Schur's first lemma for finitely generated continuous groups of $SU(d)$ 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.
 A: 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 $S$.
