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What is an example of a surjective homomorphism $B(H)\to\mathbb C$, where $B(H)$ is the set of bounded linear operators on a Hilbert space $H$, and $\mathbb C$ is the complex numbers.

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One silly question (I'm sorry, I don't know much about hilbert spaces)- Why are you sure that such homomorphisms exists for any $H$? e.g. if $B(H)$ is a simple algebra then you can't have any homomorphisms (By Schur's lemma). Is it always the case that $B(H)$ is non-simple? – kneidell Feb 15 '14 at 15:34
sorry i don't mean B(H), I mean a commutative unital subspace of B(H). @kneidell – user108605 Feb 15 '14 at 16:09
so what you're looking for is an example of a surjective homomorphism from some commutative unital subspace of $B(H)$ to $\mathbb C$? Isn't $\mathbb C$ such a subspace (under the appropriate injection)? – kneidell Feb 15 '14 at 17:18
I want an explicit map mate @kneidell. – user108605 Feb 15 '14 at 18:26
maybe one doesnt exist – user108605 Feb 15 '14 at 18:30


Let $v\in H$ be any non-zero vector. What can you say about the map $$\varphi\mapsto \varphi(v):B(H)\to \mathbb C?$$

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If we call the map f, then f(TS)=TS(v)=T(S(v)) which does not equal f(T)f(S)=T(v)S(v) so not a homomorphism @kneidell – user108605 Feb 15 '14 at 10:06
@user108605 you are right. I didn't realize you were asking for algebra homomorphisms, my mistake. I'll try to think of something else. – kneidell Feb 15 '14 at 15:30

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