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The torus $\mathbb{T}^n$, the sphere $\mathbb{S}^n$ and the hyperbolic space $\mathbb{H}^n$ admit metrics of constant (sectional) curvature $0, 1, -1$ respectively. Do they afford metrics of constant curvature in $\{0, 1, -1\}$ for a different value? (e.g. does the torus admit a metric of curvature $-1$, or $1$?)

In the 2-dimensional case, the Gauss-Bonnet theorem answers the question negatively, but what about higher dimensions?

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It is a standard result in Riemannian geometry that if you have $M$ with constant sectional curvature $k$ then there is a Riemannian covering map from $M_k\ to M$, where $M_k$ is the sphere, Euclidean space or hyperbolic space. Some topology should finish the proof that the answer is negative. –  Otis Chodosh Jun 7 '13 at 2:57
Sorry, to clarify, $M_k$ is the model with constant curvature $k$ (i.e. in your list, except Euclidean space rather than the torus, because its simply connected) –  Otis Chodosh Jun 7 '13 at 2:58
Unless I am very much mistaken, hyperbolic space $\mathbb H^n$ is diffeomorphic to euclidean space $\mathbb E^n$. The former is modeled as the half space $\mathbb R_{>0} \times \mathbb R^{n-1}$ with metric scaling as $x_1^{-2}$, where $x_1$ is the first coordinate; the diffeomorphism is $x_1 \mapsto \log x_1$. –  Theo Johnson-Freyd Jun 7 '13 at 4:44
Thats true: one hyperbolic model is $\frac{dr^2}{1+r^2} + r^2 g_{S^{n-1}}$. This works in dimension $2$-by the way. –  Otis Chodosh Jun 7 '13 at 5:26
So, sorry (I was distracted by the replacement of euclidean space by the torus) the answer should be that of your list, only hyperbolic space admits a constant sectional curvature metric of the other kind. The others do not. If you had euclidean space instead of the torus, this would also admit a hyperbolic metric. I think this answers the question. Nothing is different in dimension 2 vs higher dimensions for this question. –  Otis Chodosh Jun 7 '13 at 5:29

1 Answer 1

up vote 4 down vote accepted

The right thing to do is to restrict to compact manifolds of dimension $n\ge 2$. Then the answer is: If $M$ admits a metric of (possibly variable) sectional curvature $K<0$, or $K=0$ or $K>0$, then it does not admit a metric of curvature $K'$ with $K'<0, K'=0$ of $K'>0$, unless $KK'>0$ or $K=K'=0$.

The proof is elementary Riemannian geometry, most of which you can find, say, in do Carmo's book "Riemannian geometry". For instance, it follows from Cartan-Hadamard theorem that a manifold with metric of nonpositive curvature has contractible universal cover. This takes care of $K>0$ (and $K'>0$).

Consider a manifold $M$ admitting a flat metric. Then $\pi_1(M)$ contains $Z^2$, since $M$ is covered by the $n$-torus. On the other hand, fundamental groups of compact manifolds of negative curvature cannot contain $Z^2$. Hence, you cannot have a compact manifold which admits both a flat metric and a metric of negative curvature. Actually, a bit more follows from the flat torus theorem: Every metric of nonpositive curvature on the $n$-torus has to be flat.

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For $K>0$ you could have cited the Mayer-Vietoris theorem to justify that the universal cover of $M$ is not contractible. In the other cases, the growth of fundamental group seems a bit more elementary to me than the inclusion of $\mathbb{Z}^2$. –  Benoît Kloeckner Jun 7 '13 at 9:13
@Benoît: True on both; when I started typing the answer, I first referred to Meyer's theorem, but then somehow settled on Cartan-Hadamard. For growth: I tried to restrict to proofs for which one reference (do Carmo) sufficed. –  Misha Jun 7 '13 at 12:11

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