Why is the definition of von Neumann trace independent of the choice of the Hilbert space?

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?

• 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? – Matthew Daws Feb 3 at 12:17
• @MatthewDaws: Thank you, I added your correct understanding to the question. – Meisam Soleimani Malekan Feb 3 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!