# [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 Bright Glow
2 years, 5 months ago

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