# Are irreducible subgroups Zariski-dense?

A subgroup $H$ of an algebraic group $G$ is said to be Zariski-dense if its Zariski closure is all of $G$ (or alternatively, if every polynomial which vanishes on all elements of $H$ vanishes identically).

My question: Is any irreducible subgroup of $SL(2,\mathbb{C})$ Zariski-dense? (It's easy to see that the converse is true).

Remark: The original motivation for this question is that I have seen stated in two papers that a subgroup of $SL(2, \mathbb{C})$ is Zariski dense if and only if its natural action on $\mathbb{P}^1$ has no fixed points (which is equivalent to this subgroup being irreducible). However, as explained in the answers and comments below, this statement is in general false.

• This doesn't seem true to me. $SL_2(\mathbb{C})$ has lots of finite irreducible subgroups, for example $\widetilde{A}_5$, which are certainly not Zariski dense. Am I misunderstanding the meaning of irreducible here? – Qiaochu Yuan Dec 4 '17 at 19:26
• What does it mean for a group to be irreducible? – R. van Dobben de Bruyn Dec 4 '17 at 20:43
• Perhaps the correct hypothesis should be that a subgroup $G$ of $\textbf{SL}_2(\mathbb{C})$ is Zariski dense if and only if every finite index subgroup $H$ of $G$ acts irreducibly on $\mathbb{C}^{\oplus 2}$. – Jason Starr Dec 4 '17 at 21:24
• A subgroup of $SL_2$ is Zariski-dense iff it has no finite orbit on $\mathbb{P}^1$. For a subgroup with Zariski-connected closure, this is equivalent to the absence of fixed point. But suitable finite subgroups (as already mentioned), or the group of monomial matrices (normalizer of diagonal matrices), are irreducible. – YCor Dec 4 '17 at 23:06
• Concerning the unfortunate juxtaposition of the terms "irreducible" and "Zariski-dense", it should be clarified that in traditional algebraic geometry the first term has a topological meaning unrelated to its use here associated with irreducible linear representations. – Jim Humphreys Dec 6 '17 at 2:04

To summarize the discussion. Let $G$ be a subgroup of $\mathrm{SL}_2(\mathbf{C})$, $H$ its Zariski closure.

Proposition. Equivalent statements:

(i) $G$ acts irreducibly on $\mathbf{C}^2$;

(ii) $H$ acts irreducibly on $\mathbf{C}^2$;

(iii) $G$ fixes no point on $\mathbb{P}^1_\mathbf{C}$;

(iv) $H$ fixes no point on $\mathbb{P}^1_\mathbf{C}$;

(v) One of the following holds:

$\quad$(a) $H$ (and hence $G$) is finite and non-abelian;

$\quad$(b) $H$ is conjugate to the monomial group, made up of diagonal matrices and anti-diagonal matrices with determinant 1;

$\quad$(c) $H=\mathrm{SL}_2(\mathbf{C})$ (i.e., $G$ is Zariski-dense).

Proof. The equivalence between (i),(ii),(iii),(iv) is trivial (although that iii/iv implies i/ii is specific to dimension 2).

For a finite group in dimension 2 in characteristic zero, non-irreducibility implies that the action is diagonalizable; hence (a) implies (i). The only fixed points by the group of diagonal matrices with determinant 1 are the two coordinate axes; then are switched by the monomial group, hence (b) implies (iv), and (c) implies (iv) follows (and is clear anyway).

Conversely, suppose that none of (a),(b),(c) holds. If $G$ is finite, this means that $G$ is abelian, its irreducible representations have dimension 1 and the negation of (i) follows. Otherwise, we discuss on the Lie algebra of $H$. If it is conjugate to the upper unipotent or upper triangular subalgebra, then the corresponding connected group fixes a unique point, which is then fixed by $H$ and we obtain the negation of (ii). Since (c) does not hold, it is then conjugate to the subalgebra of diagonal matrices. Hence the Zariski connected component of $H$ consists of the group $D$ diagonal matrices with determinant 1; it has index two in its normalizer (monomial matrices). Since (b) does not hold, we deduce that $H=D$, and again is abelian and does not act irreducibly. $\Box$

I believe that a subgroup of $SL(2, C)$ (viewed as a complex algebraic group, it is different if it is viewed as a real group) is Zariski dense if and only if it is non-elementary, so in other words it has more than two limit points in $\overline{\mathbb{H}^3}$ One fixed point in the interior of $\mathbb{H}^3$ corresponds to the finite case (as in Qiaochu's comment), one fixed point on the boundary is a purely parabolic group, two fixed points (on the sphere at infinity) corresponds to an invariant geodesic.

• A Zariski dense subgroup has no fixed point at all in $\overline{\mathbb{H}^3}$. – YCor Dec 4 '17 at 23:01
• @YCor limit points, fixed. – Igor Rivin Dec 5 '17 at 0:26
• Also an invariant geodesic means 2 limits points at infinitely, but not always fixed: e.g. the monomial matrices switch the two limit points. – YCor Dec 5 '17 at 9:10