Does left-invertible imply invertible in full group C*-algebras (discrete case)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:26:23Z http://mathoverflow.net/feeds/question/78948 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78948/does-left-invertible-imply-invertible-in-full-group-c-algebras-discrete-case Does left-invertible imply invertible in full group C*-algebras (discrete case)? Yemon Choi 2011-10-24T03:07:12Z 2011-10-28T07:31:56Z <p>The following question/problem has been bugging me on and off for some time now: so I thought it might be worth broaching here on MO, as a case of "ask the experts".</p> <p>Let $G$ be a discrete group. Kaplansky observed that since the group von Neumann algebra $VN(G)$ is a finite von Neumann algebra, each left-invertible element in $VN(G)$ is invertible. A proof is outlined in</p> <blockquote> <p>M.S. Montgomery, Left and right inverses in group algebras, Bull. AMS 75 (1969)</p> </blockquote> <p>(Well, she actually states a weaker result, but inspection shows that her argument extends to give what we claim. See also my remarks on this <a href="http://mathoverflow.net/questions/18508/is-a-left-invertible-element-of-a-group-ring-also-right-invertible/18509#18509" rel="nofollow">previous MO answer</a>.) The basic idea is to exploit the <strong>faithful</strong> trace $T\mapsto \langle T\delta_e,\delta_e\rangle$ and how it behaves on idempotents: for if $ab=I$, then $ba$ is an idempotent.</p> <p>In particular, each left-invertible element of the reduced group $C^*$-algebra is invertible.</p> <p><strong>Question.</strong> What can we say for the <strong>full</strong> group $C^*$-algebra? Is every left-invertible element in <code>$C^*(G)$</code> automatically invertible?</p> <p>Some basic observations:</p> <ul> <li><p>The case where $G$ is the free group on two generators follows from a result of M-D Choi [no relation] who showed that $C^*({\mathbb F}_2)$ embeds into a direct product of matrix algebras.</p></li> <li><p>More generally, if <code>$C^*(G)$</code> has a faithful trace then one can use the same argument as for the reduced <code>$C^*$</code>-algebra to get a positive answer.</p></li> <li><p>If $C^*(G)$ has no non-trivial projections then $ab=I$ implies $ba=I$. (I think this was known to be true for $G={\mathbb F}_2$ but I've forgotten the reference at present.)</p></li> <li><p>There are examples of $G$ where <code>$C^*(G)$</code> has no faithful trace; these can be found in work of Bekka and Louvet, and come from exploiting Property (T).</p></li> </ul> <blockquote> <p>Bekka, M. B.(F-METZ-MM); Louvet, N.(CH-NCH) Some properties of <code>$C^*$</code>-algebras associated to discrete linear groups. $C^*-algebras (Münster, 1999), 1–22, Springer, Berlin, 2000. </p> </blockquote> http://mathoverflow.net/questions/78948/does-left-invertible-imply-invertible-in-full-group-c-algebras-discrete-case/79008#79008 Answer by Ken Dykema for Does left-invertible imply invertible in full group C*-algebras (discrete case)? Ken Dykema 2011-10-24T18:29:45Z 2011-10-24T18:29:45Z <p>That's a nice question. I don't know the answer for arbitrary groups, but this finiteness property (left invertible implies invertible in the full group C*-algebra) is known for more groups. M.D. Choi's result was generalized by [Exel and Loring, Internat. J. Math. 1992]. We say that a C*-algebra if residually finite dimensional (RFD) if it has a separating family of finite dimensional representations. RFD algebras have this finiteness property and finite groups, abelian groups, etc., have RFD full group C*-algebras. Exel and Loring show that unital full free products of RFD C*-algebras are RFD. So if$C^*(G_i)$are RFD (i=1,2), then so is$C^*(G_1*G_2)$. A broader class of C*-algebras than the RFD ones are the MF algebras of [Blackadar and Kirchberg, Math. Ann., 1997]. In MF algebras, all left invertibles are invertible. Recently, [Hadwin, Q. Li, J. Shen, Canad. J. Math. 2011] showed that unital full free products of MF C*-algebras are MF.</p> http://mathoverflow.net/questions/78948/does-left-invertible-imply-invertible-in-full-group-c-algebras-discrete-case/79254#79254 Answer by Andreas Thom for Does left-invertible imply invertible in full group C*-algebras (discrete case)? Andreas Thom 2011-10-27T10:57:53Z 2011-10-28T07:31:56Z <p>There is an alternative argument for the free group; not using that free groups are residually finite-dimensional.</p> <p>Let$\pi$be a faithful representation of$C^{\ast}(F)$on a Hilbert space$H$. Then, as$U(H)$is connected,$\pi$can be deformed to the trivial representation in the point-norm topology, i.e. there exists a family of unitary representations$\pi_t$for$t \in [0,1]$, such that$t \mapsto \pi_t(a)$is norm-continuous for each$a \in C^{\ast}(F)$,$\pi_0=\pi$and$\pi_1(g)=1_H$for all$ g \in F$.</p> <p>Now, if$ab=1$in$C^{\ast}(F)$, then$\pi_t(ba)$is a continuous path of projections ending at$1_H$. Hence,$\pi_0(ba)=1_H$and$ba=1$, as$\pi$was faithful.</p> <p><strong>EDIT:</strong> The same argument works if the$C^{\ast}$-algebra embeds into some contractible algebra (i.e. homotopy equivalent to$\mathbb C$). However, even though many reasonable toplogical spaces are quotients of contractible topological spaces, only few reasonable$C^{\ast}\$-algebras have this property. There is a close relationship with the concept of quasi-diagonality, which appeared in the work of Voiculescu.</p>