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.



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


