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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.

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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

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 $\mathrm{End}(A)$.

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