Jacobson-Morozov theorem

Jacobson-Morozov theorem for a semisimple algebraic group $G$ (presumably I am working over algebraically closed field) states that: given a unipotent u, there exists a homomorphism $\phi$ from $SL_2$ to $G$ such that the matrix $\begin{pmatrix} 1&1\\0&1\end{pmatrix}$ maps to $u$.

What I want to understand is that when this map is injective and when it has kernel $\pm 1$? Do we know a criteria on $u$ which determines this?

• @YCor thanks for the idea. I want to achieve as much PSL_2 as I can. Is there any general result? – Anupam Singh Jul 3 '15 at 6:37
• @YCor Can you describe how you get $PSL_2$ in $PGL_4$? That nilpotent has a Jordan block of length 2 in the adjoint representation, which shouldn't be possible if it were the $E$ in a $PSL_2$ representation. – Ben Webster Jul 3 '15 at 9:33

To each unipotent element $u\in G$ one assigns its weighted Dynkin diagram which is basically a map $\Delta\colon\, \Pi\rightarrow \{0,1,2\}$ where $\Pi$ is a basis of simple roots of the root system of $G$. From $\Delta$ one can read off the characters on ${\mathfrak g}={\rm Lie}(G)$ of a maximal torus $T_0$ of an ${\rm SL}_2$ subgroup containing $u$ (all such subgroups are $G$-conjugate in characteristic $0$ by a classical result of Kostant, as David Stewart pointed out in his comment). If $\Delta(\alpha)=1$ for some $\alpha\in \Pi$ then $\mathfrak g$ has a faithful composition factor for the adjoint action of ${\rm SL}_2$. If $\Delta(\Pi)\subseteq \{0,2\}$ then all weights of $T_0$ on $\mathfrak g$ are even which implies that $-1\in{\rm SL}_2$ acts trivially on $\mathfrak g$. So the case of ${\rm PSL}_2$ occurs if and only if $\Delta(\Pi)\subseteq \{0,2\}$. In this case $u$ is sometimes referred to as even. All one has to do now (for $G$ exceptional) is to open Carter's book on groups of Lie type and examine the list of Dynkin labels on pp. 401--406. For $G$ classical it is explained in Carter's book how to construct $\Delta$ from the partition associated with $u$. So it fairly straightforward to figure out which unipotent elements are even for $G$ classical. Although any even unipotent element of $G$ has to be Richardson, there are plenty of them around. For example, all distinguished unipotent elements of $G$ are even.

• This is the kind of efficient approach I was looking for, not requiring elaborate computations of Jordan blocks. I'm still a bit surprised that the dichotomy between the rank 1 subgroups is not written explicitly in the literature, but the old methods of Dynkin and Kostant are the easiest to use. – Jim Humphreys Jul 14 '15 at 13:51
• @AlexanderPremet Thank you very much. It is useful. – Anupam Singh Jul 15 '15 at 8:34

I'm going to go a little bit out on a limb here and contradict @YCor's comment above; if I make a mistake, presumably someone will point it out.

If I take a $\mathfrak{sl}_2$ representation, you can figure out whether it integrates to one of $SL_2$ or $PSL_2$ just be looking at how $E=\begin{bmatrix}0 & 1\\ 0&0\end{bmatrix}$ acts: in an representation of $PSL_2$, it has only odd length Jordan blocks; in a faithful representation of $SL_2$ there will be at least one Jordan block of even length. Note that exponentiation doesn't change Jordan type, so the same is true for how $u$ acts in the group action.

Thus, it's easy to check which situation you're in by taking any faithful representation of $G$, and computing the Jordan type of $u$ acting on it: if there are all odd parts, then you have $PSL_2$; if you have any even parts, it must be $SL_2$.

• Nice. It's interesting that for any choice of faithful representation of the target Lie group you get one way to check. – YCor Jul 3 '15 at 18:11

The underlying question here is of interest, though I'm not sure how explicitly it has been answered in the literature. However, your header is misleading: the lemma/theorem of Jacobson-Morozov just concerns the embedding of (nonzero) nilpotent elements of a semisimple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ into copies of the unique simple 3-dimensional Lie algebra $\mathfrak{sl}_2$. (The history of the independent work done by Jacobson and Morozov is rather complicated, but anyway there is a modern treatment in N. Bourbaki, Groupes et algebres de Lie, Chap. VIII, $\S11$, among other places. Work of Kostant and others refined the ideas further.)

It then follows from Jacobson-Morozov along with some basic Lie theory that a corresponding Lie group homomorphism $\phi:\mathrm{SL}_2 \rightarrow G$ exists, as in the question. Here $u$ is a nontrivial unipotent element obtained from the given nilpotent element in $\mathfrak{g} = \mathrm{Lie}(G)$ by something like exponentiation. Your problem is to distinguish between embeddings of $\mathrm{SL}_2$ and $\mathrm{PSL}_2$ (equal to $\mathrm{PGL}_2$ over an algebraically closed field).

Actually, one only needs an algebraically closed field of characteristic 0 here, since semisimple Lie groups are the same as semisimple algebraic groups, while there has been careful study of prime characteristic analogues by many people to carry over some of the ideas. Moreover, the study of simple algebraic groups or Lie algebras is usually sufficient in characteristic 0.

The group $G$ has a faithful (rational) representation, though for $G$ of type $E_8$ (a useful test case since the abstract group is simple) the smallest dimension possible is 248 (adjoint representation). In all characteristics the Jordan block decomposition of each unipotent element in such a representation for each exceptional Lie type has been heroically worked out by R. Lawther in Comm. Algebra 23 (1995), 4125-4156. Since there 69 nontrivial unipotent classes for type $E_8$, the resulting tables are nontrivial to compute. Typically there are some Jordan blocks of odd size (or 1), but these seem to be compatible with representations of $\mathrm{SL}_2$, which has irreducible representations of all possible dimensions $\geq 1$. It's true that if some Jordan block is of even size, this is the only possible type of subgroup.

• Ben is right. If you have a rep of $SL_2$ which is a sum of $n_i$-dimensional irreducibles $V_{n_i}$, $1\le i\le k$, then the unipotent of $SL_2$ acts with Jordan blocks of size $n_i$, $1\le i\le k$. If all $n_i$ are odd then each $SL_2\to GL(V_{n_i})$ has $-1$ in the kernel, so the whole rep factors through $PGL_2$, while otherwise (i.e. if at least one block has even size) one of this maps is faithful so the whole rep is faithful (i.e. doesn't factor). The point is that although $SL_2$ has irreducibles in all dimensions, it does not have faithful irreducibles in all dimensions. – YCor Jul 3 '15 at 18:34
• @YCor: Thanks for the comment. I've done a bit of editing but will also need to think over the strategy Ben uses when $\phi$ involves more than one summand. It seems awkward to have to consult tables like those of Lawther (or to make up one's own for classical types), but maybe there is no alternative. – Jim Humphreys Jul 3 '15 at 21:00
• Possibly this criterion is not very practical; at least it shows hat for a unipotent element, if we perform the Jacobson-Morozov theorem, the kernel of the resulting homomorphism from $SL_2$ (which is trivial, $\pm 1$, or all of $SL_2$) only depends on the unipotent element and not on the choice of homomorphism. – YCor Jul 3 '15 at 21:14
• @YCor. The independence of the unipotent element is in fact clear, since by Kostant's work, all $\mathfrak{sl}_2$ subalgebras containing a given nilpotent element $e$ are conjugate by an element of the centraliser $G_e$ of $e$. (This works over $\mathbb{C}$ and fields of sufficiently large characteristic.) – David Stewart Jul 5 '15 at 10:48
• clear? every theorem is a clear consequence of any stronger theorem :) Ben's argument shows that it follows just from the classification of $\mathfrak{sl}_2$-representations – YCor Jul 5 '15 at 19:38