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David Stewart
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[EDIT2 (oops made mistake)][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]

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

But verily I get your proposed surjection for $p=3$ and(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. BTW The computerThis must have a cohomological interpretation---probably some version of the $7$-dimensional module for $G_2$ is not having anyturning up.]

[EDIT4 According to GAP, the simply connected version of it when calculating $F_4$ 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.]

[EDIT2 (oops made mistake)]: 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]

And verily I get your proposed surjection for $p=3$ and $p=5$.

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 derivation, but clearly these are not linearly equivalent in whatever appropriate sense. BTW The computer is not having any of it when calculating $F_4$.]

[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]

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.]

Sorry for all these edits.
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David Stewart
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[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)>]

And verily I get your proposed surjection for $p=3$ and $p=5$.

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 derivation, but clearly these are not linearly equivalent in whatever appropriate sense. BTW The computer is not having any of it when calculating $F_4$.]

[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)>]

And verily I get your proposed surjection for $p=3$ and $p=5$.

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 derivation, but clearly these are not linearly equivalent in whatever appropriate sense. BTW The computer is not having any of it when calculating $F_4$.]

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David Stewart
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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.

[EDIT[EDIT2 (oops made mistake)]: I just computed the $G_2$$G_2<\mathfrak{gl}_{14}$, $p=2$ example in GAP and tellsit's telling me that it is its own normaliser. So maybe the answer two your question is always yes. At least for the exceptionals you could checkdimension $22$ and its centraliser is dimension $1$ ??? Leave this computationally. But there ought to be a more conceptual reason.]with me]

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.

[EDIT: I just computed the $G_2$, $p=2$ example in GAP and tells me that it is its own normaliser. So maybe the answer two your question is always yes. At least for the exceptionals you could check this computationally. But there ought to be a more conceptual reason.]

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 (oops made mistake)]: 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]

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