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

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