Lifting of a ucp map with values in a von Neumann algebra ultraproduct of matrix algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T17:17:00Z http://mathoverflow.net/feeds/question/69144 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69144/lifting-of-a-ucp-map-with-values-in-a-von-neumann-algebra-ultraproduct-of-matrix Lifting of a ucp map with values in a von Neumann algebra ultraproduct of matrix algebras Mikael de la Salle 2011-06-29T20:05:09Z 2011-07-01T18:15:41Z <p>Let $u:A \to \prod_{\mathcal U} M_n$ be a unital completely positive map (ucp) from a unital separable $C^*$algebra into the von Neumann algebra ultraprodut $\prod_{\mathcal U} M_n$. </p> <p>Here $\mathcal U$ is an ultrafilter on $\mathbb N$ and $\prod_{\mathcal U} M_n$ is the quotient of <code>$B=\{(x_n)_{n \in \mathbb N}, x_n \in M_n(\mathbb C) , \sup_n \|x_n\|&lt;\infty\}$</code> by the ideal <code>$I_{\mathcal U}=\{(x_n)_n, \lim_{\mathcal U} Tr(x_n^* x_n)/n = 0\}$</code>.</p> <blockquote> <p>Does there exist a ucp lifting of $u$, i.e. a sequence $u_n:A \to M_n(\mathbb C)$ of ucp maps such that $u=q \circ (u_n)$, where $q:B\to B/I_{\mathcal U}$ is the quotient map?</p> </blockquote> <p>If not, is $u$ locally liftable? That is: given a finite dimensional operator system $E\subset A$ (= a subspace $E$ of $A$ containing $1$ and stable under $a \mapsto a^*$), does there exist a ucp lifting of the restriction of $u$ to $E$?</p> <p>Some comments: as in <a href="http://mathoverflow.net/questions/68367/" rel="nofollow">my related question</a>, I know that the answer is no in general if one replaces the von Neumann algebra ultraproduct by the $C^*$-algebra ultraproduct. But I hope that again, the situation might be much simpler in the von Neumann algebra setting. (I even have the feeling that I have known the answer to this question, but that I have forgotten it).</p> http://mathoverflow.net/questions/69144/lifting-of-a-ucp-map-with-values-in-a-von-neumann-algebra-ultraproduct-of-matrix/69284#69284 Answer by Andreas Thom for Lifting of a ucp map with values in a von Neumann algebra ultraproduct of matrix algebras Andreas Thom 2011-07-01T18:15:41Z 2011-07-01T18:15:41Z <p>The answer is no, in general there is no lifting. A lifting exists if the $C^{\ast}$-algebra has the so-called lifting property (LP), and local liftings exist if it has the local lifting property (LLP).</p> <p>I constructed in </p> <p>Andreas Thom, <em>Examples of hyperlinear groups without factorization property</em>, Groups Geom. Dyn. 4, no. 1 (2010) 195-208.</p> <p>an example of a group $G$, such that the universal group $C^{\ast}$-algebra of $G$ does not have the LLP. The idea is that $G$ is hyperlinear, but cannot have Kirchberg's factorization property. The hyperlinearity is shown by a concrete construction of micro-states, Kirchberg's factorization property has to fail since $G$ has property (T), but is not residually finite. Note that Kirchberg showed that Kazhdan groups with factorization property are residually finite. See also</p> <p>Narutaka Ozawa, <em>About the QWEP conjecture</em>, Internat. J. Math. 15 (2004), no. 5, 501– 530.</p> <p>where all these concepts are explained.</p>