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In trying to prove that the answer to the title is "no", I was led to the following problem (which I think is equivalent to the question asked in the title, but can be stated independently). If someone can see how to solve it, that would be great, but I'd also welcome pointers to specific results or papers that might be helpful. $\newcommand{\VN}{{\rm VN}} \newcommand{\SU}{{\rm SU}} \newcommand{\U}{{\rm U}}$

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 We already know that abelian groups cannot contain infinite cb-Helson subsets, see the mentioned preprint for details.

Update. Here are two related facts which do not answer my question, but which may help to put it in context. In some sense, both of them show that if we ask for slightly less then the question has a positive answer, whereas my feeling is that the question stated should have a negative answer.

Result 1. For any $n$, we can find $g_1,\dots, g_n\in \SU(2)$ such that $$ {\left\Vert \sum_{i=1}^n a_i \lambda(g_i) \right\Vert}_{\VN(\SU(2)} \geq \frac{n}{2} \quad\hbox{ for all $a_1,\dots, a_n \in \U(1)$.} $$

This can be done by fixing a maximal torus in $\SU(2)$ and taking $\{g_1,\dots, g_n\}$ to be a subset of that torus which is an independent set in the sense of classical Fourier analysis: see e.g. Rudin's Fourier analysis on groups, Theorem 5.6.7. (Any Helson subset of the torus would give similar examples.)

Result 2. $\newcommand{\CFinf}{C^\ast({\mathbb F}_\infty)}$ For any $n$, we can find unitaries $v_1,\dots, v_n \in \VN(\SU(2))$ such that

$$ {\left\Vert \sum_{i=1}^n a_i \otimes v_i \right\Vert}_{M_k\otimes \VN(\SU(2)} = n \quad\hbox{for all $k$ and all $a_1,\dots, a_n\in \U(k)$} $$

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.

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