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How to prove the following:

Show that there is no metric on $S^{2}$ having curvature bounded above by $0$ and no metric on surface of genus $g$ which is bounded below by $0$.

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Do you know the Gauss–Bonnet theorem? –  Zhen Lin Apr 24 '12 at 10:30
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Another way of seeing this (except for the torus) is Milnor's theorem on fundamental group growth.$^1$

Let $M$ be a complete, compact Riemannian manifold. Let $\gamma$ denote the growth function of the fundamental group $\pi_1M$.

If the mean curvature of $M$ is everywhere positive, then $\gamma(s)$ is bounded above by a polynomial function of $s$.

If the sectional curvatures of $M$ are everywhere negative, then $\gamma(s)$ is bounded below by an exponential function of $s$.

To apply this, observe that $S^2$ has trivial fundamental group, hence does not admit any metric of negative curvature. A little computation shows that $\pi_1\Sigma_g$ has exponential growth, for $g\geq 2$, so does not admit any metric of positive curvature.

$^1$: http://intlpress.com/JDG/archive/pdf%20image%20on%20text/1968/2-1-1ocr.pdf

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Expanding on Zhen Lin's comment, by the Gauss-Bonnet theorem, on a closed $2$-dimensional Riemannian manifold $M$, we have that $$\frac{1}{2\pi}\int_M K ~dA = \chi(M),$$ where $\chi(M)$ is the Euler characteristic of $M$. We have that $\chi(S^2) = 2$, and if $\Sigma_g$ denotes the surface of genus $g$, then $\chi(\Sigma_g) = 2 - 2g$. Hence $$\frac{1}{2\pi}\int_{S^2} K_{S^2} ~dA = 2$$ implies that $K_{S^2}$ must be somewhere positive and $$\frac{1}{2\pi}\int_{\Sigma_g} K_{\Sigma_g} ~dA = 2 - 2g$$ where $g \geq 1$ implies that $K_{\Sigma_g}$ cannot be everywhere positive (I assume your question means that for the case of the torus the curvature cannot be strictly bounded below by $0$, since the torus admits a metric of everywhere zero Gaussian curvature).

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Isn't $\int_M KdA = 2\pi\chi(M)$? –  Neal Apr 24 '12 at 20:58
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Yes, I forgot to put that factor in somehow. Thanks. –  Henry T. Horton Apr 24 '12 at 21:00
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