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Mohan Ramachandran
Mohan Ramachandran
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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 arewould 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.

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 are 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.

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.

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Mohan Ramachandran
Mohan Ramachandran
  • 4.1k
  • 24
  • 19

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 KahlerRiemannian metric h with positivenonnegative Ricci curvature then by the Bochner Weitzenbock formula (M ,h) would have no nonzero all L^2 harmonic one forms are 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.In case of Riemann surfaces note that Ricci curvature coincides with Scalar curvature .

If M had a complete Kahler metric h with positive Ricci curvature then by the Bochner Weitzenbock formula (M ,h) would have no nonzero L^2 harmonic one forms . Now L^2 condition on one forms is a conformally invariant condition. However the disc has many nonzero L^2 harmonic one forms .This leads to a contradiction.In case of Riemann surfaces note that Ricci curvature coincides with Scalar curvature .

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 are 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.

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Mohan Ramachandran
Mohan Ramachandran
  • 4.1k
  • 24
  • 19

If M had a complete Kahler metric h with positive Ricci curvature then by the Bochner Weitzenbock formula (M ,h) would have no nonzero L^2 harmonic one forms . Now L^2 condition on one forms is a conformally invariant condition. However the disc has many nonzero L^2 harmonic one forms .This leads to a contradiction.In case of Riemann surfaces note that Ricci curvature coincides with Scalar curvature .