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update 2 contains more information on exponentials of adjoints over finite fields
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Russ Woodroofe
Russ Woodroofe
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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)$sl(3,3)$ 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".

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".

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".

added insight from further computations in clearly marked section at beginning; minor updates and clarifications elsewhere
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Russ Woodroofe
Russ Woodroofe
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Computing the inner automorphism group of a finite Lie algebra

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.

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.

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);

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

Computing inner automorphism group of a finite Lie algebra

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

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 $\mod p$.

An inefficient-but-working way to compute 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);

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

Computing the inner automorphism group of a finite Lie algebra

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.

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.

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);

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.

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Russ Woodroofe
Russ Woodroofe
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Computing 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.

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 $\mod p$.

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 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. I'd like to be able to deal with an arbitrary (finite) Lie algebra, however.