Regular element of a Lie algebra#Definition

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.{{cite book |first=Mark R. |last=Sepanski |title=Compact Lie Groups |url=https://books.google.com/books?id=F3NgD_25OOsC&pg=PA156 |date=2006 |publisher=Springer |isbn=978-0-387-30263-8 |pages=156}}

Definition

Let \mathfrak{g} be a finite-dimensional Lie algebra over an infinite field.Editorial note: the definition of a regular element over a finite field is unclear. 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}).{{harvnb|Bourbaki|1981|loc=Ch. VII, § 2.2. Definition 2.}} 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),{{harvnb|Serre|2001|loc=Ch. III, § 1. Proposition 1.}} but over \mathbb{R}, it is only a finite union of connected open sets.{{harvnb|Serre|2001|loc=Ch. III, § 6.}}

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.This is a consequence of the binomial-ish formula for ad. Note that \dim \mathfrak{g}^0(x) is the same as the (algebraic) multiplicityRecall that the geometric multiplicity of an eigenvalue of an endomorphism is the dimension of the eigenspace while the algebraic multiplicity of it is the dimension of the generalized eigenspace. 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 {{slink||Definition}}. 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.{{harvnb|Bourbaki|1981|loc=Ch. VII, § 2.3. Theorem 1.}} 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}),{{harvnb|Bourbaki|1981|loc=Ch. VII, § 3.3. Theorem 2.}}

  • 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 \}.{{harvnb|Procesi|2007|loc=Ch. 10, § 3.2.}} This is because: for r = \dim \mathfrak h,

  • For each h \in \mathfrak{h}, the characteristic polynomial of \operatorname{ad}(h) is {{nowrap|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).

Notes

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References

  • {{citation|last=Bourbaki|first=N.|series=Éléments de Mathématique|title=Groupes et Algèbres de Lie|publisher=Hermann|year=1981}}
  • {{Citation | last1=Fulton | first1=William | author1-link=William Fulton (mathematician) | last2=Harris | first2=Joe | author2-link=Joe Harris (mathematician) | title=Representation Theory, A First Course | publisher=Springer-Verlag | location=Berlin, New York | series=Graduate Texts in Mathematics | isbn=978-0-387-97495-8 |mr=1153249 | year=1991 | volume=129}}
  • {{citation |author-link=Claudio Procesi |first=Claudio |last=Procesi |title=Lie Groups: an approach through invariants and representation |publisher=Springer |year=2007 |isbn=9780387260402 }}
  • {{citation |first=Jean-Pierre |last=Serre |title=Complex Semisimple Lie Algebras |publisher=Springer |year=2001 |isbn=3-5406-7827-1 }}

Category:Lie groups

Category:Lie algebras