4
$\begingroup$

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.

$\endgroup$
  • 4
    $\begingroup$ 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? $\endgroup$ – Qiaochu Yuan Dec 4 '17 at 19:26
  • 5
    $\begingroup$ What does it mean for a group to be irreducible? $\endgroup$ – R. van Dobben de Bruyn Dec 4 '17 at 20:43
  • 3
    $\begingroup$ 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 $\mathbb{C}^{\oplus 2}$. $\endgroup$ – Jason Starr Dec 4 '17 at 21:24
  • 4
    $\begingroup$ 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. $\endgroup$ – YCor Dec 4 '17 at 23:06
  • 3
    $\begingroup$ 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. $\endgroup$ – Jim Humphreys Dec 6 '17 at 2:04
4
$\begingroup$

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. $\Box$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A Zariski dense subgroup has no fixed point at all in $\overline{\mathbb{H}^3}$. $\endgroup$ – YCor Dec 4 '17 at 23:01
  • $\begingroup$ @YCor limit points, fixed. $\endgroup$ – Igor Rivin Dec 5 '17 at 0:26
  • 1
    $\begingroup$ Also an invariant geodesic means 2 limits points at infinitely, but not always fixed: e.g. the monomial matrices switch the two limit points. $\endgroup$ – YCor Dec 5 '17 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.