Let $A$ be a von Neumann algebra and let $H$ be a (separable) Hilbert space.
It is known (see e.g., Section IV, Thm. 5.5 of Takesaki I) that there exists a Hilbert space $K$ such that $A \subset \mathbb{B}(K)$ such that any normal $\*$-homomorphism $\varphi : A \to \mathbb{B}(H)$ can be written as
$$ \varphi(a) = v^\* a v,$$
where $v: H \to K$ is a partial isometry with $v^\*v = \mathrm{id}\_H$ and $v v^\* \in A^\prime \subseteq \mathbb{B}(K)$.


We are led to consider the sets $\mathrm{Hom}(A, \mathrm{B}(K))$ with its u-topology (defined by seminorms $\varphi \mapsto \|\omega \circ \varphi\|\_{A\_\*}$, where $\omega \in \mathrm{B}(H)\_\*$ is an element of the predual of $\mathrm{B}(H)$) and
$$V(A, H) := \{v \in \mathrm{B}(H, K) \mid v^\*v = \mathrm{id}\_H, vv^\* \in A^\prime\}$$
with the strong operator topology.
By the result stated above, the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is surjective.


**Q: Is this map a Serre fibration?**


If not, what are the problems here, and can we assume something on $A$ or change the topologies somehow to ensure this?


Edit: So really, what I am interested in is the question whether the map above admits some kind local sections (this is precisely what the definition of Serre fibration is about). In other words, I would like to know if a family of representations (parametrized by some suitably nice space) can be lifted to a family of implementing partial isometries, at least locally. This would be a family version of the result of Takesaki cited above.




---


So maybe let me write something about what I tried and where I failed.


(1) Let us first prove that the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is continuous.
To this end, let $\omega \in \mathrm{B}(H)\_\*$ be positive, which can be written as $\omega(A) = \mathrm{tr}(WA)$ for some positive trace-class operator $W = \sum\_{i=1}^\infty w\_i e\_i \otimes e\_i^\*$.
Let $v\_n \to v$ be a convergent sequence in $V(A, H)$ and let $\varphi\_n$, $\varphi$ be the corresponding sequences of homomorphisms $A \to \mathrm{B}(H)$.
We have to show that $\mathrm{sup}\_{\|a\| \leq 1} |\omega \circ \varphi\_n(a) - \omega \circ \varphi(a)|$ converges to zero.
If $\|a\| \leq 1$, then
$$|\omega \circ \varphi\_n(a) - \omega \circ \varphi(a)| = |\mathrm{tr}(W v\_n^\* av\_n - Wv^\*av)| \\ 
\leq \sum\_{i=1}^\infty w\_i |\langle e\_i, v\_n^\* a v\_n e\_i - v^\*ave\_i\rangle|\\
\leq \sum\_{i=1}^\infty w\_i \Bigl(|\langle (v\_n-v)e\_i, a v\_n e\_i\rangle| + |\langle ae\_i, (v\_n - v) e\_i\rangle|\Bigr)\\
\leq 2 \sum\_{i=1}^\infty w\_i \|(v\_n-v)e\_i\|,
$$
which converges to zero.


(2) The fibers are as follows: It is not hard to see that if two elements $v\_1, v\_2 \in V(A, H)$ induces the same homomorphism $\varphi : A \to \mathrm{B}(H)$, then $v\_1 = wv\_2$ for a partial isometry $w \in A^\prime$ with $ww^\* = v\_1 v\_1^\*$, $w^\*w = v\_2 v\_2^\*$.
Conversely, given $v$ implementing $\varphi$, then replacing it by $wv$ for a partial isometry $w \in A^\prime$ with $w^\*w = vv^\*$ gives another element of $V(A, H)$ implementing the same $\varphi$.


(3) So what I tried was the following: Fix a projection $p \in A^\prime$, and look at the subset $V\_p(A, H)$ of all $v$ such that $vv^\* = p$, and let $\mathrm{Hom}\_p(A, \mathrm{B}(H))$ be the set of those homomorphisms that are implemented by such a $v$.
Then after fixing a basepoint $v\_0 \in V\_p(A, H)$, then by (2), we have identifications
$$ V\_p(A, H) \approx \mathrm{U}(pH)$$
given by sending $u \in \mathrm{U}(pH)$ to $uv\_0$, and two elements $u\_1$, $u\_2$ correspond to the same element of $\mathrm{Hom}(A, \mathrm{B}(H)$ if and only if $u\_2 u\_1^\* \in \mathrm{U}(A^\prime)$.
Hence there is a continuous bijection
$$\mathrm{U}(pKp)/(\mathrm{U}(pKp)\cap \mathrm{U}(A^\prime)) \to \mathrm{Hom}(A, \mathrm{B}(H)).$$
However, it is now clear to me how to show that this is a homeomorphism.
Also, I am not sure how to attack this if we don't fix $p$ in advance.



Families of representations of von Neumann algebras
**Theorem.**
The map $V(A,H)\to\operatorname{Hom}(A,\mathbb{B}(H))$ is open.


We write $\omega\_{\xi,\eta}$ for the linear functional $x\mapsto \langle x\xi,\eta\rangle$
and $\omega\_\xi$ for $\omega\_{\xi,\xi}$.
It is an elementary fact that if $\omega\_\xi=\omega\_{\eta}$ on a von Neumann algebra $A$,
then $a\xi\mapsto a\eta$ extends to a partial isometry $u\in A'$ such that $u\xi=\eta$.
We can trim $u^\*u$ a little, if necessary, and make it satisfy $1-u^\*u \sim 1-uu^\*$ (Murray--von Neumann equivalence),
at the cost of $\|u\xi-\eta\|<\epsilon/2$, where $\epsilon>0$ is arbitrary small.
Then $u$ extends to a unitary element in $A'$, still denoted by $u$,
which satisfies $\|u\xi-\eta\|<\epsilon$.
The following perturbation lemma is well-known and follows from
the theory of standard form (see [Takesaki, Section IX]).


**Lemma.**
For any $\epsilon>0$, there is $\delta>0$ which satisfies the following.
For any von Neumann algebra $A\subset B(K)$ and any unit vectors
$\xi,\eta\in K$, if $\|(\omega\_\xi-\omega\_\eta)|\_A\|<\delta$,  

then there is a unitary element $u\in A'$ such that
$\|u\xi - \eta\|<\epsilon$.


We postpone the proof of this lemma and prove the theorem.
Let $v\_0\in V(A,H)$ and an SOT neighborhood
$$G=\{ v\in V(A,H) : \forall i\ \|(v-v\_0)\xi\_i\|<\epsilon\}$$
be given.
Here $\xi\_1,\ldots,\xi\_n\in H$ are unit vectors and
$\epsilon>0$.
Take $\delta>0$ from the Lemma for $n^{-1/2}\epsilon$.
Now suppose $v\in V(A,H)$ is such that
$$\| (\omega\_{\xi\_i,\xi\_j}\circ\operatorname{Ad}\_v 
- \omega\_{\xi\_i,\xi\_j}\circ\operatorname{Ad}\_{v\_0})|\_A \|<\delta/n$$
for all $i,j$.
We consider the unit vector
$\xi=n^{-1/2}\left[\begin{smallmatrix} \xi\_1 & \cdots & \xi\_n\end{smallmatrix}\right]^T\in H^n$ and view $\mathbb{B}(H^n)=\mathbb{M}\_n\otimes\mathbb{B}(H)$.
Then
$$\|(\omega\_{(1\otimes v)\xi} - \omega\_{(1\otimes v\_0)\xi})|\_{\mathbb{M}\_n\otimes A}\|<\delta.$$
Thus by Lemma, one finds a unitary element
$u\in A'\cong (\mathbb{M}\_n\otimes A)'\cap\mathbb{B}(H^n)$ such that
$\|(1\otimes u)(1\otimes v)\xi - (1\otimes v\_0)\xi\|<n^{-1/2}\epsilon$.
This implies that $uv\in G$, which finishes the proof.


**Proof of Lemma.**
We may assume $K = p(L^2A \otimes \ell\_2)$, where $L^2A$
is a standard representation of $A$ and
$p\in (A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell\_2)$.
Fix a unit vector $\delta\_0\in\ell\_2$.
There are unique vectors $|\xi|$ and $|\eta|$ in the positive cone $(L^2A)\_+$
such that
$\omega\_\xi=\omega\_{|\xi| \otimes\delta\_0}$ and
$\omega\_\eta=\omega\_{|\eta| \otimes\delta\_0}$ on $A \otimes \mathbb{C}1$
(see [Takesaki, Theorem IX.1.2.(iv)]).
There are partial isometries $v$ and $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell\_2)$ such that $\xi=v|\xi|$ and $\eta=w|\eta|$.
By the generalized Powers--Stormer inequality ([Takesaki, Theorem IX.1.2]),
one has $\| |\xi| - |\eta| \|^2 \le \|(\omega\_\xi-\omega\_\eta)|\_A\|$.
Hence $t:=wv^\*\in p(A\otimes \mathbb{C}1)'p = A'\cap \mathbb{B}(K)$ satisfies
$$\| t \xi - \eta \| = \| w(v^\*\xi-w^\*\eta)\|\approx 0.$$
Let $t=u|t|$ be the polar decomposition. Since $\|t\|\le1$ and
$\|t\xi\|\approx\|\eta\|=1$, one has $|t|\xi \approx \xi$ and $u\xi\approx\eta$.
We can further replace the partial isometry $u\in A'$ with a
unitary element without affecting $u\xi$ much.


