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I would like to find a reference for the following fact: every finite dimensional complex representation of a reductive Lie algebra is semisimple.

edited Apr 30 2010 at 14:07

José Figueroa-O'Farrill
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asked Apr 30 2010 at 8:18

Michele Torielli
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Note that the result in question is short enough to be mentioned in the title. Generic titles are not very helpful. – François G. Dorais♦ Apr 30 2010 at 8:24

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Does "semisimple" for representations mean "completely reducible"? If so this is false as the only Lie algebras for which all finite-dimensional representations are completely reducible are the semisimple Lie algebras. – Robin Chapman Apr 30 2010 at 8:45

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Robin is correct. Note that wikipedia states this result (and is therefore incorrect): en.wikipedia.org/wiki/… – Pete L. Clark Apr 30 2010 at 9:48

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In his textbook, Humphreys sets as an exercise that a finite-dimensional representation of a reductive Lie algebra $L$ is completely reducible if every element in the centre of $L$ acts as a semisimple endomorphism. – Robin Chapman Apr 30 2010 at 10:33

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I edited the title -- if the question is going to remain open, then at the very least it ought to have a title which conveys some information. – José Figueroa-O'Farrill Apr 30 2010 at 14:07

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Jim Humphreys · Accepted Answer · 2010-04-30 10:44:04Z UTC

The statement is false. The standard definition of "reductive" for a finite dimensional Lie algebra $\mathfrak{g}$ over an arbitrary field of characteristic 0 is given in a number of equivalent ways by Bourbaki in Chapter 1 (1960) of their treatise on Lie groups and Lie algebras: section 6, no. 4-5. By definition, $\mathfrak{g}$ is reductive provided its adjoint representation is semisimple (= completely reducible). Typical equivalent conditions: the derived algebra is semisimple; or $\mathfrak{g}$ is the direct sum of a semisimple and an abelian Lie algebra; or the solvable radical equals the center. As a consequence, a finite dimensional representation of a reductive Lie algebra is semisimple iff the center acts by semisimple endomorphisms. (An abelian Lie algebra need not be represented in that way.)

Some of this is set up as an exercise at the end of Section 6 in my Springer graduate text (1972); see also Proposition 19.1.

José Figueroa-O'Farrill · Answer 2 · 2010-04-30 11:31:30Z UTC

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In many applications, a (real) reductive Lie algebra arises as the Lie algebra of a compact Lie group. In this case, and if the representation integrates to one of the group, then it is fully reducible by a version of Weyl's unitary trick. Basically every finite-dimensional module is unitarisable and every submodule has a complementary submodule: namely, its perpendicular complement.

answered Apr 30 2010 at 11:31

José Figueroa-O'Farrill
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This reinforces the fact that the notion of "reductive" for a Lie algebra in characteristic 0 has no intrinsic interest, unless you study the Lie algebra of a Lie (or algebraic) group and relate their representations carefully. For affine algebraic groups in any characteristic the notion of "reductive" group is more interesting because of Chevalley's Jordan decomposition and its preservation under rational representations. (But in characteristic $p$ you lose the connection with complete reducibility: a reductive group is "geometrically" reductive but rarely "linearly" reductive.) – Jim Humphreys Apr 30 2010 at 12:24

In my case, I'm studying exactly the Lie algebra of an algebraic group.Can you please your statement more precise? what happens in this type of cases? – Michele Torielli May 6 2010 at 10:21

Victor Protsak · Answer 3 · 2010-05-01 00:58:20Z UTC

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To complement Jim's answer, there is a thorough discussion of complete reducibility for reductive Lie algebras (with proofs, but only in char=0) in Sections 1.6 and 1.7 of Dixmier's "Enveloping algebras", which I found much less intimidating then reading Bourbaki.

answered May 1 2010 at 0:58

Victor Protsak
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Dixmier is certainly a useful reference (though Bourbaki is also available in English). One caution: the North-Holland translation of Dixmier (translator unidentified) was republished without change by AMS in 1996 but listed some misprints at the end; in fact this doesn't include all of the misprints introduced in the translation. I encouraged the AMS to acquire this out-of-print book for the GSM series but wasn't aware then of the extent of misprints. On the other hand, Dixmier added some updates to the English version. – Jim Humphreys May 1 2010 at 14:19

Homology · Answer 4 · 2010-04-30 09:55:58Z UTC

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You can find the proof in Serre's "Lie Algebras and Lie Groups", in chapter "Semisimple Lie Algebras", section "Complete Reducibility"

answered Apr 30 2010 at 9:55

Homology
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so, in your opinion, the result is it true? – Michele Torielli Apr 30 2010 at 9:58

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@Michele: The result that you want is false. The standard example of non-semisimple representation of an abelian Lie algebra is $\begin{pmatrix} 0 & \mathbb{C} \cr 0 & 0 \end{pmatrix}$. Serre gives precise conditions on when complete reducibility holds in Theorem VI.5.1 of the stated reference. – damiano Apr 30 2010 at 10:11

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No, it is true for semisimple Lie algebras, as said above. If you want to apply this to a reductive Lie algebra in a particular case, you need to make sure that the center of the Lie algebra is mapped to semisimple endomorphisms. – Homology Apr 30 2010 at 10:43

On the full reducibility of representations of reductive Lie algebras

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