Is the Real Projective Plane's Gaussian Curvature Always Positive?

[Euge]

Here is this week's POTW:

-----
Prove that a compact differentiable surface homeomorphic to the real projective plane has a point at which the Gaussian curvature is positive.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week’s problem. You can read my solution below.

Spoiler

Let $\Sigma$ be the compact differentiable surface. Being homeomorphic to the real projective plane $\Bbb RP^2$, it has the same Euler characteristic as $\Bbb RP^2$. The projective plane has a CW-complex structure with one 0-cell, one 1-cell and one 2-cell. Hence, its Euler characteristic is $1 - 1 + 1 = 1$. By the Gauss-Bonnet theorem, the total integral of the Gaussian curvature $K$ is $\Sigma$ is $2\pi$, so by the mean value theorem, $K$ is positive at some point on $\Sigma$.