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If a regular surface is covered by a surface patch all of whose $u$- and $v$-coordinate curves are geodesics, then it has Gaussian curvature zero.

How can I show the above statement? Could you give any hint?

Edit: Using the following formula: Let $S$ be an oriented regular surface and $\sigma:U\subset\Bbb R^2\to V\subset S$ be a surface patch. Let $\gamma:I\to V$ be a regular curve which can be expressed as $\gamma(t) = \sigma(u(t),v(t))$ for some smooth functions $u,v:I\to\Bbb R$. Then $\gamma$ is geodesic if and only if $u''+(u')^2\Gamma_{11}^1+2(u'v')\Gamma_{12}^1+(v')^2\Gamma_{22}^1 = 0$ and $v''+(u')^2\Gamma_{11}^2+2(u'v')\Gamma_{12}^2+(v')^2\Gamma_{22}^2 = 0$.

From this, if we consider $\gamma$ as a coordinate curve (letting $u(t)=t$ and $v(t)=t$ resp.) then I get $\Gamma_{11}^1 =\Gamma_{11}^2 = \Gamma_{22}^1=\Gamma_{22}^2 = 0$. I don't know how to get further. I'm in introductory course in differential geometry by the way. So I want some explanation in elementary way.

The formula I wrote is in p.290 of this textbook.

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  • $\begingroup$ What formula are you using for Guassian curvature? Where in the calculation have you got stuck or confused? $\endgroup$
    – Nick
    Dec 5 '20 at 17:47
  • $\begingroup$ @Nick In fact, this exercise comes from the section that introduces Christoffel symbols. I think any formula would be allowed. Using the fact that the coordinate curves are geodesics, I found that $\Gamma_{11}^1 = \Gamma_{11}^2 = \Gamma_{22}^1 = \Gamma_{22}^2 =0$. And I don't know how to go further $\endgroup$
    – love_sodam
    Dec 5 '20 at 17:53
  • $\begingroup$ Maybe this will help: en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry ? If you have verified that the Christoffel symbols are zero, then the curvature tensor itself is identically zero. $\endgroup$
    – Nick
    Dec 5 '20 at 18:32
  • $\begingroup$ @Nick But not all of the symbols are zero. Only four of them I think. $\endgroup$
    – love_sodam
    Dec 5 '20 at 18:37
  • $\begingroup$ How about this formula, then: en.wikipedia.org/wiki/… ? Since the coordinate lines are geodesics, $\nabla_1 e_1 = 0$, etc... Maybe you can fill in the rest of the details. $\endgroup$
    – Nick
    Dec 5 '20 at 18:43
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I believe this is false. Locally, on any surface whatsoever you can always find coordinates so that the $u$-curves and $v$-curves are geodesics. It is a standard result from differential equations that on a surface I can find local coordinates whose coordinate curves are tangent to two given linearly independent vector fields.

Choose a frame $X_0,Y_0$ (unit vectors) at $p_0$. Take the geodesics in those tangent directions. Call them $C$ and $\Gamma$. Take the tangent vectors to be $X$ and $Y$ respectively along those curves. Now choose $Y$ smoothly along $C$ and $X$ smoothly along $\Gamma$ extending these, maintaining linear independence. Now follow geodesics from points of $C$ in the direction of $Y$ and follow geodesics from points of $\Gamma$ in the direction of $X$. Define the vector fields $X$ and $Y$ in the obvious way in a small neighborhood of $p_0$. We have constructed these so that they are tangent to geodesics at each point.

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