Here is Deflnition 1.5 of Hilbert module in "L^2-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002, by W. Lück:

A Hilbert $\mathcal N(G)$-module $V$ is a Hilbert space $V$ together with a linear isometric $G$-action such that there exists a Hilbert space $H$ and an isometric linear $G$-embedding of $V$ into the tensor product of Hilbert spaces $H\bar\otimes\ell^2(G)$ with the obvious $G$-action.

($\mathcal N(G)$ is the group von Neumann algebra of a group $G$).

My question is about the Hilbert space $H$ in the definition. What is important about inserting this Hilbert space in this definition?

  • 1
    $\begingroup$ Warning: the phrase "Hilbert module" gets used in two different contexts to mean two completely different things $\endgroup$ – Yemon Choi Jan 27 '19 at 12:34
  • 3
    $\begingroup$ Modules of the form $H\otimes \ell^2(G)$ are "free" in some sense -- as ${\mathcal N}(G)$-modules they correspond to amplifying the canonical ${\mathcal N}(G)$-module $\ell^2(G)$, as you can check by fixing an orthonormal basis of $H$. So Lueck's definition seems to be that he wants to consider the closed submodules of these "free modules" $\endgroup$ – Yemon Choi Jan 27 '19 at 12:37

Consider the structure theory for normal $*$-homomorphisms of a von Neumann algebra $M$. Namely, if $M\subseteq B(H)$, and $M\rightarrow B(K)$ is a normal $*$-homomorphism then, up to unitary conjugation, we may suppose that there is another Hilbert space $H'$ so that $K$ is an invariant (for the $M$ action) subspace of $H\otimes H'$ with $M$ acting as $M\otimes 1$.

So applying to $\mathcal N(G)\subseteq B(\ell^2(G))$ we obtain subspaces of $H \otimes \ell^2(G)$, with $\mathcal N(G)$ acting as $1\otimes\mathcal N(G)$. As $\mathcal N(G)$ is generated by $\{\lambda(g) : g\in G\}$ the translation operators on $\ell^2(G)$, it is not hard to see that $V\subseteq H\otimes \ell^2(G)$ is $\mathcal N(G)$-invariant if and only if it is $G$-invariant, for the translation action on $\ell^2(G)$.

Thus, your definition seems to capture exactly the spaces $V$ with $\mathcal N(G)\rightarrow B(V)$ a normal unital $*$-homomorphism. Again, given such a map, we need only know the image of $\lambda(g)$, as $g$ varies, to reconstruct the whole homomorphism; that is, we only need the $g$ action on $V$. The definition nicely captures exactly which $G$-spaces which can occur, without having to talk about the von Neumann algebra theory of $\mathcal N(G)$. Finally, this would indeed seem to be a reasonable notion of what an "$\mathcal N(G)$-module" should be (which, as Yemon suggests, is different from the notion of a "Hilbert $C^*$-module").

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.