Maximal subalgebras of Cartan type in the exceptional Lie algebras

Abstract

In this paper, we initiate the study of the maximal subalgebras of exceptional simple classical Lie algebras $\mathfrak {g}$ over algebraically closed fields k of positive characteristic p, such that the prime characteristic is good for $\mathfrak {g}$. We deal with what is surely the most unnatural case; that is, where the maximal subalgebra in question is a simple subalgebra of non-classical type. We show that only the first Witt algebra can occur as a subalgebra of $\mathfrak {g}$ and give an explicit classification of when it is maximal in $\mathfrak {g}$.

Appendix: GAP calculations

We have two jobs to perform in this section, both of which use GAP intensively. The first is to find the composition factors of $\mathfrak {g}|W$ in the case that W contains a nilpotent element of type $A_{p-1}$ representing $\partial$. For the other, recall that there are twelve remaining cases of $(\mathfrak {g},p,\mathcal {O})$ in Proposition 3.10 for which we must check whether a p-subalgebra isomorphic to $W_1$ with $\partial$ represented by a nilpotent element of type $\mathcal {O}$ normalises a non-trivial abelian subalgebra of $\mathfrak {g}$. First, we give an introduction to the methods we use here and have used at other points in the paper.

1.1 Introduction

We use the routines included in the standard GAP4 distribution for computing with Lie algebras. See [31, Manual, Chapter 64] for complete details.

A very straightforward use of GAP we employ is to use its database of root systems to evaluate a cocharacter on a root, thereby finding the weight of the cocharacter on a corresponding root vector for which we have written a function findweights. For example, if $\mathfrak {g}=E_6$, $\mathcal {O}=A_4$ then [17] notates an associated cocharacter in terms of certain elements of a maximal torus as $\tau :{\begin{matrix}2&{}2&{}2&{}-6&{}0&{} \\ &{}&{}2\end{matrix}}$. Thus we may compute

Thus one concludes that $\dim \mathfrak {g}(4)=9$ and a basis of $\mathfrak {g}(4)$ is

Very frequently, we will wish to compute the Lie bracket of two expressions of the form $\sum _\alpha \lambda _\alpha e_\alpha +\sum _\beta \mu _\beta h_\beta$, where the $e_\alpha$ are root vectors in $\mathfrak {g}$, and the $h_\beta$ are elements from a basis of a maximal torus defining the root system of $\mathfrak {g}$. The $\lambda _\alpha$ and $\mu _\beta$ are treated as scalar indeterminates. For example, if x is an expression of the above form, and y is some fixed element, then calculating [x, y] and insisting that it is zero will put conditions amongst the $\lambda _\alpha$ and $\mu _\beta$. (Such calculations can of course be done by hand, since the bracket of any pair of basis elements is given by structure constants, which can be deduced from the root system but this becomes unwieldy for large expressions.) To do this in GAP, we set up a polynomial ring R in a large enough number (e.g. $\dim \mathfrak {g}$) indeterminates and then work with $\mathfrak {g}(R)$. Note that in GAP, the “canonical” basis B of a simple (classical) Lie algebra $\mathfrak {g}$ in GAP is arranged so that (i) the last ${{\mathrm{rk}}}\mathfrak {g}$ elements are a basis for a maximal torus; (ii) the first $|\Phi ^+|$ elements are positive root vectors; (iii) the first ${{\mathrm{rk}}}\mathfrak {g}$ elements are simple root vectors; (iv) if $r\le |\Phi ^+|$ then the $r+|\Phi ^+|$th element of B is a root vector corresponding to the negative of the root corresponding to the rth element of B. The simple root vectors are normally in the Bourbaki ordering; the exception is type $F_4$, where one needs to apply a permutation.

For example if $\mathfrak {g}=E_8$, and e is a nilpotent element of type $A_4$, then the following calculates the bracket of e with a general element of the maximal torus $\mu _1h_1+\dots +\mu _8h_8\in \mathfrak {t}$.

We have also implemented a routine which will make substitutions in general elements in order to make a certain expression be zero. For example, if one wanted to calculate $\mathfrak {c}_\mathfrak {t}(e)$, one could insist that the last expression in the above output be zero. Thus we might choose to calculate the substitution into $\mathtt{y}$ of $\mathtt{x\_243}=\mathtt{2}*\mathtt{x\_241}$, $\mathtt{x\_244}=\mathtt{2}*\mathtt{x\_242}$ and so on, a process which this algorithm automates.

1.2 Composition factors of $\mathfrak {g}|W$ when $\partial \in W$ is nilpotent of type $A_{p-1}$

Let us first calculate the composition factors of the restrictions $[\mathfrak {g}|W]$ for W containing a nilpotent element e of type $A_{p-1}$ representing $\partial$. For this, we use the algorithm described in Proposition 2.3. The data required are a grading $\mathfrak {g}=\bigoplus \mathfrak {g}(i)$ and the list of the weights $\ell _i$ with multiplicities of $X\partial$ on each $\mathfrak {g}(i)$. In these cases, we have many choices for a toral element h representing $X\partial$. However, by Proposition 3.3(ii), we have that that it is of the form $H+H_0$ where $H\in {{\mathrm{Lie}}}(\tau (\mathbb {G}_m))$ and $H_0\in \mathfrak {g}_e(0)\cap {{\mathrm{im}}}{{\mathrm{ad}}}e$. We find H by deriving the cocharacter $\tau$ given in [17, p33] and insisting that it has the correct weight $\mathtt{[H,e] = [X\partial ,e] = -e}$.

In the cases where e is of type $A_{p-1}$, H and $H_0$ commute, and H is toral, so that $H_0$ is also toral. Examining Table 1, ${{\mathrm{im}}}{{\mathrm{ad}}}e \cap \mathfrak {g}_e(0)$ is the Lie algebra of a connected reductive algebraic group of rank 1 so that $H_0$ is conjugate to a scalar multiple of some fixed element. We produce this element in GAP by first taking a generic element $\mathtt{w}$ in the $-2$ weight space for $\tau$, and then considering $\mathtt{[e,w]}$. This is a generic element in ${{\mathrm{im}}}{{\mathrm{ad}}}e \cap \mathfrak {g}(0)$, and insisting that it commutes with $\mathtt{e}$ and lies in the standard maximal torus fixes a choice of $H_0$. We may now write $X\partial = H + \lambda H_0$ with some scalar $\lambda$. (Note also that if $\{\alpha _1,\dots , \alpha _{p-1}\}$ are simple roots for the Levi subalgebra of type $A_{p-1}$ then $H_0$ lies in the centre of the corresponding $\mathfrak {sl}_{p}$ so one can construct the element $H_0$ as $h_{\alpha _1}+2h_{\alpha _2}+\cdots +(p-1)h_{\alpha _{p-1}}$.) As $\lambda H_0$ is toral we must have $\lambda \in \mathbb {F}_p$.

See Table 4 for our choices of H and $H_0$.

Table 4 Choices of H and $H_0$

Table 5 Composition factors of subalgebras $W\cong W_1$ containing a nilpotent element of type e of type $A_{p-1}$, where $X\partial = H + \lambda H_0$

Proposition 4.2

For the various choices of $\lambda \in \mathbb {F}_p$, Table 5 lists the possible composition factors of $\mathfrak {g}|W$ where W contains a nilpotent element e of type $A_{p-1}$ representing $\partial$, and a toral element $H+\lambda H_0$ representing $X\partial$, for H and $H_0$ in Table 4.

1.3 The remaining cases of Lemma 3.10

The cases are:

$$\begin{aligned} (\mathfrak {g},p,e)&= (F_4,5,B_2),\ (F_4,7,C_3),\ (E_6,5,A_3),\ (E_6,7,A_5),\ (E_7,5,A_3),\nonumber \\&\qquad (E_7,7,(A_5)'),\ (E_8,7,A_5),\ (E_8,13,D_7), \end{aligned}$$

(1)

as well as the cases where $\mathcal {O}= A_{p-1}$:

$$\begin{aligned} (\mathfrak {g},p,e) = (E_6,5,A_4),\ (E_7,5,A_4),\ (E_7,7,A_6),\ (E_8,7,A_6). \end{aligned}$$

(2)

For the purposes of computation in GAP, we will need elements representing e, H and $H_0$ for these cases, determined as in “Composition factors of $\mathfrak {g}|W$ when $\partial \in W$ is nilpotent of type $A_{p-1}$” section of Appendix. In the canonical basis in GAP, these are given in Table 6.

Table 6 Choices of e, H and $H_0$

The rest of this “Appendix” is dedicated to finishing the proof of Theorem 1.1(v). We check directly in GAP to see if a non-regular subalgebra must fix a nonzero vector $v \in \mathfrak {g}$. In most cases, we may find such a v and are done. Choose one of the twelve cases above $(\mathfrak {g},p,e)$ above. Our strategy is as follows:

• Set up a simple Lie algebra g in GAP of the same type of $\mathfrak {g}$ over the ring of polynomials $\mathbb {Q}[\mathtt{x\_1},\ldots , \mathtt{x\_{\dim \mathfrak {g}}}]$ as described in “Introduction” section of Appendix. Let B be its Chevalley basis.

• From the tables in [17], set e to be the nilpotent representative in the orbit $\mathcal {O}_e$ expressed in terms of the elements B[i] and set T to be an array whose entries are the coefficients of the cocharacter $\tau$ associated to e in [17]. By the choice of cocharacter in [17], we have that each element B[i] is a weight vector for $\tau$.

• Organise the set of vectors $\{\mathtt{B[i]}:1\le i\le \dim \mathfrak {g}\}$ into weight spaces for $\tau$.

• Set $\mathtt{Xd}$ to be one of the choices $H + \lambda H_0$ as given in Table 6 representing $X\partial$ and corresponding to $\mathtt{e}$. Note that, by Proposition 3.3(iii), we have $H_0=0$ if e is not of type $A_{p-1}$.

• Next we produce an element $\mathtt{f}$ that is a candidate for $\frac{1}{2} X^2 \partial$:

  • Let f be a generic element in $\mathtt{g}$. By a generic element, we mean an element of the form $\mathtt{f}:=\sum _{i}\mathtt{x\_i.B[i]}$.

  • We ensure that $\mathtt{[e,f]=Xd}$ and $\mathtt{[Xd,f]=f}$ by considering linear relations amongst the $\mathtt{x\_i}$ resulting from these equations and substituting them in the coefficients of $\mathtt{f}$.

• The putative subalgebra W contains, additionally, the element $X^3\partial$ and, moreover W is generated by $X^3\partial$ and $\partial$. We perform a similar routine to the above to find an arbitrary element $\mathtt{ff}$ representing $\frac{1}{6}X^3\partial$ on which $X\partial$ has the correct weight, i.e. $\mathtt{[Xd,}\mathtt{ff}] = 2\mathtt{ff}$. By substituting relations in both $\mathtt{f}$ and $\mathtt{ff}$ we force the relation $\mathtt{[e,ff]=f}$. (Note that this puts many constraints on $\mathtt{ff}$ but we do not attempt to guarantee that we have $\langle \mathtt{ff,e}\rangle \cong W$; indeed, this will rarely be true.)

• We look for a vector $\mathtt{v} \ne \mathtt{0}$ in $\mathtt{g}$ which is killed by $\mathtt{e}$ and $\mathtt{ff}$. Since W is generated by these elements, this will guarantee that $\mathtt{v}$ is a fixed vector for W. Specifically:

  • We form a generic element $\mathtt{v}$ from the basis vectors.

  • We compute $\mathtt{[v,e]}$. Forcing this to be zero puts constraints on the coefficients of $\mathtt{v}$.

  • We compute $\mathtt{[Xd,v]}$ and set this to be zero, putting more constraints on the coefficients. (This ensures in fact that $\mathtt{v}\in \mathfrak {g}_e(0)$.)

  • Now consider the expression $\mathtt{[ff,v]} \in \mathtt{g}$. Suppose that $\mathtt{x}\_{\mathtt{i}_\mathtt{1}},\dots ,\mathtt{x}\_{\mathtt{i}_\mathtt{r}}$ are the indeterminates occurring in $\mathtt v$. Now it turns out that the coefficients of $\mathtt{[ff,v]}$ in the basis $\mathtt B$ are all linear expressions in the $\mathtt{x}\_{\mathtt{i}_\mathtt{1}},\dots ,\mathtt{x}\_{\mathtt{i}_\mathtt{r}}$. Thus there is a matrix A whose entries are polynomials in the coefficients of $\mathtt{ff}$ and with $A \cdot \mathtt{(x}\_{\mathtt{i}_\mathtt{1}},\dots ,\mathtt{x}\_{\mathtt{i}_\mathtt{r}})^t = 0$ if and only if $\mathtt{[ff,v] = 0}$.

• We proceed with doing row-reductions on A. If the rank of A is strictly smaller than r, we are done: $\mathtt{v}$ may be chosen to satisfy $\mathtt{[ff,v]=0}$. This deals with all but the following cases:

  $$\begin{aligned} (\mathfrak {g},p,e) =&\,(E_6,7,A_5),\ (E_7,7,(A_5)'),\ (E_8,7,A_5),\ (E_8,13,D_7), \\&(E_7,7,A_6)\text { for }\lambda =1,2,3,4,5,6 \\&(E_8,7,A_6)\text { for }\lambda =2,5. \end{aligned}$$

• Let $a*b*c$ represent the element [[a, b], c]. We go on to consider the elements $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}$ and $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{ff}$, which both must vanish for $p=7$. To see this, observe that since $\mathtt{ff}$ represents $X^3\partial$ and $\mathtt{f}$ represents $X^2\partial$, we must have $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}$ (respectively, $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{ff}$) a multiple of $X^{3+1+1+1+1}\partial$ (respectively  a multiple of $X^{3+1+1+2}\partial$), which is zero. (For the $p=13$ case, we consider the element $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{ff}*\mathtt{f}*\mathtt{f}$.) If there still are linear substitutions that may be read off from the coefficients of these elements, we perform them on both $\mathtt{f}$ and $\mathtt{ff}$. We next try to show that the remaining relations in $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f=0=ff}*\mathtt{f}*\mathtt{f}*\mathtt{ff}$ again force the rank of A to be strictly less than r, by row-reducing A one step at a time and trying to substitute the above relations.

• In the cases $(E_7,7,(A_5)')$ and $(E_8,7,A_5)$, we also use the following technique to reduce the number of indeterminates in $\mathtt{f}$ and $\mathtt{ff}$: We consider root elements $y \in {{\mathrm{Lie}}}(C_e)$ with respect to the roots of $C_e$, the reductive part of the centraliser. As ${{\mathrm{ad}}}(y)$ is nilpotent (in fact, ${{\mathrm{ad}}}(y)^4=0$ in all these cases), we obtain automorphisms $s_y(t) = \exp (t\cdot {{\mathrm{ad}}}(y))$ for $t \in k$ of $\mathfrak {g}$, and $s_y(t)$ satisfies $s_y(t)(e) =e$ and $s_y(t)(H)=H$. Thus we may replace the pair $\mathtt{(f,ff)}$ by any pair $(s_y(t)(\mathtt{f}),s_y(t)(\mathtt{ff}))$. Choosing t and y suitably, we may use this to kill a number of coefficients in $\mathtt{(f,ff)}$.

• After these steps, we succeed in finding fixed vectors for all but the following cases:

  $$\begin{aligned}&(E_7,7,A_6),\lambda =2,\mathtt{x\_63}*\mathtt{x\_108=0},\mathtt{x\_63} \ne \mathtt{0} \text { or } \mathtt{x\_108} \ne \mathtt{0} \text { and} \\&(E_7,7,A_6),\lambda =5,\mathtt{x\_41}*\mathtt{x\_116=0},\mathtt{x\_41} \ne \mathtt{0} \text { or } \mathtt{x\_116} \ne 0. \end{aligned}$$

• In the above two cases, we get that $\mathtt{[ff,v]=0}$ implies $\mathtt{v=0}$. This means that any subalgebras W which arise from these cases do not fix a nonzero vector in $\mathfrak {g}$. However, let us not insist that $\mathtt{[ff,v]=0}$ but that $\mathtt{v}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff = 0}$, by substituting linear equations in v. There still exist nonzero v satisfying this relation. In the cases $(E_7,7,A_6),\lambda =2,\mathtt{x\_63=0,x\_108}\ne \mathtt{0}$ and $\lambda =5,\mathtt{x\_41}\ne \mathtt{0,x\_116=0}$ we check with GAP that the subspace spanned by $\mathtt{v}*\mathtt{f,v}*\mathtt{f}*\mathtt{f},\dots ,\mathtt{v}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}$ is an at most 6-dimensional abelian subalgebra of $\mathtt{g}$ normalised by W, and we may choose $\mathtt{v}$ such that it is nonzero. In the cases $(E_7,7,A_6),\lambda =2,\mathtt{x\_63}\ne \mathtt{0, x\_108=0}$ and $\lambda =5,\mathtt{x\_41=0,x\_116}\ne \mathtt{0}$, we consider the subspace spanned by $\mathtt{v}*\mathtt{ff,v}*\mathtt{ff}*\mathtt{ff}*\mathtt{e,v}*\mathtt{ff}*\mathtt{ff,v}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{e,v}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff,v}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{e,v}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}*\mathtt{ff}$, which again is an abelian subalgebra of $\mathtt{g}$, now at most seven-dimensional. It is again normalised by W, and we may assume it nonzero.

From the above calculations, it follows that W normalises a non-trivial abelian subalgebra in all cases.

Example 4.3

Let us illustrate our procedure by giving a detailed example of our calculations in the case $(\mathfrak {g},p,\mathcal {O})=(E_8,7,A_6)$. The other cases are similar and easier.

Let us first suppose that $\lambda =1$. We have:

We start with completely generic $\mathtt{f}$ and $\mathtt{ff}$ and $\mathtt{v}$. So for instance $\mathtt{f} = \sum _\mathtt{i} \mathtt{x\_i B[i]}$. Now to ensure $\mathtt{[e,f]=Xd}$, we consider the difference $\mathtt{[e,f]-Xd}$.

Since the coefficient of $\mathtt{v.2}$ must vanish, we have for example, that $\mathtt{x\_244 = -5x\_242}$ and may repeat linear substitutions of this form until the whole expression vanishes. Similarly, we manage to ensure after multiple substitutions that $\mathtt{[Xd,f]=f, [Xd,ff]=2ff,[e,ff]=f, [e,v]=0}$ and $\mathtt{[Xd,v] = 0}$. By way of illustration, we have

with similar expressions for $\mathtt{ff}$ and $\mathtt{v}$.

Now $\mathtt{v}$ features the indeterminates $\mathtt{[ 1, 100, 121, 123, 129, 185, 241, 248 ]}$ and hence $r=8$ is the number of indeterminates. From $\mathtt{[ff,v]}$,

we calculate the matrix A as described in the steps above to be

After row-reductions, we obtain

and this clearly has rank less than or equal to 7. As $7<8=r$, we are done.

By contrast to this, for $\lambda = 2$, we have $r=7$ but we obtain a matrix A which, even after row-reductions, may have rank 7. But now considering $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}$:

we see that there still are some linear substitutions in $\mathtt{f}$ and $\mathtt{ff}$ which are possible, for example, from the coefficient of $\mathtt{v.2}$ we have

$$\begin{aligned}&\mathtt{x\_42=-1/6}*(\mathtt{2}*\mathtt{x\_84}^\mathtt{2}*\mathtt{x\_234+4}*\mathtt{x\_62}*\mathtt{x\_155+4}*\mathtt{x\_67}\\&\quad *\mathtt{x\_159+6}*\mathtt{x\_84}*\mathtt{x\_178}). \end{aligned}$$

After performing these, we obtain

We know that this must be zero. So let us first suppose that $\mathtt{x\_{234}} \ne \mathtt{0}$. Then the vanishing of the last expression implies $\mathtt{x\_62=x\_67=x\_84=x\_112=x\_113 = 0}$ and this is enough to ensure that the rank of A is at most 4. Thus we must have $\mathtt{x\_{234} = 0}$. Then the expression for $\mathtt{ff}*\mathtt{f}*\mathtt{f}*\mathtt{f}*\mathtt{f}$ implies that

$$\begin{aligned}&\mathtt{x\_84}*\mathtt{x\_112}*\mathtt{x\_155+6}*\mathtt{x\_84}*\mathtt{x\_113}*\mathtt{x\_159+x\_62}*\mathtt{x\_113}\nonumber \\&\quad +\mathtt{x\_67}*\mathtt{x\_112 = 0}. \end{aligned}$$

(3)

Now we start applying row-reductions to A while distinguishing further subcases. For instance, we may consider the subcase $\mathtt{x\_159, x\_155,x\_113,x\_112} \ne \mathtt{0}$ and take $\mathtt{x\_159}$ as the pivot entry for our first row-reduction in A. After further row-reductions, we end up with the following matrix:

We may recognise multiples of the left hand side of (3) above in this matrix in its sixth column. This means that $\mathtt{Ared[i][6] = 0}$ for all $i \ge 6$ and hence A has rank at most 6 so we are done. Continuing in this way, for each subcase we may row-reduce A to a matrix $A'$, where we may substitute Eq. (3) to deduce that the rank of $A'$ is strictly less than $r=7$. The cases for the remaining choices of $\lambda$ are dealt with similarly.

Finally, we give an example for the use of automorphisms of the form $\exp (t\cdot {{\mathrm{ad}}}(y))$, as mentioned in the steps above. We consider the case $(E_8,7,A_5)$. After the first steps few steps, we have expressions for $\mathtt{f}$ and $\mathtt{ff}$, where for example $\mathtt{f}$ is given as follows.

Here $C_e$ is of type $G_2A_1$. The root $y = \mathtt{B[8]}$ belongs to $C_e$. The corresponding automorphisms $\mathtt{s1} := s_y(t)=\exp (t\cdot {{\mathrm{ad}}}(y))$, $\mathtt{s2} :=s_z(t)=\exp (t\cdot {{\mathrm{ad}}}(z))$ have the following effect on $\mathtt{f}$:

We see that if the coefficient $\mathtt{x\_39}$ is nonzero in $\mathtt{f}$, we may assume that $\mathtt{x\_47}$ is also nonzero after applying $\mathtt{s1}$ with $\mathtt{t}$ chosen such that $\mathtt{x\_39}*\mathtt{t+x\_47} \ne \mathtt{0}$. But then we may apply $\mathtt{s2}$ to ensure that $\mathtt{x\_39=0}$. Thus we may assume from the outset that $\mathtt{x\_39=0}$.

We have a total of 14 further root elements $y \in {{\mathrm{Lie}}}(C_e)$ to formulate automorphisms to use for this process. Applying them to reduce dependency followed by an application of the methods from before, we again succeed in showing that the rank of the corresponding matrix A is less than r in this case.