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Let $g$ be a semi-simple Lie algebra, and $0 \to I \to h \to g \to 0$ an Abelian extension of $g$. Then $g$ acts on $I$. Considering $g$ under the adjoint action, when is there a $g$-module isomorphism between $g$ and the k-th exterior power $\Lambda^k(I)$ for some $k$? Only when $g = so(n)$, $k=2$, $k = n-2$?

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The wording of the question is not quite clear to me, including the expression "its standard representation". It would help to spell out the special case mentioned in the last line "Only when ... ?" –  Jim Humphreys Sep 15 '10 at 0:21
As an example of this situation, every element of the orthogonal Lie algebra $so(n)$ can be written as an $n \times n$ matrix, which then acts on ${\bf R}^n$ via matrix multiplication on vectors. Here $I = {\bf R}^n$ is "the standard representation." It can be easily checked that as $so(n)$-modules, $so(n) \simeq I^2$. Additionally, $I$ can be viewed as an Abelian Lie algebra (all brackets zero), and $h$ contains all linear combinations of elements in $so(n)$ with elements in $I$. –  J Lodder Sep 16 '10 at 21:23
Above, $I^2$ denotes $I^{\wedge 2}$. –  J Lodder Sep 16 '10 at 21:32
This much clarification helps, though I'm still confused about the complicated formulation in terms of an extension (split, nonsplit?) and what that has to do with your motivation. What does h have to do with anything? The first case is that of a split extension, where you are just considering the "standard representation" in your sense along with its exterior powers. But if the given Lie algebra is of type $E_8$, its "standard representation" is the adjoint representation; then the suggestion by Bugs Bunny comes into play, but no other exterior power. What is the motivation? –  Jim Humphreys Sep 16 '10 at 22:05

3 Answers 3

Here is at least a partial answer to the question, to supplement some comments I already made. The essential case is that of a simple Lie algebra over $\mathbb{C}$. For each simple type there is a "natural" irreducible representation as well as the (irreducible) adjoint representation; these coincide just for type $E_8$. Many sources (such as Chapter 8 of Bourbaki's Groupes et algebres de Lie) specify the dimensions. For types $A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2$, these are respectively: $n+1, 2n+1, 2n, 2n, 27, 56, 248, 26, 7$ and $n^2-2n, 2n^2 +n, 2n^2+n, 2n^2-n, 78, 133, 248, 52, 14$.

As indicated in the question, the second (or complementary) exterior power of the natural module agrees with the adjoint module for types $B_n, D_n$. But dimension comparison seems to rule out such coincidences in other cases. In fact, higher exterior powers of the natural representation are usually not even irreducible. (Fundamental representations overlap here somewhat, but require case-by-case discussion as done in Bourbaki.) Much is known classically about dimensions of irreducibles as well as decomposition of symmetric and exterior powers, but it can take a lot of work to make the details explicit for each simple type.

Probably the narrow question here can be studied for classical types (the Lie algebras or associated simply connected compact Lie groups) in a concrete way, but ultimately the "correct" approach requires comparison of highest weights of the various irreducible representations involved. For this one should check the "planches" at the end of Bourbaki's Chapter 6 for the way the highest root is expressed in terms of fundamental weights, etc. I'm not sure whether any single source gives a concise account of both the concrete and abstract representation theory: standard, adjoint, and fundamental representations, along with a description of the exterior powers of the standard module.

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Higher exterior powers of the "natural representation" are also irreducible for the orthogonal Lie algebra, i.e. for B and D types. –  Victor Protsak Sep 19 '10 at 23:00
Yes, I shouldn't have focused on type A alone, so I edited that. The point is just that one can't take for granted anything about irreducibility of higher exterior or symmetric powers. –  Jim Humphreys Sep 20 '10 at 10:47
Among the exceptional Lie algebras, the only one with dimension being a binomial coefficient $\binom {n}{k}$ for $n \geq 2$ is $E_6$ with $78 = \binom{13}{2}$. However, the smallest non-trivial rep. of $E_6$ has dimension 27, which rules out 13. Among the classical Lie algebras, as $sl(n)$-modules, $sl(n)$ is essentially iso. to $I^{\otimes 2}$ (with the trivial rep. deleted), and as $sp(n)$-modules, $sp(n)$ is iso. to the second symmetric power of $I$. For $k \geq 2$ and $g$ complex, simple, the only solutions to the question are from the $so(n)$ family. –  J Lodder Sep 27 '10 at 0:02
Above, "$\binom{n}{k}$ for $n \geq 2$" should be "$\binom{n}{k}$ for $2 \leq k \leq n-2$." –  J Lodder Sep 27 '10 at 0:12

No (2nd question). Take $k=1$ and $I=g$ as a $g$-module and the extension is trivial.

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Since $I$ is Abelian and $g$ semi-simple, $I$ could never be isomorphic to $g$, unless $I = g = h = 0$.

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Here I is a vector space with an action of g, possibly nontrivial. For any module, you can form a split extension (though it's much less trivial to get nonsplit ones). But the question really seems to be about modules and their exterior powers. –  Jim Humphreys Sep 18 '10 at 22:02
One more upvote and you'll have caught that wascally wabbit! –  Cam McLeman Sep 21 '10 at 19:12
While in a lighter vein, the name is actually Elmer Fudd (or for purists, Elmer J. Fudd). –  Jim Humphreys Sep 23 '10 at 18:35

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