How can one "see" the Hopf fibration in the space of lattices in the plane?

Question

This question is inspired from Etienne Ghys's talk on Knots and Dynamics from ICM 2006.

The map $L \mapsto (G_4(L), G_6(L))$ gives a bijection between all lattices $L\subset \mathbb{C}$ (including the degenerate ones) and $\mathbb{C}^2 - \{0\}$. Here $G_4$ and $G_6$ are the Eisenstein series of a lattice,

$$ G_n (L) = \sum_{\omega \in L, \omega \neq 0} \frac{1}{\omega^n}. $$

By scaling the lattice $L \mapsto t L$ we can arrange that these two numbers satisfy $|z_1|^2 + |z_2|^2 = 1$; in other words we have that

$$ \{ \mbox{lattices in } \mathbb{C} \mbox{ up to rescaling} \} \cong S^3. $$

Now $S^3$ carries a nice action of $S^1$ given by sending $(z_1, z_2) \mapsto (e^{i \theta}z_1, e^{i \theta}z_2)$, the quotient being $S^2$; this is the Hopf fibration $S^3 \rightarrow S^2$.

Since the collection of lattices up to rescaling identifies with $S^3$ in such a nice way, it is tempting to try and see this action of $S^1$ at the level of lattices. It would be nice if it were to correspond to rotation of the lattice! But alas, it does not. Firstly, it can't, because the action of $S^1$ on $S^3$ is free, while rotating a lattice might `click' it back into itself before one has rotated a full rotation. In fact we see that if we rotate the lattice via $L\mapsto e^{i\theta} L$, we find that the invariants change as $$ (G_4(L), G_6(L)) \mapsto (e^{-4i\theta}G_4(L), e^{-6i\theta}G_6(L)) $$ which is not the behaviour we are looking for.

So what does the action of $S^1$ on $S^3$ correspond to in the space of lattices up to rescaling? In other words, what is $L'$ in terms of $L$ if

$$ (G_4(L'), G_6(L')) = (e^{i \theta} G_4(L), e^{i \theta} G_6(L))? $$

Answer 1

The circle action $(z,w) \mapsto (e^{2i\theta}z, e^{3i\theta}w)$ has all orbits, except two, isotopic to the trefoil.

I'm guessing that the exceptional orbits correspond to the square lattice and the hexagonal lattice. Further, I'm thinking that the degenerate lattices land on a single orbit, proving that the "space of lattices" is identical to the trefoil complement.

Ah - looking at Ghys' lecture, I see that he says all of this. Well, it is still pretty cool.

Edit: However, this tells you that there are Hopf orbits that cross the set of degenerate lattices exactly twice. This makes a nice answer to your question unlikely?