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Write $X=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Commuting with all $\mathrm{diag}(U_1,U_2)$ requiresFor $n_1,n_2\ge 2$, the defining complex representations of each factor are irreducible. Schur lemma gives $A=z{I}_{n_1}$ and $D=w{I}_{n_2}$. The two block vector spaces are inequivalent representations of the product group, even when $n_1=n_2$: the first factor acts nontrivially on just the first block. Thus their off-diagonal intertwiners vanish, $B=C=0$. Unitarity and the unit determinant give

The assertion that this full group is always a circle needs correction. Put $d=\gcd (n_1,n_2)$, $a=n_1/d$, $b=n_2/d$. The continuous group homomorphismmaps this centralizer onto the $d$th roots of unity. Its kernel is the identity component of a Lie group, parametrized injectively by $e^{i\theta }\mapsto (e^{ib\theta },e^{-ia\theta })$. Coprimality of $a,b$ proves injectivity; a Bezout identity proves that this parametrization exhausts the kernel. Extending the primitive integer row $(a,b)$ to a determinant-one integer matrix gives coordinates in which the constraint is simply $t^d=1$. ConsequentlyFor a concrete counterexample, take $n_1=n_2=2$. The constraint is $(zw{)}^2=1$, giving two separate circles $zw=1$ and $zw=-1$. The element $\mathrm{diag}(I_2,-I_2)$ lies in the second component, so this centralizer cannot be isomorphic as a Lie group to the connected circle group.

For the intended specialization of this centralizer of two special unitary blocks, $d=\gcd (2,3)=1$ and therefore the full centralizer really is a circle group:We use this primitive integer U(1) charge normalization below. If exactly one $n_i$ equals $1$ and the other is at least $2$, the same block-scalar formula still holds. If $n_1=n_2=1$, both factors are trivial and the centralizer is all of SU(2), another exception to the printed claim.