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

Let S be a surface of revolution in R3\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 α:IR3\alpha: I \rightarrow \mathbb{R}^3, where uIu \in I is α(u)=(f(u),0,g(u)).\alpha(u) = (f(u), 0, g(u)). Then by rotating this planar curve in the z-axis, we have the parametrisation

φ(u,v)=(f(u)cosv,f(u)sinv,g(u))\varphi(u,v) = (f(u)\cos v, f(u) \sin v, g(u)) of SS.

The coefficients of the first fundamental form of this parametrisation are E=1,F=0,G=f2(u).E=1, F=0, G= f^2(u). Since we let the generating curve be parametrised by unit speed curve, φ\varphi does not depend on gg. We can express the rotations by angle θ\theta about the z-axis in the standard basis of R3\mathbb{R}^3:

Rz(θ)=[cosθsinθ0sinθcosθ0001]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 φθ(u,v)=(Rz(θ)φ)(u,v)=(f(u)cos(v+θ), f(u)sin(v+θ), g(u)).\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, φθ\varphi_{\theta} are

Eθ=<φθu,φθu>=(f(u))2(cos2(v+θ)+sin2(v+θ))+(g(u))2=(f(u))2+(g(u))2=1E_{\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 Fθ=<φθu,φθv>=(f(u))2cos(v+θ)sin(v+θ)+(f(u))2cos(v+θ)sin(v+θ)+0=0F_{\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 Gθ=<φθv,φθv>=f2(u)sin2(v+θ)+f2(u)cos2(v+θ)=f2(u)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)

Eθ=1,Fθ=0,Gθ=f2(u).\therefore E_{\theta} = 1, F_{\theta} = 0, G_{\theta} = f^2(u).

Since Eθ=E,Fθ=F,Gθ=GE_{\theta} = E, F_{\theta} = F, G_{\theta} = G, we see that the rotations of S about its axis by θ\theta are local isometries to S.

Note by Bright Glow
3 years, 2 months ago

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