The Unapologetic Mathematician

Mathematics for the interested outsider

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.

November 16, 2010 - Posted by | Algebra, Group theory, Representation Theory

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