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**](http://dx.doi.org/10.1016/j.jalgebra.2005.09.008), J. Algebra 294, No. 1, 1-18 (2005). [ZBL1085.17003](https://zbmath.org/?q=an:1085.17003).


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.



Computing the inner automorphism group of a finite Lie algebra
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](https://www.sciencedirect.com/science/article/pii/1385725882900385) 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.


