Let $\widehat{F_2}$ be the pro-$\ell$ completion of the free group of rank 2, where $\ell$ is some prime.

Every outer automorphism of $F_2$ induces an outer automorphism of $\widehat{F_2}$, hence an injection $Out(F_2)\rightarrow Out(\widehat{F_2})$.

Let $\alpha\in Out(\widehat{F_2})$ be an outer automorphism. Consider the three properties:

(A) $\alpha$ has finite order.

(B) $\alpha$ does NOT lie in the closure $\overline{Out(F_2)} $ in $Out(\widehat{F_2})$.

(C) $\alpha$ is of determinant 1 under the abelianization map $Out(\widehat{F_2})\rightarrow GL_2(\mathbb{Z}_\ell)$ (does this map admit a section?)

Are there any outer automorphisms which satisfy properties $(A),(B),(C)$?

Are there any outer automorphisms which satisfy $(A)$ and $(B)$?

I'm trying to get a grip on the outer automorphisms of $Out(\widehat{F_2})$ which don't come from "limits" of outer automorphisms of $F_2$. These automorphisms seem very mysterious to me.

  • $\begingroup$ Have you thought about $\mathrm{Out}(\widehat{\mathbb{Z}})$? $\endgroup$
    – HJRW
    Jul 24 '15 at 16:10
  • $\begingroup$ @HJRW Certainly $\widehat{\mathbb{Z}}$ has plenty of automorphisms of finite order not coming from $\mathbb{Z}$, but that's basically why I included (C). I'm not sure if there's a stronger connection between $\widehat{\mathbb{Z}}$ and my situation.. $\endgroup$
    – Will Chen
    Jul 24 '15 at 16:39
  • $\begingroup$ If we're consistent with the notation, $\widehat{\mathbb{Z}}$ denotes the pro-$\ell$-completion of $\mathbb{Z}$, which is $\mathbb{Z}_\ell$. Its automorphism group (as a topological group) is just $\mathbb{Z}_\ell^\times$. $\endgroup$
    – YCor
    Jul 25 '15 at 17:10

If an automorphism fails (C), then obviously it satisfies (B). An (A)+(B) example easily extends from rank 1 to ranks 2: take a root of unity $\zeta$ in $\mathbb Z_p^\times$ that is not $\pm1$ (requiring $p\geqslant 5$). If the generators of the free group are $x$ and $y$, then there is an automorphism of the $p$-adic completion given by $x\mapsto \zeta x$ and $y\mapsto y$. It obviously has finite order, but it has nontrivial determinant, so it is not in the closure of $Out(F)$.

Define $SOut$ as the subgroup of $Out$ with determinant $1$. I believe that $SOut(F)$ is dense in $SOut(\hat F)$, so there is nothing satisfying (B)+(C), let alone all three conditions. I will not attempt to show that, but only that all torsion is conjugate into the closure, which is awfully close. Specifically, I will show that an $\ell$-Sylow subgroup is contained in the closure; and I will suggest that there is no $p$-torsion ($p\geqslant 5$). (Added at last minute: that only covers prime-power torsion and does not show that torsion of order with multiple prime factors lifts from $SL_2(\mathbb Z_p)$ to $SOut(\hat F)$, let alone to the closure of $SOut(F)$.) The key point is that $\hat F$ is a pro-p-group and thus pro-nilpotent. That makes it much easier to analyze than if it were, say, the 2,3-completion and pro-soluble, let alone the full completion with simple composition factors. The length $n$ nilpotent quotient of a group is a characteristic quotient, so an automorphism of a group induces a automorphism of its length $n$ nilpotent quotient. The group of automorphisms of a pro-nilpotent group is the inverse limit of the automorphism groups of the length $n$ quotients. (The transition maps in this inverse limit need not be surjective, but are in the free case, as indicated below.)

The automorphism group of a nilpotent group is easy to understand. If $G_n$ is a length $n$ nilpotent group, $G_{n-1}$ its maximal length $n-1$ quotient and $Z$ the kernel, then the kernel of $Aut(G_n)\to Aut(G_{n-1})$ consists of automorphisms that differ from the identity by shearing into the kernel: $\phi$ so that $\phi(g)g^{-1}\in Z$. Using the centrality of $Z$, the map $g\mapsto \phi(g)g^{-1}$ is a homomorphism $G_n\to Z$. A useful observation is $\phi(g)g^{-1}=g^{-1}\phi(g)$. The kernel is isomorphic to to the group of homomorphisms $G_n^{ab}\to Z$. $Aut(G_n)$ need not surject to $Aut(G_{n-1})$, but a similar analysis shows that if an automorphism of $G_{n-1}$ lifts to an endomorphism of $G_n$, it has an inverse. By induction that yields a nice clean statement: an endomorphism of a nilpotent group is an automorphism if and only if the induced endomorphism of the abelianization is an automorphism.

Applied to $\hat F$, this shows that the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is a torsion-free pro-$p$ group. Also, the map is surjective because freeness makes it easy to lift endomorphisms, which are then automorphisms. (Similarly, it is easy to lift automorphisms of the length $n$ nilpotent quotient to endomorphisms of the free group, but it takes work to lift them to automorphisms, which is why I do not do it. That would imply that $SOut(F)$ is dense in $SOut(\hat F)$. That would answer your question, but the points about Sylow subgroups would still be interesting.)

Thus the kernel $Aut(\hat F)\to GL_2(\mathbb Z_p)$ is built out of composition factors of the form $Hom(A,B)$, where $A$ and $B$ are composition factors of $\hat F$, so the kernel is a pro-$p$ group. More careful consideration shows that $A$ and $B$ are torsion free, hence so the kernel. The kernel $Out(\hat F)\to GL_2(\mathbb Z_p)$ is a quotient of the prior kernel by $\hat F$, so also a pro-$p$ group. I believe that directly applying an analogous stage-by-stage analysis shows that it is torsion-free. The kernel of $GL_2(\mathbb Z_p)\to GL_2(\mathbb F_p)$ is also a pro-$p$-group, torsion-free unless $p=2$. Extension by a $p$-group cannot change the prime-to-$p$ Sylow subgroups. For each $\ell$ prime to $p$, the $\ell$-Sylow subgroups of $Aut(\hat F)$, $Out(\hat F)$, $GL_2(\mathbb Z_p)$, and $GL_2(\mathbb F_p)$ are isomorphic by the quotient map. Similarly, the $\ell$-Sylow subgroups of $SAut(\hat F)$, $SOut(\hat F)$, $SL_2(\mathbb Z_p)$, and $SL_2(\mathbb F_p)$ are isomorphic. Since $SOut(F)=SL_2(\mathbb Z)$ surjects to $SL_2(\mathbb F_p)$, its closures in $SAut(\hat F)$, $SOut(\hat F)$, and $SL_2(\mathbb Z_p)$ must contain full $\ell$-Sylow subgroups. In particular, all $\ell$-power torsion is conjugate into Sylow subgroups and thus into the closure of $SOut(F)$. And I claim that there is no $p$-power torsion (except for $p=2,3$, where all the torsion in $GL_2(\mathbb Z_p)$ is conjugate into $GL_2(\mathbb Z)=Out(F)$).

All that applies to any rank free group ($Out(F)$ always surjects to $GL(\mathbb Z)$, but is no longer a isomorphic), except that there are more possibilities for $p$-power torsion in $GL_n(\mathbb Z_p)$. I think it is all conjugate into $GL_n(\mathbb Z)$ (though I’m more sure about the fields $\mathbb Q_p$ and $\mathbb Q$). I don’t know how to tell if it lifts to $Out(\hat F)$, but if it does, it probably lifts to $Out(F)$. Anyhow, I claim without proof that $SOut(F)$ is dense in $SOut(\hat F)$.


EDIT: THE FOLLOWING IS WRONG (I'm not deleting because the comments at least are very useful)

So the answer to both questions seems to be Yes, at least for "most" odd primes $\ell$.

First, it's a fact that the map $Out(\widehat{F_2})\rightarrow GL_2(\mathbb{Z}_\ell)$ is surjective (Proposition 4.5.4 (b) in Ribes/Zalesskii's book "Profinite Groups").

Second, note that torsion elements of $SL_2(\mathbb{Z})$ have order 2,3,4, or 6, and so its closure, being a profinite group, necessarily an inverse limit of quotients of $SL_2(\mathbb{Z})$, must have the same property.

Lastly, the group $SL_2(\mathbb{Z}_\ell)\subset GL_2(\mathbb{Z}_\ell)$ often has torsion of order not in $\{2,3,4,6\}$. For example, take any element $\alpha\in\mathbb{Z}_\ell$ such that $\alpha\notin\mathbb{Q}$ and $\mathbb{Q}(\alpha)$ is totally real, and such that $\alpha^2 < 4$. This is easy for $\ell\equiv \pm 1 (8)$, since you can just take $\alpha = \sqrt{2}$. If $\ell\equiv\pm 1(12)$, you can pick $\alpha = \sqrt{3}$. Perhaps a more detailed analysis can cover the other cases, though according to YCor it may very well be possible that $SL_2(\mathbb{Z}_2)$ only has torsion of order 2,3,4,6.

Anyway, assuming we have such an $\alpha$, the polynomial $f(x) = x^2 + \alpha x + 1$ then has complex conjugate roots $\beta,\overline{\beta}$, and hence $|\beta| = |\overline{\beta}| = 1$. Since $\mathbb{Q}(\alpha,\beta)$ is a CM-field, all elements of absolute value 1 are roots of unity, so any matrix in $SL_2(\mathbb{Z}_\ell)$ with characteristic polynomial $f(x) = x^2 + \alpha x + 1$ has finite order. Since $f(x)$ must also be the minimal polynomial of such a matrix and $\alpha\notin\mathbb{Q}$, we see that its order cannot be 2, 3, 4, or 6, and thus the preimage of such matrices in $Out(\widehat{F_2})$ cannot lie in $\overline{SL_2(\mathbb{Z})}$.

I'd still welcome any remarks or references either covering the other cases or just more generally pertaining to this question...

EDIT: As pointed out by Will Sawin, of course $SL_2(\mathbb{Z}_\ell)$ contains the diagonal matrices with diagonal entries $\zeta_{\ell-1},\zeta_{\ell-1}^{-1}$, which of course have order $\ell-1$, so the answer to both questions is Yes for all $\ell \ge 11$.

  • $\begingroup$ I don't think $SL_2(\mathbf{Z}_2)$ has any element of finite order not in $\{1,2,3,4,6\}$. $\endgroup$
    – YCor
    Jul 25 '15 at 8:04
  • $\begingroup$ @YCor hmm interesting you're probably right. I certainly can't prove otherwise... do you have a reference/justification? $\endgroup$
    – Will Chen
    Jul 25 '15 at 14:40
  • $\begingroup$ Counting the cardinals of $SL_2(Z/2^nZ)$, which is $3.2^k$ for some $k=k(n)$, rules out all order except $2^k$ or $3.2^k$, so we have to exclude 8 and 12. So we have to see that no quadratic extension $L$ of $K=\mathbf{Q}_2$ contains a primitive 8th or 12th root of unity. Note that $i\notin K$. So $L=K[i]$. If $a+ib$ is a square root of $i$, then we see that $a^2=b^2$, so it is $a(1\pm i)$, and $a^2=1/2$. Since $K$ has no square roots of 2, contradiction. Also, if $a+ib$ is a primitive 12th root, then $a^2-b^2=1/2$ and $a^2+b^2=1$, so $2a^2=3/2$, but contradicts that 3 is not a square in $K$. $\endgroup$
    – YCor
    Jul 25 '15 at 17:08
  • $\begingroup$ The argument with CN fields doesn't work. All $l-1$st roots of unity are in$\mathbb Z_l$ so you can use those. $\endgroup$
    – Will Sawin
    Jul 25 '15 at 18:10
  • $\begingroup$ @WillSawin Oh you mean use the diagonal matrix with entries $\zeta_{\ell-1}, \zeta_{\ell-1}^{-1}$? Hah of course. But why doesn't the argument with CM fields work? $\endgroup$
    – Will Chen
    Jul 25 '15 at 18:22

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