# Countable union of non Zariski-dense homomorphisms

Let $F_k$ be a free group in $k>1$ letters, and $G$ a semi-simple algebraic group defined over reals $\mathbb{R}$. Consider the representation variety Hom$(F_k,G(\mathbb{R}))$. The points of this variety are the homomorphisms $\phi: F_k \to G(\mathbb{R})$. Consider the set of homomorphisms with non Zariski-dense image, i.e. those homomorphisms $\overline{\phi(F_k)}$ is a proper subgroup of $G(\mathbb{R})$, where the closure is considered in Zariski topology. Denote the subset of such homomorphisms by $\mathcal{F}$. My question is whether $\mathcal{F}$ is Zariksi-dense in Hom$(F_k,G(\mathbb{R}))$ ?

• For $k=1$ and for $G=\textbf{SL}_{2,\mathbb{R}}$, it appears to me that your subset of $\text{Hom}(F_1,G)=G$ contains an open subset of the set of real points that is Zariski dense. An element of $\textbf{SL}_{2,\mathbb{R}}$ that is diagonalizable over $\mathbb{R}$ is contained in a maximal torus, and this is a proper closed subgroup of $G$. Feb 17 '17 at 13:43
• @JasonStarr for $k=1$ every homomrphism has non-Zariski dense image, so the answer is obviously "yes". In fact, the answer is "yes" in general, as the set of Zariski dense homomorphisms is Zariski open - no need to go to a countable union. Feb 17 '17 at 14:00
• @UriBader. Perhaps you and I are reading the question differently. According to what I read, the OP wants there to exist a countable collection of closed subvarieties of $G$, each of which is a proper subset of $G$, such that $\mathcal{F}$ equals the union of these proper subsets. Yet, for $k=1$, the subset $\mathcal{F}$ appears to equal all of $G$. So $\mathcal{F}$ cannot equal a countable union of closed subsets of $G$ that are proper subsets. Feb 17 '17 at 14:46
• Sorry for the confusion. Perhaps I should write $G(\mathbb{R})$. I agree with @Jason Starr. Feb 17 '17 at 14:55
• Please note the change: $k > 1$. Feb 17 '17 at 14:59

Yes: actually $\mathcal{F}$ is Zariski-closed in $G^k$. (And since $\mathcal{F}\neq G^k$ as soon as $k\ge 1$ and $G\neq\{1\}$, we deduce in this case that $\mathcal{F}$ is not Zariski-dense.)

All this can be performed over an algebraic closure, so in the following I never suppose anything to be defined over the reals.

Let $(V,\pi)$ be an irreducible representation of $G$ (of dimension $d_V$). Let $U_V$ be the set of $(g_1,\dots,g_k)\in G^k$ acting irreducibly on $V$. Then it means that the subalgebra generated by $\pi(g_1),\dots,\pi(g_k)$ contains $d_V^2$ linearly independent elements. This is a Zariski open condition.

By Chevalley's theorem, every proper subgroup is a point stabilizer in some representation. We perform this with every maximal Zariski-closed subgroup: these are finitely up to conjugacy (indeed they are either parabolic, or reductive; in the second case this means the normalizer of a semisimple subgroup, and there are finitely many semisimple subgroups up to conjugacy.) We thus get representations $V_1,\dots,V_n$ corresponding to maximal subgroups $M_1,\dots,M_n$. A subgroup contained in a conjugate of $M_i$ is non-irreducible on $V_i$. Hence if $(g_1,\dots,g_k)$ is in $U=\bigcap U_{V_i}$, then it generates a Zariski-dense subgroup and conversely being outside $U_{V_i}$ implies failure of Zariski-density. Thus $\mathcal{F}$ is the complement of $U$, and thus is Zariski-closed.

For $k\ge 2$ and $G\neq 1$ there are indeed Zariski-dense 2-generated subgroups and in this case $\mathcal{F}\neq G^k$, so $\mathcal{F}$ is not Zariski-dense.

• "reductive also if I remember correctly" is the problem I faced. In fact much can be said. Maximal proper closed subgroups are either parabolic or normalizer of semisimple subgroups or nomalizer of maximal torus. Parablic is fine due to compactness ( I am being vague here .I hope you don't mind.) As for the latter two types, I am stuck. Because, for instance, the union of conjugates of maximal torus is dense, for semisimple elements are dense in $G$. Feb 17 '17 at 18:18
• @user49908 every reductive subgroup that is maximal is the normalizer of its derived subgroup, a semisimple subgroup. A semisimple group has finitely many conjugacy classes of semisimple subgroups (I don't have a definitive argument but I'm pretty sure).
– YCor
Feb 17 '17 at 18:37
• Yes. Right. Thats a theorem due to Richardon. Feb 17 '17 at 18:38