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Edit: According to comment conversations we revise the question.

Let $G$ be a group. We consider the following subset of $G$: $$\{g\in G \mid e^{\lambda_g} \in \mathbb{C}\lambda (G)\},$$ where $\lambda_g\in C^*_{\text{red}}(G)$ is the left regular representation of $g$.

Under which conditions is this subset a subgroup of $G$? What is a description of this subset or subgroup?

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    $\begingroup$ It seems to me that $e^{\rho_g}$ and $\mathbb CG$ aren't even comparable; the former is an operator (on $\mathbb CG$, I guess?), whereas the latter is a set of scalar-valued functions. What does it mean to ask if the former lies in a subset of the latter? $\endgroup$
    – LSpice
    Commented Oct 31, 2019 at 19:06
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    $\begingroup$ $\lambda$ and $\rho$ are the Greek analogues to the Latin letters L and R, corresponding to Left and Right. $\endgroup$
    – YCor
    Commented Oct 31, 2019 at 19:27
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    $\begingroup$ As LSpice mentioned, it's not clear how the element $\exp(\rho_g)$ could make sense as an element of the reduced $C^*$-algebra. Nevertheless the element $\exp(g)$ is a well-defined element of the reduced $C^*$-algebra. But it's straightforward that it belongs to $\mathbf{C}G$ if and only if $g$ has finite order. $\endgroup$
    – YCor
    Commented Oct 31, 2019 at 19:30
  • $\begingroup$ @LSpice Yes you are right. thank you for your helpful comment. But i thought that $\mathbb{C}G$ is implicitly used or its image under representation. I revise the question. $\endgroup$
    – Ali Taghavi
    Commented Oct 31, 2019 at 20:59
  • $\begingroup$ @YCor Thanks for your helpful comment. So i should replace $\rho_g$ by $\lambda_g$. So $exp(g)$ and $exp(\lambda_g$ are count as the same thing.. According to your comment I realized that we get the torsion subgroup in the Abelian case. $\endgroup$
    – Ali Taghavi
    Commented Oct 31, 2019 at 21:05

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Let $(\delta_g)_{g\in G}$ be the canonical basis of $\mathbf{C}G$. Define $\exp(g)=\sum\frac1{n!}\delta_g^n$ as an element of the reduced $C^*$-algebra.

I claim that $\exp(g)\in\mathbf{C}G$ iff $g$ has finite order.

It's clear if $g$ has finite order. For the converse, I claim that more generally that for every summable sequence $(c_g)_{g\in G}$, the element $\sum_g c_g\delta_g$ has norm $\ge \left(\sum_g|c_g|^2\right)^{1/2}$; in particular it belongs to $\mathbf{C}G$ (if and) only if $(c_g)$ is finitely supported.

Indeed, for $\varepsilon>0$ (say $<1$), there exists a finite subset $I$ of $G$ such that $\sum_{g\in J}|c_g|\le\varepsilon$ for every finite subset $J$ of $G$ disjoint of $I$. Hence $\left\|\sum_g c_g\delta_g\right\|\ge \left\|\sum_{g\in I} c_g\delta_g\right\|-\varepsilon$, and also $\left\|\sum_{g\in I} c_g\delta_g\right\|_2\ge \left\|\sum_{g} c_g\delta_g\right\|_2-\varepsilon$.

Evaluation at $\delta_1\in\ell_2(G)$ shows that the $C^*$-norm on $\mathbf{C}G$ is $\ge$ than the $\ell^2$ norm. Hence $$\left\|\sum_g c_g\delta_g\right\|\ge \left\|\sum_{g\in I} c_g\delta_g\right\|_2-\varepsilon\ge \left\|\sum_{g} c_g\delta_g\right\|_2-2\varepsilon;$$ since this holds for all $\varepsilon>0$ one gets $\|x\|\ge \|x\|_2$ for $x=\sum_gc_g\delta_g$, which is the desired inequality.

For the conclusion, if $\sum c_g\delta_g=\sum c'_g\delta_g$ with $(c'_g)$ finitely supported, one has $\sum (c_g-c'_g)\delta_g=0$, and hence $\sum_g|c_g-c'_g|^2=0$, so $c_g=c'_g$ for all $g$. Hence $(c_g)$ is finitely supported as well.

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  • $\begingroup$ Very interesting proof, Thank you ! $\endgroup$
    – Ali Taghavi
    Commented Nov 1, 2019 at 10:44
  • $\begingroup$ It seems that we can generalize easily your argument to prove that for a torsion free group $G$ we have always the following: "For map $exp;C^*_{\text{red} G} \to C^*_{\text{red} G}$ one has $exp(\mathbb{C}G\cap \mathbb{C}G=\{1\}$. Are you agree?. On the other hand for a projection $p\in \mathbb{C}G$ we have obviously $exp(p) \in \mathbb{C}G$. So an affirmative answer to Kaplansky conjecture. $\endgroup$
    – Ali Taghavi
    Commented Nov 1, 2019 at 12:13
  • $\begingroup$ Sorry for my typos in previous comment. I revise the typos: According to the method of your proof, Is it true that for a torsion free group $G$ the exponential map $exp$ on $C^*_{red} (G)$ satisfy $exp(\mathbb{C}G) \cap \mathbb{C}G=\{1]$? In this case we have affirmative answer to Kaplansky conjecture because i $\mathbb{C}G$ has an idempotent $p$ then $exp(p) \in \mathbb{C}G$. $\endgroup$
    – Ali Taghavi
    Commented Nov 1, 2019 at 13:04
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    $\begingroup$ No, the result just says $exp^{-1}(CG)\cap G=\{1\}$. Clearly $\exp^{-1}(CG)\cap CG$ contains the scalars $C$. I don't know if $\exp^{-1}(CG)\cap CG=C$, for $G$ torsion-free, is obviously false, or if it's a reasonable conjecture, but obviously it doesn't have any such trivial proof since otherwise Kaplansky's conjecture (implied by it) would have been settled for long. $\endgroup$
    – YCor
    Commented Nov 1, 2019 at 13:22

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