Let $H$ and $K$ be Hilbert spaces, and let $u$ be a partial isometry in $\mathcal{B}(H \otimes K)$ between projections $p\_0 = u^\ast u$ and $p\_1 = u u^\ast$ such that $p\_0, p\_1 \leq 1 \otimes (1-q)$ for some projection $q \in \mathcal B(K)$ equivalent to $1 \in \mathcal B (K)$. Does $p\_1 u (a \otimes 1)p\_0 = p\_1 (a \otimes 1)u p\_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ imply that $u$ can be extended to a unitary operator in the commutant $(\mathcal B(H)\otimes 1)' = 1 \otimes \mathcal B(K)$?


I've verified the implication for several simple examples, but I'm having trouble proving it in full generality.


Update: I've reworded the question for clarity thanks to Yemon's comments. By a unitary operator extending $u$, I mean a unitary operator that agrees with $u$ on $p\_0 (\mathcal H \otimes \mathcal K)$.


Update: Here's one example.
Let $K$ be infinite dimensional with basis $e\_1,e\_2,\ldots$, and $H$ be $n$-dimensional with bases $\xi\_1, \ldots, \xi\_n$ and $\zeta\_1, \ldots, \zeta\_n$. Let $\xi = \frac{1}{\sqrt n}\sum\_i \xi\_i \otimes e\_i$ and $\zeta = \frac{1}{\sqrt n}\sum\_I \zeta\_i \otimes e\_i$. Define $u\colon\eta \mapsto \langle \eta, \xi \rangle \zeta$. Note that the assumptions in the first sentence are satisfied with $q$ the projection on the span of $\{e\_{n+1}, e\_{n+2}, \ldots \}$. The condition that $p\_1 u (a \otimes 1)p\_0 = p\_1 (a \otimes 1)u p\_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ is here equivalent to $\langle u (a \otimes 1) \xi, \zeta\rangle = \langle (a \otimes 1) u \xi , \zeta \rangle $ for all $a \in \mathcal B(H)$, which is in turn equivalent to $\langle(a \otimes 1)\xi, \xi \rangle = \langle (a \otimes 1) \zeta, \zeta \rangle$ for all $a \in \mathcal B(H)$. The latter equation holds because both sides are equal to the trace of $a$. Thus, all conditions are satisfied.


To see that the conclusion holds, let $K\_n$ be the span of $e\_1,\ldots,e\_n$, and let $c\_{ij}$ be the unitary $n \times n$ matrix such that $\zeta\_i = \sum\_j c\_{ij} \xi\_j$. Then $$u \xi = \zeta = \frac{1}{\sqrt n}\sum\_i \zeta\_i \otimes e\_i = \frac{1}{\sqrt n} \sum\_{ij} c\_{ij}\xi\_j \otimes e\_i = \frac{1}{\sqrt n}\sum\_j \xi\_j \otimes we\_j = (1 \otimes w) \xi,$$ where $w\colon K\_n \rightarrow K\_n$ is the unitary defined by $we\_j = \sum\_i c\_{ij}e\_i$. We extend $w$ to all of $K$ by setting $\tilde{w} = w \oplus 1$. Then $1 \otimes \tilde w$ is a unitary operator in $1 \otimes B(K)$ that extends $u$.


I realized that the other examples I was thinking about are trivial.



When can a partial isometry $u$ in $\mathcal B(H \otimes K)$ be extended to a unitary in $1 \otimes \mathcal B(K)$?
If I get it right this time, the following Zorn argument proves the implication.


Denote by $q\_0=\bigvee\_{v\in \mathcal{U}(H)} (v\otimes 1)p\_0(v^\ast\otimes 1)$ and similarly $q\_1=\bigvee\_{v\in \mathcal{U}(H)} (v\otimes 1)p\_1(v^\ast\otimes 1)$. Apply Zorn's lemma to the set $I=\left\{w\in \mathcal{B}(H\otimes K)\left\vert \begin{array}{l}\,\,w\text{ is a partial isometry, }\\w^\ast w\leq q\_0,\\ww^\ast\leq q\_1,\\w\text{ extends u, and }\\\!\!\!ww^\ast(v\otimes 1)w=w(v\otimes 1)ww^\ast\text{ for all }v\in\mathcal{U}(H)\end{array}\right\}\right.$


ordered by extension: $w\_1\leq w\_2$ iff $w\_2 w\_1^\ast=w\_1 w\_1^\ast$ (in other words: $w\_2$ has larger initial support and agrees with $w\_1$ on the initial support of $w\_1$)


A maximal element $w\in I$ must have $w^\ast w=q\_0$, for otherwise there would have been a unitary $v\in\mathcal{B}(H)$ such that $f=(v^\ast\otimes 1)w^\ast w(v\otimes 1)\wedge (q\_0-w^\ast w)\not=0$. Then $(v^\ast\otimes 1)w(v\otimes 1)f + w$ is a partial isometry in $I$ that extends $w$. Since $1-q\_0$ and $1-q\_1$ are equivalent projections in $1\otimes \mathcal{B}(K)$, we can extend $w$ to a unitary.


