How does the direct product of groups apply to matrices (of Lie groups)?

Question

So having two groups $H$ and $G$ the direct product $H \times G$ is the set of ordered pairs $(h,g)$, where $h\in H$ and $g\in G$, and multiplication is carried out componentwise, i.e. $(h_1,g_1)(h_2,g_2)=(h_1 h_2,g_1g_2)$, such that the direct product is again a group.

As for Matrix Lie groups, the elements can be represented by matrices which act on vectors for example. How is this action on vectors carried out in the direct product? Take as an example the direct product of $SU(2) \times U(1)$ (which is not quite, but almost isomorphic to $U(2)$). Using the defintion I would write down pairs of matrices $(S,U), S\in SU(2), U \in U(1)$. But how do these pairs act on vectors now? And vectors of which dimension? If both groups are represented in the fundamental representation, do these pairs act on three dimensional vectors $(s_1,s_2,u)^t$?

Answer 1

If we're looking at complex representations, then every irreducible representation of $G\times H$ is the tensor product of an irrep of $G$ and an irrep of $H$, see e.g. this math.SE question. So the fundamental representation of $\mathrm{SU}(2)\times\mathrm{U}(1)$ is two-dimensional and $(s,u)\in\mathrm{SU}(2)\times\mathrm{U}(1)$ acts simply with the $s$ acting on the two-dimensional vector as in the fundamental of $\mathrm{SU}(2)$ and the $u$ multiplying the vector with a phase.

Note that the action you likely had in mind when you asked about three-dimensional vectors $(s_1,s_2,u)^T$ where the $\mathrm{SU}(2)$ acts on the first two elements and the $\mathrm{U}(1)$ on the third is not an irreducible representation of $\mathrm{SU}(2)\times\mathrm{U}(1)$ - the two subspaces spanned by vectors of the form $(s_1,s_2,0)^T$ and $(0,0,u)^T$ are irreducible subrepresentations.