Let $\hat{M}$ be a branched double cover of $M$. Is there a way to lift a Riemannian metric $g$ on $M$ to get a smooth Riemannian metric $\hat{g}$ on $\hat{M}$. Moreover, if $g$ has nonnegative sectional curvature, does $\hat{g}$ also have nonnegative sectional curvature provided that such a lift $\hat{g}$ exists?
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2 Answers
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Basically, the answer is 'no'. Consider the case of a compact Riemann surface $M$ of genus $g>1$ branched over the Riemann sphere $\mathbb{CP}^1$ (which has genus $g=0$) by a nonconstant holomorphic map $h:M\to \mathbb{CP}^1$. This doesn't induce a canonical smooth metric on $M$, even if you start with a constant positive curvature metric on $\mathbb{CP}^1$. Moreover, it is known that there is no metric on $M$ of nonnegative sectional curvature, let alone a natural one.
answered Apr 21, 2014 at 17:20
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In this setting, the "natural" thing to do is to pull-back your metric of choice on the base $M,$ obtaining a metric on $\hat{M}$ which will have singularities (i.e. failure of positive definiteness) along the branching locus. On the complement of these singularities the covering projection is (by definition) a local isometry, and thus curvature properties of the base metric will remain on that part of $\hat{M}.$ In the case of Riemann surfaces presented as branched covers over $\mathbb{CP}^1,$ these pulled back "metrics" will have conical singularities at the ramification points of the covering map. I don't know the history at all, but I would be interested to know if the original interest in metrics with conical singularities arose from these examples; it seems feasible.
answered Apr 21, 2014 at 18:05
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$\begingroup$ I think it more likely that the interest in conical singularities arose by going in the other direction: passing to some kind of orbifold quotient space $M \to Q$. Under such a quotient, the cone singularities have angle $<2\pi$ and therefore have infinitesmal negative curvature, which is good; under a covering the cone singularities have angle $>2 \pi$ and therefore have infinitesmal positive curvature, which is not so good. $\endgroup$ Commented Apr 21, 2014 at 20:42
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$\begingroup$ This kind of distinction can be seen in Thurston's orbifold theorem, for example. $\endgroup$ Commented Apr 21, 2014 at 20:44