4
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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 $2=\epsilon(g-u)\leq\|g-u\|$. $\endgroup$ – Alain Valette Oct 24 '19 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? $\endgroup$ – Ali Taghavi Oct 24 '19 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 $\mathbb{C}G$? $\endgroup$ – Ali Taghavi Oct 24 '19 at 16:56
  • 2
    $\begingroup$ 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. $\endgroup$ – Alain Valette Oct 25 '19 at 20:26
  • 1
    $\begingroup$ @AliTaghavi could you edit your question according to the discussion? $\endgroup$ – YCor Oct 26 '19 at 14:50
5
+400
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you very much for this very interesting answer. $\endgroup$ – Ali Taghavi Oct 26 '19 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.