What is the metric (distance function) on the 1-sphere and 2-sphere?

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SUMMARY

The distance function on the 1-sphere (S1) is defined by the round metric \(\mathbf{g} = d\theta^2\), while for the 2-sphere (S2), it is given by \(\mathbf{g} = \sin^2\theta\,d\phi^2 + d\theta^2\). The parametric equations for the 1-sphere are \(x = \cos(\theta)\) and \(y = \sin(\theta)\), leading to the expression \(ds^2 = d\theta^2\). For the 2-sphere, spherical coordinates are utilized with \(x = \cos(\theta)\sin(\phi)\), \(y = \sin(\theta)\sin(\phi)\), and \(z = \cos(\phi)\), allowing for the calculation of the metric using the Jacobian matrix, where \(\mathbf{g} = J^T J\).

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Hi i only know the distance function in euclidean space as the standard definition involving the sqrt of the dot product of the vector connecting two points.

What is the metric (distance function) in S1 or S2 which is the 1 sphere and 2sphere respectively.
 
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The round metric for the unit 1-sphere is (using radians), \mathbf{g} = d\theta^2, and for the unit 2-sphere, \mathbf{g} = \sin^2\theta\,d\phi^2 + d\theta^2, I think, working off of memory. It should be simple to add in the r term for arbitrary euclidean 1 and 2-spheres

In general, if J is the jacobian matrix of the manifold:
\mathbf{g} = J^T J
 
That's right. The 1-sphere (unit circle for us pedestrians) has parametric equations x= cos(\theta), y= sin(\theta) so that dx= -sin(\theta)d\theta and dy= cos(\theta) so that ds^2= dx^2+ dy^2= (sin^2(\theta)+ cos^2(\theta))d\theta^2= d\theta^2.

For the 2-sphere (the surface of the unit sphere), we can take x= cos(\theta)sin(\phi), y= sin(\theta)sin(\phi), z= cos(\phi) (spherical coordinates with \rho= 1). Then dx= -sin(\theta)sin(\phi)+ cos(\theta)cos(\phi), dy= cos(\theta)sin(\phi)+ sin(\theta)cos(\phi), and dz= -sin(\phi)d\phi.

dx^2= sin^2(\theta)sin^2(\phi)d\theta^2- 2sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ cos^2(\theta)cos^2(\phi)d\phi^2
 

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