Frankel 2.3 Problems

(1)

Let $$F : M^n \to W^r$$ and $$G : W^r \to V^s$$ be smooth maps Let $$x$$, $$y$$, and $$z$$ be local coordinates near $$p \in M^n$$, $$F(p) \in W^r$$, and $$G(F(p)) \in V^s$$, respectively. Consider the composite map $$G \circ F : M \to V$$,

(i)     Using the coordinate basis $$\partial / \partial x$$, we may investigate the differential $$(G \circ F)_*$$.

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \frac{\partial z^k}{\partial x^i} \frac{\partial}{\partial z^k}.$$

By the chain rule,

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r \frac{\partial z^k}{\partial y^j}\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial z^k}.$$

Since

$$F_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{j=1}^r \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j},$$

we see that

$$(G \circ F)_*\left(\frac{\partial}{\partial x^i}\right) = \sum_{k=1}^s \sum_{j=1}^r F_*^j\left(\frac{\partial}{\partial x^i}\right)\frac{\partial z^k}{\partial y^j}\frac{\partial}{\partial z^k},$$

and conclude

$$(G \circ F)_* = G_* \circ F_*.$$

(ii)   Using the covector coordinate basis $$dx$$, we may investigate the pull-back $$(G \circ F)^*$$.

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \frac{\partial z^k}{\partial x^i} dx^i.$$

By the chain rule,

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} \frac{\partial y^j}{\partial x^i} dx^i.$$

Since

$$G^*(dz^k) = \sum_{j=1}^r \frac{\partial z^k}{\partial y^j} dy^j,$$

we see that

$$(G \circ F)^*(dz^k) = \sum_{i=1}^n \sum_{j=1}^r G_j^*(dz^k) \frac{\partial y^j}{\partial x^i} dx^i.$$

and conclude that

$$(G \circ F)^* = F^* \circ G^*.$$