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Let $L$ be a lie algebra and $H$ an abelian subalgebra of $L$ such that each element of $h \in H$ is diagonalizable under the adjoint representation. So there exists a basis of common eigenvectors for $ad(h)$ for all $h \in H$ and non zero functionals $\lambda_1 \ldots \lambda_n \in H^*$ \begin{gather} L=C_L(H) \oplus \bigoplus_{i=1}^n L_{\lambda_i} \end{gather} With $C_L(H)$ is the centralizer of $H$ and $L_\lambda =\{x \in L: [h,x]=\lambda(h)x, \forall h\in H\} $. Now if $H$ is maximal among the abelian lie sub-algebras that act diagonalizable on $L$ via the adjoint representation, the functionals $\lambda_i$ are a root system for $H^*$ and $H$ is self-centralizing. What can we say if $H$ is not maximal?

For examble, if we don´t know that $H$ is maximal but we find out that the $\lambda_i$ are a root system for $H^*$, can we say that the decomposition of $L$ above is a root decomposition and $H$ self-centralizing? In general, are there some ways to obtain information about how $H$ is embedded in $L$ from the eigenvalues $\lambda_i$?

Lorban
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1 Answers1

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I restrict to the case that we have an algebraically closed base field of characteristic 0, and $L$ is semisimple. I call the subalgebras $H$ you are interested in -- abelian subalgebras whose elements are ad-diagonalisable -- "split toral", as they very much resemble split tori in algebraic groups.

Remark first of all that $H= \lbrace 0 \rbrace$ should always be kept in mind as extreme case for some (counter)examples.

Good news: We always have a weight decomposition \begin{gather} L=C_L(H) \oplus \bigoplus_{i=1}^n L_{\lambda_i} \end{gather}

with $\lambda_i$ the non-zero weights of $H$ in $L$, and one can show (e.g. using the Killing form) that:

i) The functionals $\lambda_i$ span $H^*$.

ii) If $\lambda_i$ is a non-zero weight, then so is $-\lambda_i$.

(But see what happens in the case $H = 0$?)

Bad news: Remark first that if $H$ is split toral but not maximal split toral, its centraliser must be strictly bigger than itself, since of course every abelian algebra that contains it centralises it, and "non-maximal" means there is a bigger one. Further, each of the following three cases can happen for $H$ split toral but not maximal split toral:

1) The $\lambda_i$ form a reduced root system.

Example: $L = \mathfrak{sl}_4(\mathbb{C})$, $H = \lbrace \pmatrix{x & 0 & 0 & 0\\ 0 & y & 0 & 0\\ 0 & 0 & -y & 0\\ 0 & 0 & 0 & -x} : x, y \in \mathbb{C} \rbrace$;

calling $\lambda_1$ the functional sending $\pmatrix{x & 0 & 0 & 0\\ 0 & y & 0 & 0\\ 0 & 0 & -y & 0\\ 0 & 0 & 0 & -x}$ to $2y$

and $\lambda_2$ the one sending $\pmatrix{x & 0 & 0 & 0\\ 0 & y & 0 & 0\\ 0 & 0 & -y & 0\\ 0 & 0 & 0 & -x}$ to $x-y$,

then all the non-zero weights are:

\begin{gather} \lambda_1, \lambda_2, \lambda_1 + \lambda_2, \lambda_1 + 2\lambda_2 \text{ and their negatives.} \end{gather}

They form a root system of type $B_2$. (Notice that the root spaces $L_\lambda$ are two-dimensional for $\lambda \in \lbrace \pm \lambda_2, \pm (\lambda_1+\lambda_2) \rbrace$.)

2) The $\lambda_i$ form a non-reduced root system.

Example: $L = \mathfrak{sl}_3(\mathbb{C})$, $H = \lbrace \pmatrix{x & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -x } : x, \in \mathbb{C} \rbrace$;

calling $\lambda$ the functional sending $\pmatrix{x & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -x }$ to $x$,

then all the non-zero weights are $\lbrace \pm \lambda, \pm 2 \lambda \rbrace$. They form a non-reduced root system of type $BC_1$. (Notice that the root spaces of $ \pm\lambda$ are two-dimensional each.)

3) The $\lambda_i$ do not form a root system at all.

Example: Well, $H = \lbrace 0 \rbrace$.

Different, fancier example: $L = \mathfrak{sl}_4(\mathbb{C})$, $H = \lbrace \pmatrix{3x & 0 & 0 & 0\\ 0 & x & 0 & 0\\ 0 & 0 & -3x & 0\\ 0 & 0 & 0 & -x} : x \in \mathbb{C} \rbrace$;

calling $\lambda$ the functional sending $\pmatrix{3x & 0 & 0 & 0\\ 0 & x & 0 & 0\\ 0 & 0 & -3x & 0\\ 0 & 0 & 0 & -x}$ to $2x$,

then all the non-zero weights are: $\pm \lambda, \pm 2\lambda, \pm 3\lambda$ which can never be a root system. Notice that each of their root spaces (maybe better called weight spaces now) are two-dimensional.

(Final remark: Over non-algebraically closed fields $k$ (but still assuming $char(k) = 0$), even maximal split toral algebras are not necessarily self-centralising, their centraliser contains the so-called "anisotropic kernel" of $L$. But their non-zero weights still define a (possibly non-reduced) root system, the "rational" or "relative" root system.)

Torsten Schoeneberg
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