A lattice in $\mathbb C$ is an abelian subgroup of $\mathbb C$ isomorphic to $\mathbb Z^2$ that generates $\mathbb C$ over $\mathbb R$. Define on the space $M$ of lattices in $\mathbb C$ a structure of smooth manifold as follows. Define a transitive action of $GL(2,\mathbb R)$ on $M$, find a stabilizer of a certain $x \in M$ and verify that it is a closed Lie subgroup. What is the dimension of $M$.
I do not really see how the action need to be defined. Any explanation or clarification of this would be helpful.
Considering the basis $\{1,i\}$ of $\mathbb{C}$, a lattice $\Gamma$ can be obtained as the free abelian group $\left\langle\{A_\Gamma(1),A_\Gamma(i)\} \right\rangle_{\mathbb{Z}}$ generated by the images of $1$ and $i$ through a matrix $A_\Gamma\in GL_2(\mathbb{R})$. The stabiliser of a lattice $\Gamma$ is the group $ GL_2(\mathbb{Z})$. The latter group acts on $GL_2(\mathbb{R})$ via the left action $$ GL_2(\mathbb{Z})\times GL_2(\mathbb{R})\rightarrow GL_2(\mathbb{R}) \qquad (Z,A)\mapsto Z\,A. $$ The obtained quotient $GL_2(\mathbb{R})/GL_2(\mathbb{Z})$ is said "set of the invariant geometric lattices" and it is the $M$ you are looking for. It is a manifold of dimension $n^2$ giving in our case $\dim(M)=4$.