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!

  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

$\text{SL}_n(\mathbb{Z})$ 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 $\text{SL}_3(\mathbb{Z})$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.