Let $A$ and $B$ be $W^{*}$-algebras, let $X$ be an $A-B$-equivalence bimodule (according to the definition given in "Morita equivalence for $C^*$-algebras and $W^*$-algebras" by Rieffel, link:http://dmitripavlov.org/scans/rieffel.pdf ), the conjugate equivalence bimodule of $X$, written $\widetilde{X}$, is the space $X$ but with conjugate operations of $A, B$ and the complex numbers, that is,

$b\widetilde{x} = \widetilde{xb^*}$, $\widetilde{x}a = \widetilde{a^*x}$,

and with inner products

$\langle \widetilde{x},\widetilde{y} \rangle_B = \langle x,y \rangle_B$, etc.

for $x,y \in X, b \in B, a \in A$.

Rieffel says that this $\widetilde{X}$ is a $B-A$-equivalence bimodule, the proof is apparently verified by routine computations, thing is, I was curious about this:

If $\widetilde{X}$ is a $B-A$-equivalence bimodule, shouldn't $\langle -,- \rangle_B: \widetilde{X} \times \widetilde{X} \rightarrow B$ satisfy

$ \langle b\widetilde{x},\widetilde{y} \rangle_B = b \langle \widetilde{x},\widetilde{y} \rangle_B$

or something similar to condition (3) of definition 3.1 on page 63 in Rieffel's paper?

(which says that $\langle -,- \rangle_B: X \times X \rightarrow B$ should satisfy

$\langle x,yb \rangle_B = \langle x,y \rangle_Bb$

for all $x,y \in B$, $b \in B$).

Thanks :)

  • 1
    $\begingroup$ Did you ever resolve this question? or are you still looking for an answer? $\endgroup$ – Yemon Choi May 4 '15 at 14:11

I do not see the problem here. It satisfies the property you think it should. Explicitly: $$ \langle b\tilde x,\tilde y\rangle_B = \langle \widetilde{ x b^*} ,\tilde{y}\rangle_B = \langle x b^*,y\rangle_B = {\langle y, x b^*\rangle_B}^* = ( \langle y,x\rangle_B b^*)^* = b \langle x,y\rangle_B = b \langle \tilde{x},\tilde{y}\rangle_B,$$ where in the 3rd equality we use that $\langle x,y\rangle_B^* = \langle y,x\rangle_B$.


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