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