$\DeclareMathOperator\GL{GL}$Let $\Gamma$ be a finitely generated group. We know that there is an affine scheme $R$ defined over $\mathbb{Q}$ such that $R(k)=\mathrm{Hom}(\Gamma, \GL_N(k))$ for any field extension $k$ of $\mathbb{Q}$. My question is, does there exist a Zariski dense open set $R'$ of the $\mathbb{Q}$-scheme $R$ such that for every field extension $k$ of $\mathbb{Q}$ and for every point $\rho:\Gamma\to \GL_N(k)$ in $R'(k)$, $\rho$ is a reductive representation ?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Does "reductive" mean "completely reducible"? $\endgroup$– LSpiceCommented Feb 14, 2023 at 19:37
-
$\begingroup$ @LSpice yes, "reductive" mean "completely reducible". From the answer by YCor below, subsets of reductive representations might not be Zariski open in the representation variety. $\endgroup$– Higgs-BosonCommented Feb 15, 2023 at 23:56
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
5
No. For instance let $\Gamma$ be the Heisenberg group (two generators $x,y$ commuting with their commutator $z$), and $N=3$. Then I think the set of semisimple representations is not dense.
It follows from the following slight refinement of Jordan's theorem (which I'm pretty sure is true and well-known): for every $N$ there exists $r_N$ such that every Zariski-closed subgroup of $\mathrm{GL}_N(\mathbf{C})$ with abelian connected component, has an abelian subgroup of index $\le r_N$.
-
$\begingroup$ Thanks for your counterexample. I note that there is a theorem which says that irreducible reprentations are Zariski dense in the space of representations $R$ of $\Gamma$. Does your counter-example mean that there is some irreducible component $R'$ in $R$ such that every representation in $R'(\mathbb{C})$ is not reductive? $\endgroup$ Commented Feb 14, 2023 at 18:21
-
3$\begingroup$ @Higgs-Boson: This theorem is known only for some very special classes of groups, free groups and surface groups. $\endgroup$ Commented Feb 14, 2023 at 22:23
-
1$\begingroup$ @Higgs-Boson yes, the counterexample says that every irreducible component containing every injective homomorphism $\Gamma\to\mathrm{GL}_N$ (for every f.g. virtually solvable group $\Gamma$) has an open dense subset consisting of non-semisimple representations. $\endgroup$– YCorCommented Feb 15, 2023 at 0:31
-
-
2$\begingroup$ @LSpice yes, this is correct, thanks for noticing. To restate, fix a finitely generated virtually solvable group. Then in every irreducible component of $\mathrm{Hom}(\Gamma,\mathrm{GL}_N)$ containing a representation $\rho$ with non-virtually-abelian image, there is a dense open subset of such representations. $\endgroup$– YCorCommented Feb 16, 2023 at 10:06