# Canonical field of Hilbert spaces in Dixmier; Plancherel Theorem

I'm working through the proof of Plancherel's Theorem in $C^{*}$-algebras by Dixmier, section 18.8. For the most part, I'm happy with it, although I have one problem. From Dixmier, I have the proof that

$$L^{2}(G)\cong\int_{\widehat{G}}^{\oplus}{(K(\pi)\otimes\overline{K}(\pi))\ d\mu(\pi)},$$

where $\pi\mapsto K(\pi)$ denotes the canonical field of Hilbert spaces. That is, if for $n\in\Bbb{N}\cup\{\infty\}$, we denote by $\widehat{G}_{n}$ the set of elements of dimension $N$ in $\widehat{G}$, then $\pi\mapsto K(\pi)$ reduces on $\widehat{G}_{n}$ to the standard $n$-dimensional Hilbert space.

I got into reading this proof from Principles of Harmonic Analysis by Anton Deitmar, which gives the statement of the equivalent part of the Plancherel Theorem as

$$L^{2}(G)\cong\int_{\widehat{G}}^{\oplus}{\mathrm{HS}(V_{\pi})\ d\mu(\pi)},$$

and says that the proof is in Dixmier.

Then my question is, how does one see that these two are indeed equivalent? Clearly, it suffices to show that

$$\mathrm{HS}(V_{\pi})\cong K(\pi)\otimes\overline{K}(\pi),$$

but I simply have no idea how to go about showing that. I suspect that I'm just being stupid and missing something which makes it particularly obvious, which probably stems from a lack of understanding of the canonical field of Hilbert spaces over $\widehat{G}$. I'm reasonably happy with the definition, but I don't really know how to make use of it here. The only real attempt I've been able to come up with is based on the idea of writing $\widehat{G}=\bigoplus\widehat{G}_{n}$, and then

$$\int_{\widehat{G}}^{\oplus}{(K(\pi)\otimes\overline{K}(\pi))\ d\mu(\pi)}=\bigoplus_{i}\int_{\widehat{G}_{n}}^{\oplus}(K(\pi)\otimes\overline{K}(\pi))\ d\mu(\pi)=\bigoplus_{i}\int_{\widehat{G}_{n}}^{\oplus}{(H_{i}\otimes\overline{H_{i}})\ d\mu(\pi)},$$

where $H_{i}$ is the standard $n$-dimensional Hilbert space. I'm aware that for finite $n$, this is just $\Bbb{C}^{n}$, although I am at a complete loss as to what Dixmier means by it for $H_{\infty}$ (if he means anything by it, he might not intend for you to look at it this way...). Even ignoring that issue, I'm at a loss as to where I should go after that (if what I wrote is even true, I'm not convinced you can split up the direct integral into a direct sum of integrals the way I did).

Any help either with the proof of the equivalence of the two Theorems, or my interpretation of $\pi\mapsto K(\pi)$ would be much appreciated!

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Also asked on MO. Please provide links if you ask the same question on different sites. –  user50183 Nov 21 '12 at 5:45

On the off chance anyone comes across this and is wondering the same thing, it turned that I was just missing a pretty simple fact that made it more or less trivial. For Hilbert spaces $V,W$, $\mathrm{End}(V,W)\cong V^{*}\oplus W$, so in particular, $\mathrm{HS}(V_{\pi})=\mathrm{End}(V_{\pi},V_{\pi})\cong V_{\pi}\otimes\overline{V_{\pi}}$. Then the result is more or less immediate following (roughly) the same approach as I sketched in the question:

$$L^{2}(G)\cong\int_{\widehat{G}}^{\oplus}{(K(\pi)\otimes\overline{K(\pi)})\ d\mu(\pi)}$$

$$\cong\bigoplus_{n\in\Bbb{N}\cup\{\infty\}}\int_{\widehat{G}_{n}}^{\oplus}{(K(\pi)\otimes\overline{K(\pi)})\ d\mu(\pi)}$$

$$\cong\bigoplus_{n\in\Bbb{N}\cup\{\infty\}}\int_{\widehat{G}_{n}}^{\oplus}{(V_{\pi}\otimes\overline{V_{\pi}})\ d\mu(\pi)}$$

$$\cong\bigoplus_{n\in\Bbb{N}\cup\{\infty\}}\int_{\widehat{G}_{n}}^{\oplus}{\mathrm{HS}(V_{\pi})\ d\mu(\pi)}\cong \int_{\widehat{G}}^{\oplus}{\mathrm{HS}(V_{\pi})\ d\mu(\pi)}.$$

Actually, looking at this now that I've written it, I don't even think you need to bother with the direct sum part.

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