Product of derivations Let $A$ be an algebra over a field $K$. If $D_1$ and $D_2$ are derivations of $A$, show that $D_1 \circ D_2$ is not necessarily a derivation (it is if $D_1$ or $D_2 = 0$), but $D_1 \circ D_2−D_2 \circ D_1$ is always a derivation of $A$.
It's a problem from Tu, Loring W's An introduction to manifolds. Obviously, the composition of linear maps is linear, but how can I check the Leibniz rule?
Meanwhile, why it defines to be $D_1 \circ D_2−D_2 \circ D_1$? Are there any motivations or backgrounds?
 A: 1. For the first part (why $D_1 \circ D_2$ is not a derivation), just try to come up with a counter-example.
For example, let $A = K[x,y]$ be polynomials in two variables. The partial derivative operators $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are derivations of $A$ (by the standard product rule from calculus). But their composition $\frac{\partial}{\partial y} \frac{\partial}{\partial x}$ is not. Just plug in something simple, like $xy$:
$$ \frac{\partial}{\partial y} \frac{\partial}{\partial x} xy = 1 $$
But on the other hand,
$$ \left( \frac{\partial}{\partial y} \frac{\partial}{\partial x} x \right) \cdot y + x \cdot \left( \frac{\partial}{\partial y} \frac{\partial}{\partial x} y \right) = 0 $$
So it does not satisfy the Leibniz rule.
2. For the second part (why $D_1 \circ D_2 - D_2 \circ D_1$ is a derivation), you can just compute "by-hand", and check it satisfies the Leibniz rule. Let $a,b \in A$, and evaluate:
$$
D_1(D_2(ab))-D_2(D_1(ab)) = D_1(D_2(a) \cdot b + a \cdot D_2(b)) - D_2( D_1(a) \cdot b + a \cdot D_1(b)) $$
Here we just used the Leibniz rule for $D_2$ and $D_1$. Now since $D_1$ and $D_2$ are linear, you can "distribute", so this gives
$$ D_1(D_2(a) \cdot b) + D_1(a \cdot D_2(b)) - D_2(D_1(a) \cdot b) - D_2(a \cdot D_1(b)) $$
Now just use the Liebniz identity again for each of the four terms, and a bunch of stuff should cancel. What is left over should give the Leibniz identity for $D_1 \circ D_2 - D_2 \circ D_1$.
3. As for why this is the "correct" definition: you said this comes from a book about manifolds. If the derivations are thought of as vector fields on a manifold, then this "commutator" $XY-YX$ corresponds to the Lie derivative.
