In "Another subexponential-time quantum algorithm for the dihedral hidden subgroup problem", Kuperberg writes that $\mathbb{C}[G]$ has a "Burnside decomposition" of

$$\mathbb{C}[G]\cong \bigoplus_{V} V^*\otimes V$$

where each $V$ is an irreducible representation.

He defines $\mathbb{C}[G]$ as a Hilbert space with an orthonormal basis identified by $G$, a finite group. So $\mathbb{C}[G]$ is the span of $\{\vert g\rangle :g\in G\}$ for orthonormal vectors $\vert g\rangle$.

This means $\mathbb{C}[G]$ works naturally as the representation space of either the left-regular or right-regular representation of $G$. Thus, it makes sense to me that, letting $V_i$ be the representation spaces of different representations, we would have

$$\mathbb{C}[G]=\bigoplus_{i} V_i.$$

But I don't see why the tensor product appears. Even if we consider $B(\mathbb{C}[G])$, the set of operators on $\mathbb{C}[G]$, to get the density matrices on this Hilbert space, we would end up with vectors of the form $v_j\otimes v_i$, where $v_j\in V_j$ and $v_i\in V_i$ for $i\neq j$.

Alternatively, he may be identifying $\mathbb{C}[G]$ with the left- or right-regular representation itself, and then decomposing it into the representations, not the representation spaces. But this doesn't make sense to me either; $V$ is finite-dimensional, so $V^*\otimes V\cong M_n$ for some $n$, so this decomposition would imply that every irreducible representation of $G$ is isomorphic to $M_n$ for some $n$, which I don't think is true.

So what is this decomposition?


The second part of each tensor product serves as a multiplicity space. It might be more satisfying to write it as a full decomposition like your second one.

So you have $\bigoplus V_\lambda \otimes V_\lambda^*$ and you want it to look more like your second equation.

What would happen is each $V_\lambda$ would show up as a direct summand $d_\lambda$ times where $d_\lambda$ is the dimension of $V_\lambda^*$.

The tensor product there is as vector spaces with one of the factors being a bookkeeping device to say how many times that irreducible is a summand. It is not tensoring as representations that have to be broken down by further C-G coefficients.

More detail possible, but for a first pass just leave it at that. Don't take it beyond just conveniently keeping track of multiplicities.

Maschke's theorem

  • $\begingroup$ Wouldn't that imply that $d_\lambda$ is also the dimension if $V_\lambda$? Does every irreducible representation really appear a number of times equal to its own dimension? $\endgroup$ – Sam Jaques Apr 15 at 14:45
  • $\begingroup$ Have you seen the sum of squares formula in character tables? $\endgroup$ – AHusain Apr 15 at 15:01

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