Let $F_k$ be a free group in $k>1$ letters, and $G$ a semisimple 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 Zariskidense 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 Zariksidense in Hom$(F_k,G(\mathbb{R}))$ ?

$\begingroup$ 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 $G$. $\endgroup$– Jason StarrFeb 17, 2017 at 13:43

$\begingroup$ @JasonStarr for $k=1$ every homomrphism has nonZariski 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. $\endgroup$– Uri BaderFeb 17, 2017 at 14:00

$\begingroup$ @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. $\endgroup$– Jason StarrFeb 17, 2017 at 14:46

$\begingroup$ Sorry for the confusion. Perhaps I should write $G(\mathbb{R})$. I agree with @Jason Starr. $\endgroup$– VanyaFeb 17, 2017 at 14:55

1$\begingroup$ Please note the change: $k > 1 $. $\endgroup$– VanyaFeb 17, 2017 at 14:59
1 Answer
Yes: actually $\mathcal{F}$ is Zariskiclosed 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 Zariskidense.)
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 Zariskiclosed 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 nonirreducible on $V_i$. Hence if $(g_1,\dots,g_k)$ is in $U=\bigcap U_{V_i}$, then it generates a Zariskidense subgroup and conversely being outside $U_{V_i}$ implies failure of Zariskidensity. Thus $\mathcal{F}$ is the complement of $U$, and thus is Zariskiclosed.
For $k\ge 2$ and $G\neq 1$ there are indeed Zariskidense 2generated subgroups and in this case $\mathcal{F}\neq G^k$, so $\mathcal{F}$ is not Zariskidense.

$\begingroup$ "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 $G$. $\endgroup$– VanyaFeb 17, 2017 at 18:18

$\begingroup$ @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). $\endgroup$– YCorFeb 17, 2017 at 18:37

$\begingroup$ Yes. Right. Thats a theorem due to Richardon. $\endgroup$– VanyaFeb 17, 2017 at 18:38