Pull-back property $f^*(dx^i)=df^i$

Question

I know that this question has been asked already, but I feel that the proof I have looked at is longer than what it should be and would like to see if this may be simplified a little. The definition I am using is:

$$f^*(w)(p)(v_1,..,v_k)=w(f(p))((Df(p)\cdot v_1),...,(Df(p)\cdot v_k))$$ where $w$ is a $k$-form and each $v_i\in T_pM$.

When trying to prove that:

$$f^*(dx^i)(p)(v)=df^i$$

where $p\in \mathbb{R}^m$, $v\in T_pM$, and $f: \mathbb{R}^m\rightarrow \mathbb{R}^n$.

we know $f^*(dx^i)(p)(v)=dx^i(f(p))(Df(p)v)$

We also have that since $\frac{\partial}{\partial y_i}$ generates the tangent space, then we can express $v$ as a linear combination of these elements, that is $v=v_1\frac{\partial}{\partial y_1}+...+v_m\frac{\partial}{\partial y_m}$. I feel confused in the notation when trying to expand this into a linear combination using the $dx_i$.

I feel a little stuck on the understanding of this notation, any help is much appreciated.

Answer 1

Your notation is pretty clear, but let me try to be slightly more pedantic. First I hope you know that $x^i$ in this context means the projection onto $i^{th}$ factor. So, $x^i: \Bbb{R}^n \to \Bbb{R}$ is the function defined by \begin{align} x^i(q^1, \dots, q^n) = q^i \end{align}

Next, even though the tangent space $T_p(\Bbb{R}^m)$ is canonically isomorphic to $\Bbb{R}^m$, for any $v\in \Bbb{R}^m$, I'll put a little subscript $v_p$ to denote the element of $T_p(\Bbb{R}^m)$ which corresponds to $v$ under the isomorphism, just so that we are clear about which space we are talking about.

Now, you wish to prove that for $f: \Bbb{R}^m \to \Bbb{R}^n$ and $x^i : \Bbb{R}^n \to \Bbb{R}$, we have \begin{align} f^*(dx^i) = d(x^i \circ f) \equiv d(f^i) \end{align} where I use $\equiv$ to mean they are exactly the same thing written in different notation. As a notational remark, I'll put the point of derivative evaluation in subscript position simply to avoid a whole lot of brackets... so I'll write $Df_p$ rather than $Df(p)$, and I'll also write $df_p$ rather than $df(p)$.

Now, the actual proof: for any point $p \in \Bbb{R}^m$ and any tangent vector $v_p \in T_p(\Bbb{R}^m)$, we have \begin{align} \left(f^*(dx^i) \right)_p(v_p) &:= dx^i_{f(p)} \big( f_{*p}(v_p) \big) \\ &:= dx^i_{f(p)}\big( [Df_p(v)]_{f(p)} \big) \tag{definition of $f_{*p}$} \\ &:= D(x^i)_{f(p)} \big( Df_p(v) \big) \\ &= D(x^i \circ f)_p(v) \tag{chain rule} \\ &:= d(x^i \circ f)_p(v_p) \end{align} Since the point $p$ and the tangent vector $v_p$ were arbitrary, we have shown that $f^*(dx^i) = d(x^i \circ f)$, as desired.

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So as you can see, 4 of the 5 equal signs were true simply by definition of $d$ and $^*$. The only "proof" part of it comes from the (standard multivariable) chain rule in the $4^{th}$ equal sign.

Answer 2

At every point $p$, the tangential space $T_p\Bbb R^m$ is isomorphic to $\Bbb R^m$ via $\frac{\partial}{\partial x_i}\mapsto e_i=(0,0,\ldots,0,1,0,\ldots,0)$. Now we have \begin{align*} f^*(\operatorname{dx}^i)(p)\left(\frac{\partial}{\partial x_j}\right)&=\operatorname{dx}^i(f(p))\left(Df(p)\cdot \frac{\partial}{\partial x_j}\right)=\operatorname{dx}^i(f(p))\left(J_f(p)e_j\right)=J_{x^i}(f(p))J_f(p)e_j\\&=J_{x^i\circ f}(p)e_j=J_{f^i}(p)e_j=\operatorname{d}(f^i)(p)e_j=\operatorname{d}(f^i)(p)\left(\frac{\partial}{\partial x_j}\right) \end{align*} where $J_f(p)$ denotes the Jacobian. Since the $\frac{\partial}{\partial x_j}, 1\leq j\leq m$ span the whole tangential space $T_p\Bbb R^m$ at $p$ we are done.