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$\DeclareMathOperator\SL{SL}$I want to ask if there exists a version of the Jacobson–Morozov theorem for integer matrices. A first approximation would ask: given an integral unipotent matrix $m \in \SL_n(\mathbb{Z})$ does there exist a homomorphism $\SL_2(\mathbb{Z}) \to \SL_n(\mathbb{Z})$ sending $$\begin{pmatrix} 1&1\\0&1\end{pmatrix} \to m?$$

Of course this quickly fails, if we take $n=2$ and $$m = \begin{pmatrix} 1&2\\0&1\end{pmatrix}$$ it's not hard to see that the map from $\SL_2(\mathbb{C}) \to \SL_2(\mathbb{C})$ we get from the Jacobson–Morozov theorem does not send $\SL_2(\mathbb{Z})$ to itself.

To remedy this instead of looking for a true homomorphism $\SL_2(\mathbb{Z}) \to \SL_n(\mathbb{Z})$ we can instead look for a "virtual homomorphism" by passing to a principal congruence subgroup $\Gamma_2(k) \subseteq \SL_2(\mathbb{Z})$ for some $k$, and looking for a map $\Gamma_2(k) \to \SL_n(\mathbb{Z})$ sending $$\begin{pmatrix} 1&k\\0&1\end{pmatrix} \to m$$ and ask if such a map always exists. In the previous example, we can take $k=2$ and just the identity map on $\Gamma_2(k)$ is such a homomorphism. When $n=2$ there is a classification of unipotent conjugacy classes, and indeed any unipotent matrix in $\SL_2(\mathbb{Z})$ virtually extends to a homomorphism in this way. My question is: does an integer unipotent matrix in $SL_n(\mathbb{Z})$ always extend to virtual copy of $SL_2$ in this sense?

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    $\begingroup$ This is a good question. Why not ask it for integer Chevalley groups in general? (I don't know the answer even for $\operatorname{SL}_n(\mathbb Z)$.) By the way, I think one shouldn't speak of the map coming from Jacobson–Morozov; it is well defined only up to conjugation by $\operatorname C_G(m)$. $\endgroup$
    – LSpice
    Commented May 20, 2020 at 22:36
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    $\begingroup$ Yes, I'd certainly be interested in the question for other Chevalley groups. I'd also be interested in replacing $\mathbb{Z}$ by a ring of integers $\mathcal{O}$, replacing the single matrix $m$ by a unipotent embedding of the additive group with some constraints. You are right that I shouldn't really say "the" map in general, but in the context I used it above I believe it's unique. $\endgroup$
    – Nate
    Commented May 20, 2020 at 22:47
  • $\begingroup$ $\DeclareMathOperator\Int{Int}\DeclareMathOperator\SL{SL}$Even for $\SL_2$ it's not unique; you can use any $\Int\begin{pmatrix} 2 & b \\ 0 & 1 \end{pmatrix}$. (I agree that none of them preserves $\SL_2(\mathbb Z)$, though.) $\endgroup$
    – LSpice
    Commented May 20, 2020 at 22:57
  • $\begingroup$ As far as I remember the group $\Gamma_2(2)$ is freely generated by $\left(\begin{array}{cc}1&2\\0&1\end{array}\right)$ and its transpose, so the required homomorphism is easy to construct. Or do I miss something? $\endgroup$
    – Victor Ostrik
    Commented May 20, 2020 at 23:47
  • $\begingroup$ Hmm... good point. I think perhaps what I actually want to know is if there exists an embedding say of $SL_2(\mathbb{C})$ into $SL_n(\mathbb{C})$, such that the restriction to the congruence subgroup behaves as specified. $\endgroup$
    – Nate
    Commented May 21, 2020 at 14:29

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The answer is yes if you replace $m$ by a power of $m$. Let $\rho: SL_2 \rightarrow SL_n$ be a homomorphism defined over $\mathbb Q$ such that the unipotent element $\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right)$ maps to the fixed unipotent matrix $m$. This means that there exists a congruence subgroup $\Gamma$ os $SL_2(\mathbb Z)$ which maps into $SL_n(\mathbb Z)$ under $\rho$ and so that a generator of the upper triangular unipotent matrices in $\Gamma$ goes into a power of $m$.

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