Product of subgroups of $SU(8)$ algebraic set?

Question

Consider the special unitary group SU(8) acting on $\mathbb{C}^8\stackrel{\sim}{=}(\mathbb{C}^2)^{\otimes 3}$.

In particular, I am interested in the two subgroups $G_1=\mathrm{id}_{\mathbb{C}^2}\otimes SU(4)$ and $G_2=SU(4)\otimes \mathrm{id}_{\mathbb{C}^2}$. The product $G_1G_2$ is not itself a subgroup and I know that it is a semialgebraic set by the Tarski-Seidenberg theorem and closed in the Lie-group topology. What I would like to know is whether $G_1G_2$ is a (real) algebraic set. I suspect that this is not the case but I cannot prove it.

Answer 1

Yes, $G_1G_2\subset\mathrm{SU}(8)$ is an algebraic set. Here is the argument:

Let $G_1{\times}G_2$ act on $\mathrm{SU}(8)\subset\mathrm{End}(\mathbb{C}^8)\simeq\mathbb{C}^{64}$ by the rule $(g_1,g_2)\cdot h = g_1hg_2^{-1}$.

Then $G_1G_2\subset \mathrm{SU}(8)\subset\mathrm{End}(\mathbb{C}^8)\simeq\mathbb{C}^{64}$ is the orbit of $I_8$ under this action. In particular, it is a smooth submanifold of $\mathrm{SU}(8)$ (of dimension $\dim(G_1)+\dim(G_2)-\dim(G_1{\cap}G_2) = 27$).

Since $G_1{\times}G_2$ is compact, we know that the algebra of $G_1{\times}G_2$-invariant polynomials on $\mathbb{C}^{64}$ separates $G_1{\times}G_2$-orbits, so there exists a finite set $\rho_1,\rho_2,\cdots,\rho_m$ of $G_1{\times}G_2$-invariant (real-valued) polynomials on $\mathbb{C}^{64}$ such that $$ G_1G_2 = \{h\in \mathbb{C}^{64}\ |\ \rho_1(h) = \cdots =\rho_m(h) = 0\ \}. $$ Thus, $G_1G_2$ is an algebraic set.