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 ${\displaystyle X\in {\mathfrak {g}}}$ is regular if its centralizer in ${\displaystyle {\mathfrak {g}}}$ has dimension equal to the rank of ${\displaystyle {\mathfrak {g}}}$, which in turn equals the dimension of some Cartan subalgebra ${\displaystyle {\mathfrak {h}}}$ (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 ${\displaystyle g\in G}$ a Lie group is regular if its centralizer has dimension equal to the rank of ${\displaystyle G}$.

In the specific case of ${\displaystyle {\mathfrak {gl}}_{n}(\mathbb {k} )}$, the Lie algebra of ${\displaystyle n\times n}$ matrices over an algebraically closed field ${\displaystyle \mathbb {k} }$(such as the complex numbers), a regular element ${\displaystyle M}$ 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 ${\displaystyle n}$ evaluated at the matrix ${\displaystyle M}$, and therefore the centralizer has dimension ${\displaystyle n}$ (which equals the rank of ${\displaystyle {\mathfrak {gl}}_{n}}$, but is not necessarily an algebraic torus).

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

For a connected compact Lie group ${\displaystyle G}$, the regular elements form an open dense subset, made up of ${\displaystyle G}$-conjugacy classes of the elements in a maximal torus ${\displaystyle T}$ which are regular in ${\displaystyle G}$. The regular elements of ${\displaystyle T}$ are themselves explicitly given as the complement of a set in ${\displaystyle T}$, a set of codimension-one subtori corresponding to the root system of ${\displaystyle G}$. Similarly, in the Lie algebra ${\displaystyle {\mathfrak {g}}}$ of ${\displaystyle G}$, the regular elements form an open dense subset which can be described explicitly as adjoint ${\displaystyle G}$-orbits of regular elements of the Lie algebra of ${\displaystyle T}$, the elements outside the hyperplanes corresponding to the root system.^[1]

Let ${\displaystyle {\mathfrak {g}}}$ be a finite-dimensional Lie algebra over an infinite field.^[2] For each ${\displaystyle x\in {\mathfrak {g}}}$, let

${\displaystyle p_{x}(t)=\det(t-\operatorname {ad} (x))=\sum _{i=0}^{\dim {\mathfrak {g}}}a_{i}(x)t^{i}}$

be the characteristic polynomial of the adjoint endomorphism ${\displaystyle \operatorname {ad} (x):y\mapsto [x,y]}$ of ${\displaystyle {\mathfrak {g}}}$. Then, by definition, the rank of ${\displaystyle {\mathfrak {g}}}$ is the least integer ${\displaystyle r}$ such that ${\displaystyle a_{r}(x)\neq 0}$ for some ${\displaystyle x\in {\mathfrak {g}}}$ and is denoted by ${\displaystyle \operatorname {rk} ({\mathfrak {g}})}$.^[3] For example, since ${\displaystyle a_{\dim {\mathfrak {g}}}(x)=1}$ for every x, ${\displaystyle {\mathfrak {g}}}$ is nilpotent (i.e., each ${\displaystyle \operatorname {ad} (x)}$ is nilpotent by Engel's theorem) if and only if ${\displaystyle \operatorname {rk} ({\mathfrak {g}})=\dim {\mathfrak {g}}}$.

Let ${\displaystyle {\mathfrak {g}}_{\text{reg}}=\{x\in {\mathfrak {g}}|a_{\operatorname {rk} ({\mathfrak {g}})}(x)\neq 0\}}$. By definition, a regular element of ${\displaystyle {\mathfrak {g}}}$ is an element of the set ${\displaystyle {\mathfrak {g}}_{\text{reg}}}$.^[3] Since ${\displaystyle a_{\operatorname {rk} ({\mathfrak {g}})}}$ is a polynomial function on ${\displaystyle {\mathfrak {g}}}$, with respect to the Zariski topology, the set ${\displaystyle {\mathfrak {g}}_{\text{reg}}}$ is an open subset of ${\displaystyle {\mathfrak {g}}}$.

Over ${\displaystyle \mathbb {C} }$, ${\displaystyle {\mathfrak {g}}_{\text{reg}}}$ is a connected set (with respect to the usual topology),^[4] but over ${\displaystyle \mathbb {R} }$, it is only a finite union of connected open sets.^[5]

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 ${\displaystyle x\in {\mathfrak {g}}}$, let

${\displaystyle {\mathfrak {g}}^{0}(x)=\sum _{n\geq 0}\ker(\operatorname {ad} (x)^{n}:{\mathfrak {g}}\to {\mathfrak {g}})}$

be the generalized eigenspace of ${\displaystyle \operatorname {ad} (x)}$ for eigenvalue zero. It is a subalgebra of ${\displaystyle {\mathfrak {g}}}$.^[6] Note that ${\displaystyle \dim {\mathfrak {g}}^{0}(x)}$ is the same as the (algebraic) multiplicity^[7] of zero as an eigenvalue of ${\displaystyle \operatorname {ad} (x)}$; i.e., the least integer m such that ${\displaystyle a_{m}(x)\neq 0}$ in the notation in § Definition. Thus, ${\displaystyle \operatorname {rk} ({\mathfrak {g}})\leq \dim {\mathfrak {g}}^{0}(x)}$ and the equality holds if and only if ${\displaystyle x}$ is a regular element.^[3]

The statement is then that if ${\displaystyle x}$ is a regular element, then ${\displaystyle {\mathfrak {g}}^{0}(x)}$ is a Cartan subalgebra.^[8] Thus, ${\displaystyle \operatorname {rk} ({\mathfrak {g}})}$ is the dimension of at least some Cartan subalgebra; in fact, ${\displaystyle \operatorname {rk} ({\mathfrak {g}})}$ is the minimum dimension of a Cartan subalgebra. More strongly, over a field of characteristic zero (e.g., ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$),^[9]

• every Cartan subalgebra of ${\displaystyle {\mathfrak {g}}}$ has the same dimension; thus, ${\displaystyle \operatorname {rk} ({\mathfrak {g}})}$ is the dimension of an arbitrary Cartan subalgebra,
• an element x of ${\displaystyle {\mathfrak {g}}}$ is regular if and only if ${\displaystyle {\mathfrak {g}}^{0}(x)}$ is a Cartan subalgebra, and
• every Cartan subalgebra is of the form ${\displaystyle {\mathfrak {g}}^{0}(x)}$ for some regular element ${\displaystyle x\in {\mathfrak {g}}}$.

For a Cartan subalgebra ${\displaystyle {\mathfrak {h}}}$ of a complex semisimple Lie algebra ${\displaystyle {\mathfrak {g}}}$ with the root system ${\displaystyle \Phi }$, an element of ${\displaystyle {\mathfrak {h}}}$ is regular if and only if it is not in the union of hyperplanes ${\textstyle \bigcup _{\alpha \in \Phi }\{h\in {\mathfrak {h}}\mid \alpha (h)=0\}}$.^[10] This is because: for ${\displaystyle r=\dim {\mathfrak {h}}}$,

• For each ${\displaystyle h\in {\mathfrak {h}}}$, the characteristic polynomial of ${\displaystyle \operatorname {ad} (h)}$ is ${\textstyle t^{r}\left(t^{\dim {\mathfrak {g}}-r}-\sum _{\alpha \in \Phi }\alpha (h)t^{\dim {\mathfrak {g}}-r-1}+\cdots \pm \prod _{\alpha \in \Phi }\alpha (h)\right)}$.

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