Does it exist a Lie algebra $\mathfrak{g}$ that is reductive but if we consider the inclusion of Lie agebras $\mathfrak{g} \subset \mathfrak{h}$ then $\mathfrak{g}$ is not reductive in $\mathfrak{h}?$
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1$\begingroup$ I'm not familiar with the definition of when a Lie subalgebra is reductive in a Lie algebra. Please can you remind us of it? $\endgroup$– Robin ChapmanCommented May 2, 2010 at 12:33
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1$\begingroup$ I found it in Diximier :"enveloping algebras". $\mathfrak{g}$ is reductive in $\mathfrak{h}$ if the adjoint representation of $\mathfrak{g}$ is semisimple in $\mathfrak{h}$. $\endgroup$– Michele TorielliCommented May 2, 2010 at 12:39
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2$\begingroup$ If you read Dixmier further, you'll see that if g is reductive in h, a semisimple module over h restricts to a semisimple module over g. Thus if g does not act semisimply on V, it cannot be reductive in gl(V). Now let g be one-dimensional, spanned by a non-semisimple endomomorphism... $\endgroup$– Victor ProtsakCommented May 2, 2010 at 23:42
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2$\begingroup$ By the way, in representation theory nearly always $h\subset g$, not the other way around. $\endgroup$– Victor ProtsakCommented May 2, 2010 at 23:48
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2 Answers
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A reductive Lie algebra $L$ is the direct sum of a semisimple Lie algebra $L_1$ and an abelian Lie algebra $L_2$. Let's consider the case where $L_2$ is one-dimensional. We can embed $L$ into a larger Lie algebra $L=L_1\oplus L_2'$ by embedding $L_2$ into $L_2'$. Let $L_2'$ be the two-dimensional Lie subalgebra $$\left(\begin{array}{cc} *& *\\\ 0& 0 \end{array} \right)$$ of $\mathfrak{gl}(\mathbf{C})$ and $$L_2=\left(\begin{array}{cc} 0& *\\\ 0& 0 \end{array} \right).$$ Then $L_2$ does not act semisimply on $L_2'$, so $L$ does not act semisimply on $L'$.
answered May 2, 2010 at 12:58
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$\begingroup$ An abelian algebra is reductive by definition, so $L_1$ is not needed. $\endgroup$ Commented May 2, 2010 at 23:27
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As remarked by Jim Humphreys in a comment to my answer to a previous question, the notion of reductive for a Lie algebra (in characteristic zero) has no intrinsic interest, which means that the answer to this question has to be positive.
Here is one possible construction. Let $\mathfrak{g} = \mathfrak{s} \oplus \mathfrak{z}$ be a reductive Lie algebra, where $\mathfrak{s}$ is semisimple and $\mathfrak{z}$ is the centre of $\mathfrak{g}$. Consider a representation $V$ of $\mathfrak{g}$ where $\mathfrak{z}$ does not act semisimply. Now let $\mathfrak{h}$ be the semidirect product $\mathfrak{g} \ltimes V$, with $V$ abelian.
answered May 2, 2010 at 12:59
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2$\begingroup$ Isn't "yes" a positive answer? $\endgroup$– Ben Webster ♦Commented May 2, 2010 at 15:36
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$\begingroup$ Oops. Must have inverted the question in my mind. Thanks! I've edited it now. $\endgroup$ Commented May 2, 2010 at 17:28