## Inverse Galois Problem

My favourite group as far as the inverse Galois problem goes is $G = \mathrm{SL}_2(\mathbf{F}_p)$. This is not known to be a Galois group over $\mathbf{Q}$ for any $p > 13$, the difficulty of course being that is must correspond to an even Galois representation. A more tractable case is $G = \mathrm{PSL}_2(\mathbf{F}_p)$, and this was recently answered by David Zywina here. Here is a more elementary version of that construction. Suppose that $\pi$ is a classical modular form of weight three with coefficients in $\mathbf{Z}[\sqrt{-1}]$ and quadratic Nebentypus character $\chi$. Note that there is an isomorphism $\pi^c:= \overline{\pi} \simeq \pi^{\vee} \otimes \| \cdot \|^2 \chi$. For all primes $v$ in $\mathbf{Q}(i)$, one obtains a representation:

$\varrho = \rho \otimes \epsilon^{-1}: \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \rightarrow \mathrm{GL}_2(\mathbf{F}_v).$

with determinant $\chi$. There are two cases, depending on whether $v|p$ is split or not. If $p \equiv 1 \mod 4$ splits, then, assuming $\pi$ is not CM, the image of $\varrho$ restricted to the kernel of $\chi$ is $\mathrm{SL}_2(\mathbf{F}_p)$ for sufficiently large $p$ which can be explicitly determined in any specific case. Thus the image of $\varrho$ is $\mathrm{SL}_2(\mathbf{F}_p)$ plus the image of complex conjugation:

$\left( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right).$

Since $p \equiv 1 \mod 4$, there exists an element $\alpha \in \mathbf{F}_p$ of square $-1$, and hence an element in $\mathrm{SL}_2(\mathbf{F}_p)$ equal to

$\left( \begin{matrix} \alpha & 0 \\ 0 & - \alpha \end{matrix} \right).$

Hence the image of $\varrho$ contains a scalar element of determinant $-1$, and thus it has projective image $\mathrm{PSL}_2(\mathbf{F}_p)$.

If $p \equiv -1 \mod 4$, then, from the isomorphism $\pi^c \simeq \pi^{\vee} \otimes \| \cdot \|^2 \chi$, there is an isomorphism $\varrho^c \simeq \varrho \otimes \chi,$ where $\varrho^c$ is the Galois conjugate induced by complex conjugation. It follows that the *projective* image of $\varrho$ lands in $\mathrm{PGL}_2(\mathbf{F}_p)$. The image of $\varrho$ is thus, for sufficiently large $p$, a subgroup of $\mathbf{F}^{\times}_{p^2} \mathrm{GL}_2(\mathbf{F}_p)$ with projective image containing $\mathrm{PSL}_2(\mathbf{F}_p)$. We first observe that this implies that $\varrho$ contains $\mathrm{SL}_2(\mathbf{F}_p$). It suffices to show that it contains all the transvections; yet the lift of any transvection in $\mathrm{PSL}_2(\mathbf{F}_p)$ is a transvection of order $p$ times a scalar of order prime to $p$, which one can remove by taking an appropriate power. Since the determinant of $\varrho$ is $\chi$, this leaves only the following three possibilities for the image of $\varrho$:

1. The subgroup of $\mathrm{GL}_2(\mathbf{F}_p)$ of matrices with determinant $\pm 1$.
2. The previous subgroup together with the the scalar element $I$ with $I^2 = -1$.
3. The group $\mathrm{SL}_2(\mathbf{F}_p)$ together with $I$.

The third group does not have a non-scalar element of order $2$ correponding to complex conjugation, and the first has traces which do not generate $\mathbf{F}_{p^2}$. Hence the image must be the second, which has projective image $\mathrm{PSL}_2(\mathbf{F}_p)$.

To conclude the argument, it suffices to show that there exists such a $\pi$. Consulting William Stein’s tables, one may take

$f = q + 4i \cdot q^3 + 2 \cdot q^5 - 8i \cdot q^7 + \ldots \in S_3(\Gamma_1(32),\chi),$

for a quadratic $\chi$ where $i^2 = -1$. Since $a_3, a_5, a_7 \ne 0$, this form does not have CM by $\mathbf{Q}(\sqrt{-1})$ or $\mathbf{Q}(\sqrt{-2})$, so $\mathrm{PSL}_2(\mathbf{F}_p)$ is a Galois group for sufficiently large $p$, which one could compute exactly if one wanted. My impression from the notation in William Stein’s tables is that the fixed field of the kernel of $\chi$ is $\mathbf{Q}(\sqrt{-1})$, so this is presumably the same family of examples that arises in Zywina. Other examples (in the range of William’s tables) are as follows:

$g = q + 2 i \cdot q^2 - 4 \cdot q^4 + (3 - 4 i) \cdot q^5 + \ldots \in S_3(\Gamma_1(20)),$
$h = q + 3 i \cdot q^2 - 5 \cdot q^4 - 3 i \cdot q^5 + \ldots \in S_3(\Gamma_1(27))$

Note that this argument requires slightly more than pure thought; it was key that there existed a non-CM form with coefficient field $\mathbf{Q}(\sqrt{-1})$, and there is no *a priori* reason why there should exist any such form. For example, suppose one wanted to generalize this argument to to $\mathrm{PSp}_4(\mathbf{F}_p)$. Then one would want to look for a non-endoscopic Siegel cusp form of weight $(a,b)$ where (edit) $2a+b$ is odd with Hecke eigenvalues in $\mathbf{Q}(\sqrt{-1})$ and quadratic Nebentypus character. Possibly such things exist but perhaps they don’t!

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### 9 Responses to Inverse Galois Problem

1. Sigismond says:

Very nice. Is it generally believed that PSL_2(F_{p^2}) is out of reach?

2. Believed by whom? I’ve never really thought about these problems. My first impression is that it should be relatively easy to prove it for various sets of positive density, but one runs into problems with the fact that given a modular form with coefficients over a field $K$, it’s hard to prevent the existence of primes $p$ which split completely in $K$.

3. DZ says:

Dear GR,

Yes, your cusp form of weight 3 and level 32 gives rise to exactly the same representations as in my paper! (I found this post when doing a literature search for a note I am finishing up.)

Amusingly, your cusp form of weight 3 and level 27 shows up at the end of Serre’s 1987 Duke paper. He shows that the mod 7 representation attached to it produces PSL_2(F_7) as a Galois group over Q (unsurprisingly, the key is that the image contains a scalar matrix with determinant -1). Serre was actually giving an example of his conjecture (he started with the PSL_2(F_7)-extension and then found the form), so he overlooked that this cusp form also produces PSL_2(F_p)-extensions for all p>5 !

-DZ

• Ah, then I should have entitled this post “a proof that Serre missed…”

4. Scott says:

By the way, his last name is Zywina, not Zwyina. If it were Zwyina, he wouldn’t be the alphabetically last mathematician in human history.

• Oops, thanks! (now corrected)

5. zetahype says:

Thanks for the exposition. Could you please elaborate on the line
“If $p \equiv -1 \mod 4$, then, from the isomorphism $\pi^c \simeq \pi^{\vee} \otimes \| \cdot \|^2 \chi$, there is an isomorphism $\varrho^c \simeq \varrho \otimes \chi$, where $\varrho^c$ is the Galois conjugate induced by complex conjugation. It follows that the *projective* image of $\varrho$ lands in $\mathrm{PGL}_2(\mathbf{F}_p)$
Could you explain how the existence of an “inner twist” by c implies that the projective image lands in PGL_2(F_p)? Where does the congruence class of p mod 4 play a role?
Thanks

• The assumption that $p \equiv -1 \mod 4$ means that the residue field of the coefficient ring is $\mathbf{F}_p(i) = \mathbf{F}_{p^2}$ (the case when $p \equiv 1 \mod 4$ is easier and was dealt with previously). Moreover, if $p \equiv 1 \mod 4$, then there are two primes above $p$ in $\mathbf{Z}[\sqrt{-1}]$, and so there is no Galois action on the coefficient field. When $p \equiv -1 \mod 4$, complex conjugation on the coefficients induces the automorphism $c$ of $\mathbf{F}_{p^2}$.

Two representations $V$ and $W$ correspond to the same projective representation if and only if $V \simeq W \otimes \chi$ for some character $\chi$. The representation $\varrho$ a priori lands in $\mathrm{GL}_2(\mathbf{F}_{p^2})$, and the projective representation lands in $\mathrm{PGL}_2(\mathbf{F}_{p^2})$. The condition that $V$ actually arises from a $\mathrm{GL}_2(\mathbf{F}_p)$ representation is that $V \simeq V^c$. The condition that the projective representation lands in $\mathrm{PGL}_2(\mathbf{F}_{p})$ is that $V \simeq V^c \otimes \chi$ for some character $\chi$.