Timeline for $\mathrm{Out}(F_{2})=\mathrm{GL}_{2}(\mathbb{Z})$, but $F_{2}=\Gamma(2)$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2021 at 7:23 | comment | added | Alex Gavrilov | What is usually denoted by $\Gamma(2)$ (at least, in areas I am familiar with) is a subgroup of $PSL_2(\mathbb{Z})$ rather then $SL_2(\mathbb{Z})$. (And yes, it is isomorphic to $F_2$.) | |
| Jul 29, 2021 at 1:38 | comment | added | verret | @SeanEberhard Indeed, for any group $G$, every automorphism can be realised as conjugation by an element of an overgroup, simply via the Cayley embedding of $G$ in $Sym(G)$. | |
| Jul 28, 2021 at 23:44 | comment | added | Will Sawin | You can take the universal family of punctured elliptic curves over the moduli space of elliptic curves, $\mathcal M_{1,2} \to \mathcal M_{1,1}$. An issue is that the isomorphism $F_2 \cong P\Gamma(2)$ represents it as the fundamental group of $\mathbb P^1$ punctured at $3$ points, not an elliptic curve punctured at one point, and the $GL_2(\mathbb Z)$-action isn't visible there. | |
| Jul 28, 2021 at 23:27 | review | Close votes | |||
| Aug 2, 2021 at 3:06 | |||||
| Jul 28, 2021 at 22:33 | comment | added | Sean Eberhard | @SamHopkins This happens with all simple groups, because $H$ is naturally a subgroup of $\mathrm{Aut}(H)$ whenever $Z(H) = 1$. | |
| Jul 28, 2021 at 22:33 | comment | added | Will Chen | $\Gamma(2)$ is actually not free. For example it contains $-I$ (this is a common mistake). However there is a subgroup of $\Gamma(2)$ which is free, which is sometimes called the Sanov subgroup. | |
| Jul 28, 2021 at 21:47 | comment | added | Andy Putman | (in other words, your first point is just a red herring and has nothing to do with the action you're looking for [which is not an action at all anyway, but an outer action, so you should be suspicious if you see something that looks like an actual action]) | |
| Jul 28, 2021 at 21:45 | comment | added | Andy Putman | The group $Out(F_n)$ acts on the abelianization $F_n^{\text{ab}} = \mathbb{Z}^n$ of $F_n$. This gives a homomorphism $Out(F_n) \rightarrow Out(\mathbb{Z}^n) = Aut(\mathbb{Z}^n) = GL(n,\mathbb{Z})$. For $n=2$, this is an isomorphism, but for higher $n$ it is surjective but not injective (its kernel is called the "Torelli subgroup" of $Out(F_n)$). | |
| Jul 28, 2021 at 21:41 | history | edited | GTA | CC BY-SA 4.0 |
added 182 characters in body
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| Jul 28, 2021 at 21:41 | comment | added | Carl-Fredrik Nyberg Brodda | @SamHopkins Ah, that's a good point. I guess the question is asking for slightly more than just a description of $\operatorname{Out}(F_2)$. | |
| Jul 28, 2021 at 21:37 | comment | added | Sam Hopkins | @Carl-FredrikNybergBrodda: Of course your point is overall correct, but may be worth noting that we can have $H\subseteq G$ and conjugation by certain elements of $G$ (not in $H$) give outer automorphisms of $H$. This happens with some of the simple Lie groups, I believe. | |
| Jul 28, 2021 at 21:33 | comment | added | Carl-Fredrik Nyberg Brodda | The action of outer automorphisms is never by conjugation (by definition, otherwise it would be inner!). You can find a detailed account of automorphisms of free groups in Lyndon-Schupp, Chapter I, §4, including $\operatorname{Out}(F_2)$. For example Proposition 4.5 gives explicit generators for $\operatorname{Out}(F_2)$ and the generators of $\operatorname{GL}_2(\mathbb{Z})$ they are mapped (isomorphically) to. | |
| Jul 28, 2021 at 21:15 | history | asked | GTA | CC BY-SA 4.0 |