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

Let L be a finite-dimensional complex semisimple lie algebra, then ad(L)=Der(L). (Der is short for derivation). In order to show that ad(L)=Der(L), the proof I followed proves that that the prependicular space to ad(L) is zero. (Perpendicular with respect to killing form of Der(L).) I understand the proof of this. But why is this condition sufficient to conclude ad(L) = Der(L)

If we knew that Der(L) is the direct sum of ad(L) and P:=Perp(ad(L)) then we would be done. And non-degeneracy of the killing form restricted to ad(L) tells us that it is sufficient to show Dim(Der(L)) = Dim(ad(L)) + Dim(P). But I am failing to see how we have that.

share|improve this question
2  
This is a homework problem from a first course in Lie algebras, so I think it's not appropriate here. –  Robert Bryant Jan 16 '13 at 20:15
1  
Whether or not it's homework-related, I agree the question is far from research level (try math.stackexchange.com?). It's a standard theorem found in many books with similar proofs, attributed by Jacobson to Zassenhaus. Not conceptually transparent but a fairly elementary consequence of nondegeneracy of the Killing form. –  Jim Humphreys Jan 17 '13 at 1:30
add comment

1 Answer

If $U$ is a vector subspace of a finite-dimensional vector space $V$ such that the perpendicular space of $U$ with respect to some symmetric bilinear form on $V$ (which does not need to be nondegenerate) is zero, then $U=V$. This is very easy to check (assume the contrary, then notice that $\dim U < \dim V$, and thus the perpendicular space of $U$ has codimension $\leq \dim U$, which means it cannot be zero). You don't need direct sum decompositions.

share|improve this answer
    
Where have you got Dim(V/Perp(U)) ≤ Dim(U). –  rustyracketman Jan 16 '13 at 18:03
    
The perpendicular space of $U$ is given by the condition that it be perpendicular to all elements in a (fixed) basis of $U$. These are $\dim U$ linear equations, and thus the vector space of their joint solutions has codimension $\leq \dim U$. –  darij grinberg Jan 16 '13 at 19:26
add comment

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.