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.



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



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> **Proposition.** Equivalent statements:
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> (i) $G$ acts irreducibly on $\mathbf{C}^2$;
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> (ii) $H$ acts irreducibly on $\mathbf{C}^2$;
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> (iii) $G$ fixes no point on $\mathbb{P}^1\_\mathbf{C}$;
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> (iv) $H$ fixes no point on $\mathbb{P}^1\_\mathbf{C}$;
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> (v) One of the following holds:
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> $\quad$(a) $H$ (and hence $G$) is finite and non-abelian;
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> $\quad$(b) $H$ is conjugate to the monomial group, made up of diagonal matrices and anti-diagonal matrices with determinant 1;
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> $\quad$(c) $H=\mathrm{SL}\_2(\mathbf{C})$ (i.e., $G$ is Zariski-dense).
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> 
> 


*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. $\Box$


