1

Given two smooth manifolds with Riemannian metric $(X,g)$ and $(Y,h)$ and a smooth map $f: X \to Y$ I understand that we define $\phi$ to be an isometry if $f^* g = h$.

I thought I understood this definition but when trying to calculate it with an easy example (case where the manifolds are the same) I'm having difficulty - it should be straightforward and I'm probably missing something obvious!

I'm looking at a 1-parameter family of maps $f_t: S^2 \to S^2$ with $f_t(\theta,\phi) = (\theta + t, \phi)$ in local (spherical polar coordinates) where the spheres are given the same spherical polar metric $g = d\theta^2 + \sin^2(\theta)d\phi^2$.

Now I understand the condition for this to be an isometry is $f_t^*g(f_t(\theta,\phi)) = g(\theta,\phi)$ but we have $g(f_t(\theta,\phi)) = d\theta^2 + \sin^2(\theta+t)d\phi^2$ and then pulling this back (using the fact that $f^* (\omega dx) = f(w)d(f(w))$ generally) gives $f_t^*g(f_t(\theta,\phi)) = d\theta^2 + \sin^2(\theta+2t)d\phi^2$

It would be very helpful to understand where I am going wrong here.

Wooster
  • 3,775

1 Answers1

0

First a small note: The condition for an isometry is $f^{*}h = g$; metrics pull back under smooth mappings.

The point with your example is, if you write the round metric $$ g = d\theta^{2} + \sin^{2}\theta\, d\phi^{2}, $$ you're using physicists' convention for the angle names: $\theta$ is colatitude and $\phi$ is longitude. The proposed coordinate translation is not, therefore, an isometry. The rotation mapping you presumably intended, $$ f_{t}(\theta, \phi) = (\theta, \phi + t), $$ is easily checked to be an isometry for each $t$.

Andrew D. Hwang
  • 78,195
  • Ah yes of course, that was a bit stupid - thanks for the fast answer. In the example you gave we have $g(f_t(\theta,\phi)) = g(\theta,\phi)$ is there an easy example where that is not the case but $f_t$ is still an isometry? – Wooster Apr 02 '16 at 13:15
  • There are rotations of the sphere besides translations in $\phi$, but they're quite messy to write out in terms of $(\theta, \phi)$. If you want to try, I'd recommend embedding the sphere, rotating $\mathbf{R}^{3}$ about the $y$-axis, then recovering spherical coordinates:\begin{align}(\theta,\phi) &\mapsto (x, y, z) = (\cos\phi \sin\theta, \sin\phi \sin\theta, \cos\theta) \ &\mapsto (x', y', z') = (x\cos t - z\sin t, y, z\cos t + x\sin t) \ &\mapsto (\theta', \phi') = (\arccos z', \operatorname{Atan2}(y', x')).\end{align}You see what I mean.... :) – Andrew D. Hwang Apr 02 '16 at 13:45
  • Great, thanks! If we have a metric in local coordinates $g = g(x,y)dxdy$ and we want to check if $\phi^* g = g$ then am I right in thinking that we have to calculate $\phi^* g(\phi(x,y))$ i.e. $\phi^* (g(\phi(x,y))dxdy) = g(\phi(x,y)^2)d(x(\phi))d(y(\phi))$? It seems odd to have a $\phi^2$ there? – Wooster Apr 03 '16 at 11:06
  • Not quite: If $x$ are local coordinates on $X$, and if $u$, $v$ are tangent vectors to $X$ at $x$, then$$(\phi^{*}h)(x)(u, v) = h(\phi(x))(D\phi(x)(u), D\phi(x)(v));$$in words, use the differential $D\phi(x)$ to push $u$ and $v$ to tangent vectors to $Y$ at $\phi(x)$, then take their inner product with respect to $h$. – Andrew D. Hwang Apr 03 '16 at 23:44