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?



Why is the definition of von Neumann trace independent of the choice of the Hilbert space?
Let $V, H, f$ as in the question. As I pointed out in my [answer to your other question](https://mathoverflow.net/a/321875/406) 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>


