For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For $E\_7$, for example, it is $27$, and is the radical of the $E\_6$ parabolic.


What about in characteristic $p$ for $p>0$? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by $29$ or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic $0$.


I don't need the exact bound, just something close will do. (I have a Lie algebra $L$ with an abelian subalgebra of dimension, say, $43$, and I want to know that $L$ cannot be embedded in $E\_7$, and in particular is not $E\_7$. But I'm in characteristic $7$, for example, or worse, $2$ or $3$.)



Maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic
Let $\ell\ge0$ be the characteristic of the (algebraically closed) ground field. Let $G$ be semisimple with Lie algebra $\mathfrak g$.


First, the maximal dimension of an abelian subalgebra of $\mathfrak g$ can increase for small $\ell$. Let, e.g., $\mathfrak g=sl(2)$ and $\ell=2$. Then the Borel subalgebra is abelian, hence the maximal dimension is $2$ instead of $1$. The same holds for $\mathfrak g=pgl(2)$ and $\ell=2$ where $\langle e,f\rangle$ is abelian.


Secondly, this is a phenomenon of very small characteristics. The following statement is not the most general one but covers most cases:


Assume that $G$ is of adjoint type and $\ell>3$. Then the dimension of a maximal abelian subalgebra is the same as in characteristic $0$.


Proof: Key is the following deformation argument: Being an abelian subalgebra is a closed condition in the Grassmannian of $\mathfrak g$. In other words, there is a closed subscheme $A\_d\subseteq Gr\_d(\mathfrak g)$ classifying abelian subalgebras of dimension $d$. The group $G$ acts on $A\_d$ by conjugation. Assume that $A\_d\ne\emptyset$. Since $A\_d$ is projective, any Borel subgroup $B$ of $G$ has a fixed point. This shows, that if $\mathfrak g$ contains an abelian subalgebra $\mathfrak a$ of dimension $d$ then it will also contain one which is normalized by $B$. Assume from now on that this is the case.


Then, in particular, $\mathfrak a$ will be normalized by a maximal torus $T\subseteq B$. Let $\mathfrak g=\mathfrak t\oplus\bigoplus\_\alpha\mathfrak g\_\alpha$ be the root space decomposition. Then any $T$-stable subspace, so also $\mathfrak a$, has the form
$$
\mathfrak a=\mathfrak a\_0\oplus\bigoplus\_{\alpha\in S}\mathfrak g\_\alpha
$$
where $S$ is a set of roots.


Since $\mathfrak a$ is ablian, its subset of semisimple elements $\mathfrak a\_0$ inside $\mathfrak b$ is also normalized by $B$. Hence $\mathfrak a\_0$ consists of fixed points of $B$ and therefore of $G$. So $\mathfrak a\_0$ sits in the schematic center of $\mathfrak g$. Because $G$ is of adjoint type we infer $\mathfrak a\_0=0$.


Thus everything depends on $S$. I argue that the condition on $S$ making $\mathfrak a=\bigoplus\_{\alpha\in S}\mathfrak g\_\alpha$ abelian is independent of $\ell$, proving our assertion.


For this let $e\_\alpha\in\mathfrak g\_\alpha$ be a Chevalley generator. Then $[e\_\alpha,e\_\beta]$ must be zero for all $\alpha,\beta\in S$. If $\alpha+\beta=0$ then $h\_\alpha=[e\_\alpha,e\_\beta]\ne0$ since $\ell\ne2$. So this case must not occur. If $\gamma=\alpha+\beta$ is a root then a famous formula of Chevalley asserts
$$
[e\_\alpha,e\_\beta]=\pm N\_{\alpha\beta}e\_\gamma\quad\text{with }N\_{\alpha\beta}\in\{1,2,3\}.
$$
Because $\ell>3$, by assumption, we get $[e\_\alpha,e\_\beta]\ne0$. So this case must not occur either. In the remaining cases we have $[e\_\alpha,e\_\beta]=0$. The condition on $S$ is therefore: $\mathfrak a$ is an abelian subalgebra if and only if $\alpha+\beta$ is not zero and not a root for all $\alpha,\beta\in S$. This condition is clearly characteristic free.


