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Let $A \subseteq \mathcal B(\mathcal H)$ be a unital C*-algebra in its universal representation. The GNS representation $\pi_\mu\colon A \rightarrow \mathcal B(\mathcal H_\mu)$ with base state $\mu$ extends uniquely to a normal $\ast$-homomorphism $\pi_\mu''\colon A'' \to \mathcal B(\mathcal H_\mu)$. Since $A''$ is a von Neumann algebra, there exists a unique projection $p\in A''$ such that $\mathbf{ker} \\ \pi_\mu''\ = A''p$. Is $p$ the least upper bound of operators $a \in A$ such that $\mu(a) = 0$ and $0 \leq a \leq 1$?

EDIT: Jonas provided a simple counterexample. For non-commutative C*-algebras, $\mu(a)=0$ of course does not imply that $\pi_\mu(a)=0$. The naive question is therefore whether $p$ is the least upper bound of operators $a\in A$ such that $0 \leq a \leq 1$ and $\mu(c^\ast a c)=0$ for all $c \in A$. More generally, I would be grateful for any such "intrinsic" characterization of the projection $p$.

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3 Answers

up vote 3 down vote accepted

So, it seems like the new question is: Is $\ker\pi_\mu \subseteq A$ dense in $\ker\pi_\mu'' \subseteq A''$. As we're talking about the universal representation, $A''=A^{**}$, the bidual of A.

So, suppose that $\mu$ is a faithful state on A, so $\ker\pi_\mu=\{0\}$. I don't think it's necessary that $\ker \pi_\mu''=\{0\}$: this is equivalent to $\pi_\mu'':A^{**}=A''\rightarrow B(H_\mu)$ being injective, and hence an isomorphism onto its range.

For example, let $G$ be an infinite discrete group, let $A=C^*_r(G)$, and let $\mu$ be the canonical trace on $A$. Then $\pi_\mu(A)'' = VN(G)$ the group von Neumann algebra (as $H_\mu$ can be identified with $\ell^2(G)$), and the predual of $VN(G)$ is $A(G)$ the Fourier algebra. The dual of A is $B(G)$ the Fourier-Stieljtes algebra; as $G$ is not compact, $A(G) \subsetneq B(G)$. Then $A^{**} = B(G)^* = W^*(G)$ (in common notation) and so the map $\pi_\mu'':W^*(G) \rightarrow VN(G)$ is the quotient induced by the adjoint of the inclusion $A(G) \rightarrow B(G)$. In particular, it's not injective.

I could, of course, have misunderstood the question...

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It may be instructive to note that MD's example has nothing to do with (non)commutativity: take $G$ to be the group of integers, so that $C^*(G)=C(T)$; then $\mu$ is just "integration round the circle" and $VN(G)=L^\infty(T)$, and we're observing that the natural map from $C(T)^{**}$* onto $L^\infty(T)$ is not injective. –  Yemon Choi Apr 20 '10 at 18:09
    
(That should have been $C(T)^{**}$ in the last line, obviously) –  Yemon Choi Apr 20 '10 at 19:19
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Actually, probably the easiest example is to let $A=C[0,1]$ and to let $\mu$ be Lebesgue measure. Then the observation simply becomes that $C[0,1]^{**}\rightarrow L^\infty[0,1]$ is a proper quotient map, which follows as the inclusion $L^1[0,1]\rightarrow C[0,1]=M[0,1]$ is not a surjection, because not all regular measures on [0,1] are absolutely continuous with respect to Lebesgue measure. –  Matthew Daws Apr 21 '10 at 9:30
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Answer to original, pre-edit question:

No, because in general $p$ need not be an upper bound for that set.

For example, suppose $A$ is the image of the universal representation of the algebra of 2-by-2 complex matrices $M_2$, and to hopefully reduce confusion I want to explicitly mention an isomorphism $\phi:M_2\to A$. Consider the vector state $\tau$ on $M_2$ induced by $(0,1)\in\mathbb{C}^2$ and let $\mu = \tau\circ\phi^{-1}$ be the corresponding state on $A$. Then $a=\phi\left(\left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right)\right)\in A$ satisfies $\mu(a)=0$ and $0\leq a\leq1$. However, $a\nleq0$, and $p$ in this case is 0 because $\mathbf{ker} \ \pi_\mu'' = \mathbf{ker} \ \pi_\mu = \{0\}$.


Addressing new question:

As you indicated, the problem with the original question is that $\mu(a)=0$ does not typically imply that $\pi_\mu(a)=0$. Your new condition takes care of this: If $x_\mu\in\mathcal{H}_\mu$ is the GNS vector for $\pi_\mu$ and $\pi_\mu(a)\neq0$, then because the numerical radius of $\pi_\mu(a)$ is nonzero and $x_\mu$ is cyclic for $\pi_\mu$, there is a $c\in A$ such that $\langle\pi_\mu(a)(\pi_\mu(c)x_\mu),\pi_\mu(c)x_\mu\rangle\neq0.$ That is, $\mu(c^*ac)\neq0$.

(The question thus reduces to whether $p=\sup\{a\in\mathbf{ker} \ \pi_\mu: 0\leq a\leq1\}$, and thus by Kaplansky's density theorem to whether $\mathbf{ker} \ \pi_\mu$ is dense in $\mathbf{ker} \ \pi_\mu''$. I don't have an argument for why that is true.)

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Jonas, thank you for your counterexample. As I explain in my edit, and as your counterexample amply shows, the question was incorrectly formulated. I hesitate to mark your answer as accepted because I hope to get an answer to the correct version of the question. I don't know what MO etiquette dictates in this situation. I certainly plan to mark this answer correct eventually, and can do so now if you'd like. –  Andre Apr 19 '10 at 10:04
    
Andre, I like how you've handled it so far. I suspected this wasn't quite what you wanted. Actually, I'm going to edit my answer to remark on the updated question. You shouldn't feel any obligation to accept this answer, particularly if you are interested in waiting to see if a more thorough and enlightening answer is forthcoming. –  Jonas Meyer Apr 19 '10 at 10:11
    
I don't see why $A\cap \mathbf{ker} \pi_\mu$ should be dense in $\mathbf{ker} \pi_\mu$. This is in fact precisely the difficulty that brought me to MO with this question. Perhaps I should have said so earlier, but I have an aversion to presenting questions mid-dissection because it makes them less attractive, and because it leads readers down the same path that got me stuck in the first place. Anyway, maybe I'm just suffering from a mental block and it's obvious after all. –  Andre Apr 19 '10 at 21:28
    
Andre, thank you for your comment, it is a good question. (You meant $\mathbf{ker} \pi_\mu''$.) Actually it isn't obvious to me either, and I should have said that I wasn't sure about this. I was speculating on the last part, which is why I said "addressing" rather than "answering", but it was my mistake not to make this clear; I should have just asked about it. I'm sorry. I don't have more to say right now, but I may think about this further. –  Jonas Meyer Apr 19 '10 at 23:56
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I am coming on this problem two years later and trying to remember things I used to know twenty years ago, but my answer (to Jonas' re-stated question) is that $\ker \pi_{\mu}$ is almost never dense in $\ker\pi_{\mu}''$. The universal representation is the direct sum of all the GNS representations and the kernel of $\pi_{\mu}''$ is everything in $A^{**}$ that arises from other GNS representations. So if $\mu$ is a faithful state, $\ker\pi_{\mu}=0$ but $\ker\pi_{\mu}''$ is generally enormous. The projection $p$ is the central cover of the representation $\pi_{\mu}$ and is discussed in the (older?) standard books on C*-algebras.

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