Why is the space of 2D lattices topologically 4D, if Gram matrices are 3D?

Question

There are several good articles on the space of 2D lattices, usually represented using Eisenstein series and the invariants $G_4$ and $G_6$.

Here are some:

• http://www.stevejtrettel.com/the-space-of-lattices.html
• http://www.josleys.com/articles/ams_article/Lorenz3.htm
• https://golem.ph.utexas.edu/category/2014/04/the_modular_flow_on_the_space.html

All these pages show that the space of lattices is 4-dimensional, and if degenerate lattices are included, it is isomorphic to $\Bbb C^2$, with the degenerate lattices being a trefoil knot. This seems very well known and a basic part of the theory of modular forms.

The reasoning in the first article is that the space of lattices is isomorphic to $GL(2,\Bbb R)/GL(2,\Bbb Z)$. Since $GL(2,\Bbb R)$ is a 4D manifold and $GL(2,\Bbb Z)$ is a discrete subgroup, the quotient is also 4D.

My question: Lattices can also be represented with positive-definite Gram matrices. These are $2 \times 2$ symmetric matrices, and hence are only 3D. For any such matrix $G$, if $M$ is a unimodular matrix, the matrix $M' G M$ represents the same lattice, so we can take the quotient and equate all such matrices. Wouldn't we then have a 3D space? Is it somehow isomorphic to the 4D one?