If $L$ is a Lie algebra over an algebraic closed field $K$ of characteristic zero, then all Cartan subalgebras are conjugated. Hence, they have all the same dimension. If $K$ is not algebraic closed but still of characteristic zero, then let $F$ the algebraic closure of $K$ and $H$ a Cartan subalgebra of $L$. Then $H\otimes F$ is a Cartan subalgebra of $L\otimes F$ with $dim_K(H)=dim_F(H\otimes F)$. Thus, the dimension of all Cartan subalgebras in characteristic zero is identical.

In the modular case this statement is wrong. There are several examples in modular Lie algebras such that the dimension of the Cartan subalgebras is not identical.

In the literature I found no examples of a modular Lie algebra based on an associative finite-dimensional unital algebra such that there exist Cartan subalgebras of different dimension.

My question is: Let $A$ be an associative finite-dimensional unital algebra over a field $K$ of characteristic $p$. Is the dimension of the Cartan subalgebras of the associated (resricted) Lie algebra based on $A$ identical?

Remark: I could prove that for $p\ne 2$ and $A$ being solvable this statement is true. In addition, for separable $A$ its true, too.

  • $\begingroup$ What is the definition of "Cartan subalgebra" for an associative finite-dimensional unital algebra? (I only know it in the context of Lie algebras of reductive groups, where the dimension is independent of the CSA.) $\endgroup$ – LSpice Sep 27 '16 at 19:15
  • $\begingroup$ For an associative algebra $A$ the associated Lie algebra $A_{Lie}$ is defined by the multiplication $a\circ b:=ab-ba$ for all $a,b\in A$. $A_{Lie}$ is an Lie algebra, and hence Cartan subalgebras are regarded in my question for $A_{Lie}$. $\endgroup$ – Sven Wirsing Sep 27 '16 at 19:18
  • $\begingroup$ Sorry, I guess I meant: what is the definition of a Cartan subalgebra of an abstract Lie algebra? $\endgroup$ – LSpice Sep 27 '16 at 19:19
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    $\begingroup$ If $L$ is a Lie algebra, then a subalgebra $H$ of $L$ is a Cartan subalgebra if and only if $H$ is nilpotent and self-normalizing ($N_L(H)=H$). $\endgroup$ – Sven Wirsing Sep 27 '16 at 19:20
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    $\begingroup$ cross-posted: math.stackexchange.com/questions/1943988/… $\endgroup$ – Sven Wirsing Sep 27 '16 at 19:42

By a theorem of Premet (MR) (which was later on again proven by Farnsteiner (article MR)) the following theorem holds for a restricted Lie algebra over an algebraically closed field of characteristic p: if a torus has maximal dimension, then the CSA based on this torus has dimension the rank of the Lie algebra.

In particular, if all tori are of the same dimension, then the CSAs are of the same dimension, too.

This theorem can be used for this situation to prove uniquenessalso for modular Lie algebras based on associative algebras:

Let $A$ be an associative finite-dimensional $K$-algebra over a field $K$ of positive characteristics and $L$ be an algebraic closed superfield of $K$. If $H$ is a CSA of $A_{Lie}$, then $H\otimes L$ is a CSA of $(A\otimes L)_{Lie}$. In addition, $dim_K(H)=dim_L(H\otimes L)$ is valid. The $L$-algebra $A\otimes L$ is associative and finite-dimensional and possesses a separable radical factor algebra (because $L$ is algebraically closed and the factor algebra is isomorphic to diect product of full matrix algebras over $L$). In view of the theorem of Premet/Farnsteiner we have to prove that the dimension of the maximal tori of such algebras is unique. For this we start with the observation that the maximal tori of $A$ are mapped to maximal tori of $A/J(A)$ ($J(A)$ the Jacobson radical of $A$) by $T -> (T+J(A))/J(A)$. The dimension of the maximal tori does not change because the intersection of $T$ with $J(A)$ is zero. Thus, we have to calculate the dimension of maximal tori for separable associative algebras. These are direct products of separable simple associative algebras. Maximal tori of direct products are exactly the direct products of maximal tori of the components. We have reduced the question to simple separable associative algebras. Here we can prove that maximal tori and CSA are identical and the dimension is the product of $dim_K(Z(A))$ with the index of the central-simple algebra $A$ over its center.

  • $\begingroup$ The references are useful, but I don't think you've answered the basic question asked (which may in fact be rather subtle). Once a restricted Lie algebra is cut loose from the theory of affine algebraic groups, it gets very hard to study its structure in classical ways. Even the notion of "rank" is difficult to define, and there might be maximal "tori" of varying dimensions. Probably the problem here is to see what if anything extra comes from the assumption that there is an underlying associative algebra structure. P.S. This "answer" should be added to the original question, I guess. $\endgroup$ – Jim Humphreys Oct 8 '16 at 13:39
  • $\begingroup$ I have added the proof in my answer for an associative algebra which was my original question. $\endgroup$ – Sven Wirsing Oct 10 '16 at 6:27

I thought about this topic again and find the following solution proposal: As described in the first solution we can assume that $K$ is algebraically closed, because Cartan subalgebras and base field extensions are compatible and preserve the dimension -- as well as the nilpotency and solvable class ! -- too. The factor algebra by the nilradical is isomorphic to a direct product of full matrix algebras over L. Let (by using the theorem of Wedderburn-Malcev) $T$ be a unital subalgebra of $A$ and $I_1,...,I_r$ simple ideals of $T$ and $n_1,\cdots , n_r\in \mathbb{N}$ such that $T=I_1 \oplus \cdots \oplus I_r$ and $I_i\cong L^{n_i\times n_i}$ are valid for all $1\le i \le r$. The Cartan subalgebras are the centralizers of the maximal tori of $A$. We will proove that all maximal tori are conjugated. Thus, all Cartan subalgebras are conjugated, too, and the theorem is proven. For all $1\le i\le r$ let $e_i$ the unit of $I_i$. $T$ is a unital subalgebra of $A$ and $1_A=1_T=e_1+ \cdots +e_r$ is valid. Let $X$ be a maximal torus of $A$. Then $X$ can be conjugated into $T$ (by a generalized version of the Wedderburn-Malcev theorem.). $T$ is separable and hence maximal tori and CSA are identical and they are the direct sum of the CSAs of the simple components of $T$. Hence for all $1\le i\le r$ there exists a CSA $X_i$ of $ I_i$ with $X=X_1+\cdots + X_r$. By a theorem of Jacobson the CSAs of full matrix algebras over algebraically closed fields are exactly the conjugates of the subalgebra of diagonal matrices. For all $1\le i\le r$ let $D_i$ be a subalgebra of $I_i$ isomorphic to the subalgebra of diagonal matrices of $L^{n_i\times n_i}$. We conclude that for all $1\le i\le r$ there exists a unit $g_i$ in $I_i$ such that $X_i^{g_i}=D_i$ is valid. The element $g:=g_1+\cdots + g_r$ is a unit of $A$ (Let $h_i\in I_i$ with $g_i \cdot h_i=e_i$ for all $1\le i\le r$. Then $g\cdot (h_1+\cdots + h_r)=e_1+\cdots +e_r=1_A$ because the ideals $I_1,\cdots ,I_r$ are direct). By the directness of the ideals decomposing $T$ we conclude $X^g=X_1^{g_1}+ \cdots +X_r^{g_r}=D_1+\cdots +D_r$.


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