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I want to derive the round metric $g=d\theta^{\,2}+\sin\left(\theta\right)^2d\phi^{\,2}$ but I cannot get the correct answer. I know that the metric in cartesian coordinates is $g=dx^2+dy^2$. I've used the formula $dx=\frac{dx}{d\theta}d\theta+\frac{dx}{d\phi}d\phi$ and $dy=\frac{dy}{d\theta}d\theta+\frac{dy}{d\phi}d\phi$. From

$x=\sin\left(\theta\right)\cos\left(\phi\right)$ and $y=\sin\left(\theta\right)\sin\left(\phi\right)$,

I find that

$\frac{dx}{d\theta}=\cos\left(\theta\right)\cos\left(\phi\right)$,

$\frac{dx}{d\phi}=-\sin\left(\theta\right)\sin\left(\phi\right)$,

$\frac{dy}{d\theta}=\cos\left(\theta\right)\sin\left(\phi\right)$,

$\frac{dy}{d\phi}=\sin\left(\theta\right)\cos\left(\phi\right)$.

This leads to $g=dx^2+dy^2=\cos\left(\theta\right)^2d\theta^{\,2}+\sin\left(\theta\right)^2d\phi^{\,2}$ which is not correct.

edit: metric was typed incorrectly

user331373
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    Your formula for the round metric is incorrect (the coefficient $\sin \theta$ should be squared).

    Embed the sphere using spherical coordinates in the usual way, and compute the pullback of $\bar{g} = dx^2 + dy^2 + dz^2$ to the sphere, which amounts to replacing $x, y, z$ with their expressions in spherical coordinates.

     – Travis Willse Apr 14 '16 at 11:53
  • So that means I calculate $dx=\frac{dx}{d\theta}d\theta+\frac{dx}{d\phi}d\phi$ and similarly for $dy$ and $dz$ and then put them into $\bar{g}=dx^2+dy^2+dz^2$? – user331373 Apr 14 '16 at 12:19
  • Yes, that's the idea: The pullback by the inclusion $\iota$ is $\iota^* \bar g$ is $\iota^* (dx^2 + \cdots ) = d(\iota^* x)^2 + \cdots = d(x \circ \iota)^2 + \cdots$. – Travis Willse Apr 14 '16 at 12:38
  • So that means the metric for the sphere that I wrote is not right as it is missing a $dz^2$ term. It seems strange that I only need $d\theta$ and $d\phi$ to describe the metric in spherical coordinates but I need all of $dx$, $dy$ and $dz$ to do the same. I always thought that the number of those differentials I needed was equal to the dimension of the manifold. – user331373 Apr 14 '16 at 12:45
  • You are missing the $z$-coordinate in your spherical coordinates. You are correct in that you only need two differential 1-forms (squared) to describe the metric since the manifold is $2$-dimensional (The metric is a $2$-tensor on a $2$-dimensional vector space). – Sak Apr 14 '16 at 22:36
  • @user331373 the sphere a 2-dimensional manifold embedded in 3-dimensional space. Hence the metric in the ambient space has three differentials, but when you pull it back to the 2-sphere there are only two. – ziggurism Oct 01 '23 at 21:43

2 Answers2

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calculate

$$g_{\theta\theta}=\frac {\partial x}{\partial \theta }\frac {\partial x}{\partial \theta }+\frac {\partial y}{\partial \theta }\frac {\partial y}{\partial \theta }+\frac {\partial z}{\partial \theta }\frac {\partial z}{\partial \theta }$$

$$g_{\theta\phi}=\frac {\partial y}{\partial \theta }\frac {\partial y}{\partial \phi}+\frac {\partial x}{\partial \theta }\frac {\partial x}{\partial \phi }+\frac {\partial z}{\partial \theta }\frac {\partial z}{\partial \phi }$$

$$g_{\phi\theta}=\frac {\partial x}{\partial \phi}\frac {\partial x}{\partial \theta }+\frac {\partial y}{\partial \phi }\frac {\partial y}{\partial \theta }+\frac {\partial z}{\partial \phi }\frac {\partial z}{\partial \theta }$$

$$g_{\phi\phi}=\frac {\partial y}{\partial \phi }\frac {\partial y}{\partial \phi }+\frac {\partial x}{\partial \phi }\frac {\partial x}{\partial \phi}+\frac {\partial z}{\partial \phi }\frac {\partial z}{\partial \phi }$$ now plug it in the matrix $$g=\begin{pmatrix}g_{\theta \theta }&&g_{\theta \phi}\\g_{ \phi \theta} &&g_{\phi\phi}\end{pmatrix}$$

Nebo Alex
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Let $\Psi(\theta, \phi): (\sin\left(\theta\right)\cos\left(\phi\right), \sin\left(\theta\right)\sin\left(\phi\right), \cos(\theta) )$. Then the coefficients of the metric are given by (The dot denotes the dot product, and using the Pythagorean identity $\sin^2\alpha + \cos^2\alpha=1$):

$\frac{\partial\Psi}{\partial \theta}\cdot\frac{\partial\Psi}{\partial\theta}=\left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)\cdot \left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)= 1$,

$\frac{\partial\Psi}{\partial \theta}\cdot\frac{\partial\Psi}{\partial \phi}=\left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)\cdot \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)=0$,

$\frac{\partial\Psi}{\partial\phi}\cdot\frac{\partial\Psi}{\partial\phi}= \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)\cdot \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)=\sin^2\theta$

Noah Schweber
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Sak
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