Rank Of A Lie Algebra

In mathematics, a regular element of a Lie algebra or Lie group is an element whose centralizer has dimension as small as possible.
For example, in a complex semisimple Lie algebra, an element $X \in \mathfrak$ is regular if its centralizer in $\mathfrak$ has dimension equal to the rank of $\mathfrak$, which in turn equals the dimension of some Cartan subalgebra $\mathfrak$ (note that in earlier papers, an element of a complex semisimple Lie algebra was termed regular if it is semisimple and the kernel of its adjoint representation is a Cartan subalgebra).
An element $g \in G$ a Lie group is regular if its centralizer has dimension equal to the rank of $G$.

Basic case

In the specific case of $\mathfrak_n(\mathbb)$, the Lie algebra of $n \times n$ matrices over an algebraically closed field $\mathbb$(such as the complex numbers), a regular element $M$ is an element whose Jordan normal form contains a single Jordan block for each eigenvalue (in other words, the geometric multiplicity of each eigenvalue is 1).
The centralizer of a regular element is the set of polynomials of degree less than $n$ evaluated at the matrix $M$, and therefore the centralizer has dimension $n$ (which equals the rank of $\mathfrak_n$, but is not necessarily an algebraic torus).

If the matrix $M$ is diagonalisable, then it is regular if and only if there are $n$ different eigenvalues. To see this, notice that $M$ will commute with any matrix $P$ that stabilises each of its eigenspaces. If there are $n$ different eigenvalues, then this happens only if $P$ is diagonalisable on the same basis as $M$; in fact $P$ is a linear combination of the first $n$ powers of $M$, and the centralizer is an algebraic torus of complex dimension $n$ (real dimension $2n$); since this is the smallest possible dimension of a centralizer, the matrix $M$ is regular. However if there are equal eigenvalues, then the centralizer is the product of the general linear groups of the eigenspaces of $M$, and has strictly larger dimension, so that $M$ is not regular.

For a connected compact Lie group $G$, the regular elements form an open dense subset, made up of $G$-conjugacy classes of the elements in a maximal torus $T$ which are regular in $G$. The regular elements of $T$ are themselves explicitly given as the complement of a set in $T$, a set of codimension-one subtori corresponding to the root system of $G$. Similarly, in the Lie algebra $\mathfrak$ of $G$, the regular elements form an open dense subset which can be described explicitly as adjoint $G$-orbits of regular elements of the Lie algebra of $T$, the elements outside the hyperplanes corresponding to the root system.

Definition

Let $\mathfrak$ be a finite-dimensional Lie algebra over an infinite field. For each $x \in \mathfrak$, let
:$p_x(t) = \det(t - \operatorname(x)) = \sum_^ a_i(x) t^i$
be the characteristic polynomial of the adjoint endomorphism \operatorname(x) : y \mapsto , y/math> of $\mathfrak g$. Then, by definition, the rank of $\mathfrak$ is the least integer $r$ such that $a_r(x) \ne 0$ for some $x \in \mathfrak g$ and is denoted by $\operatorname(\mathfrak)$. For example, since $a_(x) = 1$ for every ''x'', $\mathfrak g$ is nilpotent (i.e., each $\operatorname(x)$ is nilpotent by Engel's theorem) if and only if $\operatorname(\mathfrak) = \dim \mathfrak g$.

Let $\mathfrak_ = \$. By definition, a regular element of $\mathfrak$ is an element of the set $\mathfrak_$. Since $a_$ is a polynomial function on $\mathfrak$, with respect to the Zariski topology, the set $\mathfrak_$ is an open subset of $\mathfrak$.

Over $\mathbb$, $\mathfrak_$ is a connected set (with respect to the usual topology), but over $\mathbb$, it is only a finite union of connected open sets.

A Cartan subalgebra and a regular element

Over an infinite field, a regular element can be used to construct a Cartan subalgebra, a self-normalizing nilpotent subalgebra. Over a field of characteristic zero, this approach constructs all the Cartan subalgebras.

Given an element $x \in \mathfrak$, let
:$\mathfrak^0(x) = \bigcup_ \ker(\operatorname(x)^n : \mathfrak \to \mathfrak)$
be the generalized eigenspace of $\operatorname(x)$ for eigenvalue zero. It is a subalgebra of $\mathfrak g$. Note that $\dim \mathfrak^0(x)$ is the same as the (algebraic) multiplicity of zero as an eigenvalue of $\operatorname(x)$; i.e., the least integer ''m'' such that $a_m(x) \ne 0$ in the notation in . Thus, $\operatorname(\mathfrak g) \le \dim \mathfrak^0(x)$ and the equality holds if and only if $x$ is a regular element.

The statement is then that if $x$ is a regular element, then $\mathfrak^0(x)$ is a Cartan subalgebra. Thus, $\operatorname(\mathfrak g)$ is the dimension of at least some Cartan subalgebra; in fact, $\operatorname(\mathfrak g)$ is the minimum dimension of a Cartan subalgebra. More strongly, over a field of characteristic zero (e.g., $\mathbb$ or $\mathbb$),
*every Cartan subalgebra of $\mathfrak$ has the same dimension; thus, $\operatorname(\mathfrak g)$ is the dimension of an arbitrary Cartan subalgebra,
*an element ''x'' of $\mathfrak g$ is regular if and only if $\mathfrak^0(x)$ is a Cartan subalgebra, and
*every Cartan subalgebra is of the form $\mathfrak^0(x)$ for some regular element $x \in \mathfrak g$.

A regular element in a Cartan subalgebra of a complex semisimple Lie algebra

For a Cartan subalgebra $\mathfrak h$ of a complex semisimple Lie algebra $\mathfrak g$ with the root system $\Phi$, an element of $\mathfrak h$ is regular if and only if it is not in the union of hyperplanes $\bigcup_ \$. This is because: for $r = \dim \mathfrak h$,
*For each $h \in \mathfrak$, the characteristic polynomial of $\operatorname(h)$ is

This characterization is sometimes taken as the definition of a regular element (especially when only regular elements in Cartan subalgebras are of interest).

Notes

References

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*{{citation , first=Jean-Pierre , last=Serre , title=Complex Semisimple Lie Algebras , publisher=Springer , year=2001 , isbn=3-5406-7827-1

Lie groups
Lie algebras