Replacement for Lie-algebra complements

Question

All groups are linear algebraic over some fixed field $k$.

I believe that it is true that, in characteristic $0$, if $G'$ is a reductive subgroup of $G$, then there is a $G'$-invariant complement to $\operatorname{Lie}(G')$ in $\operatorname{Lie}(G)$. I guess this is nearly a consequence of complete reductivity, but I'm not sure how to ensure rationality of the complement if some isotypic component for the adjoint component of $G'$ on $\operatorname{Lie}(G)$ is neither contained in, nor disjoint from, $\operatorname{Lie}(G')$.

In positive characteristic, this can fail. I think, but haven't checked, that the adjoint embedding $\operatorname{PGL}_2 \to \operatorname{GL}(\mathfrak{pgl}_2)$ in characteristic 2 is a counterexample.

Now, there are some phenomena that behave well in zero or large characteristic, but that fail in small characteristic, like the existence of Jacobson–Morosov triples. (EDIT: @JimHumphreys points out that this example may be misleading.) In this setting, we have McNinch's theory of optimal $\operatorname{SL}_2$-homomorphisms, of which it is my understanding that they often provide a substitute for JM triples that allow the use of many of the classical techniques.

My question is whether there is some similar substitute for the existence of complements to reductive Lie subalgebras in small positive characteristic. A specific example that would work for me is if it were known that, whenever $G'$ is a reductive subgroup of $G$, then $\operatorname{Lie}(G')$ surjects onto $\mathrm N_{\operatorname{Lie}(G)}(\operatorname{Lie}(G'))/\mathrm C_{\operatorname{Lie}(G)}(\operatorname{Lie}(G'))$. (This is a consequence of, but hopefully strictly weaker than, the existence of a complement.) In case it makes the problem any easier, it would suffice for me to be able to handle the case where $G'$ is adjoint, $G$ is $\operatorname{GL}(\operatorname{Lie}(G'))$, and the map $G' \to G$ is the adjoint embedding.

EDIT: @JimHumphreys points out that it may be unclear exactly what I am asking. First, my reference to "small positive characteristic" is misleading; I am looking for something that works in all characteristics. At least part of the question is "what is the 'right' replacement?" (to which one can imagine that there are many possible answers). One awkward, but sufficient for my purposes, replacement is:

If $G'$ is an adjoint group, $G = \operatorname{GL}(\operatorname{Lie}(G'))$, and $G' \to G$ is the adjoint embedding, then $\operatorname{Lie}(G')$ surjects onto $\mathrm N_{\operatorname{Lie}(G)}(\operatorname{Lie}(G'))/\mathrm C_{\operatorname{Lie}(G)}(\operatorname{Lie}(G'))$.

and so a proof of this (or counterexample) would certainly constitute an answer.

Answer 1

Let me give this a try:

It seems you are really asking about whether there are any outer derivations of $Lie(G')$ in $Lie(G)$. [EDIT] This does not, unfortunately amount to exactly the question of whether outer derivations exist, because they may not be realised inside the normaliser of $Lie(G')$ in $Lie(G)$ (a good example is to take $G'$ is a torus; then the normaliser in $\mathfrak{gl}(Lie(G'))$ is itself as $G'$ is then a maximal torus of $GL(Lie(G'))$). Of course there are no outer derivations in characteristic $0$ for the Lie algebras of semisimple groups. In positive characteristic the situation is rather different. For classical simple Lie algebras (i.e. ones coming from algebraic groups---there are plenty more simple Lie algebras in char $p$ like the Witt algebras $W(n;\underline{m})$), there are no outer derivations in very good characteristic because the Killing form is non-degenerate. (You could read Seligman Modular Lie Algebras, p112.) So if you can reduce to this case then I suppose you're alright.

Now the Killing form is degenerate for $\mathfrak{psl}_{lp^r}$ (or indeed $\mathfrak{sl}_{lp^r}$) but you are ruling out these on the basis you are looking at the Lie algebra of an adjoint group and verily $\mathfrak{pgl}_{lp^r}$ has no outer derivations. However! There will possibly be an issue for $G_2$ in characteristic $2$ since the Lie algebra is isomorphic (qua Lie algebra) to $\mathfrak{psl}_4$ (amazing but true--e.g. count dimensions). Then this will have extra derivations and I would think this may give rise to a counterexample. Possibly also $F_4$ in characteristic $3$, $E_8$ in characteristic $5$, $\dots$ though I would take a small bet in favour of there are none except for $G_2$ in exceptional type. I would not be surprised to learn that there is a paper where someone has established exactly the outer derivations in all characteristics and for all isogeny types, but unfortunately I can't point you to one.

[EDIT2]: I just computed the $G_2<\mathfrak{gl}_{14}$, $p=2$ example in GAP and it's telling me that its normaliser is dimension $22$ and its centraliser is dimension $1$ ??? Leave this with me]

[EDIT3: Here's the code:

gl:=FullMatrixLieAlgebra(GF(2),14);
gap> g:=SimpleLieAlgebra("G",2,GF(2));
<Lie algebra of dimension 14 over GF(2)>
gap> mats:=List(Basis(g),i->AdjointMatrix(Basis(g),i));;
gap> Bgl:=Basis(gl);
CanonicalBasis( <Lie algebra over GF(2), with 27 generators> )
gap> l:=[];              
[  ]
gap> for i in [1..14] do
> temp:=0*Bgl[1];
> for j in [1..14] do
> for k in [1..14] do
> temp:=temp+mats[i][j][k]*Bgl[(j-1)*14+k];
> od;
> od;
> Append(l,[temp]);
> od;
gap> h:=Subalgebra(gl,l);
<Lie algebra over GF(2), with 14 generators>
gap> Dimension(h);
14
gap> LieSolvableRadical(h);                     
<Lie algebra of dimension 0 over GF(2)>
gap> LieNormaliser(gl,h);  
<Lie algebra of dimension 22 over GF(2)>
gap> LieCentralizer(gl,h);
<Lie algebra of dimension 1 over GF(2)>]

So your proposed surjection breaks for $G_2$ in characteristic $2$.

But verily I get your proposed surjection for $p=3$ (and $p=5$ which is a good prime but just to make sure).

I even checked to make sure that $\mathfrak{pgl}_4$ has no outer derivations when $p=2$ and this does seem to be correct. I would have expected there to be just one extra outer derivation for $\mathfrak{psl}_4=Lie(G_2)$, but clearly these are not linearly equivalent in whatever appropriate sense. This must have a cohomological interpretation---probably some version of the $7$-dimensional module for $G_2$ is turning up.]

[EDIT4 According to GAP, the simply connected version of $F_4$ has no outer derivations in char 2 (where it is not simple) or 3 (where it is simple so adjoint and s.c. are isomorphic) and $E_6$ has no outer derivations in char 2 or 3 (again, it's simple so adj=sc), but I would guess it might take a day or two to complete $E_7$ and $E_8$. If you want, you can email me and I can do these for you.]

Answer 2

Probably the recent paper by Herpel and Stewart here helps to settle your basic question positively. Though the correspondence between subgroups of Lie groups and Lie algebras in the classical situation (or more generally for algebraic groups in characteristic 0) works quite well, it usually breaks down a lot in prime characteristic. But after going through all the work to classify semisimmple and then reductive algebraic groups and especially to understand their internal structure, it turns out that many aspects of the correspondence do work (most of the time) but usually with exceptions.

This isn't easy to predict, but for example the "smoothness" property which makes global and infinitesimal normalizers correspond is worked out for "fairly good" primes in their paper. Here the basic situation is the same as in your question: a pair consisting of a reductive group $G$ and a reductive subgroup $G'$. Denote the respective Lie algebras by $\mathfrak{g}$ and $\mathfrak{g'}$. As in characteristic 0, the Lie algebra normalizer (and centralizer) are the Lie algebras of the group normalizer (and centralizer), provided you avoid some primes for the root systems involved. Here it's easiest to start with a reductive group such as $\mathrm{GL}_n$ whose derived group is simple, to avoid complications with multiple root sytems.

In all characteristics, the automorphism group of a (say) simple algebraic group is easily understood' in particular, its identity component is just the adjoint group. Combining this with the Herpel-Stewart results in their $\S3$, I think you arrive at a positive answer to your question. This doesn't really involve representation theory, fortunately, where there are still many unknowns. In particular, it's apparently independent of thinking about the adjoint representation. But it does involve a lot of heavy lifting and has its own prerequisites.