# 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 $a_{r}(x)\neq 0$ 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 ${\mathfrak {g}}_{\text{reg}}=\{x\in {\mathfrak {g}}|a_{\operatorname {rk} ({\mathfrak {g}})}(x)\neq 0\}$. 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
    ${\mathfrak {g}}^{0}(x)=\sum _{n\geq 0}\ker(\operatorname {ad} (x)^{n}:{\mathfrak {g}}\to {\mathfrak {g}})$
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 $a_{m}(x)\neq 0$ in the notation in § Definition. Thus, $\operatorname {rk} ({\mathfrak {g}})\leq \dim {\mathfrak {g}}^{0}(x)$ 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 ⋃$_{α∈Φ}${h∈h∣α(h)=0}{\textstyle \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$^r$(t$^{dim⁡g−r}$−∑$_{α∈Φ}$α(h)t$^{dim⁡g−r−1}$+⋯±∏$_{α∈Φ}$α(h)){\textstyle 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).