📂Geometry

Pull Back in Differential Geometry

Overview

We define the pullback on a differential manifold. If differential manifolds are complex, one can think of $M = \mathbb{R}^{m}$ and $N = \mathbb{R}^{n}$.

Definition¹

Given two differential manifolds $M, N$ and a differentiable function $f : M \to N$, we can consider a function $f^{\ast}$ that maps $N$’s $k$-forms to $M$’s $k$-forms. Let $\omega$ be a $k$-form on the manifold $N$, then a $k$-form $f^{\ast}\omega$ on the manifold $M$ is defined as the pullback of $\omega$ as follows.

$$ \begin{equation} (f^{\ast}\omega)(p) (v_{1}, \dots, v_{k}) := \omega (f(p))\left( df_{p}v_{1}, \dots, df_{p}v_{k} \right),\quad v_{i} \in T_{p}M \end{equation} $$

Explanation

The name pullback implies that, contrary to $f$ mapping from $M$ to $N$, $f^{\ast}$ maps from $N$ to $M$. The definition and notation are quite complex, so let’s understand them step by step.

• $f^{\ast}$

$f^{\ast}$ is a map that sends $k$-forms of $N$ to $k$-forms of $M. Therefore, if $\omega$ is a $k$-form of $N$, then $f^{\ast}\omega = f^{\ast}(\omega)$ is a $k$-form of $M.

• $f^{\ast}\omega (p)$

A $k$-form on the manifold $M$ maps $p \in M$ to an element of $\Lambda^{k}(T_{p}^{\ast}M)$.

$$ f^{\ast}\omega : M \to \Lambda^{k}(T_{p}^{\ast}M) $$

$$ \Lambda^{k} (T_{p}^{\ast}M) := \left\{ \varphi : \underbrace{T_{p}M \times \cdots \times T_{p}M}_{k \text{ times}} \to \mathbb{R}\ | \ \varphi \text{ is k-linear alternating map} \right\} $$

In other words, $f^{\ast}\omega (p) \in \Lambda^{k} (T_{p}^{\ast}M)$ is also a function. By the definition of $\Lambda^{k} (T_{p}^{\ast}M)$, $f^{\ast}\omega (p)$ takes “$k$ tangent vectors at $p$” as variables. Thus, $(1)$ is the expression that specifically defines this function’s value. To emphasize that $f^{\ast}(p)$ itself is a function, let’s use the following notation.

$$ (f^{\ast}\omega)_{p} = f^{\ast}\omega (p) $$

• $\omega (f(p))$

Since $\omega$ is a $k$-form of $N$, it maps the point $f(p)$ of $N$ to an element of $\Lambda^{k}(T_{f(p)}^{\ast}N)$.

$$ \Lambda^{k} (T_{f(p)}^{\ast}N) := \left\{ \varphi : \underbrace{T_{f(p)}N \times \cdots \times T_{f(p)}N}_{k \text{ times}} \to \mathbb{R}\ | \ \varphi \text{ is k-linear alternating map} \right\} $$

By the definition of $\Lambda^{k} (T_{f(p)}^{\ast}N)$, $\omega (f(p))$ is also a function. $\omega (f(p))$ takes “$k$ tangent vectors at $f(p)$” as variables. Here too, to emphasize that $\omega (f(p))$ itself is a function, let’s use the following notation.

$$ \omega_{f(p)} = \omega (f(p)) $$

• $df_{p}v_{i}$

$$ df_{p} : T_{p}M \to T_{f(p)}N $$

For $f : M \to N$, the differential $df_{p}$ of $f$ is defined as above. Therefore, if $v_{i} \in T_{p}M$, then $df_{p}v_{i} = df_{p}(v_{i})$ is an element of $T_{f(p)}N$.

Now, combining these, we obtain $(1)$.

$$ (f^{\ast}\omega)_{p} (v_{1}, \dots, v_{k}) := \omega_{f(p)}\left( df_{p}v_{1}, \dots, df_{p}v_{k} \right),\quad v_{i} \in T_{p}M $$

The domains of these two functions show the following difference.

$$ \begin{align*} (f^{\ast}\omega)_{p} : && \underbrace{T_{p}M \times \cdots \times T_{p}M}_{k \text{ times}} &\to \mathbb{R} \\ \omega_{f(p)} : && \underbrace{T_{f(p)}N \times \cdots \times T_{f(p)}N}_{k \text{ times}} &\to \mathbb{R} \end{align*} $$

Think of the differential $df_{p} : T_{p}M \to T_{f(p)}N$ as bridging this difference. Hence, $df_{p}$ is also called push forward. For a $1$-form $\varphi$, the following holds true.

$$ \begin{equation} \varphi( dfv) = f^{\ast}\varphi(v) \end{equation} $$

Pullback of $0$-forms

Let’s consider $f : M \to N$ as a function defined between two differential manifolds. Let $g : N \to \mathbb{R}$ be a function (a $0$-form of $N$). The pullback $f^{\ast}g : M \to \mathbb{R}$ is defined as the following function (a $0$-form of $M$).

$$ f^{\ast}g := g \circ f $$

Coordinate Transformation

Let’s assume a function $f : \mathbb{R}^{n} \to \mathbb{R}^{m}$ is given. Let $\mathbf{x} = (x_{1}, \dots ,x_{n}) \in \mathbb{R}^{n}$, and $\mathbf{y} = (y_{1}, \dots ,y_{m}) \in \mathbb{R}^{m}$.

$$ f(x_{1}, \dots, x_{n}) = (f_{1}(\mathbf{x}), \dots, f_{m}(\mathbf{x}) )= (y_{1}, \dots ,y_{m}) $$

And let $\omega = \sum\limits_{I} a_{I} dy_{I}$ be a $k$-form on $\mathbb{R}^{m}$. Then the pullback $f^{\ast}\omega$ is as follows, based on these properties.

$$ \begin{align*} f^{\ast} \omega &= f^{\ast} \left( \sum a_{I}dy_{I} \right) \\ &= \sum f^{\ast} \left( a_{I}dy_{I} \right) \\ &= \sum f^{\ast}a_{I} f^{\ast}dy_{I} \\ &= \sum f^{\ast}a_{I} f^{\ast}(dy_{i1} \wedge \cdots \wedge dy_{ik}) \\ &= \sum f^{\ast}a_{I} (f^{\ast}dy_{i1} \wedge \cdots \wedge f^{\ast}dy_{ik}) \end{align*} $$

Here, due to $(2)$, $f^{\ast}dy_{i1}(v) = dy_{i1}(df(v)) = d(y_{i1}\circ f)(v) = df_{i1}(v)$, and $f^{\ast}a_{I} = a_{I} \circ f$, so,

$$ \begin{equation} f^{\ast} \omega = \sum a_{I}(f_{1}, \dots f_{m}) df_{i1} \wedge \cdots \wedge df_{ik} \end{equation} $$

This formula signifies coordinate transformation. Let’s see how it specifically works in the following example.

Example

Let’s assume a $1$-form $\omega$ on $\mathbb{R}^{2} \setminus \left\{ 0, 0 \right\}$ is as follows.

$$ \omega = - \dfrac{y}{x^{2} + y^{2}}dx + \dfrac{x}{x^{2} + y^{2}}dy = a_{1}dx + a_{2}dy $$

Let’s transform this $1$-form in Cartesian coordinates to polar coordinates. Let $U = \left\{ (r,\theta) : 0 \lt r, 0 \le \theta \lt 2\pi \right\}$. And let $f : U \to \mathbb{R}^{2}$ be as follows.

$$ f(r,\theta) = (r\cos\theta, r\sin\theta) = (f_{1}, f_{2}) $$

Now, let’s calculate $df_{1}, df_{2}$. Since $f_{1} = r\cos\theta, f_{2}=r\sin\theta$,

$$ \begin{align*} df_{1} &= \dfrac{\partial f_{1}}{\partial r}dr + \dfrac{\partial f_{1}}{\partial \theta}d\theta = \cos\theta dr - r \sin \theta d\theta \\ df_{2} &= \dfrac{\partial f_{2}}{\partial r}dr + \dfrac{\partial f_{2}}{\partial \theta}d\theta = \sin\theta dr + r \cos \theta d\theta \\ \end{align*} $$

Then, by $(3)$,

$$ \begin{align*} f^{\ast} \omega &= a_{1}(f_{1}, f_{2})df_{1} + a_{2}(f_{1}, f_{2})df_{2} \\ &= - \dfrac{f_{2}}{f_{1}^{2} + f_{2}^{2}}(\cos\theta dr - r \sin \theta d\theta) + \dfrac{f_{1}}{f_{1}^{2} + f_{2}^{2}}df_{2}(\sin\theta dr + r \cos \theta d\theta) \\ &= - \dfrac{r\sin\theta}{r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta}(\cos\theta dr - r \sin \theta d\theta) \\ &\quad + \dfrac{r\cos\theta}{r^{2}\cos^{2}\theta + r^{2}\sin^{2}\theta}(\sin\theta dr + r \cos \theta d\theta) \\ &= -\dfrac{\sin\theta \cos\theta}{r}dr + \sin^{2}\theta d\theta + \dfrac{\cos\theta \sin\theta}{r}dr + \cos^{2}\theta d\theta \\ &= d\theta \end{align*} $$

Therefore,

$$ \int - \dfrac{y}{x^{2} + y^{2}}dx + \dfrac{x}{x^{2} + y^{2}}dy = \int d\theta $$

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Properties

Let $M, N$ be differential manifolds of dimensions $m, n$ respectively, and let $f : M \to N$. Let $\omega, \varphi$ be $k$-forms on $N$. Let $g$ be a $0$-form on $N$. Let $\varphi_{i}$s be $1$-forms on $N. Then, the following hold true.

$$ \begin{align} f^{\ast} (\omega + \varphi) =&\ f^{\ast}\omega + f^{\ast}\varphi \tag{a} \\ f^{\ast} (g \omega) =&\ (f^{\ast}g) (f^{\ast}\omega) \tag{b} \\ f^{\ast} (\varphi_{1} \wedge \cdots \wedge \varphi_{k}) =&\ f^{\ast}(\varphi_{1}) \wedge \cdots \wedge f^{\ast}(\varphi_{k}) \tag{c} \end{align} $$

Here, $+$ and $\wedge$ represent the sum and wedge product of $k$-forms, respectively.

Let $\omega, \varphi$ be arbitrary forms on $N. Let $L$ be a $l$-dimensional differential manifold, and let $g : L \to N$.

$$ \begin{align*} f^{\ast}(\omega \wedge \varphi) &= (f^{\ast}\omega) \wedge (f^{\ast}\varphi) \tag{d} \\ (f \circ g)^{\ast} \omega &= g^{\ast}(f^{\ast}\omega) \tag{e} \end{align*} $$

Proof

Proof $(a)$

$$ \begin{align*} (f^{\ast}(\omega + \varphi))_{p} (v_{1}, \dots, v_{k}) =&\ (\omega + \varphi)_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\ =&\ \omega_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) + \varphi_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\ =&\ (f^{\ast} \omega)_{p}(v_{1}, \dots, v_{k}) + (f^{\ast} \varphi)_{p}(v_{1}, \dots, v_{k}) \\ =&\ \left( f^{\ast}\omega + f^{\ast}\varphi \right)_{p}(v_{1}, \dots, v_{k}) \end{align*} $$

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Proof $(b)$

Let’s define the product of a $0$-form $g$ and a $k$-form $\omega$ as follows.

$$ (g\omega)(p) = g(p) \omega (p) $$

Note that $g(p) = g_{p}$ is a scalar, and $\omega (p) = \omega_{p}$ is a function. Then,

$$ \begin{align*} (f^{\ast} (g\omega))_{p} (v_{1}, \dots, v_{k}) =&\ g\omega_{f(p)} \left( df_{p}v_{1}, \dots, df_{p}v_{k} \right) \\ =&\ g_{f(p)} \omega_{f(p)} (df_{p}v_{1}, \dots, df_{p}v_{k}) \\ =&\ g\circ f(p) \omega_{f(p)} (df_{p}v_{1}, \dots, df_{p}v_{k}) \\ =&\ (f^{\ast}g)_{p} (f^{\ast}\omega)_{p} (v_{1}, \dots, v_{k}) \end{align*} $$

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Proof $(c)$

$$ \begin{align*} (f^{\ast}\left( \varphi_{1} \wedge \cdots \wedge \varphi_{k} \right))_{p} (v_{1}, \dots, v_{k}) =&\ (\varphi_{1} \wedge \dots \wedge \varphi_{k})_{f(p)} \left( df_{1}, \dots, df_{k} \right) \\ =&\ \det [\varphi_{i}df(v_{j})] \\ =&\ \det [ f^{\ast} \varphi_{i}(v_{j})] \\ \end{align*} $$

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