# Regarding equality of norms in Hilbert bimodules

I was reading the book Elements of noncommutative geometry and in page 160 lemma 4.21 the authors state that in a Hilbert B-A bimodule $$E$$ the two norms induced by the two inner products coincide. Reading the proof I noticed something missing to justify one of the inequalities. He states that $$\|(\{s|s\}s|\{s|s\}s)\|_A\leq \|t\mapsto \{s|s\}t\|^2\|(s|s)\|_A$$ and that $$\|t\mapsto \{s|s\}t\|=\|\{s|s\}\|_B$$ Here $$\{\cdot |\cdot \}$$ denotes the inner product over $$B$$ that makes $${}_B E$$ a left Hilbert $$B$$-module and $$(\cdot | \cdot)$$ denotes the inner product over $$A$$ that makes $$E_A$$ a left Hilbert $$A$$-module.

However since $$t\mapsto \{s|s\}t$$ can be seen as a bounded operator on $$E_A$$ and on $${}_{B}E$$, where the norms could be different (in principle) then the norm of $$t\mapsto \{s|s\}t$$ isn't unique. This means that we can't mix the equation above with the inequality, since those $$\|t\mapsto \{s|s\}t\|$$ represent different norms in each case. When trying to come up with a fix for this proof I felt some extra hypothesis was missing, for example have both $$E_A$$ and $${}_{B}E$$ be full Hilbert modules. Reading Blackadar's book on operator theory, I found that in page 151 he proves the same result for full Hilbert bimodules as I suspected. Is the fullness a necessary hypothesis?

• I don't think it matters. You can replace $A$ by the closure of $(E|E)$ and $B$ by the closure of $\{E|E\}$ to obtain a full bimodule with the same norms. – MaoWao Nov 21 '18 at 14:28