(caveat: I'm not a number-theorist or Langlands-programme-er, and I don't expect to understand all the answers to this question, but I figured they might be useful to someone besides me).

I've been making videos of symmetries of Klein's $j(\tau)$:

https://math.stackexchange.com/questions/466975/elements-of-sl2-mathbbz-which-fix-roots-of-kleins-absolute-invariant-j

https://math.stackexchange.com/questions/338332/visualizing-functions-invariant-or-almost-under-modular-transformation

in an attempt to have a better visual feel of the symmetries of $\mathbf{SL}(2,\mathbb{Z})$.

I know that $\mathbf{SL}(2,\mathbb{Z})$ and $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$ are related -- for instance this paper by Ribet: http://math.berkeley.edu/~ribet/Articles/motives.pdf but that's outside of my scope of knowledge, and that $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$ is something which no one quite understands yet, so:

  • Is $\mathbf{SL}(2,\mathbb{Z})$ a subgroup of $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})$? and if so,

  • What properties does their quotient $\mathrm{Gal}(\mathbb{\bar{Q}},\mathbb{Q})/\mathbf{SL}(2,\mathbb{Z})$ have?

  • I can think of two relationships between these groups but neither of them takes this form. One of them passes through the relationship between modular forms and Galois representations and the other passes through Grothendieck-Teichmüller theory. Which one do you have in mind? – Qiaochu Yuan Aug 20 '13 at 7:54
  • @QiaochuYuan: I would be interested in descriptions of either. – deoxygerbe Aug 20 '13 at 7:57
  • Well, there are many resources (such as previous MO questions) about both of these topics. It seems better to do some independent reading and then to ask a more specific question later. – Qiaochu Yuan Aug 20 '13 at 8:01
  • Dessin d'enfants is besides Langlands program another place to look at. – Marc Palm Aug 20 '13 at 8:06
up vote 7 down vote accepted

The answer to the question as stated is no. The reason is that $\text{SL}_2(\mathbb{Z})$ contains an element of order $4$, while $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ does not. In fact the following more general claim holds.

Proposition: Let $K$ be a field and let $g \in \text{Gal}(\overline{K}/K)$ have finite order. Then $g^2 = \text{id}$.

Proof. Let $L$ be the fixed field of $g$. Then $\overline{K}/L$ is a finite extension of degree the order of $g$, but $\overline{K}$ is algebraically closed, so by the Artin-Schreier theorem $[\overline{K} : L] = 1$ or $2$. $\Box$

  • I don't get the proposition. This contradicts the inverse Galois problem, or not? – Marc Palm Aug 20 '13 at 8:05
  • 3
    @Marc: no. The inverse Galois problem is about what finite groups appear as quotients, not subgroups, of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. – Qiaochu Yuan Aug 20 '13 at 8:05
  • 2
    It might also be interesting to observe that $\mathrm{SL}_2(\mathbb{Z})$ cannot occur as a quotient of $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ either, since $\mathrm{SL}_2(\mathbb{Z})$ is not profinite and Galois groups are all profinite. Its profinite completion $\mathrm{SL}_2(\widehat{\mathbb{Z}})$, though, which is the infinite product $\prod \mathrm{SL}_2(\mathbb{Z}_p)$ over all primes $p$, occurs as quotient of the Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ since each copy does (at least for $p\geq 11$), see Thm 4 of A. Pande, Int. J. Number Th., 2011. – Filippo Alberto Edoardo Aug 21 '13 at 7:49
  • @Filippo: this isn't clear to me; what if you quotient by a non-closed subgroup? – Qiaochu Yuan Aug 22 '13 at 5:50
  • @Qiaochu You are right, I was assuming in my comment that the subgroup was closed. I have no idea about the general case. – Filippo Alberto Edoardo Aug 23 '13 at 6:45

I edit my answer due to two downvotes and the comments. It is an easy argument, that if an embedding would exist the quotient would look horrible. QY shows no embedding is possible, so to some extent my answer is negligible.

$Gal( \overline{\mathbb Q}, \mathbb Q)$ is a profinite group, in particular compact, and $SL_2(\mathbb{Z})$ is a discrete infinite group. Every embedding of an infinite discrete group in a compact group is dense, and no nice topological quotient exists. Usually, the cosets are not even measurable, look at $\mathbb{R} / \mathbb{Q}$.

You might be able to find $SL_2(\mathbb{Z})$ as a subgroup of a quotient, though. It is at least conjectured that $SL_2(\mathbb{Z}/N)$ turns up as quotients for every $N$. If they would turn up in a consistent manner, you might build the projective limit, i.e., find the profinite completion $\otimes_p SL_2(\mathbb{Z}_p)$ of $SL_2(\mathbb{Z})$ as a quotient, which has $SL_2(\mathbb{Z})$ as a discrete subgroup.

  • 2
    I don't really understand the claim being made here. Any infinite discrete group which is residually finite embeds into its profinite completion. – Qiaochu Yuan Aug 20 '13 at 8:07
  • This is a dense embedding then and no nice quotient exists. Nice = closed. – Marc Palm Aug 20 '13 at 8:08
  • It would be very nice if we could identify a dense subgroup of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ as simple as $\text{SL}_2(\mathbb{Z})$! – Qiaochu Yuan Aug 20 '13 at 8:10
  • Okay, but for me subgroups of locally compact groups should be closed by definition. Of course, your algebraic statement is much stronger. – Marc Palm Aug 20 '13 at 8:13

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