# Differentiating the pull-back of a one-form

Let $\Omega$ be an open subset of a vector space $V$ and let $\alpha\colon \Omega\to V^*$ be a one-form on $\Omega$. Assuming that $\alpha$ is differentiable, then for any $x\in \Omega$, $D\alpha (x)\colon V\to V^*$ is a linear map. Hence we can think of this derivative as a bilinear map $D\alpha(x)\colon V\times V\to \mathbb{R}$.

We say that $\alpha$ is symmetric if the bilinear form $D\alpha$ is symmetric for all $x$. I'm trying to understand the following fact from a book that I'm reading: Let $f\colon \Omega'\to \Omega$ be a smooth map from an open subset $\Omega'$ of a vector space $V'$ to $\Omega$. Then

If $\alpha$ is symmetric, then so is the pull-back $f^*\alpha$.

Here the pull-back is defined by $(f^*\alpha)(x)(v)=\alpha(f(x))Df(x) v$. $\quad$ (I)

My book proves this as follows:

$$D(f^*\alpha)(x)(v_1, v_2)=D\alpha(f(x))(Df(x) v_1, Df(x) v_2)$$ and clearly this is symmetric in $v_1$ and $v_2$ if $D\alpha(f(x))$ is symmetric.

I don't quite see where the displayed equality comes from. (Using coordinates shouldn't be a good idea because symmetry seems to be coordinate dependent.)

I believe the source of my difficulty lies in not feeling comfortable with differentiating something like $S(x)\circ T(x)$. Here $S(x)$ and $T(x)$ are both functions (after evaluation) and $S(x)\circ T(x)$ is the composition of two functions.

This is perhaps too elementary, but any help with clarification is appreciated. Thanks!

• How does your book define $D\alpha(x)$? Something like $$D\alpha(x)(v_1,v_2) = \lim_{t \to 0}\frac{1}{t}\left[\alpha(x+tv_1)(v_2)-\alpha(x)(v_2)\right]$$ perhaps? – Thurmond May 22 '16 at 19:38

## 1 Answer

Let's start by making a quick "type list" of everything we're dealing with here. I'm going to rewrite some of your equations to help me keep everything together, because you're often conflating $V$ and $V'$.

• $x$ is a point in the space $\Omega \subset V$.

• $x'$ is a point in the space $\Omega' \subset V'$.

• $v_1, v_2$ are vectors in $V$.

• $v'_1, v'_2$ are vectors in $V'$.

• $\alpha$ is a one-form on $\Omega$; that is, a map $\Omega \to V^*$, equivalently a map $\Omega \times V \to \mathbb R$.

• $D\alpha$ is the derivative of $\alpha$. So for $x \in \Omega, v \in V$, we have $$D\alpha(x)(v) = \lim_{t \to 0}\frac{1}{t}\left[\alpha(x + tv)-\alpha(x)\right],$$ which means that $D\alpha$ is a map $\Omega \times V \to V^*$, which we want to think of via currying as a map $\Omega \times V \times V \to \mathbb R$ given by $$D\alpha(x)(v_1, v_2) = \lim_{t \to 0}\frac{1}{t}\left[\alpha(x+tv_1)(v_2)-\alpha(x)(v_2)\right].$$ Remember that $\alpha$ is only linear in the second coordinate, not the first - this is why we have derivatives, to linearize it in the first coordinate!

• $f$ is a map $\Omega' \to \Omega$.

• What's the "type" of $Df(x')$? Well, $f$ is a function $\Omega' \to \Omega$, and $Df(x')$ should capture what happens when we vary $x' \in \Omega'$ infinitesimally by elements of $V'$. So $Df(x')(v')$ should measure the variation in $f(x')$ when we move $x'$ in the direction of $v'$; that is, $Df(x')(v')$ should look (up to limits and $1/t$) like $f(x'+tv') - f(x')$. But $f(x'+tv')$ and $f(x')$ are points in $\Omega$, so their difference is in $V$. Hence $Df(x')$ is a map $V' \to V$, so $Df$ is a map $\Omega' \times V' \to V$.

• $f^*\alpha$ is the pullback of $\alpha$ by $f$, a one-form on $\Omega'$; that is, a map $\Omega' \to (V')^*$ or $\Omega' \times V' \to \mathbb R$. It is given by $$(f^*\alpha)(x')(v') = \alpha(f(x'))\left[Df(x')(v')\right].$$ This expression is surprisingly subtle. $Df(x')$ is a map $V' \to V$, and $v'$ is an element of $V'$. So $Df(x')(v')$ is an element of $V$. And since $f(x)$ is an element of $\Omega$, $\alpha(f(x))$ is an element of $V^*$. So we have an element of $V^*$ and an element of $V$, thus we get something in $\mathbb R$ as desired!

• $D(f^*\alpha)$, the derivative of the pullback, is going to be a map $\Omega' \times V' \to (V')^*$, which we'll again want to think of as a map $\Omega' \times V' \times V' \to \mathbb R$.

Now that we've got everything straight, let's run through the computation that $$D(f^*\alpha)(x')(v'_1,v'_2) = D\alpha(f(x'))(Df(x')v'_1,Df(x')v'_2).$$

We have \begin{align} D(f^*\alpha)(x')(v'_1,v'_2) &= \lim_{t \to 0}\frac{1}{t}\left[(f^*\alpha)(x'+tv'_1)(v'_2) - (f^*\alpha)(x')(v'_2)\right] \\ &= \lim_{t \to 0}\frac{1}{t}\left[ \alpha(f(x'+tv'_1))(Df(x'+tv'_1)(v'_2))-\alpha(f(x'))(Df(x')(v'_2)) \right]. \end{align} Now for the other side. We have $$D\alpha(f(x'))(Df(x')v'_1,Df(x')v'_2) = \lim_{t\to0}\frac{1}{t}\left[\alpha(f(x')+tDf(x')(v'_1))(Df(x')(v'_2))-\alpha(f(x'))(Df(x')(v'_2))\right].$$

So to finish we only need to show that $$\alpha(f(x')+tDf(x')(v'_1))(Df(x')(v'_2)) \approx \alpha(f(x'+tv'_1))(Df(x'+tv'_1)(v'_2))$$ up to error of order $O(t^2)$ since that will vanish in the limit.

By Taylor's theorem, since $f$ is smooth, $f(x'+tv'_1) = f(x') + tDf(x')(v'_1) + O(t^2)$. So the first arguments of $\alpha$ are close enough for our purposes, and since $\alpha$ is $C^1$ we can get $O(t^2)$ approximation for $\alpha$ itself. Similarly, linearizing $Df$ should give us an $O(t^2)$ error as well. So up to terms that we care about (i.e., those that don't vanish when we take the derivative) the equality is true.