This is a really basic question, which draws as its source two of the pictures from the Wikipedia article about Gaussian curvature.
If it is true that the sum of the angles of a triangle on a surface of negative Gaussian curvature is less than 180 degrees (as it says on Wikipedia), and that the sum of the angles of a triangle on a surface of positive Gaussian curvature is more than 180 degrees (which I believe is the case for a sphere, I think the sum is 270 degrees), then:
Since the inside of a torus supposedly has negative Gaussian curvature, and the outside supposedly has positive Gaussian curvature, does a triangle inscribed on the inside of a torus have a sum of angles less than 180 degrees while a triangle inscribed on the outside of a torus have a sum of angles greater than 180 degrees?
By "inside" I mean "closer to the donut hole" and by "outside" I mean "away from the donut hole".
Thus an "ant walking on a torus" could tell precisely when it arrived "at the highest point" of the torus (the circle on the torus where every point has zero Gaussian curvature which divides the above two mentioned regions of positive and negative curvature) when the triangles it is drawing on the ground have a sum of angles of precisely 180 degrees?
Also does this mean that the torus essentially has both elliptic and hyperbolic geometries, depending on which side the ant is walking on?
I was thinking about this question because I was trying to think of examples of closed surfaces which have negative Gaussian curvature over an extended region, because it is not possible for the surface to be closed and have negative curvature everywhere and be embeddable in $\mathbb{R}^3$, see:
- existence of closed surface having only negative Gaussian curvature.
- https://mathoverflow.net/questions/111101/surfaces-in-mathbb-r3-with-negative-curvature-bounded-away-from-zero
- https://mathoverflow.net/questions/32597/compact-surfaces-of-negative-curvature
Best Answer
Quantitatively, if $T$ is a geodesic triangle bounding a topological disk in a surface with a Riemannian metric, the sum of the interior angles of $T$ is equal to $\pi$ (radians) plus the integral of the (Gaussian) curvature over the interior of $T$. (Sometimes this fact is called the local Gauss-Bonnet theorem, and the integral of the curvature is called the angular defect.)
When you speak of "the torus", you're equipping a particular smooth manifold (a product of circles) with a metric from a particular two-parameter family obtained by embedding (into Euclidean $3$-space) one circle factor as a circle of radius $r$ and revolving about an axis at distance $R > r$ from the center of the circle, sweeping out the other circle factor. Every metric of this type does indeed have negative curvature on "the inside" and positive curvature on "the outside", though the curvature is not constant in any region, but only along the "latitudes".
Measurements on the surface of the torus can detect the "parabolic points" (where the Gaussian curvature vanishes) in the following sense: A small geodesic triangle entirely "on the outside" has positive angular defect, while a small geodesic triangle entirely "on the inside" has negative angular defect. A small triangle that contains points of positive curvature and points of negative curvature can have positive, zero, or negative angular defect, depending on its shape. Detecting the precise locations of the parabolic curves therefore requires (i) measuring infinitely many triangles (ii) to arbitrary precision.
Because the curvature of such a torus is not constant, calling its geometry "elliptic" or "hyperbolic" would be at least mildly non-standard.
While you're correct that no closed (compact, boundaryless) surface embedded in Euclidean $3$-space has a metric of negative curvature,
(Incidentally, if it matters, a geodesic triangle on a sphere can have arbitrary angular defect between $0$ and $4\pi$, non-inclusive. Angular defect $\pi/2$ occurs for a triangle enclosing one octant, and therefore having three right angles.)