Spivak Calculus on Manifolds - Theorem 4-10

Question

Part (4) of Theorem 4-10 in Spivak's Calculus on Manifolds says the following:

If $\omega$ is a $k$-form on $\mathbb{R}^m$ and $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is differentiable, then $f^{*}(d \omega) = d(f^{*}\omega)$.

In the Proof, Spivak says that it is clear if $\omega$ is a $0$-form. I tried expanding both sides using the definitions, but I'm not getting the desired result even after a lot of effort. I suppose I'm only missing something straightforward. Could anyone please help me out? Thanks in advance.

Edit: Some of what I attempted is as follows:

\begin{align} &\ f^{*} d\omega (p) (v_p)\\ =&\ f^{*} \bigl(d\omega (f(p))\bigr)(v_p)\\ =&\ d\omega \bigl(f(p)\bigr) (f_{*}v_p)\\ =&\ d\omega \bigl(f(p)\bigr) \bigl(Df(p)(v)\bigr)_{f(p)}\\ =&\ D\omega (f(p))(Df(p)(v)) \end{align}

I also tried writing $d\omega$ as $\sum_{i=1}^n \omega_i dx^i$, so that \begin{align} f^{*} d\omega &= f^{*}\left(\sum_{i=1}^n \omega_i dx^i\right)\\ &= \sum_{i=i}^n f^{*}(\omega_i dx^i)\\ &= \sum_{i=1}^n \omega_i \circ f \cdot f^* (dx^i)\\ &= \sum_{i=1}^n \omega_i \circ f \cdot \sum_{j=1}^n D_j f^i \cdot dx^j \end{align}

Then, $d(f^*\omega)(p)(v_p) = D(f^*\omega)(p)(v)$, but I don't know how to connect this with the last line.

As I understand, a $0$-form is just a function from $\mathbb{R}^n$ to $\mathbb{R}$. The operator $d$ takes a $k$-form and converts it into a $k+1$-form.

Answer 1

Notation/definitions:

• if $V$ is a vector space, and $V^*$ its dual, in the following write $$\langle \omega,v \rangle = \omega (v),$$ where $\omega \in V^*$, and $v\in V$.

• Suppose $g\colon M \to \mathbb R$ is a (nice) function: the $1$-form $dg$ is the unique element in the cotangent-space, (i.e., the dual vector space to the tangent space), such that $$ \langle dg , v \rangle = v (g),$$ where $v$ is an arbitrary tangent vector (dropping 'at a point $p$' and corresponding subscripts for legibility), and $v(g)$ denotes the $v$-directional derivative of $g$, i.e., if $$ \gamma: \mathbb R \to M$$ is a path such that $\gamma ( 0 ) = p$, and $\gamma'(0) = v$, then $$\langle dg, v \rangle= {d\over dt}\Bigg\rvert_{t=0}\, g \circ \gamma.$$

• if $f\colon M \to N$, and $v$ and $\gamma$ are as in the preceding, then $f_*v = (f\circ \gamma)'(0)$, i.e., $$(f_*v)\,(g) = {d\over dt}\Bigg\rvert_{t=0}\, g \circ (f\circ \gamma) =v ( g \circ f ).$$
• if $\omega$ is a $1$-form, then $\langle f^*\omega,v\rangle= \langle\omega, f_*v\rangle$.
• if $g$ is $0$-form, i.e., a function, then $f^* g = g \circ f$.

To answer your question:

On the one hand, $$ \langle f^*dg, v\rangle= \langle dg, f_*v\rangle = {d\over dt}\Bigg\rvert_{t=0}\, g \circ (f\circ \gamma) = v (\, g \circ f \,).$$ On the other, $$ \langle d (f^*g), v \rangle = \langle d (g \circ f), v\rangle = {d\over dt}\Bigg\rvert_{t=0}\, (g \circ f) \circ \gamma = v (\, g \circ f \,).$$

Therefore $f^*dg = d (f^* g)$.

Answer 2

I am posting my own answer, using the notation and theorems in Spivak, just to clarify my understanding.

Suppose $\omega$ is a $0$-form on $\mathbb{R}^m$, that is, $\omega : \mathbb{R}^m \to \mathbb{R}$ is a $C^\infty$ function. Then, $$ d\omega = \sum_{\alpha=1}^m D_\alpha \omega \cdot dx^\alpha \qquad \text{(by Theorem 4-7)} $$ is a $1$-form on $\mathbb{R}^m$. Let $\, f : \mathbb{R}^n \to \mathbb{R}^m$ be a differentiable function. $\, f^*(d\omega)$ is thus a $1$-form on $\mathbb{R}^n$. Let $p \in \mathbb{R}^n$. Then, $$ \begin{align*} f^*(d\omega)(p) &= f^* \left( \sum_{\alpha=1}^m D_\alpha \omega \cdot dx^\alpha \right) (p) \\ &= \sum_{\alpha=1}^m f^* (D_\alpha \omega \cdot dx^\alpha) (p) &&\text{(by Theorem 4-8(2))}\\ &= \sum_{\alpha=1}^m [(D_\alpha (\omega) \circ f) \cdot f^*(dx^\alpha)](p) &&\text{(by Theorem 4-8(3))}\\ &= \sum_{\alpha=1}^m \left[ (D_\alpha(\omega) \circ f) \cdot \sum_{\beta=1}^n D_\beta \;\! f^\alpha \cdot dx^\beta \right](p) &&\text{(by Theorem 4-8(1))}\\ &= \sum_{\beta=1}^n \left[ \sum_{\alpha=1}^m D_\alpha \omega(f(p)) \cdot D_\beta \;\! f^\alpha(p) \right] dx^\beta(p)\\ &= \sum_{\beta=1}^n D_\beta(\omega \circ f)(p) \cdot dx^\beta(p) &&\text{(by Theorem 2-9)}\\ &= \left(\sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta \right) (p). \end{align*} $$ Hence, $$ f^*(d\omega) = \sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta. $$ On the other hand, $\, f^* \omega$ is a $0$-form on $\mathbb{R}^n$ given by $$ f^* \omega(p) = \omega(f(p)) = \omega \circ f(p) $$ for all $p \in \mathbb{R}^n$. That is, $$ f^* \omega = \omega \circ f. $$ Therefore, $d(f^* \omega)$ is a $1$-form on $\mathbb{R}^n$, and we have $$ \begin{align*} d(f^* \omega) &= \sum_{\beta=1}^n D_\beta(f^* \omega) \cdot dx^\beta &&\text{(by Theorem 4-7)}\\ &= \sum_{\beta=1}^n D_\beta(\omega \circ f) \cdot dx^\beta. \end{align*} $$ Thus, if $\omega$ is a $0$-form on $\mathbb{R}^m$ and $\, f : \mathbb{R}^n \to \mathbb{R}^m$ is differentiable, then $$ f^*(d\omega)=d(f^*\omega). $$