Varieties of Elementary Subalgebras of Maximal Dimension for Modular Lie Algebras

Abstract

Motivated by questions in modular representation theory, Carlson, Friedlander, and the first author introduced the varieties $\mathbb E(r, \mathfrak g)$ of r-dimensional abelian p-nilpotent subalgebras of a p-restricted Lie algebra $\mathfrak g$ in [8]. In this paper, we identify the varieties $\mathbb E(r, \mathfrak g)$ for a reductive restricted Lie algebra $\mathfrak g$ and r the maximal dimension of an abelian p-nilpotent subalgebra of $\mathfrak g$.

To Dave Benson on the occasion of his 60th birthday

Appendix A. Maximal Sets of Commuting Roots

In the Appendix we give some details for the description of the maximal subsets of commuting positive roots found in Table 2 and the stabilizers of certain ideals in Table 3. Note that the maximal subsets of commuting positive roots are computed in Malcev [17] (except that he skips the proof for $E_8$) so we give details here only for completeness of our arguments.

We begin with the computation of maximal subsets of commuting positive roots. For type E, $D_n$ when $n < 7$, and $B_n$ when $n < 5$ we use a computer program which we have made available online [25]. For the remaining types we provide the following arguments. In several cases we start by looking at commutative subsets of the entire root system $\Phi$, not just the positive roots. In this case, $\Gamma \subset \Phi$ is a commutative subset if for any $\alpha , \beta \in \Gamma$, $\alpha + \beta$ is not a root and $\beta \not = -\alpha$.

1.1 A.1 Type $A_n$

One can check, as in Grantcharov and Serganova [12], that sending $J \subseteq \left\{ 1, 2, \ldots , n + 1\right\}$ to the set of roots $\left\{ \epsilon _i - \epsilon _j \ | \ i \in J, j \notin J\right\}$ yields a bijection between proper nontrivial subsets of $\left\{ 1, 2, \ldots , n + 1\right\}$ and sets of inclusion maximal commutative subsets of $\Phi$. As J gets sent to a set of size $|J|(n + 1 - |J|)$ we see that this set is of maximal order when $n = 2m$ and $|J| = m, m + 1$ or when $n = 2m + 1$ and $|J| = m + 1$. It is a set of positive roots if and only if $J < \left\{ 1, \ldots , n + 1\right\} \setminus J$, thus in type $A_{2m}$ we have $J = \left\{ 1, \ldots , m\right\}$ or $\left\{ 1, \ldots , m + 1\right\}$ yielding $\Phi ^\mathrm {rad}_{m}$ and $\Phi ^\mathrm {rad}_{m + 1}$, respectively, and in type $A_{2m + 1}$ we have $\left\{ 1, \ldots , m + 1\right\}$ yielding $\Phi ^\mathrm {rad}_{m + 1}$.

1.2 A.2 Type $B_n$

We assume $n \ge 5$. The set $R = \left\{ \epsilon _i \ | \ 1 \le i \le n\right\}$ is an inclusion maximal set of non-commuting roots so any maximal set of commuting roots in $\Phi ^+$ consists of a maximal set of commuting roots in $\Phi ^+ \setminus R$ together with at most one element from R. Observe that $\Psi = \Phi \setminus \pm R$ is a root system of type $D_n$ with simple roots $\left\{ \alpha _1, \ldots , \alpha _{n - 1}, \alpha _{n - 1} + 2\alpha _n\right\}$. The maximal set $\Psi ^\mathrm {rad}_n$ can commute with any $\epsilon _i$ and yields $S_i$. The maximal set $\Psi ^\mathrm {rad}_{n - 1}$ commutes with $\epsilon _i$ when $i < n$ and yields $S_i^*$.

1.3 A.3 Type $C_n$

Sending $J \subseteq \left\{ 1, \ldots , n\right\}$ to the set

$$\begin{aligned} \phi (J) = \left\{ \epsilon _i + \epsilon _{i'}, \epsilon _i - \epsilon _j, -\epsilon _j - \epsilon _{j'} \ | \ i, i' \in J \ \text {and} \ j, j' \notin J\right\} \end{aligned}$$

gives a bijection $\phi$ between the power set of $\left\{ 1, \ldots , n\right\}$ and inclusion maximal unipotent commuting subsets of $\Phi$. Among those subsets J satisfying $|J| = m$, the number of positive roots in $\phi (J)$ attains a maximum of $\frac{1}{2}m(m + 1) + m(n - m)$ when $J < \left\{ 1, \ldots , n\right\} \setminus J$ and this maximum value for a given m attains a maximum of $\frac{1}{2}n(n + 1)$ when $m = n$. Thus we take the positive roots of $\phi (\left\{ 1, \ldots , n\right\} )$ and get $\Phi ^\mathrm {rad}_{n}$.

1.4 A.4 Type $D_n$

We assume $n \ge 7$.

Lemma A.1

Let $\Phi$ be type $D_n$. If $R \subseteq \Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ is an inclusion maximal set of commuting roots which contains $\epsilon _1 - \epsilon _2$ then $R = \Phi ^\mathrm {rad}_{1}$ has order $2n - 2$. If it does not contain $\epsilon _1 - \epsilon _2$ then it consists of the root $\epsilon _1 + \epsilon _2$ together with one choice of root from each of the sets $\left\{ \epsilon _1 + \epsilon _r, \epsilon _2 - \epsilon _r\right\} _{2 < r \le n}$ and $\left\{ \epsilon _1 - \epsilon _r, \epsilon _2 + \epsilon _r\right\} _{2 < r \le n}$, and hence has order $2n - 3$.

Proof

We have $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2} = \left\{ \epsilon _1 \pm \epsilon _i, \epsilon _2 \pm \epsilon _j \ | \ 2 \le i \le n \ \text {and} \ 3 \le j \le n\right\}$. If $\epsilon _1 - \epsilon _2$ is contained in our maximal set then the roots $\epsilon _2 \pm \epsilon _j$ are not, so the set contains at most the roots $\epsilon _1 \pm \epsilon _i$, i.e., the roots of $\Phi ^\mathrm {rad}_{1}$. These indeed commute and there are $2n -2$ of them. If $\epsilon _1 - \epsilon _2$ is not contained in our maximal set then note that $\epsilon _1 + \epsilon _2$ is the longest root and therefore is contained in any inclusion maximal set of commuting roots. The remaining roots form the sets of non-commuting pairs given in the statement. One sees that roots from distinct pairs commute and there are $2n - 4$ such pairs.    $\square$

Observe that $m(\Phi ) \ge |\Phi ^\mathrm {rad}_{n}| = \frac{1}{2}n(n - 1)$. Also $\Phi ^\mathrm {rad}_{1}$ is inclusion maximal and of smaller order so no element of $\mathrm {Max}(\Phi )$ contains $\Phi ^\mathrm {rad}_{1}$. Now $\Psi = \Phi \setminus \pm \Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ is a root system of type $D_{n - 2}$ with simple roots $\left\{ \alpha _3, \ldots , \alpha _n\right\}$. Every set of commuting roots in $\Phi ^+$ is the union of sets of commuting roots from $\Psi ^+$ and $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$. By the lemma above the set of commuting roots from $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ can have at most $2n - 3$ elements and by induction the set from $\Psi ^+$ can have at most $\frac{1}{2}(n - 2)(n - 3)$. These sum to the order of $\Phi ^\mathrm {rad}_{n}$ so a maximal set of commuting roots must be the union of a maximal set from $\Psi ^+$ and a set of order $2n - 3$ from $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$.

Now it suffices to take $R \in \mathrm {Max}(\Psi )$ and check which roots in $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ it commutes with. If $R = \Psi ^\mathrm {rad}_n$ then $\epsilon _i + \epsilon _n \in R$ for all $2< i < n$ so $\epsilon _1 - \epsilon _j, \epsilon _2 - \epsilon _j \notin R$ for all $2 < j \le n$. This identifies a unique inclusion maximal set of commuting roots in $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ and it’s union with $\Psi ^\mathrm {rad}_n$ is $\Phi ^\mathrm {rad}_{n}$. If $R = \Psi ^\mathrm {rad}_{n - 1}$ then $\epsilon _i - \epsilon _n \in R$ for all $2< i < n$ so $\epsilon _1 - \epsilon _i, \epsilon _2 - \epsilon _i, \epsilon _1 + \epsilon _n, \epsilon _2 + \epsilon _n \notin R$. Again this identifies the inclusion maximal set in $\Phi ^\mathrm {rad}_{\alpha _1, \alpha _2}$ and its union with $\Psi ^\mathrm {rad}_{n - 1}$ is $\Phi ^\mathrm {rad}_{n - 1}$.

1.5 A.5 Type $F_4$

Recall that the roots of $F_4$ are

$$\begin{aligned} \pm \epsilon _i, \pm \epsilon _i \pm \epsilon _j, \frac{1}{2}(\pm \epsilon _1 \pm \epsilon _2 \pm \epsilon _3 \pm \epsilon _4). \end{aligned}$$

We will denote positive roots of the last type by $\epsilon _{ijk} = \frac{1}{2}(\epsilon _1 + i\epsilon _2 + j\epsilon _3 + k\epsilon _4)$ where $i, j, k \in \left\{ \pm 1\right\}$ and will write, for example, $\epsilon _{+-+}$ instead of $\epsilon _{1, -1, 1}$.

Lemma A.2

If $\epsilon _{ijk} \ne \epsilon _{i'j'k'}$ then $\epsilon _{ijk}$ and $\epsilon _{i'j'k'}$ commute if and only if there is exactly one sign change between (i, j, k) and $(i', j', k')$. In particular, a commuting set of roots can have at most 2 roots of the form $\epsilon _{ijk}$.

Proof

Observe that the terms in $\epsilon _{ijk} + \epsilon _{i'j'k'}$ are exactly those $e_t$ for which the sign did not change (including $\epsilon _1$). If there are 1 or 2 such terms then $\epsilon _{ijk} + \epsilon _{i'j'k'}$ is a root. As the roots are distinct there cannot be 4 such terms, therefore for $\epsilon _{ijk}$ and $\epsilon _{i'j'k'}$ to commute there must be 3 such terms, hence exactly one sign change.

If (i, j, k), $(i', j', k')$, and $(i'', j'', k'')$ are mutually distinct and there is exactly one sign change from (i, j, k) to $(i', j', k')$ and $(i'', j'', k'')$ then there are 2 sign changes from $(i', j', k')$ to $(i'', j'', k'')$. This proves that 3 roots of the form (i, j, k) cannot pairwise commute.    $\square$

Now observe that the roots $\pm \epsilon _i$ and $\pm \epsilon _i \pm \epsilon _j$ give $B_4 \subseteq F_4$. As the commuting property is preserved when intersecting with a subroot system the lemma above gives that a maximal set of commuting roots in $F_4$ can have at most 9 roots: 7 from a maximal set in $B_4$ plus 2 additional roots of the form $\epsilon _{ijk}$. This maximum is indeed attained so every maximal set of commuting positive roots in $F_4$ is identified by a tripel $(C, ijk, i'j'k')$, where $C \subseteq B_4$ is a maximal set of commuting roots and $\epsilon _{ijk}$ and $\epsilon {i'j'k'}$ are the two additional roots. We now compute all the possibilities.

1.5.1 A.5.1 $C = \Phi ^\mathrm {rad}_{1}$

Every $\epsilon _{ijk}$ commutes with $C = \left\{ \epsilon _1, \epsilon _1 \pm \epsilon _i \ | \ i = 2, 3, 4\right\}$. Once (i, j, k) are chosen there are three tripels $(i', j', k')$ which differ by a single sign change. This gives 24 ordered pairs of roots $(\epsilon _{ijk}, \epsilon _{i'j'k'})$ that can be added, thus 12 possible sets of roots for this case.

1.5.2 A.5.2 $C = S_t$

We have $C = \left\{ \epsilon _t, \epsilon _i + \epsilon _j \ | \ 1 \le i < j \le 4\right\}$ and for $\epsilon _{ijk}$ to commute with $\epsilon _t$ the sign on $\epsilon _t$ must be positive. We cannot have more than one negative sign in (i, j, k) otherwise $\epsilon _{ijk}$ would not commute with some root of the form $\epsilon _i + \epsilon _j$. Thus the two roots of the form $\epsilon _{ijk}$ must be $\epsilon _{+++}$ and $\epsilon _{ijk}$ where there is exactly one negative sign in (i, j, k) and this negative sign is not on the $\epsilon _t$ term.

Thus for $C = S_1$ we get three maximal sets corresponding to the three choices for a negative sign and for $C = S_2, S_3, S_4$ we get two maximal sets each. This gives 9 possible sets of roots for this case.

1.5.3 A.5.3 $C = S^*_t$

We have $C = \left\{ \epsilon _t, \epsilon _i + \epsilon _j, \epsilon _{i'} - \epsilon _4 \ | \ 1 \le i< j< 4, 1 \le i' < 4\right\}$. Because of the $\epsilon _{i'} - \epsilon _4$ terms the only $\epsilon _{ijk}$ with a single negative that commutes with C is $\epsilon _{++-}$. As $\epsilon _{--+}$ and $\epsilon _{---}$ don’t commute with $\epsilon _2 + \epsilon _3$ we find that the two additional elements must be $\epsilon _{++-}$ and $\epsilon _{+++}$ or $\epsilon _{++-}$ and $\epsilon {ij-}$ where exactly one of i, j is negative and the negative is not on the $\epsilon _t$ term.

The first choice is valid for any t. For the second when $C = S_1$ there are two choices for the additional negative and when $C = S_2, S_3$ there is one choice for the additional negative. This gives 7 possible sets of roots for this case.

1.6 A.6 Type $G_2$

There are 3 short and 3 long positive roots. No pair of short positive roots commute so there can be at most 1 short root in a maximal commuting set. The pair of long roots $(\alpha _2, 3\alpha _1 + \alpha _2)$ does not commute so a maximal set contains the highest root $3\alpha _1 + 2\alpha _2$ together with at most 1 other long and 1 short root. With this one can check that the maximal sets are

$$\begin{aligned}&\left\{ \alpha _1, 3\alpha _1 + \alpha _2, 3\alpha _1 + 2\alpha _2\right\} , \\&\left\{ \alpha _1 + \alpha _2, 3\alpha _1 + \alpha _2, 3\alpha _1 + 2\alpha _2\right\} , \\&\left\{ \alpha _2, 2\alpha _1 + \alpha _2, 3\alpha _1 + 2\alpha _2\right\} , \\&\left\{ \alpha _2, \alpha _1 + \alpha _2, 3\alpha _1 + 2\alpha _2\right\} , \\&\left\{ 2\alpha _1 + \alpha _2, 3\alpha _1 + \alpha _2, 3\alpha _1 + 2\alpha _2\right\} . \\ \end{aligned}$$