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 $g = so(n)$, $k=2$, $k = n-2$?
3 Answers
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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.
answered Sep 19, 2010 at 19:51
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$\begingroup$ Higher exterior powers of the "natural representation" are also irreducible for the orthogonal Lie algebra, i.e. for B and D types. $\endgroup$ Commented Sep 19, 2010 at 23:00
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$\begingroup$ 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. $\endgroup$ Commented Sep 20, 2010 at 10:47
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$\begingroup$ 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. $\endgroup$– J LodderCommented Sep 27, 2010 at 0:02
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$\begingroup$ Above, "$\binom{n}{k}$ for $n \geq 2$" should be "$\binom{n}{k}$ for $2 \leq k \leq n-2$." $\endgroup$– J LodderCommented Sep 27, 2010 at 0:12
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No (2nd question). Take $k=1$ and $I=g$ as a $g$-module and the extension is trivial.
answered Sep 16, 2010 at 14:47
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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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1$\begingroup$ 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. $\endgroup$ Commented Sep 18, 2010 at 22:02
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6$\begingroup$ One more upvote and you'll have caught that wascally wabbit! $\endgroup$ Commented Sep 21, 2010 at 19:12
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$\begingroup$ While in a lighter vein, the name is actually Elmer Fudd (or for purists, Elmer J. Fudd). $\endgroup$ Commented Sep 23, 2010 at 18:35
$E_8$, 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? $\endgroup$