# How can one “see” the Hopf fibration in the space of lattices in the plane?

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

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

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I guess you mean lattices in $\mathbb C$ not in $\mathbb C^2$. – Torsten Ekedahl Oct 17 '10 at 13:17
Thanks, yes I did, I changed it now. – Bruce Bartlett Oct 17 '10 at 19:16
No, you didn't :-) – Robin Chapman Oct 17 '10 at 19:19
Line 3 - shouldn't $\mathbb{C}−0$ be $\mathbb{C}^2−0$? – Sam Nead Oct 17 '10 at 20:31
Ghys's work starts from the fact that non-degenerate lattices up to scale form the complement of a trefoil knot in $S^3$. That knot is not preserved by the Hopf $S^1$-action, which instead traces out unknots. This seems inauspicious for a nice lattice interpretation of the circle-action... – Tim Perutz Oct 17 '10 at 20:32

The circle action $(z,w) \mapsto (e^{2i\theta}z, e^{3i\theta}w)$ has all orbits, except two, isotopic to the trefoil.
Yes, the $S^1$ action on the space of lattices given by rotating the lattice is an interesting action to think about. But it's not the one I'm interested in... I'm trying to figure out the $S^1$ action on the space of lattices that corresponds to the $S^1$ action on $S^3$. We know one exists... what is it? Tim Perutz above suggests that there simply won't be a "nice" lattice interpretation... but I'm still holding out hope. – Bruce Bartlett Oct 18 '10 at 19:05