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 $3$.)


1 Answer 1


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.

  • 1
    $\begingroup$ As you mention one could make a more general statement, I guess this is clear but I want to write this down. Suppose that $\ell \geq 0$ does not divide any of the $N_{\alpha\beta}$. Then the maximal dimension of an abelian subalgebra is $n + k$, where $n$ is the maximal dimension of an abelian subalgebra in char $0$, and $k$ is the dimension of the center of $\mathfrak{g}$. For ADE types this always holds, since $N_{\alpha\beta} = \pm 1$. For types $B$, $C$, $F_4$ assuming $\ell \neq 2$ suffices. For type $G_2$ we need $\ell \neq 2,3$. $\endgroup$
    – spin
    Commented Feb 24, 2020 at 11:14
  • $\begingroup$ There is also an AMS Memoir by Ross Lawther "Maximal Abelian Sets of Roots" which seems to classify the sets of roots $S \subseteq \Phi$ with the property that for all $\alpha, \beta \in S$, we have $\alpha + \beta \neq 0$ and $\alpha + \beta \not\in \Phi$. $\endgroup$
    – spin
    Commented Feb 24, 2020 at 11:17
  • $\begingroup$ Thank you very much for this, and thanks to spin for the comments. I found two days ago, and have been meaining to update this question with, a reference for the p good case. A paper by Julia Pevtsova and Jim Stark called 'Varieties of Elementary Subalgebras of Maximal Dimension for Modular Lie Algebras' proves this, but with the requirement that the abelian subalgebra have trivial p-restriction. This seems to be needed to make sure that you can conjugate $\mathfrak a$ into $\mathfrak u$.... $\endgroup$ Commented Feb 25, 2020 at 12:43
  • $\begingroup$ ... You seem to have proved more than they did in their paper, which is interesting considering it is more or less the main result of their paper! In particular, they definitely require $p$ to be good, which for $E_8$ for example is a long way from your condition. I will have to read this proof again, as it's a little far outside my research area (I want to apply this result to work with finite groups of Lie type) but it all seems good to me. $\endgroup$ Commented Feb 25, 2020 at 12:44
  • $\begingroup$ @spin It turns out that I need this more general statement. I have $\ell=2$ and an abelian $29$-space for $E_7$. So it just sneaks over the bound. This has been incredibly useful. $\endgroup$ Commented Jul 16, 2020 at 9:54

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