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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
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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

1 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?

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This circle action is a Seifert fibration and the exceptional orbits clearly are the square and hexagonal lattices. They are circles at two ends of the sphere, and the tori in between are fibered by trefoils. So, the most important lesson is that this is another fibration which is really just as interesting as the Hopf fibration, but a tad more complicated. You even get that the quotient is a sphere. (More precisely, an orbifold decoration of a sphere.) –  Greg Kuperberg Oct 17 '10 at 21:32
    
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
    
By the way, earlier this year I got an AIMS essay student Ihechukwu Chinyere to give a computer animation of the fibers of this "modular fibration", see youtube.com/watch?v=eqeqbjec97w. –  Bruce Bartlett Oct 1 '12 at 8:06

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