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A group $G$ is co-Hopfian if every injective homomorphism $G\to G$ is bijective, i.e., if $G$ contains no proper subgroups isomorphic to $G$. My question is whether Thompson's group $T$ is co-Hopfian.

For context, Thompson's groups $F$ and $V$ are very much not co-Hopfian, roughly due to the fact that there are many copies of the unit interval inside the unit interval, and many copies of the Cantor set inside the Cantor set. However, there are no copies of the circle properly inside the circle, so the same intuition does not apply to $T$.

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    $\begingroup$ Le Boudec and Matte Bon proved (Ann ENS 2018, arxiv.org/abs/1605.01651) that every nontrivial continuous action of $T$ on the circle is semiconjugate to the standard action. This might be a step towards a positive answer. $\endgroup$
    – YCor
    Commented Oct 8, 2020 at 19:21
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    $\begingroup$ I am a bit confused with the remark that there is no copies of the circle sitting inside the circle. Since $\mathbb{T}\cong \mathbb{R}\oplus \mathbb{Q}/\mathbb{Z}$, we can choose a subgroup of $\mathbb{Q}/\mathbb{Z}$ isomorphic to itself, Is this something too naive? $\endgroup$
    – Bombyx mori
    Commented Oct 9, 2020 at 5:14
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    $\begingroup$ The inclusion of the centralizer in $T$ of a rotation through $2\pi/n$ is `close' to showing that $T$ is non-co-Hopfian: isn't this subgroup isomorphic to a (presumably non-split) central extension with kernel $C_n$ and quotient $T$? $\endgroup$
    – IJL
    Commented Oct 9, 2020 at 17:24
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    $\begingroup$ By the way I've received an email, from a reliable source, with a proof that $T$ is not co-Hopfian, but I want to be 100% sure it works before publicly declaring the problem solved. $\endgroup$
    – Matt Zaremsky
    Commented Oct 9, 2020 at 17:42
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    $\begingroup$ Besides, I'm not sure I can tell whether the subgroup of $T$ of elements whose slopes are in $4^\mathbf{Z}$ is isomorphic to $T$. $\endgroup$
    – YCor
    Commented Oct 9, 2020 at 20:25

1 Answer 1

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The answer to the question is no, $T$ is not co-Hopfian, i.e., it does contain proper subgroups isomorphic to itself. Nicolás Matte Bon explained this to me over email (he doesn't use Mathoverflow, but someone showed him this question).

Matte Bon's strategy is to look at $T$ acting on the Cantor set $C=\{0,1\}^{\mathbb{N}}$ and define a new (faithful) action of $T$ on $C$ that yields an (injective) endomorphism $T\to T$ (and then argue that it is not surjective). For non-surjectivity, the idea is that under the new action there is a proper non-empty invariant open subset (so the action is not minimal), unlike for the usual action of $T$. Constructing such an action is a bit too complicated to explain here, but it uses ideas from Section 11 of Matte Bon's paper [1]. What I can do here is explain one concrete example of a proper endomorphism $T\to T$, which I sorted out after understanding Matte Bon's general construction.

To describe the endomorphism $T\to T$, I will use the strand diagram model for elements of $T$ (and $F$ and $V$), see, e.g., Definition 2.7 of Belk-Matucci [2]. For each split or merge vertex (Definition 2.1(2)), draw a small neighborhood around the vertex, not meeting any other vertices. For a split this neighborhood has one incoming strand and two outgoing strands, and for a merge this neighborhood has two incoming strands and one outgoing strand. Now to define $T\to T$ we replace the picture inside each such neighborhood with a more complicated picture. For a split, replace the strand going from the split vertex to the right exit with the picture in Figure 2 of Belk-Matucci (the usual "$x_0$" generator). For a merge, do this same thing but flipped upside-down. (Don't change any cyclic permutations.) This defines a well defined injective endomorphism $T\to T$ (if we only allow cyclic permutations that is; if we allow all permutations then it's $V\to V$, and if we allow no permutations then it's $F\to F$). To see it's non-surjective, one can use the strand diagram method of analyzing dynamics (see, e.g., Figure 19 of Belk-Matucci) to check that if $f$ is in the image of this endomorphism and $c\in C$ starts with $11$ then $f(c)$ has a "$11$" somewhere in it. (Since $T$ certainly contains elements violating this rule, this gives non-surjectivity.)

[1] Nicolás Matte Bon, Rigidity properties of full groups of pseudogroups over the Cantor set. arXiv link

[2] James Belk, Francesco Matucci, Conjugacy and dynamics in Thompson's groups. Geometriae Dedicata 169.1 (2014) 239-261. arXiv link

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  • $\begingroup$ At least to define the endomorphism should be computer-checkable, since it consists in defining images of generators and checking they satisfy the defining relators (once a presentation of $T$ is fixed). Since generators map to nontrivial elements the endomorphism is non-constant, and hence injective by simplicity of $T$. (Then remains the non-surjectivity.) $\endgroup$
    – YCor
    Commented Oct 12, 2020 at 10:34
  • $\begingroup$ An optimistic improvement would be to use the results of Matte Bon's paper [1] to completely classify endomorphisms of $T$ (modulo post-composition by conjugation). $\endgroup$
    – YCor
    Commented Oct 12, 2020 at 10:36
  • $\begingroup$ True, $T$ has a nice small presentation with 3 generators and 6 relations (somewhere in Cannon-Floyd-Parry) so this might not be too tedious. And yeah, that could be interesting to classify the endos, and maybe not too unreasonable.... $\endgroup$
    – Matt Zaremsky
    Commented Oct 12, 2020 at 11:58

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