[Differential Geometry] Are Rotations on a Surface of Revolution Isometries?

Let S be a surface of revolution in \(\mathbb{R}^3\). Are the rotations about its axis all isometries of S?

Without loss of generality, let's assume that the axis of revolution is the z-axis, and the generating curve in the xz-plane is parametrised by a unit speed parametrised curve \(\alpha: I \rightarrow \mathbb{R}^3\), where \(u \in I\) is \[\alpha(u) = (f(u), 0, g(u)).\] Then by rotating this planar curve in the z-axis, we have the parametrisation

\[\varphi(u,v) = (f(u)\cos v, f(u) \sin v, g(u))\] of \(S\).

The coefficients of the first fundamental form of this parametrisation are \(E=1, F=0, G= f^2(u).\) Since we let the generating curve be parametrised by unit speed curve, \(\varphi\) does not depend on \(g\). We can express the rotations by angle \(\theta\) about the z-axis in the standard basis of \(\mathbb{R}^3\):

\[R_z(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1
\end{bmatrix}\]

By composing this rotation with our parametrisation, we have \[\varphi_{\theta}(u,v) = (R_z(\theta)\circ \varphi)(u,v) = (f(u)\cos(v+\theta),~ f(u)\sin(v+\theta),~ g(u)).\]

Hence the coefficients of the first fundamental form of the surface after rotation, \(\varphi_{\theta}\) are

\[E_{\theta} = \left<\varphi_{\theta u}, \varphi_{\theta u}\right> = (f'(u))^2(\cos^2(v+\theta) + \sin^2(v+\theta))+(g'(u))^2 = (f'(u))^2+(g'(u))^2 = 1\] \[F_{\theta} = \left<\varphi_{\theta u}, \varphi_{\theta v}\right> = -(f'(u))^2\cos(v+\theta)\sin(v+\theta) +(f'(u))^2\cos(v+\theta)\sin(v+\theta) + 0 = 0\] \[G_{\theta} = \left<\varphi_{\theta v}, \varphi_{\theta v}\right> = f^2(u)\sin^2(v+\theta)+f^2(u)\cos^2(v+\theta) = f^2(u)\]

\[\therefore E_{\theta} = 1, F_{\theta} = 0, G_{\theta} = f^2(u).\]

Since \(E_{\theta} = E, F_{\theta} = F, G_{\theta} = G\), we see that the rotations of S about its axis by \(\theta\) are local isometries to S.

Note by Tasha Kim
1 month ago

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