# Actions of the discrete Heisenberg group by formal power series of two variables

I am interested in faithful actions of the discrete Heisenberg group $H$ by smooth diffeomorphisms of a surface $S$, that is, 1-1 homomorphisms $\phi \colon H \to \text{Diff}^{\infty}(S)$.

We say $p \in S$ is a fixed point if, for every $g \in H$, $\phi(g)(p) = p$. In this case, picking coordinates about $p$, the derivative gives us an action $H \to GL_2(\mathbb{R})$.

Let $G$ be the following group: $G$ is the set of pairs of formal power series $(\sum_{i, j} a_{i, j}x^iy^j, \sum_{i, j} b_{i, j}x^iy^j)$ such that $a_{0,0} = b_{0,0} = 0$, and the determinant $a_{1,0}b_{0,1} - a_{0,1}b_{1,0} \neq 0$. This forms a group under composition.

If we keep track of higher derivatives for the action $\phi$ at the fixed point $p$, we get an action $\psi \colon H \to G$. We have thus turned the analytic problem of understanding homomorphisms $\phi \colon H \to \text{Diff}^{\infty}(S)$ into the algebraic problem of understanding homomorphisms $\psi \colon H \to G$.

I know an example of an injective homomorphism $\psi \colon H \to G$. Ideally, I would like to classify (up to conjugacy) all possible injective homomorphisms $\psi \colon H \to G$. Does anyone know something about the algebra of the group $G$, which might help me move in that direction?

• Just a remark: the only solvable subgroups of the mapping class group are virtually abelian. So you may assume that the image of your group in $Diff(S)$ must have center which is isotopic to the identity, since the image in $Mod(S)$ will be abelian. – Ian Agol Jul 1 '12 at 22:21
• That's true; thanks for your remark. – Kiran Parkhe Jul 10 '12 at 14:27