Highschooler here, trying to learn about the Euler characteristic, Gaussian curvature and the Gauss-Bonnet theorem linking them together.
As per the Gauss-Bonnet theorem: total curvature $= 2 \pi \times$ euler characteristic.
Here's my confusion. A square (for example a flat sheet of paper) has a Gaussian curvature of zero. But following the formula $\chi = V – E + F$, I calculate that a square's Euler characteristic is $1$.
This is because vertices $V = 4$, edges $E = 4$ and faces $F = 1$. Therefore $\chi = 4 – 4 + 1\Rightarrow \chi = 1$.
So I get the equation $0 = 2\pi 1$, i.e. $0 = 2\pi$.
Where is my mistake?
Best Answer
The first difficulty is that the version of the Gauss-Bonnet theorem you seem to be using is for compact 2-manifolds without boundary. A sphere is a compact 2-manifold without boundary. The boundary of a cube (six squares glued along their edges) is a compact 2-manifold without boundary. A square (taken to be closed since you say the vertices and edges are part of the manifold) is a compact 2-manifold with boundary.
In describing a manifold, one usually omits "without boundary". One usually includes "with boundary". The default state of a manifold is having no boundary.
There is a version of Gauss-Bonnett for compact 2-manifolds with boundary. $$ \int_M K \,\mathrm{d}A + \int_{\partial M} k_g \,\mathrm{d}s = 2 \pi \chi(M) \text{,} $$ where the first integral is of the Gaussian curvature over the surface and the second integral is the geodesic curvature on the boundary.
The closed square is homeomorphic to the closed disk. The boundary of the closed disk is a circle. The geodesic curvature of the circle boundary of a disk measures how much that curve closes up in like manner to a circle (much as the Gaussian curvature measures how much a surface closes up in like manner to the sphere). Of course, a circle closes up exactly the way one circle does, so this integral contributes $2\pi$ on the left-hand side when you study a closed disk or a closed square.
(There is a subtletly here. It is easy to conflate "extrinsic" curvature caused by a specific embedding with geodesic ("intrinsic") curvature. We can embed our circle along many revolutions of a helix, then outside the helix back to where we started. This embedding has a lot of curvature, but a circle is just a circle...)
A less critical difficulty is that a square only looks flat when you embed it a particular way. You can curl a square up into a tube -- which is not flat. You can even bend this tube around to make the ends meet -- which is again not flat.
If to a square we glue the upper and lower edges together [*] and then glue the two new circles together, we get a compact 2-manifold (without boundary). This object is a torus. Due to the gluings, all four vertices of the square have been glued into one vertex and both opposite pairs of edges of the square have been glued together. The result has one vertex, two edges, and one face, with Euler characteristic zero and total curvature zero.
This zero is what you were expecting for a flat square. It might be surprising that our embedding has to exhibit all the "curviness" of a torus to get zero Gaussian curvature -- but all that "curviness" is extrinsic curvature.
[*] We should be careful about how we do this gluing. For the first pair of edges we should glue so as to obtain an annulus not a Moebius strip. For the gluing of the circles, if we glue the same way as the first gluing, we obtain a torus. If we glue "the other way around", we get a Klein bottle. Of course, the Klein bottle with constant curvature is flat, so also has zero Gaussian curvature.