I'm working through Frankel's "The Geometry of Physics" this summer, and I'm stuck on a problem concerning the pushforward and pullback operations. The problem is stated as follows:
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 $p \epsilon M$. $F(p) \epsilon W$ and $G(F(p)) \epsilon V$, respectively. We may consider the composite map $G \circ F: M \rightarrow V$.
(i) Show, by using bases $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ $\frac{\partial}{\partial z}$, that $$(G \circ F)_* = G_* \circ F_*$$
(ii) Show, by using bases $dx$, $dy$, $dz$, that $$(G \circ F)^* = G^* \circ F^*$$
So far I've started with the fact that: $$(G \circ F)_* = (G_* \circ F) \cdot F_*$$ by the chain rule, but I'm not sure how to proceed. I feel like if I can get help with the first one the second one shouldn't be much of an issue.
Thank you.
Best Answer
The equation $(G_\star\circ F)\cdot F_\star$ doesn't make much sense to me. Namely, what is the $\cdot$.
A way to go through this is to compute the coordinate expressions of the pushforward. For a vector $X\in T_p M$ and a function $f\in C^\infty(W)$ we have $$(F_*X)f:=X(f\circ F).$$ In particular, if $X=X^\mu\left(\frac{\partial}{\partial x^\mu}\right)_p$, we obtain $$(F_*X)f:=X^\mu\frac{\partial(f\circ F)}{\partial x^\mu}(p):=X^\mu\partial_\mu(f\circ F\circ x^{-1})(x(p))=X^\mu\partial_\mu(f\circ y^{-1}\circ y\circ F\circ x^{-1})(x(p)).$$ Notice that now $f\circ y^{-1}$ is a function defined on an open subset of $\mathbb{R}^r$ and taking values on $\mathbb{R}$. Similarly $y\circ F\circ x^{-1}$ is a function defined on an open subset of $\mathbb{R}^n$ and taking values on $\mathbb{R}^r$. We can thus use the usual chain rule of multivariable calculus to obtain $$(F_*X)f=X^\mu\partial_a(f\circ y^{-1})(y(F(p)))\partial_\mu(y^a\circ F\circ x^{-1})(x(p))=X^\mu\frac{\partial f}{\partial y^a}(F(p))\frac{\partial y^a\circ F}{\partial x^\mu}(p).$$ Notice that in this computation there is a proof for the chain rule on manifolds. Letting $F_{(y)}^a:=y^a\circ F$, then we find the coordinate expression $$F_*X=X^\mu\frac{\partial F_{(y)}^a}{\partial x^\mu}(p)\left(\frac{\partial}{\partial y^a}\right)_{F(p)}$$ Applying this formula again we have for $Y\in T_{F(p)}M$ $$G_* Y=Y^a\frac{\partial G_{(z)}^A}{\partial y^a}(F(p))\left(\frac{\partial}{\partial z^A}\right)_{G(F(p))}.$$ Putting it all together, we have $$(G_* \circ F_*)X=X^\mu\frac{\partial F_{(y)}^a}{\partial x^\mu}(p)\frac{\partial G_{(z)}^A}{\partial y^a}(F(p))\left(\frac{\partial}{\partial z^A}\right)_{G(F(p))}=X^\mu\frac{\partial G_{(z)}^A \circ F}{\partial x^\mu}(p)\left(\frac{\partial}{\partial z^A}\right)_{G(F(p))},$$ using our chain rule for manifolds. However, using the coordinate expression of the pushforward but now applied to $G\circ F$ we see that indeed $$(G\circ F)_*X=(G_* \circ F_*)X.$$
The proof for the pullbacks follows exactly the same reasoning.
May I suggest the lectures of Prof. Schuller https://www.youtube.com/playlist?list=PLPH7f_7ZlzxTi6kS4vCmv4ZKm9u8g5yic ? I found them very helpful. If you would like a more compact version for GR check https://www.youtube.com/playlist?list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_ .