Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $H$ be a Hilbert space, we know that weak topology over $B(H)$, operator algebra of bounded linear operators from $H$ into $H$, is the topology generated by $\{\langle \cdot \xi,\eta\rangle:\; \xi,\eta\in H\}$.

So naturally, I think about the norm of $\langle \cdot \xi,\eta\rangle$ as a linear functional over $V$ a von Neumann subalgebra of $B(H)$. And I guess that

$$\|\langle\cdot \xi,\eta\rangle\|=\inf \{\|\xi'\|_H \|\eta'\|_H:\; s.t. \;\langle T \xi',\eta'\rangle = \langle T \xi,\eta\rangle\; \forall T\in V\}.$$

But I am not sure how can I show that. Indeed I am wondering whether this is correct or not even!

share|cite|improve this question
What does the notation $\langle\cdot\xi,\eta\rangle$ mean? – Davide Giraudo Feb 7 '12 at 9:14
@Davide: What's also commonly written as $\omega_{\xi,\eta}$; the normal functional $T\mapsto \langle T\xi, \eta \rangle$. – Matthew Daws Feb 7 '12 at 10:03
@Mahmood: As Yemon said over at MO, what you describe is the "weak operator topology" and NOT the "weak topology"-- these are two very different things. – Matthew Daws Feb 7 '12 at 10:03
up vote 0 down vote accepted

The answer is "yes". The following ideas can be found in standard texts.

The functional $\omega:T\mapsto \langle T\xi,\eta\rangle$ is normal, so it has a polar decomposition: $\omega = v |\omega|$ where $v\in V$ is a partial isometry, and $|\omega|$ is positive. $v$ is uniquely defined by the extra condition that $v^*v=s(|\omega|)$ the support projection of $|\omega|$. So $v^*\omega = v^*v|\omega| = |\omega|$. However, obviously $v^*\omega=\langle\cdot v^*\xi,\eta\rangle$.

At this point I cheat, and invoke a result from Kadison+Ringrose, Vol II, Prop 7.3.12. This says that as $|\omega|=\langle\cdot v^*\xi,\eta\rangle$ is positive, we can find $\xi'\in H$ with $|\omega|=\langle\cdot \xi',\xi'\rangle$. Clearly $\| |\omega| \| = \|\xi'\|^2$. As $e=v^*v$ is the support projection of $|\omega|$ it is central, and so we see that $|\omega| = \langle \cdot \xi',\xi'\rangle = \langle \cdot e\xi',e\xi'\rangle$. It follows that $\| |\omega| \| = \|\xi'\|^2 = \|e\xi'\|^2$ and so $e\xi'=\xi'$. So also $\|v\xi'\|=\|\xi'\|$.

Now, $\omega = \langle \cdot v\xi',\xi'\rangle$, and so $\|\omega\| = \||\omega|\| = \|\xi'\|^2 = \|v\xi'\| \|\xi'\|$. So your formula for the norm is correct, and actually the infimum is obtained.

I will admit that this was harder than I thought. I wonder if anyone else has an easier proof?

Edit: A general comment. Very often, we work with von Neumann algebras in "standard position". From your bio I see you are interested in abstract harmonic analysis, so perhaps interested in group von Neumann algebras $VN(G)$. Acting on $L^2(G)$ these are in standard position. Then every normal functional arises as $\langle \cdot \xi,\eta \rangle$ for some $\xi,\eta$. So in this case, I wouldn't need to use the reference, and the argument becomes a lot easier.

share|cite|improve this answer
Dear Matthew, Thanks. I already knew the answer for $VN(G)$. I tried to generalize this idea to use it in similar cases and I could not. The result from Kadison+Ringrose that you used was new for me which seems to be the key of the proof. – Mahmood Feb 7 '12 at 18:06
Dear Mahmood. If it is the case that you already had a proof for VN(G) then it would have been very helpful if you could have written this out in your question, and then explained that you didn't see how to generalise the proof to other algebras. That would have helped me (and others) to give a better answer (and would have made the question much more suitable for mathoverflow). – Matthew Daws Feb 8 '12 at 8:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.