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Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ with compact support, which forms an algebra under convolution. Suppose $e \in C_c(G)$ is an idempotent, so that $H = eC_c(G)e$ is an algebra with identity. Is it now true that if $f \star g = e$ in $H$, then $g \star f = e$ as well?

This may be too general to be true, so to be more specific: suppose $K < G$ is a compact subgroup such that $K\backslash G/K$ is countable, suppose $\phi : G \to \mathbf{C}^{\times}$ is a character, and suppose the idempotent $e$ is the function with support in $K$ such that $e(x) = \phi(x^{-1})/\mu(K)$ for $x \in K$. Now is it true that if $f \star g = e$ in $H$ then $g \star f = e$ as well?

[I am reading something that claims some $f$ is a unit but then checks it by checking the existence of $g$ such that $f \star g = e$. So really, the question is whether it follows by some general business that $g \star f = e$, or whether one has to do another computation to check the other direction.]

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  • $\begingroup$ Daft/ignorant question: does the countability hypothesis on the double coset space force $K$ to be open in $G$? $\endgroup$
    – Yemon Choi
    Commented Nov 19, 2009 at 1:12
  • $\begingroup$ I certainly intended compact open subgroup, at any rate. $\endgroup$
    – D. Savitt
    Commented Nov 19, 2009 at 1:59
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    $\begingroup$ Returning to this after a long gap, just in case you have any residual interest: it turns out that on any unimodular $G$, if you have a and b in C_c(G) such that $a*b=a+b$, then $b*a=a+b$. So now take $a=e-f$ and $b=e-g$, with $e,f,g$ as in your question, then this seems to answer your question in the positive. See Theorem 3.6 of arxiv.org/abs/1205.4354 $\endgroup$
    – Yemon Choi
    Commented Jul 1, 2012 at 22:26

3 Answers 3

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I don't have a solution, but here are some thoughts which might be of use or interest.

You may have seen this already, but if your group is discrete then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. I think this property is inherited by the algebra obtained when one compresses by an idempotent in $C_c(G)$.

The earliest reference I know of is somewhere in Kaplansky's Fields and Rings; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in

Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc. 75 1969 539--540. MR0238967 (39 #327)

(The proof uses the existence of a faithful tracial state on $VN(G)$, plus the fact that every idempotent in a $C^*$-algebra is similar in the algebra to a self-adjoint idempotent -- something which was not all that obvious to me the first time I saw this result.)

I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^*$-algebras of some totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments.

Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind.

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    $\begingroup$ I checked with the author of what I was reading, who confirms that it was a mistake -- they should have been checking invertibility in both directions. (Well, in fact it was only the left-invertibility of $f$ that was actually being used subsequently.) So although Yemon's comment doesn't answer the question, it may be as helpful as we're going to get. $\endgroup$
    – D. Savitt
    Commented Nov 19, 2009 at 17:07
  • $\begingroup$ For a final answer, see Yemon's second comment to the question. $\endgroup$
    – D. Savitt
    Commented Jul 3, 2012 at 23:56
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I think the general problem with infinite matrices (that are row-column-finite) arises from the fact that the shift matrix ei -> ei+1 (for i ranging over positive integers) only has a left inverse.

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  • $\begingroup$ Duh. Good call. Editing. $\endgroup$
    – D. Savitt
    Commented Nov 18, 2009 at 23:43
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Hey DLS. The finiteness assumptions you put on the matrices mean that they can be interpreted as linear maps from an infinite direct sum of copies of k to itself. Call this spae X. So now your question seems to be "if f,g:X-->X and fg=id, is gf=id?" and now it's easy to see counterexamples: e.g. imagine one of the matrices sends e_i to e_{i+1} and the other sends e_i to e_{i-1} and kills e_1.

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  • $\begingroup$ Hi! Yes, let's move on to the non-trivially-false part of the question. $\endgroup$
    – D. Savitt
    Commented Nov 18, 2009 at 23:47
  • $\begingroup$ To clarify: this was an answer to warm-up part (0) of the question, which has since been deleted. I don't know the answer to the question in its current form. $\endgroup$
    – Kevin Buzzard
    Commented Nov 19, 2009 at 21:53

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