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

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

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?

• 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 $2=\epsilon(g-u)\leq\|g-u\|$. – Alain Valette Oct 24 '19 at 6:22
• @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? – Ali Taghavi Oct 24 '19 at 16:49
• @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 $\mathbb{C}G$? – Ali Taghavi Oct 24 '19 at 16:56
• 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. – Alain Valette Oct 25 '19 at 20:26
• @AliTaghavi could you edit your question according to the discussion? – YCor Oct 26 '19 at 14:50

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.