5 updated to include info from Ozawa

Update One possible attack on the problem is to show that $C^\ast_\delta(\SU(2))$, the norm-closed subalgebra of ${\mathcal B}(L^2(G))$ that is generated by the subset $\{\lambda(g) \colon g\in \SU(2)\}$, is exact in the sense of Kirchberg/Wassermann et al. For then we could exploit known lower bounds on the exactness constant of $\hbox{MAX}(\ell_1^n)$, due I think to Pisier, to show that the question has a negative answer.

(This was the original motivation for this MO question )

However, I have not found anything in the literature to tell me that $C^\ast_\delta(\SU(2))$ is exact: it is known that this algebra quotients onto $C^\ast_r(\SU(2)_d)$, which probably is exact, but it's not clear to me if this observation helps.

Also: here

This essentially follows from combining the following results: (i) the generators of the free group ${\mathbb F}_\infty$, when viewed as elements of $\CFinf$, span an operator space $V$ which is completely isometric to $\hbox{MAX}(\ell_1)$; (ii) there is an injective unital *-homomorphism from $\CFinf$ into a product of matrix algebras; (iii) we can embed a product of matrix algebras into $\VN(\SU(2))$ completely isometrically. Strictly speaking, this construction only gives us $v_1,\dots, v_n$ in the unit ball of $\VN(\SU(2))$, but I think that if we chase the details through the construction then one can modify what it gives us to get unitaries as claimed.

One possible attack on the problem is to show that $C^\ast_\delta(\SU(2))$, the norm-closed subalgebra of ${\mathcal B}(L^2(\SU(2)))$ that is generated by the subset $\{\lambda(g) \colon g\in \SU(2)\}$, is exact in the sense of Kirchberg/Wassermann et al. For then we could exploit known lower bounds on the exactness constant of $\hbox{MAX}(\ell_1^n)$, due I think to Pisier, to show that the question has a negative answer.

However, I have not found anything in the literature to tell me that $C^\ast_\delta(\SU(2))$ is exact: it is known that this algebra quotients onto $C^\ast_r(\SU(2)_d)$, which probably is exact, but it's not clear to me if this observation helps.

Upper-date 2012-10-24: it turns out that $C^\ast_\delta(\SU(2))$ is not exact (N. Ozawa, personal communication), see the updates to this MO question.

4 fixed typo


We let $\U(k)$ be the group of unitary $k\times k$ matrices, and $\lambda: \SU(2)\to {\mathcal B}(L^2(\SU(2))$ the left regular representation.

Question. Does there exist $c\in (0,1]$ such that, for any $n$, there exist $g_1,\dots, g_n\in \SU(2)$ satisfying $$\inf_{k\geq 1} \inf \left\{ {\left\Vert \sum_{i=1}^n a_i \otimes \lambda(g_i) \right\Vert}_{M_k\otimes\VN(\SU(2))} : a_1,\dots, a_n \in \U(k)\right\} \geq cn,$$
where the norm on $M_k\otimes\VN(\SU(2))$ is the usual (spatial) one?

Remark. My feeling is that no such $c$ should exist, but this is based more on wishful thinking than anything else. I have a vague hope that one could prove no such $c$ exists by a probabilistic argument: that is, pick the $a_1,\dots, a_n$ randomly from some suitable distribution on $\U(k)$, and show that the expected value of $\left\Vert \sum_{i=1}^n a_i \otimes \lambda(g_i) \right\Vert_{M_k\otimes\VN(\SU(2)))}$ is at most $o(n)$, no matter how you choose the special elements $g_1,\dots, g_n$. However, I couldn't find a way to get decent upper bounds on the expectation of the norm, at least with the natural choice of Haar measure on $\U(k)$.

Motivation. A negative answer to this question would tell us that there are no infinite "cb-Helson subsets" of $\SU(2)$, in the terminology of http://arxiv.org/abs/1104.2953 We already know that abelian groups cannot contain infinite cb-Helson subsets, see the mentioned preprint for details.

Update One possible attack on the problem is to show that $C^\ast_\delta(\SU(2))$, the norm-closed subalgebra of ${\mathcal B}(L^2(G))$ that is generated by the subset $\{\lambda(g) \colon g\in \SU(2)\}$, is exact in the sense of Kirchberg/Wassermann et al. For then we could exploit known lower bounds on the exactness constant of $\hbox{MAX}(\ell_1^n)$, due I think to Pisier, to show that the question has a negative answer.

(This was the original motivation for this MO question )


2 tweaked title, clarified the remark
1