A Hilbert module defined 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 $G$).

After that the notion of von Neumann trace defined as:

Let $f:V\rightarrow V$ be a positive endomorphism of a Hilbert $\mathcal N(G)$-module. Choose a Hilbert space $H$, a Hilbert basis $\{b_i: i\in I\}$ for $H$, a $G$-equivariant projection $\text{pr}:H\otimes\ell^2(G)\rightarrow H\otimes\ell^2(G)$ and an isometric $G$-isomorphism $u:\text{im(pr)}\xrightarrow{\cong}V$. Let $f:H\otimes\ell^2(G)\rightarrow H\otimes\ell^2(G)$ be the positive operator given by the composition $$ \bar f:H\otimes\ell^2(G)\xrightarrow{\text{pr}}\text{im(pr)}\xrightarrow{u}V\xrightarrow{f}V\xrightarrow{u^{-1}}\text{im(pr)}\hookrightarrow H\otimes\ell^2(G)$$ Define the von Neumann trace of $f:V\rightarrow V$ by
$$ \text{tr}_{\mathcal N(G)}(f):=\sum_{i\in I}\langle f(b_i\otimes e), b_i\otimes e\rangle \quad\in[0,\infty]; $$ where $e\in G\subset\ell^2(G)$ is the unit element.

The "positive" means "positive operator" in the sense of Hilbert spaces, and "endomorphism" mean that $f$ needs to commute with the $\mathcal N(G)$-action.

The author claims after this definition that

This definition is independent of the choices of $H$, $\{b_i: i\in I\}$, $\text{pr}$ and $u$.

But only the independence of the choice $\{b_i: i\in I\}$ was proved. I could not prove the independence of the choice of $H$ in the definition above. What is the main idea to prove this claim?

  • 2
    $\begingroup$ Could you say what "positive endomorphism of a Hilbert $\mathcal N(G)$-module" means? Does "positive" mean "positive operator" in the sense of Hilbert spaces, and "endomorphism" mean that $f$ needs to commute with the $\mathcal N(G)$-action? $\endgroup$ – Matthew Daws Feb 3 '19 at 12:17
  • $\begingroup$ @MatthewDaws: Thank you, I added your correct understanding to the question. $\endgroup$ – Meisam Soleimani Malekan Feb 3 '19 at 13:37

Let $V, H, f$ as in the question. As I pointed out in my answer to your other question what is really going on (in my opinion) is that we are studying normal $*$-homomorphisms of $\newcommand{\mc}{\mathcal}\mc N(G)$. So have a normal $*$-homomorphism $\pi:\mc N(G)\rightarrow\mc B(V)$ and an isometry $v:V\rightarrow H\otimes\ell^2(G)$ with $$ v\pi(x) = (1\otimes x)v \qquad (x\in \mc N(G)), $$ where $\mc N(g)$ acts on $\ell^2(G)$ in the canonical way.

Set $M = \mathbb C\otimes\mc N(G)$ a von Neumann algebra on $H\otimes\ell^2(G)$. Let $\newcommand{\pr}{\operatorname{pr}}\pr$ be the projection onto the range of $v$, so that $\pr = vv^*$. Then $\pr \in M'$, the commutant, as the image of $v$ is $M$-invariant. Notice that $\pi(x) \mapsto v\pi(x)v^* \in \pr M \pr$ is an isomorphism, where $\pr M \pr$ is the induced von Neumann algebra. This has commutant $\pr M' \pr$ (the reduced von Neumann algebra). A simple calculation shows that $\pi(\mc N(G))' \ni y \mapsto vyv^*$ is an isomorphism between $\pi(\mc N(G))'$ and $\pr M'\pr$.

In the notation of the original question, $\overline f$ is exactly the image of $f\in \pi(\mc N(G))'$ in $\pr M'\pr$. Notice that $$ y\mapsto \sum_i (y(b_i\otimes e)|b_i\otimes e) $$ induces a faithful semi-finite trace $T$ on $M' = \mc B(H) \overline\otimes \mc N(G)'$. It is in fact the tensor product of the canonical traces on $\mc B(H)$ and $\mc N(G)'$. The restriction of this trace to $\pr M' \pr \cong \pi(\mc N(G))'$ is exactly what we are interested in: we need to show that it does not depend on $H$ and $v$ (that it does not depend on the choice of $(b_i)$ is fairly clear in this picture).

Suppose $v':V\rightarrow H'\otimes\ell^2(G)$ is another isometry intertwining the $\mc N(G)$ actions. By enlarging $H$ if necessary, we may suppose that $H=H'$. Then $u = vv'^*$ is a partial isometry, in $M'$, with initial projection $\pr'$ and final projection $\pr$. Then $y \mapsto u^*yu$ is an isomorphism between $\pr M' \pr$ and $\pr' M' \pr'$, so it suffices to show that $T(y) = T(u^*yu)$ for positive $y \in M'$. But this is clear as $T$ is a trace!

Edit: For background reading, chapter 8 of the "preprint book" by Popa and Anantharaman available here is very good: http://www.math.ucla.edu/~popa/Books/IIun-v13.pdf

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