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Given a first fundamental form, i.e. $$\frac{(du)^2 + (dv)^2}{u^2 + v^2}.$$ How can I calculate the Gaussian curvature $K$? I do not really know how to approach the problem, since the formulas for the Gaussian curvature involve the second fundamental form. Is there a way to calculate the second fundamental form out of the first one?

Edit. Here is the theorema egregium as stated in Riemannian Manifolds by John M. Lee:

Let $M \subseteq \mathbb{R}^3$ be a $2$-dimensional submanifold and $g$ the induced metric on $M$. For any $p\in M$ and any basis $(X,Y)$ for $T_pM$, the Gaussian curvature of $M$ at $p$ is given by $$K= \frac{Rm(X,Y,Y,X)}{|X|^2|Y|^2-\langle X,Y\rangle^2}$$ Therefore the Gaussian curvature is an isometry invariant of $(M,g)$.

Where $Rm$ denotes the Riemann curvature tensor and $$Rm(X,Y,Z,W) = \langle R(X,Y)Z,W\rangle$$ where $R$ is the Riemann curvature endomorphism.

TheGeekGreek
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  • You should look at Gauss's Theorem Egregium. – John Hughes Dec 15 '16 at 15:56
  • @JohnHughes Ah thanks, then I am on the right way. I was not entirely sure if I have to use it. Somehow I do not quite see yet how to apply it. I am using the one stated in the book Riemannian manifolds by John M. Lee (will post shortly). There is a formula in terms of the basis elements of the tangential space. – TheGeekGreek Dec 15 '16 at 16:05
  • Right -- the key thing is that Rm can be expressed entirely in terms of derivatives of the metric $g$; this version of the theorem obscures that slightly. – John Hughes Dec 15 '16 at 18:09
  • @JohnHughes I looked at the previous pages and found nothing which is very enlightening for me...I think the reason is, that I do not have a really strong background on tensor analysis (second year of bachelor studies). Would you mind to write a short answer? I mean a bit more guidance since I think then I will also more understand what the book is saying... – TheGeekGreek Dec 15 '16 at 20:11
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    This statement of Theorema Egregium is overkill, and not particularly enlightening for what you're trying to do. You should fish up the version in Do Carmo's Differential Geometry of Curves and Surfaces. There is an explicit formula there given for the curvature only in terms of quantities involving the metric. Of course, if you unpack the statement you have here, you will find that they are equivalent statements. – A. Thomas Yerger May 31 '17 at 22:07

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You have $g(u, v)$ as a matrix with entries $g_{ij}(u, v)$ that are functions of $u, v$. You can compute the inverse of this matrix (it's diagonal, after all!), and call its entries $g^{ij}(u, v)$.

Define $$ \Gamma^\ell_{ij} (u, v) = \frac{1}{2} \sum_{k=1}^2 g^{kl} \left ( \frac{\partial g_{ik}}{\partial u^j} - \frac{\partial g_{ij}}{\partial u^k} + \frac{\partial g_{kj}}{\partial u^i} \right ) $$ where $u^1$ means $u$, and $u^2$ means $v$, and everything on the right hand side is actually a function of $u$ and $v$.

Now define $$ R^\ell_{ijk} = \frac{\partial \Gamma_{ik}^\ell}{\partial u^j} - \frac{\partial \Gamma_{ij}^\ell}{\partial u^k} + \sum_{p = 1,2} \left (\Gamma_{ik}^p \Gamma_{pj}^\ell - \Gamma_{ij}^p \Gamma_{pk}^\ell \right) $$

(I'm not making this up!) Also: everything above is a function of $u$ and $v$, of course.

Then the theorem egregium says that the curvature $K$ is given by $$ K(u, v) = \frac{1}{g(u, v)} \sum_{\ell = 1,2} R^\ell_{121} g_{\ell 2} $$ where $g(u, v)$ in this formula denotes the determinant of the matrix of $g_{ij}$s.

Now I'm not man enough to do all those calculations and derivatives for you, but the vast majority of the terms you compute will simply be zero (yay!). I copied these formulas from Millman and Parker's Elements of Differential Geometry; I mention that so that you can double-check that I got the indices correct. :(

John Hughes
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  • Very nice! Thanks a lot! No, the calculations I will do by myself of course. I think that the book by Lee is just a little bit too abstract for my understanding of the subject at the moment. +1 – TheGeekGreek Dec 15 '16 at 21:14