Kernel projections in the universal representation. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:50:51Z http://mathoverflow.net/feeds/question/21820 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21820/kernel-projections-in-the-universal-representation Kernel projections in the universal representation. Andre 2010-04-19T06:39:06Z 2012-04-07T21:18:30Z <p>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 $0 \leq a \leq 1$?</p> <p>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 $p$.</p> http://mathoverflow.net/questions/21820/kernel-projections-in-the-universal-representation/21826#21826 Answer by Jonas Meyer for Kernel projections in the universal representation. Jonas Meyer 2010-04-19T08:53:39Z 2010-04-20T00:00:21Z <h3>Answer to original, pre-edit question:</h3> <p>No, because in general $p$ need not be an upper bound for that set.</p> <p>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 <code>$a=\phi\left(\left( \begin{matrix} 1 &amp; 0 \\ 0 &amp; 0 \end{matrix} \right)\right)\in A$</code> satisfies $\mu(a)=0$ and $0\leq a\leq1$. However, $a\nleq0$, and $p$ in this case is 0 because <code>$\mathbf{ker} \ \pi_\mu'' = \mathbf{ker} \ \pi_\mu = \{0\}$</code>.</p> <hr> <h3>Addressing new question:</h3> <p>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 <code>$x_\mu\in\mathcal{H}_\mu$</code> is the GNS vector for <code>$\pi_\mu$</code> and <code>$\pi_\mu(a)\neq0$</code>, then because the numerical radius of <code>$\pi_\mu(a)$</code> is nonzero and <code>$x_\mu$</code> is cyclic for <code>$\pi_\mu$</code>, there is a $c\in A$ such that <code>$\langle\pi_\mu(a)(\pi_\mu(c)x_\mu),\pi_\mu(c)x_\mu\rangle\neq0.$</code> That is, <code>$\mu(c^*ac)\neq0$</code>.</p> <p>(The question thus reduces to whether <code>$p=\sup\{a\in\mathbf{ker} \ \pi_\mu: 0\leq a\leq1\}$</code>, and thus by Kaplansky's density theorem to whether <code>$\mathbf{ker} \ \pi_\mu$</code> is dense in <code>$\mathbf{ker} \ \pi_\mu''$</code>. I don't have an argument for why that is true.)</p> http://mathoverflow.net/questions/21820/kernel-projections-in-the-universal-representation/21970#21970 Answer by Matthew Daws for Kernel projections in the universal representation. Matthew Daws 2010-04-20T15:16:11Z 2010-04-20T15:16:11Z <p>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, <code>$A''=A^{**}$</code>, the bidual of A.</p> <p>So, suppose that $\mu$ is a <em>faithful</em> 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.</p> <p>For example, let $G$ be an infinite discrete group, let <code>$A=C^*_r(G)$</code>, 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 <code>$A^{**} = B(G)^* = W^*(G)$</code> (in common notation) and so the map <code>$\pi_\mu'':W^*(G) \rightarrow VN(G)$</code> is the quotient induced by the adjoint of the inclusion $A(G) \rightarrow B(G)$. In particular, it's not injective.</p> <p>I could, of course, have misunderstood the question...</p> http://mathoverflow.net/questions/21820/kernel-projections-in-the-universal-representation/93466#93466 Answer by Douglas Somerset for Kernel projections in the universal representation. Douglas Somerset 2012-04-07T21:18:30Z 2012-04-07T21:18:30Z <p>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.</p>