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Since any $C^*$-algebra can be represented as an algebra of bounded operators $\mathcal{B(H)}$ on a Hilbert space $\mathcal{H}$, the spectral theorem applies to all $C^*$-algebras:

($*$) $A=\int_{\sigma(A)}\lambda d\nu(\lambda)$ where $A$ is a bounded self-adjoint and $d\nu$ is a $\mathcal{P(B(H))}$-valued measure.

This can be generalized to self-adjoint unbounded operators, or restricted to self-adjoint compact ones.

But I read a lot of comments in books or papers (for example p.9 here: http://arxiv.org/abs/quant-ph/0601158 ) that later it was proven that the spectral theorem is valid for any von neumann algebra $\mathfrak{M}$, with $d\nu$ a $\mathcal{P}\mathfrak{(M)}$-valued measure.

1) Since a von Neumann algebra is also a $C^*$-algebra, what new information does this "new" spectral theorem give?

2) Is the topology of convergence leading to equality in ($*$) the only difference?

3) If so can you explain why the von Neumann version is not presented as the most general one for bounded operators (since convergence in weak topology implies the norm one)?

Any further clarifications on the different versions of the spectral theorem are most welcome!

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8  
The new information it gives is that the spectral projections $\nu(E)$ always lie in the von Neumann algebra generated by $A$, though they will not in general lie in the C*-algebra generated by $A$. –  Mike Jury Aug 23 '13 at 19:03
    
Great, thanks for the answer Mike. Could you please state the vNA version of the theorem in complete form or give me a reference where the point you made appears clearly? I can't find any. –  Issam Ibnouhsein Aug 23 '13 at 19:38
1  
Convergence in the weak topology does not imply convergence in norm. –  MTS Aug 23 '13 at 19:39
    
You are right MTS, I was confused it's the other way around. I can't figure out how to cross the third question. –  Issam Ibnouhsein Aug 23 '13 at 19:50
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You may want to compare the continuous functional calculus and the Borel functional calculus. –  Michael Aug 24 '13 at 14:42

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