Let (M,g) be a Poincare disk. Does there exists a complete Kahler metric h, such that the scalar curvature of h is positive?
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2 Answers
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Take any smooth function $f$ on the disk so that $f$ goes to infinity at the boundary and then our metric must be of the form $g=e^f(dx^2+dy^2)$. Every metric on any oriented surface is Kaehler, in this case for the usual complex structure. The curvature is the Laplacian of $f$. To get positive scalar curvature, you need to get the Laplacian to be negative. But $f$ needs to still grow to infinity as we approach the boundary, which is impossible by the maximum principle applied to concentric circles around the origin.
answered Nov 11, 2016 at 8:52
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11$\begingroup$ Actually, if you use polar coordinates and write your $f(x^2+y^2) = h(r)$ (where $r^2 = x^2+y^2$), you'll see that the positivity of the scalar (i.e., Gauss) curvature is equivalent to $rh''(r) + h'(r) <0$ for $0<r<1$ and $h''(0)<0$ (since $h$ is an even function of $r$, we automatically have $h'(0)=0$). This easily implies that $h$ is a decreasing function of $r$ when $0<r<1$ (just integrate twice), so your metric cannot be complete. $\endgroup$ Commented Nov 11, 2016 at 10:27
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2$\begingroup$ In fact, if you correct my sign of the Laplacian in the curvature, you can see that a conformal factor needs to have negative Laplacian and still grow to infinity along the unit disk, which is impossible by the maximum principle applied to concentric circles around the origin. $\endgroup$ Commented Nov 11, 2016 at 11:02
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2$\begingroup$ May be such a metric does not exist. Robert Bryant already showed that if the metric exists can not be rotationally invariant. Now since SO(2) is compact and preserves the complex structure of the disk you can "average" a non rotational example to get a rotational invariant one. $\endgroup$ Commented Nov 22, 2016 at 15:40
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The answer to your question is no.In case of surfaces the scalar curvature and Ricci curvature coincide with Gauss curvature. If an n dimensional M had a complete Riemannian metric h with nonnegative Ricci curvature then by the Bochner Weitzenbock formula all L^2 harmonic one forms would be parallel.By an observation of Calabi M has infinite volume. Therefore all L^2 harmonic forms are zero. Now L^2 condition on one forms is a conformally invariant condition on a Riemann surface. However the disc has many nonzero L^2 harmonic one forms .This leads to a contradiction.
answered Dec 2, 2016 at 14:47