# Pullback Calculation

If we define the 2-form $\omega=\frac{1}{r^3}(x_1dx_2\wedge dx_3+x_2dx_3\wedge dx_1+x_3dx_1\wedge dx_2)$ with $r=\sqrt{x_1^2+x_2^2+x_3^2}$

If we now define $x(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ for $\theta\in(0,\pi)$ and $\phi\in (0,2\pi)$

I now want to show that the pullback of this map is:

$x^*\omega=\sin\theta d\theta\wedge d\phi$

Now my definition of the pullback of a map $c:[a,b]\rightarrow D$ is $\alpha(c'(t))dt$ but am unsure as to how to proceed? I can see that this curve $x$ is a parametrization of the sphere in $\mathbb{R}^3$ but after that I am a bit lost?

Thanks for any help

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simply plug the expressions for $x_{1,2,3}$ in terms of $\theta,\phi$ into $\omega$ and calculate the resulting form. –  O.L. May 22 '13 at 9:36
@O.L. sorry I am confused as to how I evalute $x'(t)$ what it is in terms of $\theta$ and $\phi$? –  hmmmm May 22 '13 at 9:54
@hmmmm: What O.L. means is this: You should plug in $x_1 = \sin \theta \cos \phi$ and $x_2 = \sin\theta \sin \phi$, and so on. Then, for example, $$dx_1 = d(\sin\theta \cos\phi) = d(\sin\theta)\cos\phi + \sin\theta \,d(\cos\phi) = \cos\theta \cos \phi\, d\theta - \sin\theta\sin\phi \,d\phi.$$ Similarly for the others. –  Jesse Madnick May 22 '13 at 10:07
I guess $c(t)$ appeared in a one-dimensional example which is not very much related to your question. Here, instead of a curve mapped to some space, you have a surface (2D sphere). In a sense, the analog of $t$ here is the pair $(\theta,\phi)$, and the analog of $c$ is given by the triple of functions $x_{1,2,3}(\theta,\phi)$. –  O.L. May 22 '13 at 10:10
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## 1 Answer

A more general equation for how to compute pullbacks is as follows: If $f: M \to N$ is a smooth map of smooth manifolds and $\omega$ is an $n$-form on $N$, written in local coordinates as $$\omega = \sum_I \omega_I dy^{i_1} \wedge \cdots \wedge dy^{i_n}$$ then $f^*\omega$ can be written in induced coordinates as $$f^*\omega = \sum_I (\omega_I \circ f) d(y^{i_1} \circ f) \wedge \cdots \wedge d(y^{i_n} \circ f)$$ where you think of $y^i$ as the function which picks out the $i^{th}$ coordinate of $f$.

Your $\omega$ is given by $$\omega = \omega_{23} dx_2 \wedge dx_3 + \omega_{31} dx_3 \wedge dx_1 + \omega_{12} dx_1 \wedge dx_2$$ where $\omega_{23} = \frac{x_1}{r^3}, \omega_{31} = \frac{x_2}{r^3}$ andd $\omega_{12} = \frac{x_3}{r^3}$. So now you just use the formula: I will do the $\omega_{23} dx_2 \wedge dx_3$ term as an example:

$$f^*(\omega_{23} dx_2 \wedge dx_3) = (\omega_{23} \circ f) d(x_2 \circ f) \wedge d(x_3 \circ f)$$

Now \begin{align*} \omega_{23} \circ f &= \frac{\sin\theta \cos\phi}{\sqrt{\sin^2\theta\cos^2\phi + \sin^2\theta \sin^2\phi + \cos^2\phi}^3} = \sin\theta\cos\phi \\ d(x_2\circ f) &= d(\sin\theta\sin\phi) = \cos\theta \sin\phi d\theta + \sin\theta \cos\phi d\phi \\ d(x_3\circ f) &= d(\cos\theta) = -\sin\theta d\theta. \end{align*} Putting it all together, we get \begin{align*} (\omega_{23} \circ f) d(x_2 \circ f) \wedge d(x_3 \circ f) &= (\sin\theta\cos\phi)(\cos\theta\sin\phi d\theta + \sin\theta\cos\phi d\phi) \wedge (-\sin\theta d\theta ) \\ &= \sin^3\theta\cos^2\phi d\theta \wedge d\phi \end{align*}

(Of course, modulo mistakes in my computations!)

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