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I'm reading Terry Gannon's Moonshine Beyond the Monster, and in section 2.4.3 he hints at (but does not explicitly describe) a way to extend the action of $SL_2(\mathbb{Z})$ on modular forms to an action of the braid group $B_3$. Here is what he says about this action:

First, we lift modular forms $f : \mathbb{H} \to \mathbb{C}$ to functions $\phi_f : SL_2(\mathbb{R}) \to \mathbb{C}$ as follows: let

$$\phi_f \left( \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right] \right) = f \left( \frac{ai + b}{ci + d} \right) (ci + d)^{-k}.$$

Thinking of $f$ as a function on $SL_2(\mathbb{R})$ invariant under $SO_2(\mathbb{R})$, we have now exchanged invariance under $SO_2(\mathbb{R})$ for invariance under $SL_2(\mathbb{Z})$. ($SO_2(\mathbb{R})$ now acts by the character corresponding to $k$.) In moduli space terms, an element $g \in SL_2(\mathbb{R})$ can be identified with the elliptic curve $\mathbb{C}/\Lambda$ where $\Lambda$ has basis the first and second columns (say) of $g$, and $\phi_f$ is a function on this space invariant under change of basis but covariant under rotation.

Second, $SL_2(\mathbb{R})$ admits a universal cover $\widetilde{SL_2(\mathbb{R})}$ in which the universal central extension $B_3$ of $SL_2(\mathbb{Z})$ sits as a discrete subgroup. Unfortunately, Gannon doesn't give an explicit description of this universal cover (presumably because it's somewhat complicated).

Question: What is a good explicit description of this universal cover and of how $B_3$ sits in it (hence of how it acts on modular forms)? In particular, does it have a moduli-theoretic interpretation related to the description of $B_3$ as the fundamental group of the space $C_3$ of unordered triplets of distinct points in $\mathbb{C}$? (These triplets $(a, b, c$) can, of course, be identified with elliptic curves $y^2 = 4(x - a)(x - b)(x - c)$.)

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4 Answers 4

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You can think of the space of positively oriented covolume-one bases of $\mathbb{R}^2$ as a torsor under $SL_2(\mathbb{R})$, i.e., it is a manifold with a simply transitive action of the group. If you choose a preferred basepoint, such as $(\mathbf{i},\mathbf{j})$, you get an identification with the group. You can think of elements of the universal cover as positively oriented area-one bases equipped with a homotopy class of paths in $\mathbb{R}^2 - \{0\}$ from $\mathbf{i}$ to the first element of the basis. There is a reasonably straightforward composition law that involves multiplying matrices and composing paths.

Gannon's description of lifting to $SL_2(\mathbb{R})$ implies the lifts of even weight modular forms to $\widetilde{SL_2(\mathbb{R})}$ are invariant under the action of $B_3$. In particular, classical modular forms are rather boring from the perspective of the braid group. In order to detect the central extension, you need to consider modular forms of fractional weight. When the weight is not an integer, you don't get an action of $SO(2)$, but instead, an action of the universal cover $\mathbb{R}$. The resulting action of $B_3$ is necessarily nontrivial, since the restriction to the center is by a nontrivial character of $\mathbb{Z}$. I don't know many explicit constructions of fractional weight forms, other than half-integer weight forms like $\eta$ and theta functions, and vector-valued forms constructed from them. However, you can generate a family of examples by choosing powers of the cusp form $\Delta$, which admits a logarithm since it is globally regular and nonvanishing.

My understanding of the explicit relationship to configurations of points and elliptic curves is the following: Given a path of triples of distinct points $(a_1(t),a_2(t),a_3(t))$, we get a path on the space of elliptic curves of the form $y^2 = (x-a_1(t))(x-a_2(t))(x-a_3(t))$, but this will throw away an action of real translations and dilations (irrelevant) together with the central extension and the circle action (important). If we just look at the isomorphism types (i.e., the $j$-invariants) of the curves, we get a path through the quotient of the upper half plane by $SL_2(\mathbb{Z})$. We need to choose a discrete structure to remove the quotient by the center, and a one dimensional continuous structure to promote our space to three dimensions. To retain the angular information that we lost by passing to elliptic curve isomorphism, we fix a tangent direction at infinity to remove rotational symmetry. This tangent direction is manifested when we choose our discrete structure: a homotopy class of nonintersecting paths from the three points to infinity, because we demand that the paths asymptotically approach infinity in that direction. The elliptic curve is a double cover of the complex projective line, ramified at the three points and infinity. We can choose once and for all a uniform convention for lifting the three paths to primitive homology cycles, such that any pair generates $H_1$, and one cycle is the sum of the other two, so those two form a preferred basis. To get the parametrization of $\widetilde{SL_2(\mathbb{R})}$ from the first paragraph, we choose a preferred basepoint configuration of three points with paths from infinity and asymptotic direction, and consider a triple $(a_1(t),a_2(t),a_3(t))$ that starts at the basepoint. By uniqueness of homotopy lifting, we get a family of tangent vectors at infinity together with a family of elliptic curves with oriented bases of homology. By rescaling the bases in $\mathbb{R}^2$ to have unit covolume, we get the parametrization in the first paragraph.

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    $\begingroup$ There is a straightforward way to pass between the description of $\widetilde{SL_2(\mathbb{R})}$ in my first paragraph and Robin Chapman's description, but Robin's description is not only more concise, but lends itself to nice descriptions of finite-order covers like the metaplectic group (using branches of $n$th roots of $cz+d$). $\endgroup$
    – S. Carnahan
    Apr 4, 2010 at 19:49
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$\widetilde{\textit{SL}_2(\mathbb{R})}$ is not so complicated, but one of the best descriptions is just that. It has no faithful finite-dimensional representations, which makes things a little tricky. It's easy to see that $\textit{SL}_2(\mathbb{R})$ is topologically a solid torus, so it does have a central extension by $\mathbb{Z}$.

Another description of this central extension is to think of $PSL_2(\mathbb{R})$ acting on the hyperbolic plane. Then $\widetilde{SL_2(\mathbb{R})}$ is pairs $(g,\phi)$ where $g \in PSL_2(\mathbb{R})$ and $\phi:\mathbb{R}\to\mathbb{R}$ is a map that induces the map from $S^1$ to $S^1$ induced by $g$ on the circle at infinity. (Note that this goes by a quotient from $SL_2(\mathbb{R})$ and then centrally extends back again.)

For the relation to $B_3$, you can think of $PSL_2(\mathbb{Z})$ as the mapping class group of the 4-times punctured sphere, with one puncture (at infinity) distinguished and required to go to itself. To go from there to its central extension $B_3$, you remove a disk (rather than a point) around infinity, and only look at isotopies that fix that boundary, which is similar to what happened at the circle at infinity above.

More generally, all mapping class groups have a unique central extension; if there's a distinguished point, you can do it in the braid-like way (looking at isotopies that fix a boundary circle), while if there isn't, you do it by lifting the map of the surface to a map from $\mathbb{H}^2$ to itself (not an isometry) and then lifting the action on the circle at infinity to a map from $\mathbb{R}$ to itself. (These are equivalent when they are both defined.)

(I still feel like it ought to be possible to make the connection crisper.)

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I'm afraid I know nothing about $B_3$ but here is my favourite construction of $\widetilde{\mathrm{SL}}_2(\mathbb{R})$.

The elements of $\widetilde{\mathrm{SL}}_2(\mathbb{R})$ are pairs $(A,f)$ where $$A=\left(\begin{array}{cc} a&b\\\ c&d \end{array}\right)\in\mathrm{SL}_2(\mathbb{R})$$ and $f$ is a branch of $\log(cz+d)$ for $z$ in the upper half plane. We compose these as follows: $(A,f)(B,g)=(AB,h)$ where $$h(z)=f(Bz)+g(z)$$ and of course $Bz$ denotes the usual linear fractional transformation action of $B$ on $z$ in the upper half-plane.

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This is only a partial answer to your question, concerning how to visualise the universal cover of $\mathrm{SL}(2,\mathbb{R})$.

A nice picture of this group (together with a proof that it is not a matrix group) is given in Graeme Segal's lectures in the book Lectures on Lie groups and Lie algebras. Alas, the Google Books preview does not cover Segal's lectures and I don't have the book here with me in order to scan the nice picture he draws.

The bi-invariant metric on $\mathrm{SL}(2,\mathbb{R})$ is lorentzian and has constant negative sectional curvature. Its universal cover is the lorentzian analogue of hyperbolic space and in the Physics literature it goes by the name of (three-dimensional) anti de Sitter spacetime $\mathrm{AdS}_3$ and has been studied a lot.

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