4
$\begingroup$

I have been reading the paper "On the images of modular and geometric three-dimensional Galois representations" (see also the earlier arXiv preprint). But there is a detail in section 6.2 case (c) and case (d) that I do not follow, and would be very grateful, if anyone could clarify it.

The detail in cases (c) and (d) appears to be similar, so I'll just write up the situation for one of them.

Let P(G) be an irreducible subgroup contained in a maximal subgroup of PSL$(3,l)$ that leaves invariant a triangle with coordinates in $\mathbb{F}_{l}$ and makes all six permutations of its vertices (see Mitchell's paper or Bloom's paper).

Then we have an exact sequence, $1 \to R \to P(G) \to U \to 1$, with $R$ diagonal and $U$ isomorphic to a subgroup of $S_3$. We want to show that the projective image of the inertia subgroup at $l$, $P(I)$, is contained in $R$. It is claimed that $P(I)$ is cyclic with order greater than or equal to $l - 1 > 3$, and so lies in $R$, as $S_3$ does not have cyclic groups of order greater than 3.

I can see from the description of $P(I)$ in section 5 that it has cyclic quotients of order greater than or equal to $l-1>3$, but not why it is cyclic. I don't see either why this must mean it lies in $R$. For example, we have the exact sequence, $1 \to \langle i\rangle \to Q_8 \to C_2 \to 1$, where $Q_8$ is the quaternion group of order 8. Yet there are cyclic subgroups of $Q_8$ of order greater than 2, which aren't contained in $<i>$, for example $\langle j\rangle$, and $\langle k\rangle$.

1: MR2045504 11F80 Dieulefait, Luis; Vila, Núria On the images of modular and geometric three-dimensional Galois representations. Amer. J. Math. 126 (2004), no. 2, 335–361.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ It's useful to add more specific links to the article, which was published in 2004 (ams.org/mathscinet-getitem?mr=2045504) but appeared earlier in preprint form (front.math.ucdavis.edu/0111.5043). $\endgroup$
    – Jim Humphreys
    Commented Apr 1, 2017 at 19:05
  • 2
    $\begingroup$ In general there are cyclic subgroups of $P(G)$ that are not contained in $R$. For example, the maximal subgroup of ${\rm PSL}(3,8)$ with structure $7^2:S_3$ has elements of order $14$. But I am unsure of the definition of the inertia subgroup $I$. $\endgroup$
    – Derek Holt
    Commented Apr 2, 2017 at 12:50
  • $\begingroup$ There is a description of (the image of) the inertia subgroup in section 5 of Dieulefait and Vila's paper. But for this you would need to know the definition of a fundamental character. There's a nice [post here][1] explaining them. If you're happy to read about them. How did you find this example? I would like to have a look at these subgroups too, if possible. [1]:mathoverflow.net/questions/44654/… $\endgroup$
    – mdave16
    Commented Apr 2, 2017 at 13:21
  • 4
    $\begingroup$ For an example in odd characteristic, ${\rm PSL}(3,11)$ has a maximal (imprimitive) subgroup with structure $10^2:S_3$. For $P(G)$ we can take its (maximal) subgroup with structure $5^2:S_3$, which is still irreducible. It has elements of order $10$. I have spent a lot of my time during the past several years working on the structure of the maximal subgroups of the classical groups, so I have become very familiar with them. That's how I thought of these examples. $\endgroup$
    – Derek Holt
    Commented Apr 2, 2017 at 16:55
  • 1
    $\begingroup$ P.s. if you want an explicit example of the matrices in Dereks ${\rm PSL}(3,11)$ example take $s = \left(\begin{array}{rrr} 4 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 3 & 0 \end{array}\right) $. I am also worried about the argument the authors present cause they claim that R is the center of the group, which it is not. The matrix s is diagonalizable, so it could very well be in the image of one of the inertia matrices described in section 4 after choosing a different basis. There is an honest problem in the argument as written, and I would contact the authors privately to ask for clarification. $\endgroup$
    – Maarten Derickx
    Commented Dec 25, 2021 at 11:52

0

Reset to default

You must log in to answer this question.