Let G be an linear algebraic group, and let H be a zariski dense subgroup of G. Then does H have to be a irreducible subgroup of G? Here H being irreducible means that H has no nontrivial invariant subspaces. Assume that G comes with a faithful irreducible representation. For example we can take G=SL(n,R) or SO(n,R). In these cases subgroup H being irrreducible means that when H acts on R^n there is no nontrivial invariant subspace.
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1$\begingroup$ Subspaces of what? Does G come with a faithful irrep, that remains irreducible for H? In which case you want somehow to exclude G = the circle. $\endgroup$– Allen KnutsonCommented Oct 30, 2010 at 2:42
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$\begingroup$ Yes G comes with a faithful irreducible representation. For example we can take G=SL(n,R) or SO(n,R). In these cases subgroup H being irrreducible means when H acts on R^n there is no nontrivial invariant subspaces. $\endgroup$– user9552Commented Oct 30, 2010 at 4:14
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$\begingroup$ @unknown Mixing Lie group and algebraic group language is risky, as when you take matrix groups over the reals as examples: these are better viewed as the real points of certain algebraic groups (or group schemes). The Zariski topology comes into play when you look at polynomial or rational function properties, so "faithful irreducible representation" in the question should refer to a "rational" (algebraic group) representation. $\endgroup$– Jim HumphreysCommented Oct 30, 2010 at 19:12
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2 Answers
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If $G$ is a linear algebraic group, then you can recognize the coordinate algebra of $G$ as the algebra generated by all matrix entries of all algebraic representations of $G$. (In other words, every matrix entry is a function on $G$, they are all regular functions, and they generate the coordinate ring $O(G)$.) This implies that if $H \subseteq G$ is Zariski dense, then the $H$-decomposition of every algebraic representation of $G$ is exactly the same as the $G$-decomposition. Because, if $W \subseteq V$ is an $H$-subrepresentation, then this says that in a basis of $V$ that extends a basis of $W$, certain matrix entries vanish on elements of $H$; then since $H$ is Zariski dense, the same matrix entries vanish on all of $G$. In particular, every $G$-irreducible representation is also $H$-irreducible.
answered Oct 30, 2010 at 6:15
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1$\begingroup$ More simply, perhaps (without referring to coordinate algebras): with $W\subset V$ as above, the stabilizer of $W$ is an algebraic subgroup of $G$, hence equal to $G$ if it contains $H$. $\endgroup$ Commented Nov 1, 2010 at 12:13
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$\begingroup$ Sure, I guess that's another way to explain the same point. $\endgroup$ Commented Nov 1, 2010 at 21:17
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Let me add another perspective. $H = SL_2(\mathbb{Z})$ is Zariski dense in $G=SL_2(\mathbb{R})$, but the restriction of a holomorphic discrete series representation of $G$ to $H$ is never irreducible. This is OK because these are not algebraic reps of $G$. [1] is a good reference about this.
[1] B. Bekka. Square integrable representations, von Neumann algebras and an application to Gabor analysis. J. Fourier Anal. Appl., 10(4):325–349, 2004.
