# Prove the given surface has Gaussian curvature zero

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.

• What formula are you using for Guassian curvature? Where in the calculation have you got stuck or confused?
– Nick
Dec 5 '20 at 17:47
• @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 Dec 5 '20 at 17:53
• 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.
– Nick
Dec 5 '20 at 18:32
• @Nick But not all of the symbols are zero. Only four of them I think. Dec 5 '20 at 18:37
• 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.
– Nick
Dec 5 '20 at 18:43

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.