom$ space. We’ve seen that this is the multiplicity of the trivial representation in $V\otimes W$ , which we’ve also seen is the inner product $\langle\chi^\mathrm{triv},\chi\otimes\psi\rangle$ . We calculate:

$\displaystyle\begin{aligned}\langle\chi^\mathrm{triv},\chi\otimes\psi\rangle&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\overline{\chi^\mathrm{triv}(g)}\chi(g)\psi(g)\\&=\frac{1}{\lvert G\rvert}\sum\limits_{g\in G}\chi(g)\psi(g)\end{aligned}$

This may not be as straghtforward and generic a result as the last one, but it’s at least easily calculated for any given pair of modules $V$ and $W$ .

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The Unapologetic Mathematician

Mathematics for the interested outsider