0

Could you give an example of such Riemannian manifold: 1. compact; 2. non-positive sectional curvature

user360777
  • 402

2 Answers2

3

The usual example is a "flat torus": $[0,1]^2$ with the top and bottom, as well as left and right, edges identified.

There are similar examples with constant negative curvature, but they are a bit more involved to describe.

One possibility is to take two regular hyperbolic right-angled octagons and glue them together along opposite edges, producing a cylinder-like thing with two longitudinal seams where each end has four right-angled corners. You can then glue each of these edges shut by identifying parts of them in the flat-torus pattern.

      B--3--C              Q--3--R
     /       \            /       \
    4         4          5         5
   /           \        /           \
  A             D      P             S
  |             |      |             |
  1             2      2             1
  |             |      |             |
  H             E      W             T
   \           /        \           /
    7         7          8         8
     \       /            \       /
      G--6--F              V--6--U

The resulting manifold is a genus-2 surface. It has four points where four of the original octagon corners come together:

    |           |           |           |
    5           2           1           8
   Q|R         P|D         T|H         U|V
-3--+--3-   -5--+--4-   -8--+--7-   -6--+--6-
   C|B         S|A         W|E         G|F
    4           1           2           7
    |           |           |           |

A simpler construction would be to take a larger hyperbolic octagon with corner angle 45°, and simply identify opposite sides (preserving orientation). This also produces a compact genus-2 surface with constant negative curvature.

hmakholm left over Monica
  • 286,031
  • The flat torus can also be realized as $S^1 \times S^1 \subset \mathbb R^4$: https://en.wikipedia.org/wiki/Clifford_torus. – kahen Jun 03 '17 at 10:20
  • oh I didn't notice this usual example, my original intention is to find a manifold with $K<0$ – user360777 Jun 03 '17 at 10:24
  • @user360777: I've described one hyperbolic example (not particularly distinguished except the first orientable one I could come up with). – hmakholm left over Monica Jun 03 '17 at 11:07
  • @HenningMakholm i don't totally understand it, is this topologically 2-torus or Klein bottle? – user360777 Jun 03 '17 at 12:07
  • @user360777: There result is homeomorphic to a double torus. Once you have glued sides 1-1 and 2-2 together, what you have is homeomorphic to two rectangles (BCQR) and (GFVU) connected by a handle, and then you stitch each of those rectangles into a torus. – hmakholm left over Monica Jun 03 '17 at 12:11
  • There are several other possible ways to stitch the two octagons together that also produce genus-2 surfaces (including ones that are not globally isometric to this one, unless I'm much mistaken). – hmakholm left over Monica Jun 03 '17 at 12:17
  • @HenningMakholm Preissman's theorem says if M is a compact Riemannian manifold with $K<0$, the any nontrivial Abelian subgroup of $\pi_1(M)$ is infinite cyclic, so a 2-torus cannot carry such a metric. I think this glued manifold is unbound as a complete manifold then non-compact. – user360777 Jun 03 '17 at 12:24
  • @user360777: The theorem you reference excludes a 2-dimensional torus (a 2D surface of genus 1), which is something different from the double torus (a 2D surface of genus 2) that results here. – hmakholm left over Monica Jun 03 '17 at 12:26
  • @HenningMakholm it's not 2-dimensional torus, the fundamental group of double torus is $\mathbb{Z}\oplus\mathbb{Z}$ but it's not infinite cyclic – user360777 Jun 03 '17 at 12:30
  • @user360777: $\mathbb Z\oplus\mathbb Z$ is the fundamental group for a 2-dimensional torus, but not of the double torus. – hmakholm left over Monica Jun 03 '17 at 12:34
  • Also, it is impossible to get a non-compact surface by stitching together a finite number of compact patches -- that produces a quotient space, and all quotients of a compact space are compact. – hmakholm left over Monica Jun 03 '17 at 12:38
  • @HenningMakholm fundamental group is a topological invariant, so it's not about the dimension of Euclidean space the tori is embedded (like 3 or 4), but only the genus of the 2-dimensional orientable closed manifold – user360777 Jun 03 '17 at 12:45
  • @user360777: I'm not talking about "the dimension of the Euclidean space the tori is embedded in". A double torus is not homeomorphic to an ordinary 2-dimensional torus. It has different genus (2 rather than 1), and a different fundamental group! – hmakholm left over Monica Jun 03 '17 at 12:48
  • @HenningMakholm i think i got something wrong, you're right! one more question, how you deal with the boundary when gluing? i mean the octagon doesn't have a boundary – user360777 Jun 03 '17 at 12:55
  • @user360777: Why would the octagon not have a bounday? – hmakholm left over Monica Jun 03 '17 at 12:57
  • @HenningMakholm you can't define a metric on the boundary – user360777 Jun 03 '17 at 13:00
  • @user360777: Umm, how is that different from the usual stitching construction for the flat torus? To be completely formal, I think you could have a chart for each of the octagon interiors, a chart for each of the four shared corners, and a long thin chart for each of the eight seams. – hmakholm left over Monica Jun 03 '17 at 13:02
  • @HenningMakholm here you need to consider the metric, you can't define a metric on the boundary, like the Poincare's hyperbolic disk, the boundary is 'infinity'. when dealing with just topology, you don't have to consider this – user360777 Jun 03 '17 at 13:08
  • @user360777: The octagon is finite: a bounded, closed subset of the hyperbolic plane. In the disk model it does not go near the boundary of the model. – hmakholm left over Monica Jun 03 '17 at 13:14
  • @HenningMakholm i got it, thanks! it seems we can construct such a torus by gluing 2 squares similarly? – user360777 Jun 03 '17 at 13:34
  • it seems the metric at the gluing boundary is continuous but not differentiable – user360777 Jun 03 '17 at 14:28
2

Take the unit disc $D$ in $\Bbb C$ with hyperbolic metric, then factor it by a discrete group of Mobius transformations acting freely on $D$. If this group has the property that the quotient is compact, then you have an example. (All compact Riemann surfaces of genus $\ge2$ can be obtained this way.)

Angina Seng
  • 158,341