Is $SU(2)\otimes SU(3)$ a subgroup of $SU(4)$?

Question

This should not be a complex question. Why do we hear that $SU(2)\otimes SU(3)$ is a subgroup of $SU(5)$, but not $SU(4)$? The dimension of $SU(2)\otimes SU(3)$ is $3+8=11$, and dimension of $SU(4)$ is $15$, so $SU(2)\otimes SU(3)$ should also be a subgroup of $SU(4)$, shouldn't it?

Answer 1

I think that usually $\mathrm{SU}(2)\times\mathrm{SU}(3)$ is the direct product, where we can interpret its elements concretely as block-diagonal matrices with blocks from $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$, and $\mathrm{SU}(2)\otimes\mathrm{SU}(3)$ is the pointwise Kronecker product of $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$. This automatically makes $\mathrm{SU}(2)\times\mathrm{SU}(3)$ a subgroup of $\mathrm{SU}(5)$ and makes $\mathrm{SU}(2)\otimes\mathrm{SU}(3)$ a subgroup of $\mathrm{SU}(6)$. In this case it happens that they are isomorphic as Lie groups, but I don't see why we would be talking about $\otimes$ instead of $\times$. In general if $G$ and $H$ are matrix Lie groups, then the homomorphism $G\times H\to G\otimes H$ is onto with a kernel of the form $\{(zI,z^{-1}I)\mid z\in\mathbb{C}^\times\}$, which in this case is trivial since the scalar matrices in $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$ use scalar square and cube roots of unity respectively.

Dimension is a straightforward way to prove a group $H$ cannot be embedded in $G$; this happens often enough that it's a good idea to check dimensions. But if you find out $\dim H<\dim G$, you still pretty much have no idea if $H$ can be a subgroup of $H$. For instance, $S^3$ is not a subgroup of $S^1\times\cdots\times S^1$ no matter how high the dimension of the torus is, since $S^3$ is abelian while $(S^1)^n$ isn't. Put pointedly, if you were to start writing down all pairs of Lie groups (up to isomorphism) where $\dim H<\dim G$, you would soon find out even among these pairs it's still very rare for $H$ to be isomorphic to a subgroup of $G$ if $\dim H$ is not vastly smaller.

It is not possible to embed $\mathrm{SU}(2)\times\mathrm{SU}(3)$ in $\mathrm{GL}_4\mathbb{C}$. If it were, we could embed it into the maximal compact subgroup $\mathrm{U}(4)$, which we can show is impossible. The smallest complex irreps of $\mathrm{SU}(3)$ have dimensions $1,3,6$ (see the dimension formula $d(p,q)$ here), so $\mathbb{C}^4\cong\mathbb{C}^3\oplus\mathbb{C}$ as $\mathrm{SU}(3)$-reps ($\mathbb{C}^3$ being either the standard rep or its conjugate, and $\mathbb{C}$ being trivial). If the $\mathrm{SU}(2)$ subgroup of $\mathrm{U}(4)$ commutes with $\mathrm{SU}(3)$ then it must preserve this decomposition, and act trivially on $\mathbb{C}$, effectively making it a subgroup of $\mathrm{U}(3)$, but $\mathrm{SU}(2)\times\mathrm{SU}(3)$ cannot be embedded in $\mathrm{U}(3)$ because of dimensions.