Computing the inner automorphism group of a finite Lie algebra

I'm interested in writing GAP code to compute the inner automorphism group of a finite Lie algebra. (I'd like to be able to group conjugate subalgebras together.) I've had trouble finding good references on the topic, and I'm getting an unexpectedly large group in testing.

UPDATE 2:
As @Paul Levy says in his answer below, there are problems with making sense of the exponential map when $p$ is small with respect to $n$. GAP computation shows that extending to $\mathbb{Z}$ coefficients, then restricting back down to $\mathbb{F}_p$ coefficients, fails for IIRC $sl(4,\mathbb{F}_3)$.

Indeed, there are larger problems for small $p$. The image $\operatorname{exp}(g)$ may not be an automorphism (may not preserve brackets)! I found the following paper very helpful:

Mattarei, S., Artin-Hasse exponentials of derivations, J. Algebra 294, No. 1, 1-18 (2005). ZBL1085.17003.

Equations (1.1) and (2.1) of Mattarei's paper give precise measurements of how the product fails to respect exponents. A sufficient condition is for the maximal nilpotency index of elements of $\operatorname{ad} \mathfrak{g}$ to be at most $(p+1)/2$. For example, it is enough for $(p+1)/2$ to be at least the dimension of $\mathfrak{g}$. Likely one can do better.

In particular, it looks very likely that the exponential map of the adjoint does not give automorphisms for $\mathfrak{g} = sl(3,3)$, explaining the resulting large group.
END UPDATE 2

UPDATE:
Upon closer examination, I realized that $sl(3,3)$ is not simple. The nilradical, solvable radical, and center coincide as a 1-dimensonal subalgebra. Modding out yields a 7-dimensional algebra, which (partly) explains the occurrence of $G_2$.
It appears that $\mathbb{F}_3$ is an exception, as sometimes happens with small fields. I computed the inner automorphism group of $sl(3,5)$ to be $PSL(3,5)$, which is according to reasonable expectation.
In short, it appears that the answer to the question of whether the inner automorphism group of $sl(3,3)$ has order 9285337152 is "yes".

I still wish that I had better references on inner automorphisms of finite Lie algebras.
END UPDATE

Let me review what I know. I'm not an expert on Lie theory, so apologies if any of this is terribly naive.

For a complex Lie algebra, computing the inner automorphism group is well-established. The inner automorphism group is generated by the exponentials of the adjoints of the elements of the Lie algebra. Thus, $$\operatorname{Inn}\mathfrak{g} = \langle I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \cdots\rangle,$$ where $A = \operatorname{ad} x$, taken over all $x \in \mathfrak{g}$.

In passing to rational Lie algebras, one gives up convergent limits. The usual thing to do seems to be to restrict to nilpotent elements (those whose adjoints are nilpotent matrices), so that the exponential has only finitely many terms. See e.g. https://www.encyclopediaofmath.org/index.php/Inner_automorphism .

In passing to fields of prime order $p$, or more generally those of characteristic $p$, one gives up well-defined division by $p$. I understand that one gets around this by pulling the matrix entries back to $\mathbb{Z}$, computing the exponential over $\mathbb{Z}$ (again, restricting to nilpotent adjoints), then taking the result $\mod p$.
Exponentials in characteristic $p$ are mentioned briefly in e.g. the course notes at http://math.berkeley.edu/~reb/courses/261/8.pdf , though I haven't found any place where they are discussed carefully.

I implemented the latter exponentiation in GAP for fields of prime order. (Code omitted for the time being.) Since GAP wants Lie algebras to act on the left, but groups to act on the right, the correct matrix for basis $B$ is computed by:  LieInnerAutomorphismMatrix:=function(B, x) # Since GAP adjoint matrices act on the left, we transpose to act on right return ExponentialMap(TransposedMat(-AdjointMatrix(B, x))); end; 

An inefficient-but-working way to compute the inner automorphism group (the group generated by exponentials of nilpotent adjoints) is:  nilpotents:=Filtered(L, x->IsNilpotentElement(L,x)); B:=Basis(L); generators:=List(nilpotents, x->LieInnerAutomorphismMatrix(B, x)); G:=Group(generators); 

I tested my code on $sl(2,3)$, and got $PSL(2,3)$. So far so good.

Then I tested on $sl(3,3)$. I get a group of order 9285337152 (!!). The group in question is an extension of an elementary abelian 3-group by the exceptional Lie-type group $G(2,3)$. For comparison, $GL(3,3)$ has order 11232.

Incidentally, all nilpotent adjoints over $sl(3,3)$ have order at most 2, so the 'standard' matrix exponentiation formula (avoiding conversion to/from $\mathbb{Z}$) also works; it gives the same answer.

My questions are:

1. Is there any possibility that this is right, and the group of inner automorphisms of $sl(3,3)$ really has order 9285337152? If so, the group seems like it must be too big to be interesting -- what should I be computing instead?
2. Is there a good place to read about inner automorphisms and similar topics for finite Lie algebras? (Preferably something on the textbook level, or otherwise aimed at nonexperts.)

As far as Question 1 goes, I guess that if I restricted the situation further, I could compute the Chevallay group. But I'd like to be able to deal with an arbitrary (finite) Lie algebra.

• On your update 2: the exponential of a nilpotent element of $\mathfrak{g}=\mathfrak{sl}(3,3)$ is an automorphism of $\mathfrak{g}$, since all nilpotent elements have $p$-th power zero, so the ordinary exponential equals the exponential truncated at the $p$-th power, and in particular is well-defined. So I think your group of order 9285337152 will consist of automorphisms of $\mathfrak{g}$. Feb 23, 2018 at 23:48
• Unless I'm missing the point, a main point of the Mattarei paper is that the exponential in characteristic $p$ of a nilpotent element need not be an automorphism, even if the $p$-th power is 0. I haven't yet asked GAP to look for a non-automorpism in the group, however -- it might be worthwhile to do so. Mar 3, 2018 at 23:06
• I haven't looked at the Mattarei paper in detail, but they are dealing with derivations of an arbitrary derivation of an arbitrary non-associative algebra. In your case, you have an inner derivation of a Lie algebra. So I am fairly sure that (e.g. by using the adjoint representation to reduce to the case of $\mathfrak{gl}_n$) if $({\rm ad}\, x)^p=0$ then ${\rm exp}({\rm ad}\, x)$ is (well-defined and is) an automorphism of ${\mathfrak g}$ (corresponding, in the right setting, to conjugating by $e^x$). This is completely algebraic and the proof goes through as in characteristic zero. Mar 5, 2018 at 12:25
• I can confirm by computation that exponentials of adjoints with nilpotency less than $p$ (in characteristic $p$) may not be isomorphisms. (And this also appears elsewhere in the literature.) May 23, 2018 at 22:53
• Example please? May 24, 2018 at 10:51

You are interested in finite Lie algebras, which is the same thing as finite-dimensional Lie algebras over finite fields. The usual way to understand such a problem is first to consider finite-dimensional Lie algebras algebraically closed fields of characteristic $p$.

So let's assume the ground field $k$ is algebraically closed.

Hogeweij determined in this paper the automorphism group of the following Lie algebras:

• ${\mathfrak g}={\rm Lie}(G)$ where $G$ is a simple algebraic group;

• ${\mathfrak f}={\mathfrak g}/{\mathfrak z}({\mathfrak g})$ where ${\mathfrak g}={\rm Lie}(G)$ for a simply-connected simple algebraic group $G$.

Slightly confusingly, he called ${\mathfrak f}$ the classical Lie algebra of type $X_l$ (where $X_l$ is the root system). (Hogeweij remarked that the automorphism groups of most of the classical Lie algebras were earlier determined by Steinberg.) He used the notation: $\widetilde{G}$, $\widetilde{\mathfrak g}$ for the simply-connected case (called universal by Hogeweij) and $\bar{G}$, $\bar{\mathfrak g}$ for the adjoint case. There are no other possibilities except in type $A_l$ with $(l+1)$ composite or type $D_l$ with $p=2$. Note that ${\mathfrak z}({\mathfrak g})$ is trivial and $\widetilde{\mathfrak g}\cong\bar{\mathfrak g}$ except:

• when $p$ divides $(l+1)$ in type $A_l$;

• when $p=2$ in types $B_l$, $C_l$, $D_l$, $E_7$;

• when $p=3$ in type $E_6$.

The appearance of $G_2$ from your calculations is (surely) related to the following fact, well-known to specialists in positive characteristic Lie algebras: in characteristic 3 (and only in characteristic 3) a Lie algebra of type $G_2$ has an ideal, generated by the short root elements, isomorphic to $\mathfrak{psl}(3,k)$. (Note that this is a 7-dimensional Lie algebra which is not the Lie algebra of an algebraic group.) The simple algebraic group of type $G_2$ therefore acts as automorphisms of $\mathfrak{psl}(3,k)$. However, I don't understand quite how it arises in your set-up.

Indeed, Hogeweij's tables show that (in characteristic 3) the automorphism group of $\mathfrak{psl}(3,k)$ is simple of type $G_2$, but both $\mathfrak{sl}(3,k)$ and $\mathfrak{pgl}(3,k)$ have automorphism group equal to a semidirect product of ${\rm PGL}(3,k)$ and the group of diagram automorphisms (which is cyclic of order 2). So unless ${\rm PGL}(3,\overline{{\mathbb F}_3})$ somehow contains something like $G_2({\mathbb F}_3)$ (which seems unlikely to me, though I am not certain) then I can't see how your result can be correct.

This brings me to my second point. Your algorithm is doomed to fail for small $p$, since the exponential of ${\rm ad}\, x$ is only defined if $({\rm ad}\, x)^p = 0$. In general this holds for some but not all nilpotent elements - it holds for all nilpotent elements if and only if $p$ is greater than or equal to the Coxeter number. The usual construction of the Chevalley groups (if I remember correctly) uses only $\exp(\lambda\,{\rm ad}\, e_\alpha)$ where $e_\alpha$ is a simple (positive or negative) root element. The initial choice of a ${\mathbb Z}$-form ensures that these exponentials are well-defined. But I don't see any reason to expect that the group generated by the exponentials of these simple root elements will be equal to the group generated by the exponentials of all nilpotent ($p$-th power zero) elements. This probably (at least in part) explains the discrepancy between your group of order 9285337152 and the group you expected to obtain.

• Thank you! The comment on what goes wrong with extending the Chevalley construction is particularly helpful. Feb 21, 2018 at 18:26

The main problem I see with this line of questioning is the lack of a clear definition of "inner automorphism group of a finite Lie algebra". Many Lie algebras don't occur as Lie algebras of linear algebraic groups, and then the notion of "inner automorphism" is undefined. On the other hand, your examples tend to be reductions modulo a prime of $\mathbb{Z}$-forms in a Lie algebra such as $\mathfrak{sl}_n$ which does occur as the Lie algebra of a familiar matrix group. Good $\mathbb{Z}$-forms (which Chevalley first formalized for simple Lie algebras over $\mathbb{C}$) allow one to construct a finite Lie algebra for any finite field. But these are very special examples.

On the other hand, the structure of arbitrary finite Lie algebras can get extremely complicated and their automorphism groups are usually unknown. Such Lie algebras are also often remote from familiar matrix groups.

As you've realized, the otherwise well-behaved Lie algebra $\mathfrak{sl}_3$ over the field of 3 elements has a center (consisting of scalar matrices); the same is true for $\mathfrak{sl}_p$ over the field of $p$ elements for any prime $p$. There is nothing special here about the prime 3. (Also, I don't understand your added comment about $G_2$.)

Computations involving Lie algebras have been a specialty of Willem de Graaf, who has written a book on the subject here. But he tends to start with Lie algebras attached to simple algebraic groups, over fields of characteristic 0.

• Thank you for your thoughtful reply. I'll need to think a little to absorb it, but will respond immediately to a couple of things. First, unless I'm mistaken, sl(n,p) will have trivial center unless p divides n. (This is the situation in which scalar matrices have nonzero trace.) I should probably be embarrassed that I needed a GAP computation to realize that :-). Feb 13, 2018 at 8:23
• Also, let me try to explain my comment on G_2. The exceptional group G_2(3) is the 'inner automorphism group', as I compute above, of sl(3,3) mod its center. (Or so the StructureDescription command in GAP tell me.) I don't completely understand why this should be the answer, but as G_2 has a nice action on 7-dimensional space, it doesn't seem like complete nonsense. One possible story: I haven't checked carefully, but maybe sl(3,3) is cryptomorphic to a mod 3 version of the octonions. Feb 13, 2018 at 8:53
• I guess it's a matter of taste and convention as to whether inner automorphism groups are defined outside of linear algebraic groups. To me, the definition of an inner automorphism should be an automorphism arising from exponentiation (subject to suitable restrictions) of a Lie algebra element. There are still some problems with definition in this case, as I'll shortly put in an 'UPDATE2' to my post, but as long as p is large wrt n, one does get a consistent notion. Feb 21, 2018 at 18:24

I accepted @Paul Levy's answer, but meanwhile I want to add some further extended comments, references, and experimental data. Perhaps this will be useful to someone else at some point.

As Paul Levy says, in order for there to be any hope of defining an exponential of an adjoint $\operatorname{ad}x$, one must have $(\operatorname{ad}x)^p = 0$. There are then two ways one could reasonably proceed:

1. Take the truncated exponential function $\sum_{n=0}^{p-1} (\operatorname{ad}x)^n / n!$.

2. Convert the adjoint matrices to integral matrices (by taking the integer between $0$ and $p-1$ corresponding with each element of $\mathbb{Z}_p$). Exponentiate over integers/rationals. Consider the result mod $p$.

The first is discussed in the Mattarei paper referenced in the question, and elsewhere in the Lie literature. The truncated exponential of an adjoint need not be a Lie isomorphism (homomorphism), unless we assume a stronger condition such as $(\operatorname{ad}x)^{(p+1)/2}=0$.

The second is harder to find, probably for good reason. Articles on the Groupprops Wiki appears to be relevant; I've seen it elsewhere, but don't have the reference handy. There are multiple troubles in this approach. There's an obvious 'lift' to the integers only for fields of prime order. The integer matrix associated with a nilpotent matrix over $\mathbb{F}_p$ may fail to be nilpotent. And apparently the 'divided power' Leibniz condition mentioned in the Groupprops link can fail (see below).

Here's the part that may be interesting:

I implemented both approaches in GAP, and ran them on $sl(3,3)$. There are 1093 1-dimensional spaces generated by nilpotent elements, each of nilpotency 3. Of those, 157 give Lie isomorphisms with the truncated exponential approach, while 55 give Lie isomorphisms with the second 'divided powers' approach. For the divided powers approach, I only considered those $x$ for which the integer matrix associated with $\operatorname{ad}x$ was nilpotent, and one might be able to include more elements by looking for terms in the rational exponential series that are 0 mod $p$. The elements that worked with the 2nd approach were a proper subset of those that worked for the 1st. I checked for Lie isomorphism by computing the table of structure constants according to a given basis and its image.

The groups generated was $PSL(3,3)$ in both cases. This is contrary to what I posted above. (The difference arises with the isomorphism check.)

I repeated the same experiments with the quotient of $sl(3,3)$ by its center. Here, there are 364 1-subspaces with nilpotent elements. All give isomorphisms via the truncated exponential. Only 24 give isomorphisms via the divided powers approach. Both approaches give $G(2,3)$ as a group of automorphisms.

I don't know of an example where the two approaches give different groups, but expect that such examples likely exist.

Based on this experimentation, it looks to me like the best way to practically generate a large group of (inner) automorphisms of a finite Lie algebra in characteristic $p$ is to follow the 1st approach. That is, take exponentials of adjoints of elements with nilpotency at most $p$, then throw out the maps that are not homomorphisms. I wish I had a way to restrict to a small subset of the set of nilpotent elements.