Edit: According to answer and comments by Prof. Valette we edite the question.

The Kadison Kaplansky conjecture says:

Kadison-Kaplansky conjecture: If $G$ is a torsion-free discrete group then $C^*_{\mathrm{red}}(G)$ has no nontrivial projection.

It is a particular case of a more general conjecture, The Baum-Connes conjecture.

Obviously existence of a non-trivial projection $e$ for a $C^*$-algebra $A$ implies that $A$ has a self-adjoint unitary element $u=1-2e$ which is neither $1$ nor $-1$.

Our main question is the following:

Question: Let $G$ be a discrete group and $u$ be a self-adjoint unitary element of $C^*_{\mathrm{red}}(G)$ different from $1$ and $-1$. Let $g\in G \subset C^*_{\mathrm{red}}(G)$ be a nearest element to $u$ among all group elements $h\in G\subset C^*_{\mathrm{red}}(G)$. Here by "nearest" we mean the nearest according to the distance arising from the operator norm on $C^*_{\mathrm{red}}(G)$. Can one say that such a $g$ is a torsion element of $G$ which is different from the neutral element $e\in G$?

Or can one find a nearest element $g$ as above and then prove that $g$ is a nontrivial torsion element?

Does the main question has an affirmative answer at least in the abelian case?

A refinement of the question according to comment discussion: Assume that $u$ is a non trivial self adjoint unitary element of $C^*_{\text{red}} G$ and $g\in G$ is a group element such that $g$ or $-g$ minimize the quantity $|\pm h-u|,\quad h\in G$. Does this implies that $g$ is of finite order?Can one find such a $g\neq e$?

Note: By emphasising on the word "non-neutral element" one can easily check that this nearest element $\pm g$ is always non neutral for the simple case $G=\mathbb{Z}/3\mathbb{Z}$.

How can this very interesting existing answer be generalized to a group which is not necessarilly abelian?

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    $\begingroup$ Your question needs to be refined. Indeed the set of self-adjoint unitaries is invariant under $u\mapsto -u$, but the copy of $G$ in $C^*_r(G)$ is not. To illustrate what I am saying, I have the following claim: if $G$ is amenable and $u$ is a self-adjoint unitary in $C^*_r(G)$, then either $u$ or $-u$ is at distance 2 of ANY $g\in G$. Obviously we have $\|g-u\|\leq 2$, as $g$ and $u$ are unitaries. Let $\epsilon$ be the trivial rep of $G$: by amenability it is defined on $C^*_r(G)$. Changing sign if necessary, assume $\epsilon(u)=-1$. Then $2=\epsilon(g-u)\leq\|g-u\|$. $\endgroup$ Commented Oct 24, 2019 at 6:22
  • $\begingroup$ @AlainValette Thank you very much for your very helpful and perfect answer. I realize that the last part of my question about $\mathbb{Z}/3\mathbb{Z}$ is obviousely false. BTW does amenability of a group impliy that every unitary representation of G on arbitrary H has an extension to a $C^*$ morphism from $C^*_{red} G$ to $B(H)$? Are they equivalent? $\endgroup$ Commented Oct 24, 2019 at 16:49
  • $\begingroup$ @AlainValette Is there a terminology for the kernel of $\epsilon$ restricted to $\mathbb{C} G$?When G is not amenable what is a terminology for the $C^*$ algebra generated by ker $\epsilon$ restricted to $\mathbb{C}G$? $\endgroup$ Commented Oct 24, 2019 at 16:56
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    $\begingroup$ Yes, a group $G$ is amenable if and only if every unitary representation of $G$ extends to $C^*_r(G)$. The kernel of $\epsilon$ on $\mathbb{C}G$ is usually called the augmentation ideal. $\endgroup$ Commented Oct 25, 2019 at 20:26
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    $\begingroup$ @AliTaghavi could you edit your question according to the discussion? $\endgroup$
    – YCor
    Commented Oct 26, 2019 at 14:50

1 Answer 1


Let $G$ be a discrete abelian group, denote by $\epsilon$ the trivial character. Let $u\in C^*_r(G)$ be a self-adjoint unitary element such that $\epsilon(u)=1$. If $g\in G$ is such that $\|u-g\|<2$, then $g$ has finite order. Indeed, by Pontryagin duality $C^*_r(G)\simeq C(\hat{G})$, with $\hat{G}$ the Pontryagin dual of $G$. Since the Fourier transform $\hat{u}$ takes the values $\pm1$ on $\hat{G}$, and $\hat{u}$ has value 1 at the identity of $\hat{G}$, as $\hat{u}^{-1}(1)$ is clopen we see that $\hat{u}=1$ on $\hat{G}^0$, the connected component of identity of $\hat{G}$.

Observe now that, denoting by $T(G)$ the torsion subgroup of $G$, the dual of $G/T(G)$ identifies canonically with $\hat{G}^0$. If $g\in G$ has infinite order, it defines a non-zero element of $G/T(G)$, so $\hat{g}$ defines a non-trivial character of $\hat{G}^0$. Hence the image of $\hat{G}^0$ is a non-trivial, closed, connected subgroup of $\mathbb{T}$, so it is $\mathbb{T}$. So there exists $\chi\in\hat{G}^0$ such that $\chi(g)=-1$, to the effect that $|(\hat{u}-\hat{g})(\chi)|=2$, hence $\|\hat{u}-\hat{g}\|=\|u-g\|=2$. This concludes the proof.

  • $\begingroup$ Thank you very much for this very interesting answer. $\endgroup$ Commented Oct 26, 2019 at 5:36

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