Journal of Lie Theory Volume 13 (2003) 511–518 \begin{matrix}c \\ ◯ \end{matrix} 2003 Heldermann Verlag

On the Cartan Subalgebras of Lie Algebras over Small Fields

Salvatore Siciliano^{∗}

Communicated by K. Strambach

Abstract. In this note we study Cartan subalgebras of Lie algebras defined over finite fields. We prove that a possible Lie algebra of minimal dimension without Cartan subalgebras is semisimple. Subsequently, we study Cartan subalgebras of 𝔤𝔩(n, F). AMS classification: 17B50 Keywords: Lie algebras, Cartan subalgebras, finite fields.

1. Introduction

Let L be a finite-dimensional Lie algebra over a field F. A subalgebra H of L is called a Cartan subalgebra of L if it is nilpotent and self-normalizing, i.e. N_{L}(H) = H.

They play a fundamental role in the theory of Lie algebra of characteristic 0 and they represent also an important tool for studying modular Lie algebras.

We recall that, for any x ∈ L, the Fitting null component relative to ad x is defined by

L_{0}(ad x) = {y ∈ L | y(ad x)^{n} = 0 for certain n ∈ N}.

An element h ∈ L is said to be regular if, for any x ∈ L:

dim_{F} L_{0}(ad h) ≤ dim_{F} L_{0}(ad x)

In 1967, Barnes [1] showed that if dim_{F} L < |F | then a subalgebra H is a Cartan subalgebra of minimal dimension if and only if there exists a regular element x in L such that H = L_{0}(ad x).

On the contrary, when the base field has not more than dim_{F} L elements, the existence of Cartan subalgebras of finite-dimensional Lie algebras is still an open problem.

However, in some cases the hypothesis on the cardinality of F can be dropped. For example, one can demonstrate that solvable Lie algebras always

ISSN 0949–5932 / $2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag have Cartan subalgebras (when char F = 0, they are conjugate under the action of inner automorphisms, too: see [4] or [5]); restricted Lie algebras have Cartan subalgebras and they coincide with the centralizers of the maximal tori (see [5]).

∗ Supported by Dipartimento di Matematica “Ennio De Giorgi”–Università degli Studi di Lecce.

In this paper, we analyze some properties of a possible minimal counterexample of Lie algebra which does not have a Cartan subalgebra. Indeed, we reduce the general problem of the existence of the Cartan subalgebras of Lie algebras to the case of semisimple non restricted Lie algebras over a ”small” field with respect to the dimension.

In the second part, we determine the Cartan subalgebras of 𝔤𝔩(n, F) when the field F has at least n^{2} + 1 elements generalizing a result already obtained by Jacobson for algebraically closed fields.

I want to express my most sincere gratitude to professor Willem A. de Graaf for his very useful suggestions.

2. Properties of a possible minimal counterexample

For the proof of the next result we refer to [5].

Lemma 2.1. Let ℒ, \overline{ℒ} be two Lie algebras on a field F and let f : ℒ → \overline{ℒ} be a homomorphism. Then if H is a Cartan subalgebra of ℒ, f(H) is a Cartan subalgebra of f(ℒ). Also, if \overline{H} is a Cartan subalgebra of f(ℒ) and H is a Cartan subalgebra of f^{−1}(\overline{H}), then H is a Cartan subalgebra of ℒ.

Using this result, we are able to prove:

Proposition 2.2. Let ℒ be a minimal example of a finite dimensional Lie algebra over a field F without Cartan subalgebras. Then

1. F is a finite field and |F | ≤ dim_{F} ℒ;

2. ℒ is not a restricted Lie algebra;

3. ℒ is semisimple

Proof. The assertions 1. and 2. are consequence of the remarks already stated in the introduction.

In order to prove 3., suppose by contradiction that ℒ is not semisimple. Then the radical Rad ℒ is not {0} and, because Rad ℒ is solvable, there exists a positive integer k such that Rad^{(k)} ℒ ≠ {0} and Rad^{(k+1)} ℒ = {0}.

Now, set J = Rad^{(k)} ℒ; then it results J ≠ {0} and [J, J] = {0}, in other words J is a nonzero abelian ideal of ℒ.

Consequently, considered the quotient Lie algebra ℒ/J, we have that 0 ≤ dim_{F} ℒ/J < dim_{F} ℒ. By minimality of the dimension of ℒ, it follows that there exists a Cartan subalgebra \overline{H} of ℒ/J and then ℒ has a subalgebra H such that \overline{H} = H/J.

Since H/J is nilpotent and J is abelian, in particular H/J and J are solvable, also H is solvable, therefore by [4] (or [5]) this assures the existence of a Cartan subalgebra ℌ of H.

Consider now the canonical mapping

χ : ℒ → ℒ/J.

We conclude that χ is a homomorphism between the Lie algebras ℒ and ℒ/J, \overline{H} is a Cartan subalgebra of ℒ/J = χ(ℒ) and ℌ is a Cartan subalgebra of H = χ^{−1}(\overline{H}).

Hence, in view of Lemma [5] ℌ is a Cartan subalgebra of ℒ,contradicting the original assumption on ℒ.

3. Cartan subalgebras of 𝔤𝔩(n, F)

In [3], Jacobson proved that if F is an algebraically closed field, A ∈ 𝔤𝔩(n, F) is a regular element if and only if its characteristic polynomial det(λI − A) has n distinct roots. Moreover the Cartan subalgebra determined by A coincides with the centralizer C_{𝔤𝔩(n,F)}(A). We will apply the Barnes’ theorem to prove the following generalization:

Theorem 3.1. Let n ∈ N and F a field with |F | > n^{2}. Then a subalgebra H of 𝔤𝔩(n, F) is a Cartan subalgebra if and only if there exists X ∈ H having n characteristic roots in some extension of F such that H = C_{𝔤𝔩(n,F)}(X).

Proof. Denote by V the n-dimensional vector space over F.

Since the property of an element to be regular or not is unchanged on extending the base field, we can assume, without loss of generality, that the characteristic roots of any element are always in F.

Let A ∈ 𝔤𝔩(n, F), let α_{1}, α_{2}, …, α_{r} ∈ F be the distinct characteristic roots of A and let V_{α_{1}}, V_{α_{2}}, … V_{α_{r}} be the corresponding characteristic spaces.

Since the Lie algebra ℒ = F A is nilpotent and the subspaces V_{α_{i}} are invariant under ℒ we have that α_{1}, α_{2}, …, α_{r} are the weights for ℒ and

V = V_{α_{1}} ⊕ V_{α_{2}} ⊕ … ⊕ V_{α_{r}}  (1)

is the decomposition of V into the weight spaces relative to ℒ.

Moreover, for every 1 ≤ i ≤ r, we recall that the dual module V_{−α_{i}}^{∗} of V_{α_{i}} is a weight module for ℒ with the weight −α_{i}, hence (1) induces a decomposition of the dual space V^{∗} given by

V^{∗} = V_{−α_{1}}^{∗} ⊕ V_{−α_{2}}^{∗} … ⊕ V_{−α_{r}}^{∗}.

Now, if R is the representation of ℒ relative to the module V ⊗_{F} V^{∗} and Ã = AR, following the same argument used in [1] we see that the Fitting null component relative to Ã in V ⊗_{F} V^{∗} is

⨁_{i=1}^{r} V_{α_{i}} ⊗_{F} V_{−α_{i}}^{∗}.

For 1 ≤ i ≤ r set n_{i} = dim_{F} V_{α_{i}}. Then

∑_{i=1}^{r} dim_{F} (V_{α_{i}} ⊗_{F} V_{−α_{i}}^{∗}) = ∑_{i=1}^{r} ( dim_{F} V_{α_{i}} · dim_{F} V_{−α_{i}}^{∗}) = ∑_{i=1}^{r} n_{i}^{2}

and by (1) we have also n = dim_{F} V = ∑_{i=1}^{r} n_{i}.

Consider now the system of diophantine equations

\{ \begin{matrix}∑_{i=1}^{r} n_{i}^{2} = m \\ ∑_{i=1}^{r} n_{i} = n \end{matrix}  (2)

For any positive integers x_{1}, x_{2}, …, x_{r} we have

(x_{1} + x_{2} + … + x_{r})^{2} ≤ x_{1}^{2} + x_{2}^{2} + … + x_{r}^{2},

thus m is minimal if and only if r = n and n_{i} = 1 for all i.

By assumption |F | > n^{2}, in particular the field F (and obviously any extension) has at least n elements. It is well known that the module V ⊗_{F} V^{∗} is isomorphic in a natural way to 𝔤𝔩(n, F) (as a module over ℒ, the corresponding representation is the adjoint mapping). Thus the dimensionality of 𝔤𝔩(n, F)_{0}(ad A) is minimal (i.e. A is regular) if and only if A has n distinct eigenvalues.

Now, since |F | > n^{2} = dim_{F} 𝔤𝔩(n, F) and the Cartan subalgebras of 𝔤𝔩(n, F) are equidimensional (see Theorem 3.2, below), in view of Barnes’ theorem the Cartan subalgebras of 𝔤𝔩(n, F) are exactly the Fitting null component relative to the adjoint mapping of the regular elements of 𝔤𝔩(n, F). So a subalgebra H of 𝔤𝔩(n, F) is a Cartan subalgebra if and only there exists X ∈ 𝔤𝔩(n, F) whose characteristic polynomial has n distinct roots in its splitting field such that H = L_{0}(ad X) .

Then X is diagonalizable, so V admits a basis {f_{i}}_{i=1}^{n} consisting of eigenvectors of X. The centralizer C_{𝔤𝔩(n,F)}(X) contains all elements of 𝔤𝔩(n, F) whose matrix is diagonal relative to the basis {f_{i}}_{i=1}^{n}, so dim_{F} C_{𝔤𝔩(n,F)}(X) ≥ n. But C_{𝔤𝔩(n,F)}(X) ⊆ L_{0}(ad X) = H and dim_{F} L_{0}(ad X) = ∑_{i=1}^{n} n_{i} = n, therefore H = C_{𝔤𝔩(n,F)}(X).

Generally, the result fails without having any hypothesis on the cardinality of the field. If, for example, n > 2, F is a field with less than n elements and H is the subalgebra consisting of the diagonal elements of 𝔤𝔩(n, F), then H is a Cartan subalgebra of 𝔤𝔩(n, F). On the other hand, the eigenvalues of each X ∈ H are exactly the elements of the principal diagonal of its associated matrix, which are elements of F and thus X cannot have n distinct elements.

Nevertheless, a part of Theorem 3.1 can be generalized: indeed, the Cartan subalgebras of 𝔤𝔩(n, F), F an arbitrary field, are all n-dimensional.

For the definition of the concept of a torus we refer to [5].

Theorem 3.2. Let F be an arbitrary field and let H be a subalgebra of 𝔤𝔩(n, F). Then H is a Cartan subalgebra if and only if H is a maximal torus of 𝔤𝔩(n, F). Moreover, dim_{F} H = n.

Proof. In characteristic 0 the result is trivial as all Cartan subalgebras are conjugate over the algebraic closure of F.

Suppose thus F is of characteristic p > 0.

We remark that 𝔤𝔩(n, F) is obviously restricted. By [5], Theorem 4.5.17, we have that H is a Cartan subalgebra of 𝔤𝔩(n, F) if and only if H = C_{𝔤𝔩(n,F)}(T), where T is a maximal torus.

Let K be the algebraic closure of F. Then T ⊗_{F} K is a maximal torus of 𝔤𝔩(n, F) ⊗_{F} K. And T ⊗_{F} K is conjugate under the action of GL(n, K) to the maximal torus D of 𝔤𝔩(n, F) ⊗_{F} K consisting of diagonal matrices. So H ⊗_{F} K is conjugate to C_{𝔤𝔩(n,F)⊗_{F} K}(D) = D.

It follows that dim H = n, and H = T.

For n = 2 a complete determination of the Cartan subalgebras of 𝔤𝔩(2, F) can be obtained without restriction on the field by some elementary considerations. This not only yields the previous result, but establishes also a one-to-one corrispondence between the Cartan subalgebras and certain elements of 𝔤𝔩(2, F).

In the following, we identify an element x of 𝔤𝔩(2, F) with its associated matrix, {E_{ij}} is the canonical basis of 𝔤𝔩(2, F) and L_{x} denotes the Lie subalgebra generated by x and the identity matrix I.

First we show some preliminary lemmas.

Lemma 3.3. Let F a field, char F ≠ 2, and x = ( \begin{matrix}0 & a_{12} \\ a_{21} & 0 \end{matrix} ) where a_{12}, a_{21} ∈ F \ {0}. Then L_{x} is a Cartan subalgebra of 𝔤𝔩(2, F) and L_{x} = C_{𝔤𝔩(2,F)}(x).

Proof. Since the structure constants of L_{x} are 0, L_{x} is abelian and in particular nilpotent,hence it remains only to show that N_{𝔤𝔩(2,F)}(L_{x}) ⊆ L_{x}.

If y = ( \begin{matrix}b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} ) ∈ N_{𝔤𝔩(2,F)}(L_{x}), we have that

[x, y] = ( \begin{matrix}a_{12}b_{21} − a_{21}b_{12} & a_{12}(b_{22} − b_{11}) \\ a_{21}(b_{11} − b_{22}) & a_{21}b_{12} − a_{12}b_{21} \end{matrix} )  (3)

and by the assumption on y there exist h, k ∈ F such that [x, y] = kI + hx.

But tr([x, y]) = 0, so 2k = 0 and then k = 0 (as char F ≠ 2) and from (3) it follows b_{21} = \frac{a_{21}}{a_{12}} b_{12} and b_{11} = b_{22}.

For ĥ = \frac{b_{12}}{a_{12}} and k̂ = b_{11} we have k̂I + ĥx = y so y ∈ L_{x}.

Therefore L_{x} is a Cartan subalgebra of 𝔤𝔩(2, F) and it is easy to show that L_{x} coincides with C_{𝔤𝔩(2,F)}(x).

Remark 3.4. In Lemma 3.3, if we have a_{21} = 0 and a_{12} ≠ 0, then [x, E_{11}] = −x with E_{11} ∉ L_{x}, thus L_{x} is not a self-normalizing subalgebra.

Analogously if a_{12} = 0 and a_{21} ≠ 0.

If a_{12} = a_{21} = 0, L_{x} = F I and in Theorem 3.2 we have seen that L_{x} cannot be a Cartan subalgebra.

Therefore the condition a_{12}, a_{21} ≠ 0 in the Lemma is necessary.

If char F = 2 and x is as in Lemma 3.3, then [x, E_{11}] = x with E_{11} ∉ L_{x}, hence also the assumption char F ≠ 2 is essential.

Lemma 3.5. Let F be an arbitrary field and let x = ( \begin{matrix}1 & a_{12} \\ a_{21} & 0 \end{matrix} ) with a_{12}, a_{21} ∈ F such that 1 + 4a_{12}a_{21} ≠ 0. Then L_{x} is a Cartan subalgebra of 𝔤𝔩(2, F) and L_{x} = C_{𝔤𝔩(2,F)}(x).

Proof. As in Lemma 3.3, it suffices to prove that, for every y = ( \begin{matrix}b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} ) ∈ 𝔤𝔩(2, F), if [x, y] ∈ L_{x} then y ∈ L_{x}. We have

[x, y] = ( \begin{matrix}a_{12}b_{21} − a_{21}b_{12} & b_{12} + a_{12}(b_{22} − b_{11}) \\ a_{21}(b_{11} − b_{22}) − b_{21} & a_{21}b_{12} − a_{12}b_{21} \end{matrix} )  (4)

and [x, y] = kI + hx, for some h, k ∈ F.

From tr([x, y]) = 0 it follows h = −2k and combining with (4) we obtain the pair of relations

\{ \begin{matrix}2a_{12}(a_{12}b_{21} − a_{21}b_{12}) = b_{12} + a_{12}(b_{22} − b_{11}) \\ 2a_{21}(a_{12}b_{21} − a_{21}b_{12}) = a_{21}(b_{11} − b_{22}) − b_{21}. \end{matrix}  (5)

We distinguish a few cases.

In the first case, a_{12} ≠ 0, a_{21} ≠ 0. Then by (5) we have

b_{12}(1 + 4a_{12}a_{21}) = \frac{a_{12}}{a_{21}} b_{21}(1 + 4a_{12}a_{21})

and, since 1 + 4a_{12}a_{21} ≠ 0, this forces b_{12} = \frac{a_{12}}{a_{21}} b_{21} and b_{11} = b_{22} + \frac{b_{12}}{a_{12}} . For k̂ = b_{22} and ĥ = \frac{b_{12}}{a_{12}} , we have y = k̂I + ĥx ∈ L_{x}. In the second case, a_{12} = 0, a_{21} ≠ 0. Then (5) give immediately b_{12} = 0

and b_{11} = b_{22} + \frac{b_{21}}{a_{21}} , so for k̂ = b_{22} and ĥ = \frac{b_{21}}{a_{21}} we have y = k̂I + ĥx ∈ L_{x}. In the third case, a_{12} ≠ 0, a_{21} = 0. Analogous to the previous case . Fourth case, a_{12} = a_{21} = 0. In this case, L_{x} is the set of all diagonal

matrices which is a Cartan subalgebra of 𝔤𝔩(2, F).

In conclusion, N_{𝔤𝔩(2,F)}(L_{x}) = L_{x} and, clearly, L_{x} = C_{𝔤𝔩(2,F)}(x).

Remark 3.6. The requirement 1+4a_{12}a_{21} ≠ 0 is always satisfied if char F = 2. When char F ≠ 2, this condition cannot be omitted: in fact, for a_{21} = −\frac{1}{4a_{12}} , and

y = ( \begin{matrix}0 & 4a_{12}^{2} \\ 1 & 0 \end{matrix} ), then y ∉ L_{x}, while [x, y] = a_{12}(−2I + 4x) ∈ L_{x}.

Remark 3.7. Lemmas 3.3, 3.5 can also be proved using Theorem 3.2. For example, if char F ≠ 2, then 1 + 4a_{12}a_{21} ≠ 0 is equivalent to x being semisimple.

Proposition 3.8. Let F be an arbitrary field. Then the Cartan subalgebras of 𝔤𝔩(2, F) are precisely the centralizers C_{𝔤𝔩(2,F)}(x), where

if char F ≠ 2:

x = \begin{cases}( \begin{matrix}0 & a_{12} \\ a_{21} & 0 \end{matrix} ) & with a_{12}, a_{21} ∈ F \ \{0\} or \\ ( \begin{matrix}1 & \overline{a}_{12} \\ \overline{a}_{21} & 0 \end{matrix} ) & with \overline{a}_{12}, \overline{a}_{21} ∈ F and 1 + 4\overline{a}_{12}\overline{a}_{21} ≠ 0 \end{cases}

if char F = 2:

x = ( \begin{matrix}1 & c_{12} \\ c_{21} & 0 \end{matrix} ) with c_{12}, c_{21} ∈ F.

Proof. By lemmas 3.3 and 3.5, the centralizers defined in the statement are Cartan subalgebras.

Conversely, let H be a Cartan subalgebra of 𝔤𝔩(2, F). In view of Theorem 3.2, H is two-dimensional and I ∈ H, hence there exists \overline{x} = ( \begin{matrix}x_{11} & x_{12} \\ x_{21} & x_{22} \end{matrix} ) ∈ H such that \{I, \overline{x}\} is a basis for H.

If x_{11} ≠ x_{22}, setting

x = ( \begin{matrix}1 & \frac{x_{12}}{x_{11}−x_{22}} \\ \frac{x_{21}}{x_{11}−x_{22}} & 0 \end{matrix} )

we have \overline{x} = x_{22}I + (x_{11} − x_{22})x, thus H ⊆ L_{x},which together with dim_{F} H = dim_{F} L_{x} = 2 implies H = L_{x}.

If x_{11} = x_{22}, considering

x = ( \begin{matrix}0 & x_{12} \\ x_{21} & 0 \end{matrix} )

we have \overline{x} = x_{11}I + x and as above H = L_{x}.

We observe that if x_{1} and x_{2} are as in the statement of Proposition 3.8 and they are not proportional, then the respective subalgebras determined by them are evidently distinct.

Therefore there is a one-to-one correspondence between the Cartan subalgebras of 𝔤𝔩(2, F) and the non proportional elements x.

In particular

Corollary 3.9. Let F be a finite field, |F | = p^{n}. Then the number of Cartan subalgebras of 𝔤𝔩(2, F) is p^{2n}.

Proof. We compute the number of the elements of 𝔤𝔩(2, F) whose centralizers determine the Cartan subalgebras.

If char F = 2, the number of the pairs (a_{12}, a_{21}) ∈ F^{2} is p^{2n}, so thanks to Proposition 3.8 the assertion follows.

If char F ≠ 2, the number of non proportional pairs (a_{12}, a_{21}) of nonzero elements of F is p^{n} − 1. Moreover, the number of solutions (in F) for the equation 1 + 4a_{12}a_{21} = 0 is p^{n} − 1. So the number of the pairs (a_{12}, a_{21}) such that 1 + 4a_{12}a_{21} ≠ 0 is p^{2n} − (p^{n} − 1). Hence the number of the Cartan subalgebras is again p^{2n}.

References

[1] Barnes, R. E., On Cartan Subalgebras of Lie Algebras, Math. Z. 101 (1967), 350–355.

[2] Gastineau-Hills, H. M., “Homomorphisms and Formations of Soluble Lie Algebras,” M. Sc. Thesis, University of Sydney, 1967.

[3]  Jacobson, N., “Lie Algebras,” Wiley Interscience, New York London, 1962.

[4] Seligman, G. B., “Modular Lie algebras,” Springer-Verlag , New York etc., 518 Siciliano

[5] Winter, D. J., “Abstract Lie Algebras,” M. I. T. Press, Cambridge, Mass., 1972.

Salvatore Siciliano Dip. di Matematica ”E. De Giorgi” Università degli Studi di Lecce Via Provinciale Lecce-Arnesano 73100-LECCE, Italy salvatore.siciliano@unile.it

Received July 11, 2002 and in final form October 14, 2002