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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 wor$k$ed it out. The problem had a mista$k$e in the question. Bill cook edit m$a$de me realized the lecturer mean't - instead of =.2012-04-03

1 Answers 1

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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 $\mathrm{End}(A)$.