Theorem- Let $M^{n}$ a complete Riemannian manifold with constant sectional curvature equal to $K$. Then the universal cover $\bar{M}$ of $M$ , with covering metric is isometric to $H^{n}$ if $K=-1$, $\mathbb{R}^{n}$, if $K=0$ and finally $\mathbb{S}^{n}$ if $K=1$.

What information about the manifold $M$ the theorem tell me? Once universal cover $\bar{M}$ to be isometric to one of the spacial forms.


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    $\begingroup$ What do you mean what is the advantage? It's cool! $\endgroup$ – user98602 Nov 18 '16 at 1:57
  • $\begingroup$ @MikeMiller i would like to know what information about manifold $M$ the theorem tell me? $\endgroup$ – C. Junior Nov 18 '16 at 2:16
  • $\begingroup$ once the universal cover is isometric to a space forms $\endgroup$ – C. Junior Nov 18 '16 at 2:17
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    $\begingroup$ For instance, you can make explicit computations using hyperbolic/spherical trigonometry. You can replace many hard Riemannian geometry computations using linear algebra (via the Lorentzian space model of the hyperbolic space). Every triangle is uniquely (up to congruence) determined by its side-length. The fundamental group of a constant curvature space-form is linear... $\endgroup$ – Moishe Kohan Nov 18 '16 at 2:27

Here is one example. You may have heard about the 3-dimensional Poincare Conjecture (if $M$ is a compact, without boundary, simply-connected 3-dimensional manifold, then $M$ is homeomorphic to $S^3$). Here is how it was proven by Gregory Perelman, broadly speaking:

  1. $M$ is known to admit a smooth structure. Equip $M$, therefore with a randomly chosen Riemannian metric $g$. Since we know nothing about the properties of $g$, we cannot deduce any conclusions about $M$ from the existence of $g$.

  2. Deform $g$ via the Ricc flow (with surgeries) to an Einstein metric $g_0$ on $M$. (OK, not on $M$ itself but on pieces of a connected sum decomposition of $M$, but let's ignore this.)

  3. In dimension 3 each Einstein metric has constant curvature. It is easy to see that this curvature has to be positive (since $M$ is compact and simply connected), hence, after rescaling, we can assume it to be equal to $1$. Hence, by the Killing-Hopf theorem (I always thought of it is Elie Cartan's theorem...) $(M, g_0)$ is isometric to the unit 3-dimensional sphere with its standard metric. In particular, $M$ itself is diffeomorphic to $S^3$.

Now, you see how this theorem is useful and what advantage one has having a metric of constant curvature on a manifold.


For me, the Killing-Hopf theorem is interesting because it tells you that a complete connected Riemannian $n$-manifold $M$ with constant negative sectional curvature is a classifying space $B\Gamma$ and hence an Eilenberg-MacLane space $K(\Gamma,1)$ for its fundamental group $\Gamma$, which is necessarily a discrete subgroup of the orthochronous Lorentz group $O^+(n,1)$. This follows because the universal cover is contractible. From the point of view of geometric group theory, this is interesting because it means that cohomology of $M$ with local coefficients is isomorphic to group cohomology of the fundamental group $\Gamma$.


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