How Differentials Compose « Explicating Thingummyhoos

How Differentials Compose

Next problem, Frankel’s 2.3(1), part(i):

Let $F:M^n \rightarrow W^r$ and $G:W^r \rightarrow V^s$ be smooth maps. Let x, y and z be local coordinates near $p \in M$ , $F(p) \in W$ , and $G(F(p)) \in V$ , respectively. We may consider the composite map $G \circ F : M \rightarrow V$ . Show, by using bases $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ , and $\frac{\partial}{\partial z}$ , that $(G \circ F)_* = G_* \circ F_*$ .

Now, the differential of F, $F_*$ , acts like this: $F_* \left( \frac{\partial}{\partial x^j} \right) = \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \frac{\partial}{\partial F^i\left(x\right)}$ , where $x = (x^1, x^2, \cdots , x^n) \in M^n$ and $F^i(x)$ is the ith component of $F(x)$ . A similar definition holds for $G_*$ and $(G \circ F)_*$ .

The problem asks us to consider the differentials acting on the bases $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ , and $\frac{\partial}{\partial z}$ . However, it should be obvious that we need only prove that $(G \circ F)_*\left(\frac{\partial}{\partial x}\right) = G_* \circ F_*\left(\frac{\partial}{\partial x}\right)$ . The same proof would apply to the other two bases, and hence to the whole vector space that has those three bases as its basis.

Thus
$G_* \circ F_* \left( \frac{\partial}{\partial x^j} \right)$
$= G_* \left( \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \frac{\partial}{\partial F^i\left(x\right)} \right)$
$= \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} G_* \left( \frac{\partial}{\partial F^i\left(x\right)} \right)$
$= \sum_i \frac{\partial F^i\left(x\right)}{\partial x^j} \sum_k \frac{\partial G^k \left(F\left(x\right)\right)}{\partial F^i \left(x\right)} \frac{\partial}{\partial G^k\left(F\left(x\right)\right)}$
$= \sum_k \left(\sum_i \frac{\partial G^k \left(F\left(x\right)\right)}{\partial F^i \left(x\right)} \frac{\partial F^i\left(x\right)}{\partial x^j} \right) \frac{\partial}{\partial G^k\left(F\left(x\right)\right)}$
$= \sum_k \frac{\partial \left(G \circ F \right)^k \left(x\right)}{\partial x^j} \frac{\partial}{\partial G^k \left( F\left(x\right) \right)}$
$= (G \circ F)_* \left( \frac{\partial}{\partial x^j} \right)$