Regular element 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 $$X \in \mathfrak{g}$$ is regular if its centralizer in $$\mathfrak{g}$$ has dimension equal to the rank of $$\mathfrak{g}$$, which in turn equals the dimension of some Cartan subalgebra $$\mathfrak{h}$$ (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 $$\mathfrak{gl}_n(\mathbb{k})$$, the Lie algebra of $$n \times n $$ matrices over an algebraically closed field $$\mathbb{k}$$(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 $$\mathfrak{gl}_n$$, 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 $$\mathfrak{g}$$ 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 $$\mathfrak{g}$$ be a finite-dimensional Lie algebra over an infinite field. For each $$x \in \mathfrak{g}$$, let
 * $$p_x(t) = \det(t - \operatorname{ad}(x)) = \sum_{i=0}^{\dim \mathfrak{g}} a_i(x) t^i$$

be the characteristic polynomial of the adjoint endomorphism $$\operatorname{ad}(x) : y \mapsto [x, y]$$ of $$\mathfrak g$$. Then, by definition, the rank of $$\mathfrak{g}$$ is the least integer $$r$$ such that $$a_r(x) \ne 0$$ for some $$x \in \mathfrak g$$ and is denoted by $$\operatorname{rk}(\mathfrak{g})$$. For example, since $$a_{\dim \mathfrak g}(x) = 1$$ for every x, $$\mathfrak g$$ is nilpotent (i.e., each $$\operatorname{ad}(x)$$ is nilpotent by Engel's theorem) if and only if $$\operatorname{rk}(\mathfrak{g}) = \dim \mathfrak g$$.

Let $$\mathfrak{g}_{\text{reg}} = \{ x \in \mathfrak{g} | a_{\operatorname{rk}(\mathfrak{g})} (x) \ne 0 \}$$. By definition, a regular element of $$\mathfrak{g}$$ is an element of the set $$\mathfrak{g}_{\text{reg}}$$. Since $$a_{\operatorname{rk}(\mathfrak{g})}$$ is a polynomial function on $$\mathfrak{g}$$, with respect to the Zariski topology, the set $$\mathfrak{g}_{\text{reg}}$$ is an open subset of $$\mathfrak{g}$$.

Over $$\mathbb{C}$$, $$\mathfrak{g}_{\text{reg}}$$ is a connected set (with respect to the usual topology), but over $$\mathbb{R}$$, 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 $$x \in \mathfrak{g}$$, let
 * $$\mathfrak{g}^0(x) = \sum_{n \ge 0} \ker(\operatorname{ad}(x)^n : \mathfrak{g} \to \mathfrak{g})$$

be the generalized eigenspace of $$\operatorname{ad}(x)$$ for eigenvalue zero. It is a subalgebra of $$\mathfrak g$$. Note that $$\dim \mathfrak{g}^0(x)$$ is the same as the (algebraic) multiplicity of zero as an eigenvalue of $$\operatorname{ad}(x)$$; i.e., the least integer m such that $$a_m(x) \ne 0$$ in the notation in. Thus, $$\operatorname{rk}(\mathfrak g) \le \dim \mathfrak{g}^0(x)$$ 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 $$\mathfrak{g}^0(x)$$ is a Cartan subalgebra. Thus, $$\operatorname{rk}(\mathfrak g)$$ is the dimension of at least some Cartan subalgebra; in fact, $$\operatorname{rk}(\mathfrak g)$$ is the minimum dimension of a Cartan subalgebra. More strongly, over a field of characteristic zero (e.g., $$\mathbb{R}$$ or $$\mathbb{C}$$),
 * every Cartan subalgebra of $$\mathfrak{g}$$ has the same dimension; thus, $$\operatorname{rk}(\mathfrak g)$$ is the dimension of an arbitrary Cartan subalgebra,
 * an element x of $$\mathfrak g$$ is regular if and only if $$\mathfrak{g}^0(x)$$ is a Cartan subalgebra, and
 * every Cartan subalgebra is of the form $$\mathfrak{g}^0(x)$$ 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 $\bigcup_{\alpha \in \Phi} \{ h \in \mathfrak{h} \mid \alpha(h) = 0 \}$. This is because: for $$r = \dim \mathfrak h$$,
 * For each $$h \in \mathfrak{h}$$, the characteristic polynomial of $$\operatorname{ad}(h)$$ is $t^r \left(t^{\dim \mathfrak g - r} - \sum_{\alpha \in \Phi} \alpha(h) t^{\dim \mathfrak g - r - 1} + \cdots \pm \prod_{\alpha \in \Phi} \alpha(h)\right)$.

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