Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let A be an algebra over K with multiplication $(x,y) \rightarrow x \cdot y$. A linear operator D on the vector space A is called a derivation of A if $D(x \cdot y)=(Dx) \cdot y + x \cdot (Dy)$ $( \forall x, y \in A)$.

Verify that the commutator $[ D,D' ]= D \circ D'-D' \cdot D $ is a derivation when D and D' are derivations of A.

So from definitions $[ D, D' ](x \cdot y)=(DD'-D'D)(x \cdot y)=DD'(x) \cdot y - D'D(x) \cdot y + x \cdot DD'(y) - x \cdot D'D(y)$.

This is what I think you have to do.

share|improve this question
    
I've worked it out. The problem had a mistake in the question. Bill cook edit made me realized the lecturer mean't - instead of =. –  dannie Apr 3 '12 at 20:44

1 Answer 1

Yes. Just run through the definition.

$[D,D'](xy) = (D \circ D')(xy)-(D' \circ D)(xy) = D(D'(xy))-D'(D(xy))$ $=D(D'(x)y+xD'(y))-D'(D(x)y+xD(y))=D(D'(x)y)+D(xD'(y))-D'(D(x)y)-D'(xD(y))=$ $D(D'(x))y+D'(x)D(y)+D(x)D'(y)+xD(D'(y))-D'(D(x))y-D(x)D'(y)-D'(x)D(y)-xD'(D(y))=$ $D(D'(x))y+xD(D'(y))-D'(D(x))y-xD'(D(y))=$ $\left(D(D'(x))-D'(D(x))\right)y+x\left(D(D'(y))-D'(D(y))\right)=$ $[D,D'](x)y+x[D,D'](y)$

Therefore, $[D,D']$ is itself a derivation.

Thus the subspace of derivations forms a Lie subalgebra of $\mathrm{End}(A)$.

share|improve this answer

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.