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 $ucoloneta mapsto langle eta, xi rangle zeta$. 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, zetarangle = 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 $wcolon K_n rightarrow K_n$ 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.
No comments:
Post a Comment