From: Boudewijn Moonen Subject: Re: Lie algebra Date: Thu, 11 Mar 1999 10:40:20 +0100 Newsgroups: sci.math.research Keywords: Representations of Lie algebras -- reductive etc. bossy wrote: > I am reading a paper by Armand Borel and have received help from here before > (many thanks). What is the definition of a Lie algebra being > 1) completely reducible > 2) reductive in gl(n, R) > 3) reductive > > In fact the paper does define these concepts but then proceeds to use the > term "fully reducible" apparently in place of reductive without any > explanation. > > The paper is Arithmetic Subgroups of Algebraic Groups by Borel & Harish > Chandra. In particular Lemma 1.5 > > Many thanks > > Phil A) GENERALITIES As far as I can see it, the standard definitions are as follows (general references for the following are Bourbaki's Chapter I on Lie algebras and Chevalley's book on Lie groups,, in particular Volume III): Let g be a Lie algebra (G a Lie group) over the real or complex numbers. Then - a representation of g (of G) on a vector space V is called \emph{completely} or \emph{fully reducible} if every invariant subspace of V has an invariant vector space complement. This is equivalent to saying that it decomposes as a direct sum of irreducible representations - g (or G) is called \emph{completely} or \emph{fully reducible} if every finite dimensional representation is so - g is called \emph{reductive} if its adjoint representation is fully reducible - if g is a Lie subalgebra of a Lie algebra g' it is called \emph{reductive in g'} if the adjoint representation of g' restricted to g is fully reducible One has the following facts: 1) According to a famous and nontrivial theorem of Weyl a semisimple Lie algebra (group) is fully reducible; the converse also holds rather trivially (see Bourbaki) 2) The following properties for a Lie algebra are equivalent: (i) g is reductive (ii) [g,g] is semisimple (iii) g = s + a as a direct sum of a semisimple Lie algebra s and an abelian Lie algebra a (iv) the radical r(g) of g coincides with the centre c(g) of g If this is the case, s = [g,g], a = c(g). 3) From this, it is not difficult to prove: For a Lie algebra g the following properties are equivalent (i) g is reductive (ii) a finite dimensional representation V of g is fully reducible if and only if c(g) acts by semisimple endomorphisms in V. (Use the direct sum decomposition g = [g,g] + c(g) and Weyl's Theorem). 4) As a consequence one has for a Lie subalgebra of a Lie algebra g' the equivalence (i) g is reductive in g' (ii) g is reductive, and all elements of c(g) are semisimple In particular, for a Lie group G with finitely many connected components, or for a Lie algebra g, being fully reducible is equivalent to being semisimple and definitely not equivalent to being reductive; e.g the Lie algebra gl(n,R) is reductive, and the twodimensional representation - - | 0 trace X | X |---> | | | 0 0 | - - of gl(n,R) clearly is not fully reducible. B) THE BOREL--HARISH-CHANDRA PAPER The confusion regarding the relations between the notions "fully reducible" and "reductive" are possibly caused by the following two points. Point 1. In case of a Lie algebra, or Lie group, of endomorphisms of a finite dimensional vector space V - which is the case considered in the paper - there appears to be a conflicting terminology going back to the older literature. E.g. if one consults Jacobson' Lie algebra book one will find that in this case g or G sometimes is called fully reducible if V is a fully reducible g(or G)-module. So to prevent confusion one should call this property \emph{fully reducible in V}. With this terminology, a Lie algebra g is reductive iff if ad g is fully reducible in gl(g), and g is reductive in g'iff ad g is fully reducible in gl(g'). Since an endomorphism X in g is semisimple iff the endomorphism ad X in gl(g) is semisimple, we have because of 3): A Lie subalgebra g of gl(V) is reductive in gl(V) iff it is fully reducible in V. So, because of 3) and 4), the following properties are equivalent in this case: (i) G is fully reducible in V (ii) g is fully reducible in V (iii) g is reductive in gl(V) (iv) g is reductive, and c(g) consists of semisimple endomorphisms of V which is my proposal for how to interprete the lemma in 1.2 of the paper. Note that because of the example above (iv) is not equivalent to G being fully reducible in the general sense of A), since g=gl(n,R) satisfies (iv). Point 2 An additional subtlety appears in the definition of a \emph{reductive} algebraic group in 3.1 of the paper. For arbitrary (connected) Lie groups, this is done in such a way that a Lie group is reductive iff its corresponding Lie algebra is reductive. In the algebraic case, however, one makes the additional requirement that the connected component of the centre of G is a \emph{diagonizable} commutative connected group and so consists of \emph{semisimple} endomorphisms of the vector space V, where by definition G is an algebraic subgroup of GL(V). This means that an algebraic group G is \emph{defined} to be reductive in such a way that this holds iff a) its Lie algebra g is reductive (as an abstract Lie algebra) and,\emph{in addition}, b) the centre c(g) of g consists of semisimple endomorphisms of V, i.e. according to (iii) above iff g is reductive in gl(V). We further have that G is fully reducible in V iff g is fully reducible in V. Since, finally, semisimplicity of elements of G (or g) is preserved in rational representations (see e.g. Theorem 15.3 in Humphrey's book on algebraic groups), we have the following equivalences: (see 3.1 of the paper) (i) G is reductive (ii) G is fully reducible in V, i.e. V decomposes as a direct sum of irreducible rational representations of G (ii)' g is fully reducible in V, i.e. V decomposes as a direct sum of irreducible rational representations of g (iii) g is reductive in gl(V), i.e. gl(V) decomposes as a direct sum of irreducible rational representations of g under the adjoint representation of gl(V) restricted to g (iv) every rational representation of G (or g) is fully reducible If (iv) would hold for all representations of G as a Lie group, G would be semisimple; restricting it to rational representations then characterizes the reductive algebraic groups. (In the same way, the conditions a) and b) above are requirements for reductive \emph{algebraic} Lie algebras - quite a silly notion, "algebraic Lie algebra", but, alas, it is standard-.) In particular, the two-dimensional representation of gl(n,R) above does not integrate into a rational representation of GL(n,R), as one can easily check (the multiplicative group G_m and the additive group G_a are not isomorphic as algebraic groups, i.e. there does not exist an "algebraic exponential function"). Hope this helps. Best regards. -- Dr. Boudewijn Moonen Institut fuer Photogrammetrie der Universitaet Bonn Nussallee 15 D-53115 Bonn GERMANY e-mail: Boudewijn.Moonen@ipb.uni-bonn.de Tel.: GERMANY +49-228-732910 Fax.: GERMANY +49-228-732712