## Central Extensions and Weight One Forms

As mentioned in the comments to the last post, Kevin Buzzard and Alan Lauder have made an extensive computation of weight one modular forms in characteristic zero (see also here). Thinking about what that data might contain, I wondered about the following question: what are the images of the Galois representations associated to the weight one forms of type $A_5$?

Let us take a step back. Consider a projective representation

$\psi: G_{{\mathbf{Q}}} \rightarrow {\mathrm{PGL}}_2({\mathbf{C}})$

with image $A_5$, and assume that it is odd. (That is, complex conjugation has order $2.$) According to Tate, there exists a lift

$\rho: G_{{\mathbf{Q}}} \rightarrow {\mathrm{GL}}_2({\mathbf{C}}).$

This lift is unique up to twisting. Since the Schur multiplier $H_2(A_5,{\mathbf{Z}})$ of $A_5$ is ${\mathbf{Z}}/2{\mathbf{Z}}$, there is a unique minimal lift up to twist whose image is a central extension ${\widetilde{A}_5}$ by a cyclic group $\Delta$ of $2$-power order. Note that $\Delta$ is not trivial, since $A_5$ does not have any two-dimensional representations. If $|\Delta| = 2$, then the determinant of the corresponding 2-dimensional representation of ${\widetilde{A}_5}$ is trivial, which contradicts the assumption that $\psi$ is odd. (Equivalently, there is an obstruction at $\infty$ to lifting to the central extension by ${\mathbf{Z}}/2{\mathbf{Z}}.$) Hence $4$ divides $|\Delta|.$ What is the expected distribution of $\Delta$ as one runs over all odd $A_5$-extensions?

My first guess (without any prior thought or computation) was that this might obey some form of Cohen–Lenstra heuristic, suitably interpreted.

Note that the image of the determinant has order $|\Delta|/2.$ The corresponding determinant representation is a character of ${\mathbf{Q}}$ of $2$-power order. Since ${\mathbf{Q}}$ has trivial class number, the order $|\Delta|/2$ is equal to the maximal ramification degree $e_p$ of this representation over all primes $p.$

Over ${\mathbf{Q}}$, Tate’s lifting theorem has the following stronger form: one may choose a lift $\rho_p$ of $\psi_p := \psi|_{D_p}$ and insist that $\rho|_{I_p} = \rho_p | I_p$; that is, they agree on inertia. This is essentially a consequence of the fact that ${\mathbf{Q}}$ has trivial class group. For convenience, suppose that $\psi$ is unramified at $2$ and $3.$ Suppose that $\psi$ is ramified at $p.$ There are three possibilities:

1. The image of $\psi_p$ at a ramified prime $p$ is cyclic of order 2, 3, or 5.
2. The image of $\psi_p$ at a ramified prime $p$ is $D_6$ or $D_{10}.$
3. The image of $\psi_p$ at a ramified prime $p$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2{\mathbf{Z}}.$

For a fixed $p$, let $\epsilon$ denote the Teichmuller lift of the mod-$p$ cyclotomic character. (Fix an isomorphism of $\mathbf{C}$ with $\overline{\mathbf{Q}}_p$ for all $p.$)

Let us consider the three cases in turn.

In the first case, the image factors through a cyclic quotient. One may thus take $\rho_p$ to be a direct sum which, on inertia, has the shape $\chi \oplus 1$ up to twist. By comparing this to the projective representation, we see that $\chi$ has order 2, 3, or 5, and so, after finding the twist such that the determinant has $2$-power order, we see that $e_p = 1$ or $e_p = 2.$

In the second case, the lift on inertia is (up to twist) of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ Since the order of $\omega_2$ is $p^2 - 1$, the order of the ratio is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 3 or 5. It follows that $r$ is even. Yet the determinant is equal to

$\displaystyle{ \omega^{(p+1)r}_2 = \epsilon^{r}},$

Since $r$ is even, we see that, after twisting, we may take $e_p = 1.$

Finally, in the third case, the lift is of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ We now find that the order of the ratio of these characters is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 2, and the determinant is $\epsilon^r.$ If $r$ is even, then, as above, we may twist so that $e_p = 1.$ Hence, the only way that the image after minimal twist does not have $|\Delta| = 4$ is if we are in this third situation with $p \equiv 1 \mod 4$, with $r$ odd, and then (after twisting) we find that $e_p$ is the largest power of $2$ dividing $p - 1.$

(I confess that I originally forgot the fact that the third possibility could occur, and was only after noticing that this seemed to imply the inverse Galois problem was false thought a little bit more about the possibilities.)

To summarize:

Lemma Assume that $\psi$ is unramified at $2$ and $3$ and has projective image $A_5$, and a lift with image $\widetilde{A}_5$ with minimal kernel. Then order of $\Delta$ is $4$ unless there exists a prime $p \equiv 1 \mod 4$ such that the image of the decomposition group at $p$ under $\psi$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}.$ In this case, we have $\Delta$ to be twice the largest power of $2$ dividing $p -1$ for all such primes $p.$

Let $\Delta(\psi)$ denote the corresponding power of $2.$

We see that $\Delta(\psi)$ is determined by purely local phenomena. This still doesn’t quite answer what the distribution of the extension $\widetilde{A}_5$ will be. However, I imagine that Bhargava style heuristics should certainly be able to predict the ratio of $A_5$ with $\Delta(\psi) = 2^n.$ Does anyone have a sense of how easy this might be to prove? Or, much more modestly, how easy it would be to compute from these quite precise heuristics the exact predicted distribution of the central extensions of $A_5$ coming from weight one modular forms?

(I confess, it is not even obvious to me from this construction how to prove that all central extensions $\widetilde{A}_5$ occur as Galois groups — but I presume this is known, and hopefully one of my readers can provide a reference.)

(According to KB, BTW, all the $A_5$ representations with $N \le 1500$ have $\Delta = 4.$)

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