Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathfrak{h}?$

share|improve this question
1  
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? –  Robin Chapman May 2 '10 at 12:33
1  
I found it in Diximier :"enveloping algebras". $\mathfrak{g}$ is reductive in $\mathfrak{h}$ if the adjoint representation of $\mathfrak{g}$ is semisimple in $\mathfrak{h}$. –  Michele Torielli May 2 '10 at 12:39
1  
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... –  Victor Protsak May 2 '10 at 23:42
    
By the way, in representation theory nearly always $h\subset g$, not the other way around. –  Victor Protsak May 2 '10 at 23:48

2 Answers 2

up vote 2 down vote accepted

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 $L'$.

share|improve this answer
    
An abelian algebra is reductive by definition, so $L_1$ is not needed. –  Victor Protsak May 2 '10 at 23:27

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.

share|improve this answer
1  
Isn't "yes" a positive answer? –  Ben Webster May 2 '10 at 15:36
    
Oops. Must have inverted the question in my mind. Thanks! I've edited it now. –  José Figueroa-O'Farrill May 2 '10 at 17:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.