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Is it true that $SL_n(\mathbb{Z})$ is co-hopfian for $n \geq 3$? I heard about this result, but I can't find any reference. Would be grateful for any references!

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    $\begingroup$ Reminder: co-hopfian means: every injective endomorphism is surjective. $\endgroup$ – YCor May 28 '17 at 21:36
  • $\begingroup$ I bet one could make a fairly elementary argument, without using superrigidity or CSP. The upper nilpotent subgroup should map to a nilpotent subgroup, which ought to allow you to pin it down inside $SL_n\mathbb{Z}$ as a subgroup of an upper triangular group up to conjugation. Then one can probably show that the matrices in $SL_{n-1}\mathbb{Z} \ltimes \mathbb{Z}^{n-1}$ map identically, using the fact that the linear action of $SL_{n-1}\mathbb{Z}$ is determined by the action on the upper triangular $\mathbb{Z}^{n-1}$. Then these reps. can probably be assembled to show that it is a conjugation. $\endgroup$ – Ian Agol Jun 7 '17 at 22:53
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Yes it's true. Indeed, denote it by $\Gamma$. Every injective endomorphism $f$ of $\Gamma$ extends ($*$), at least in restriction to some finite index subgroup $\Lambda$, to an automorphism of $G=\mathrm{SL}_n(\mathbf{R})$; fix a Haar measure for the latter.

The automorphism group of the latter is unimodular (since it has the finite index subgroup $\mathrm{PGL}_n(\mathbf{R})$ which has no nontrivial continuous homomorphism to $\mathbf{R}$), and hence maps $\Lambda$ to a subgroup of the same covolume. So $\Lambda$ and $f(\Lambda)$ have the same index in $\Gamma$. This implies that $\Gamma$ and $f(\Gamma)$ also have the same index in $\Gamma$, that is, $f(\Gamma)=\Gamma$.

($*$) is a particular instance of Margulis' superrigidity, but maybe was previously known. Also, maybe there is a more direct approach.

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  • $\begingroup$ Thinking twice, I'm pretty sure that every injective endomorphism $f$ indeed extends to $G$ (no only in restriction to a finite index subgroup). $\endgroup$ – YCor May 28 '17 at 22:09
  • $\begingroup$ I know, that in Margulis superrigidity theorem we nead also the following condition: the image of the homomorphism $f: \Gamma \to \Gamma$ is dense in Zariski topology. Is it true in our case? Can you explain please why? $\endgroup$ – Maria Gerasimova May 28 '17 at 22:50
  • $\begingroup$ Yes we need Zariski-dense image, maybe. Indeed there's no homomorphism $\Gamma\to G$ with infinite non-Zariski-dense image. Indeed first it would be injective (or kernel of order 2) and hence the Zariski closure would virtually have a simple quotient of dimension $<\dim(G)$, and applying superrigidity at this level would yield a contradiction. $\endgroup$ – YCor May 28 '17 at 22:56
  • $\begingroup$ I have one more question... Why the extension of out injective endomorphism to $SL_n(\mathbb{R})$ will be necessary an automorphism? $\endgroup$ – Maria Gerasimova May 29 '17 at 18:31
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    $\begingroup$ If I remember correctly, Raghunathan had some earlier version of Margulis' theorem for non-uniform irreducible lattices. $\endgroup$ – Misha May 31 '17 at 13:58
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You don't need superrigidity. A completely elementary argument is given by Bob Steinberg in:

Steinberg, Robert, Some consequences of the elementary relations in $SL_n$, Finite groups - coming of age, Proc. CMS Conf., Montreal/Que. 1982, Contemp. Math. 45, 335-350 (1985). ZBL0579.20038.

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  • $\begingroup$ It's certainly simpler than Margulis' but certainly not "completely elementary": first, it relies on writing down a presentation (the "standard" K-theoretic one). Second, it relies on the congruence subgroup property (Theorem 3, for which the author indeed a proof that is simpler than any I knew before). Using this he gets the substitute for Margulis' superrigidity, namely a structural result for homomorphisms $SL_n(Z)\to GL_m(C)$ (Theorem 6). The proof takes 3 pages. Finally, one still needs some basic Lie theory (unimodularity of the automorphism group of $SL_n(R)$)... $\endgroup$ – YCor May 29 '17 at 9:53
  • $\begingroup$ @YCor Yes, "cpmpletely elementary" is an exaggeration, I agree... $\endgroup$ – Igor Rivin May 29 '17 at 16:58
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$\text{SL}_n(\mathbb{Z})$ is indeed cohopfian and evenmore, every non-trivial homomorphism $\text{SL}_n(\mathbb{Z})\to \text{SL}_n(\mathbb{Z})$ is onto, and this could be seen in a fairly elementary fashion. My answer here should be seen as an extended comment to YCor's answer. For simplicity I will deal with the case $n=3$.

The following theorem, which elementary proof will be sketched below, is a version of "super-rigidity".

Theorem: Given a non trivial homomorphism $\phi:\text{SL}_3(\mathbb{Z})\to \text{SL}_3(\mathbb{C})$ there exists a basis $v_1,v_2,v_3\in\mathbb{C}^3$ such that the image of $\phi$ could be identified with the group of orientation preserving automorphisms of $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3\in \mathbb{C}^3$.

The proof of the lemma relies only on the following fact concerning elementary matrices in $\text{SL}_3(\mathbb{Z})$. Denote $$ U_1=\{E_{1,2}(t)\mid t\in\mathbb{Z}\},~ U_2=\{E_{1,3}(t)\mid t\in\mathbb{Z}\},~U_3=\{E_{2,3}(t)\mid t\in\mathbb{Z}\},~ $$ $$ U_4=\{E_{2,1}(t)\mid t\in\mathbb{Z}\},~U_5=\{E_{3,1}(t)\mid t\in\mathbb{Z}\},~U_6=\{E_{3,2}(t)\mid t\in\mathbb{Z}\}. $$ This 6 subgroups $U_i$ (with the standing convention $i\in\mathbb{Z}/(6)$) satisfy:

  • they are all isomorphic to $\mathbb{Z}$,
  • they are all conjugated in $\text{SL}_3(\mathbb{Z})$ (by the Weyl group).
  • they sequentially commute, that is $[U_i,U_{i+1}]=1$.
  • for all $i$, $U_i$ is the commutator group of $U_{i-1}$ and $U_{i+1}$, that is $[U_{i-1},U_{i+1}]=U_i$.
  • together they generate $\text{SL}_3(\mathbb{Z})$.

It follows that for every $i$, $H_i=\langle U_{i-1},U_{i+1}\rangle$ is an isomorphic copy of the discrete Heisenberg group $H$ whose center is given by $U_i$. By the facts that the $U_i$'s generate and they are all conjugated, it must be that the non-trivial $\phi$ in the theorem is non-trivial on each of the ${U_i}$'s. Thus for every $i$, $\phi|_{H_i}$ is non-trivial on the center. This is useful because of the following lemma, which proof I leave as an exercise.

Lemma: Let $\psi:H\to \text{SL}_3(\mathbb{C})$ be a homomorphism which is non-trivial on the center. Then the image of the center is a rank-1 unipotent group (a group of operators $u$ which eigenvalues are all 1, and such that the image of $u-1$ is 1-dimensional).

Advice: when trying to follow my notation in the proof below, have in mind the case where $\phi$ is the standard representation.

Sketch of the proof of the theorem: By the lemma, for every $i$, $\phi(U_i)$ is a rank-1 unipotent group. Denote by $P_i<\mathbb{C}^3$ the invariant plane of $\phi(U_i)$. By commutation, $P_i$ is preserved under $U_{i-1}$ and $U_{i+1}$. By the fact that $[U_{i-1},U_{i+1}]$ it is easy to see that either $P_{i-1}=P_i$ or $P_{i+1}=P_i$, but not both. It follows that either $$ (1)~~P_1=P_2,~P_3=P_4,~P_5=P_6 \quad \text{or} \quad (2)~~P_1=P_6,~P_3=P_2,~P_5=P_4. $$ Note that applying inverse-transpose to $\text{SL}_3(\mathbb{Z})$ takes $U_i\to U_{i+3}$, so upon precomposing $\phi$ with inverse-transpose we may and will assume that option (2) occurs (as is the for the standard representation). So we have these three planes $P_1,P_3,P_5$ and their three lines of intersection $L'_i=P_{i-1}\cap P_{i+1}$, $i=2,4,6$. For notational convenience it makes sense to denote $L_i=L'_{2i}$, $i=1,2,3$.

Reflecting on $\phi(H_2)$ it is easy to see that $L_1$ is the image of the rank one transformations in $\phi(U_1)-1$ and $\phi(U_2)-1$. Similarly, $L_2$ is the image of the rank one transformations in $\phi(U_3)-1$ and $\phi(U_4)-1$ and $L_3$ is the image of the rank one transformations in $\phi(U_5)-1$ and $\phi(U_6)-1$. Fix $0\neq v_1\in L_1$ and set $v_2 = (\phi(E_{2,1}(1))-1)v_1$ and $v_3 = (\phi(E_{3,2}(1))-1)v_1$ it is easy to check that coordinate change $e_1\mapsto v_i$ conjgates $\text{GL}_3(\mathbb{Z})$ to the group of automorphisms of the lattice $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3$. $~~~\square$

The proof we gave above is constructive and actually gives more than demanded: for $i=1,2,3$ the lines $\mathbb{C}v_i$ are identified with the images of the rank-1 operators $\phi(E_{i,i-1}(1))-1$ and an arbitrary choice of $v_1\in \text{Im}(\phi(E_{1,3}(1))-1)$ determines uniquely the choices of $v_2$ and $v_3$ in the corresponding lines. For simplicity of formulation I did not put this into the theorem, but this is useful. For example, if $\phi:\text{SL}_3(\mathbb{Z})\to \text{SL}_3(\mathbb{Z})<\text{SL}_3(\mathbb{C})$, it follows by the construction that the lines $L_i$ are rational, and upon choosing $v_1$ with integer coordinates, the vectors $v_i$ are integer. We get that $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3< \mathbb{Z}^3$. Finally, observe that for a full rank subgroup $\Lambda<\mathbb{Z}^3$, unless $\Lambda$ is charcteristic, there exists an orientation preserving automorphism of $\Lambda$ which does not extend to $\text{SL}_3(\mathbb{Z})$. Comparing with the theorem above, we conclude that $\mathbb{Z}v_1\oplus \mathbb{Z}v_2\oplus \mathbb{Z}v_3< \mathbb{Z}^3$ is characteristic and thus $\phi$ is onto $\text{SL}_3(\mathbb{Z})$.

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