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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 $v_0 \in V_p(A, H)$, by (2), we have identifications $$ V_p(A, H) \approx \mathrm{U}(pK)$$ given by sending $u \in \mathrm{U}(pK)$ 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 $$\mathrm{U}(pK)/(\mathrm{U}(pK)\cap \mathrm{U}(A^\prime)) \to \mathrm{Hom}_p(A, \mathrm{B}(H)).$$ 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.

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    $\begingroup$ As a complete non-expert, I looked up "Serre fibration" on wikipedia, and I have to say I could not make sense of the definition, as applied in this case. Could you say a bit more about what one would need to prove, and where in particular you get stuck? $\endgroup$ Commented Jan 12, 2022 at 9:20
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    $\begingroup$ So the definition one should be looking at is found at the following link, or is there a better equivalent condition you are working with? Serre fibrations aren't that familiar to some folks working in vNas. encyclopediaofmath.org/wiki/Serre_fibration $\endgroup$
    – Jon Bannon
    Commented Jan 12, 2022 at 12:51
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    $\begingroup$ @MatthiasLudwig, if you're asking for help on something and someone asks you to define an unfamiliar term, this is not a great way to respond. BTW Matthew Daws is very active on mathoverflow and solves a lot of hard problems posed here, so you're really not doing yourself a favor by telling him "the question is perfectly sensible" and refusing to provide a definition. $\endgroup$
    – Nik Weaver
    Commented Jan 12, 2022 at 14:21
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    $\begingroup$ Sorry if I appeared blunt, this was not the intention. If Serre fibration does not mean much to you, you can take it to mean "fiber bundle" or any other kind of fibration. The definition linked by Jon Bannon is the one to use. Or the one on Wikipedia (en.wikipedia.org/wiki/Fibration). $\endgroup$ Commented Jan 12, 2022 at 21:34
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    $\begingroup$ @MatthewDaws: In the simplest case, the questions asks whether given a continuous path p:[0,1]→Hom(A,B(H)) together with a lift v∈V(A,H) of p(0), we can construct a continuous path q:[0,1]→V(A,H) that lifts p and such that q(0)=v. The general condition asks the same question in continuous families indexed by disks. $\endgroup$ Commented Jan 13, 2022 at 0:05

1 Answer 1

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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)]). Hence $v\colon (a\otimes 1)(|\xi|\otimes\delta_0)\mapsto (a\otimes1)\xi$, $a\in A$, extends to a partial isometry in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\xi=v(|\xi|\otimes\delta_0)$ and $vv^*\le p$. Likewise, there is a partial isometry $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\eta=w(|\eta|\otimes\delta_0)$ and $ww^*\le p$. 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 $$\| t \xi - \eta \| = \| w(v^*\xi-w^*\eta)\|\le\| |\xi|-|\eta| \| \approx 0.$$ 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.

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  • $\begingroup$ Thank you for this great answer! $\endgroup$ Commented Jan 17, 2022 at 17:59
  • $\begingroup$ Hi, going back to understanding the proof of the lemma in more detail, I wondered the following: How do you get the existence of these partial isometries $v$ and $w$? And what exactly are the properties that you would like? Should they satisfy $v^*v = w^*w = 1$? $\endgroup$ Commented Sep 22, 2023 at 15:15
  • $\begingroup$ @Matthias Ludewig: I wasn't active recently. In any case, it was not $|\xi|$ but $\|\xi|\otimes\delta_0$. Because $\xi$ and $|\xi|\otimes\delta_0$ are unit vectors in $K\subset L^2(A)\otimes\ell_2$, $\xi=v(|\xi|\otimes\delta_0)$ implies $v^*\xi=|\xi|\otimes\delta_0$ (which also follows from the definition). The answer is edited now. $\endgroup$ Commented Oct 7, 2023 at 12:12
  • $\begingroup$ Delete the part "$K\subset$". $\endgroup$ Commented Oct 7, 2023 at 12:21

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