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Suppose that a finite set $S$ of $d\times d$ special unitary matrices densely generates a representation $\rho$ of a continuous subgroup of $G$ of $SU(d)$. That is, for every $\epsilon>0$ and every $g\in G$, there exists a finite product of the matrices in $S$ which is $\epsilon$-close to $\rho(g)$ in operator norm. Furthermore suppose that the only matrices which commute with all elements of $S$ are constant multiples of the identity. Does these facts suffice to show that the representation $\rho$ is irreducible?

If $G$ were finite, then Schur's first lemma says that $\rho$ is irreducible if and only if the only matrices which commute with all of $S$ are constant multiples of the identity. The subtlety in the continuous case is that every matrix $\rho(g)$ is not exactly obtainable from a finite product of matrices in $S$. Hence the statement "only constant multiples of the identity commute with $S$" does not immediately give that "only constant multiples of the identity commute with $\rho(g)$ for all $g\in G$", as it does in the finite case.

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    That shows that if a matrix$A$commutes with all of S, then it commutes with all of G. I'm interested in the other direction: Suppose a matrix$A$does not commute with all of S. Does this mean A does not commute with all of G? The same continuity argument doesn't work here, because there could be a matrix B in G where A commutes with B but not with any of the finite products from S. (For instance, the commutator [$\rho (g_i)$,A] could be nonzero for all $i$ and tend to $0$ as $i \rightarrow \infty$.)2012-12-19

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It is possible to have $S$ and a matrix $B$ for which $B$ does not commute with anything in $S$, but $B$ does commute with the limit of a sequence of finite products of things in $S$. In fact, we'll find a sequence with limit $A$ for which $A$ doesn't commute with any finite products of things in $S$ (but, of course, $A$ commutes with itself).

Let $S=\{e^{2\pi i\sqrt{2}}, e^{2\pi j \sqrt{2}}\}\subseteq Sp(1) = $ unit quaternions. I claim that the closure of the group generated by $S$, denoted $\overline{S}$ is all of $Sp(1)$.

The idea is that just by looking at $e^{2\pi i\sqrt{2}}$, the closure of the smallest subgroup containing this is an $S^1 = \{e^{i\theta}\}\subseteq Sp(1)$ and likewise the other element generates another $S^1 = \{e^{j\theta}\}\subseteq Sp(1)$.

Hence, $\overline{S}$ must contain these two circles. Note that since $\overline{S}$ is a closed subgroup of $Sp(1)$, it's an embedded Lie subgroup, so we can talk about its Lie algebra. On the Lie algebra level, this implies the Lie algebra of $\overline{S}$ contains $i$ and $j$. Since it's a Lie subalgebra, it contains $[i,j] = 2k$. Since its a subspace, it must contain all combinations of $i,j,k$, i.e., it's all of $\mathfrak{sp}(1)$.

This implies $\overline{S} = Sp(1)$.

Now, it's easy to find things which don't commute with any element in $S$. For example, $k$ doesn't since $k e^{2\pi i \sqrt{2}} k^{-1} = e^{-2\pi i \sqrt{2}}$ and likewise for $e^{2\pi j \sqrt{2}}$. On the other hand, of course, $k$ commutes with itself.

If you're more stringent and want to find an element of $Sp(1)$ which doesn't commute with any finite combination of things in $S$, you can still do it (I think).

The idea is that the group generated by $S$ is countable. Each element of $Sp(1)$, with the exception of $\pm 1$ lie in a unique maximal torus. I think one can prove (but I admit I haven't gone through the details) that no finite combination of things in $S$ produces $\pm 1$. (If some finite combination of things in $S$ does make $\pm 1$, then I'd bet that changing one of the $\sqrt{2}$s to something algebraically independent from $\mathbb{Q}(\sqrt{2})$ would fix it.) Believing this, now argue as follows.

For each finite combination of things in $S$, the set of all matrices which commutes with this is an $S^1\subseteq Sp(1)$. Doing this for all finite combinations of things in $S$ gives countably many $S^1\subseteq Sp(1)$. Since there are uncountably many $S^1\subseteq Sp(1)$, at least one $S^1\subseteq Sp(1)$ has not been used. Pick any element in this $S^1$ which is not $\pm 1$. This element commutes with itself, but not with any finite combination of things in $S$.

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    Putting these answers together, we see that the above conditions do suffice to show that $\rho$ is irreducible. Indeed, suppose that the only matrices which commutes with all of $S$ are constant multiples of the identity. Then since $S\subset \rho(G)$, the only matrices which commute with all of $\rho(G)$ are constant multiples of the identity. By the answer to the second question, this suffices to show $\rho$ is irreducible. In the end, the tangential question about taking the closure of finite products of $S$ was not relevant - though it is interesting in its own right.2012-12-20