## Does left-invertible imply invertible in full group C*-algebras (discrete case)?

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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".

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

M.S. Montgomery, Left and right inverses in group algebras, Bull. AMS 75 (1969)

(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 previous MO answer.) The basic idea is to exploit the faithful 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.

In particular, each left-invertible element of the reduced group $C^*$-algebra is invertible.

Question. What can we say for the full group $C^*$-algebra? Is every left-invertible element in `$C^*(G)$` automatically invertible?

Some basic observations:

• 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.

• More generally, if `$C^*(G)$` has a faithful trace then one can use the same argument as for the reduced `$C^*$`-algebra to get a positive answer.

• 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.)

• There are examples of $G$ where `$C^*(G)$` has no faithful trace; these can be found in work of Bekka and Louvet, and come from exploiting Property (T).

Bekka, M. B.(F-METZ-MM); Louvet, N.(CH-NCH) Some properties of `$C^*$`-algebras associated to discrete linear groups. \$C^*-algebras (Münster, 1999), 1–22, Springer, Berlin, 2000.

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Would Bill Johnson know? There is an ask-johnson tag. See mathoverflow.net/questions/tagged/ask-johnson – Will Jagy Oct 24 2011 at 3:36
@Will: I think of WBJ as more of a Banach-space specialist than an operator algebraist, but it is entirely possible he might spot something I haven't here. – Yemon Choi Oct 24 2011 at 3:48
I printed out your arXiv piece 1003.1650v2. Very nice. – Will Jagy Oct 24 2011 at 4:02

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Just a remark: if a $C^*$-algebra $A$ does not have tracial states then $A^{**}$ has an isometry which is not unitary, thus $A^{**}$ does not satisfy Kaplanski condition. Little bit more is true: for every $n$ there are $n$ pairwise orthogonal isometries in $A^{**}$. – Kate Juschenko Oct 24 2011 at 9:51
Nice question, and I do not know the answer. In fact, Yemon would be hard pressed to ask a question about operator algebras to which I know the answer but he does not. – Bill Johnson Oct 24 2011 at 12:43
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## 2 Answers

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.

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Welcome to MO, Ken! – Jon Bannon Oct 25 2011 at 0:23
Thanks for the references - I will have to set aside some time to read up on these results. – Yemon Choi Oct 25 2011 at 0:24

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There is an alternative argument for the free group; not using that free groups are residually finite-dimensional.

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$g \in F$.

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.

EDIT: 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.

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I don't quite follow-- is the point about why this works for a free group that I can take a path through the unitaries for each generator, and then let these (uniquely) induce my *-rep of C^*(F)-- that we have no relations to worry about means that this always works?? – Matthew Daws Oct 27 2011 at 19:28
Thanks Andreas. Reading your answer it reminded me that I'd seen a similar outline before - I guess this kind of argument is very natural for those who've looked at K-theory of group (C*-)algebras. After doing some looking, it seems that this kind of argument was used by J. Cohen in ams.org/mathscinet-getitem?mr=546507 – Yemon Choi Oct 27 2011 at 19:29
Matthew, you are absolutely right. – Andreas Thom Oct 28 2011 at 7:27
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