A question regarding Kadison-Kaplansky idempotent conjecture (A nearest group element $g$ to a nontrivial self adjoint unitary element u )

Question

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.

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?

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.