# non compact riemannian manifolds

According to Gromoll and Meyer:

• Let M be a complete non-compact Riemannian manifold of positive sectional curvature. Then M is diffeomorphic to $\mathbb{R}^n$.

Thus, I think to classify non-compact Riemannian manifolds of positive curvature up to isometry, it suffices to seek for metrics of positive curvature on euclidean spaces. I did some search to find some results related to this problem, but I found few ones.

Therefore, I would be very thankful if give me some information, or/and introduce me some references about metrics of positive curvature on $\mathbb{R}^n$.

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You mean to classify non-compact Riemannian manifolds of positive sectional curvature, not just to classify non-compact Riemannian manifolds –  Ben McKay Aug 7 '13 at 16:59
A paraboloid is a nice example: $y=x_1^2+x_2^2+\dots+x_n^2$. Intuitively, that should be something like what they all look like, as their Ricci curvature must decay to zero when you go far away from any chosen point. –  Ben McKay Aug 7 '13 at 17:01
Cross posted to MSE: math.stackexchange.com/questions/461691/… –  Willie Wong Aug 13 '13 at 8:26
What do you mean by classifying ? Being positively curved is an open condition for metrics (with the $C^2$ topology), and the result you quote shows that non compact positively curved manifolds are classified up to diffeomorphism. Can one expet something more ? –  Thomas Richard Aug 13 '13 at 13:09
What is 'find all' ? It is possible to write a set of big and ugly formulas involving the coefficients $g_{ij}$ of the metric $g$ which will be equivalent to asking the curvature to be positive (and some other formulas saying that $g$ is complete), but I don't see why you could get anything better than that. Just look at rotationaly symmetric metrics $g=dr^2+f^2(r)d\theta^2$ on $\mathbb{R}^2$. The fact that $g$ is positively curved is equivalent to some inequality involving $f$ and its first and second derivatives. What more is there to say about those $f$ ? –  Thomas Richard Aug 13 '13 at 16:20