# Checking if a surface in $\mathbb{R^3}$ exists if the first fundamental form is given

Given the first fundamental form and the shape operator: $$g = \begin{pmatrix} 1 & 0\\ 0 & \cos^2(u) \end{pmatrix} \text{and} \,S = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},$$ I want to check, if there exists a surface $$f: \Omega \rightarrow \mathbb{R^3}$$ satisfying these conditions. We know that $$S^T = bg^{-1},$$ where $$b$$ is the second fundamental form. Thus, $$S = I = bg^{-1}$$ and $$b = g.$$ From here we find, $$E = 1, F = 0, G = \cos^2(u), e = 1, f=0, g = \cos^2(u).$$ We may also find that the mean curvature is $$H=1$$ and the Gauss curvature, $$K = 1,$$ the principal curvatures being $$K_1 = K_2 = 1.$$ How can I go from here to eventually find a parametrization of the surface?

• that may not be exactly what you want, but the surfaces of constant curvature are well known.... Commented Feb 19 at 21:30
• Thanks. I thought first, it would be a sphere, but can not find an explicit parametrization. Can you help ? Commented Feb 19 at 21:33
• It sure looks like a variant of spherical coordinates to me. Commented Feb 19 at 22:53
• Spherical coordinates \begin{align} x&=\cos\varphi\sin\theta\,,&y&=\sin\varphi\sin\theta\,,&z=\cos\theta \end{align} become with $u=\theta+\frac\pi 2$ \begin{align} x&=\cos\varphi\cos u\,,&y&=\sin\varphi\cos u\,,&-z=\sin u \end{align} which should be your parametrization. Commented Feb 19 at 23:28
• Thanks. With this proposal of change of parameters, the metric tensor is alright, but not the shape operator. I was rather looking for a formal procedure, by maybe integrating. An Ansatz seems to me rather arbitrary. Commented Feb 20 at 8:02

As we know we have in traditional spherical coordinates \begin{align} f(\varphi,\theta)&=\pmatrix{\cos\varphi\sin\theta\\\sin\varphi\sin\theta\\\cos\theta}\,,& f_\varphi&=\pmatrix{-\sin\varphi\sin\theta\\\cos\varphi\sin\theta\\0}\,,& f_\theta&=\pmatrix{\cos\varphi\cos\theta\\\sin\varphi\cos\theta\\-\sin\theta}\,,\\[2mm] E&=\langle f_\theta,f_\theta\rangle=1\,,&F&=\langle f_\varphi,f_\theta\rangle=0\,,&G&=\langle f_\varphi,f_\varphi\rangle=\sin^2\theta\,,\\[2mm] f_{\varphi\,\varphi}&=\pmatrix{-\cos\varphi\sin\theta\\-\sin\varphi\sin\theta\\0}&f_{\varphi\,\theta}&=\pmatrix{-\sin\varphi\cos\theta\\\cos\varphi\cos\theta\\0}\,,&f_{\theta\,\theta}&=\pmatrix{-\cos\varphi\sin\theta\\-\sin\varphi\sin\theta\\-\cos\theta}\,,\\[2mm] L&=-\langle f_{\theta\,\theta},\mathbf{n}\rangle=1\,,&M&=-\langle f_{\varphi\,\theta},\mathbf{n}\rangle=0\,,&N&=-\langle f_{\varphi\,\varphi},\mathbf{n}\rangle=\sin^2\theta \end{align} where we use that the normal vector is just $$\mathbf{n}=f(\varphi,f_\theta)\,.$$ Then the shape operator is $$S=\pmatrix{L&M\\M&N}\pmatrix{E&F\\F&G}^{-1}=\pmatrix{1&0\\0&1}$$ Switching to $$u=\theta+\frac\pi2$$ leads to $$\sin\theta\to-\cos u$$ and $$\cos\theta\to\sin u$$ so that in those coordinates \begin{align} f(\varphi,u)&=\pmatrix{-\cos\varphi\cos u\\-\sin\varphi\cos u\\\sin u}\,,& f_\varphi&=\pmatrix{\sin\varphi\cos u\\-\cos\varphi\cos u\\0}\,,& f_u&=\pmatrix{\cos\varphi\sin u\\\sin\varphi\sin u\\\cos u}\,,\\[2mm] E&=\langle f_u,f_u\rangle=1\,,&F&=\langle f_\varphi,f_u\rangle=0\,,&G&=\langle f_\varphi,f_\varphi\rangle=\cos^2u\,,\\[2mm] f_{\varphi\,\varphi}&=\pmatrix{\cos\varphi\cos u\\\sin\varphi\cos u\\0}&f_{\varphi\,u}&=\pmatrix{-\sin\varphi\sin u\\\cos\varphi\sin u\\0}\,,&f_{u\,u}&=\pmatrix{\cos\varphi\cos u\\\sin\varphi\cos u\\-\sin u}\,,\\[2mm] L&=-\langle f_{u\,u},\mathbf{n}\rangle=1\,,&M&=-\langle f_{\varphi\,u},\mathbf{n}\rangle=0\,,&N&=-\langle f_{\varphi\,\varphi},\mathbf{n}\rangle=\cos^2 u \end{align} and again $$S=\pmatrix{L&M\\M&N}\pmatrix{E&F\\F&G}^{-1}=\pmatrix{1&0\\0&1}\,.$$