TITLE: Why is the definition of von Neumann trace independent of the choice of the Hilbert space?
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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