This question is still wide open – all of the answers so far rely on magical calculations. I've only accepted an answer because, by bounty rules, otherwise one would be accepted automatically. I can't change the accepted answer, but it would be amazing to have more discussion on this question.
I'd like a nice proof (or a convincing demonstration), for a surface in $\mathbb R^3$, that explains why the following notions are equivalent:
1) Curvature, as defined by the area of the sphere that Gauss map traces out on a region.
1.5) The integral of the product of principal curvatures.
2) The angle defect of parallel transport about a geodesic triangle.
(This equivalence may be considered as either a part of the Theorema Egregium or a part of Gauss-Bonnet. Proving that numbers 1 and 1.5 are the same is pretty easy).
Motivation: I'm teaching a five-day class for very bright high-school students. The idea is to give them an impression of what geometry is about. However, when I looked at Spivak's proof of this, it was much more of a messy calculation than I expected. I'd like, if at all possible, something more conceptual, ideally with a nice picture attached to it.
Since this doesn't have to be a perfectly complete class, I'll be perfectly happy with a good illustration of why this is true instead of a rigorous proof, if a conceptual and rigorous proof is completely out of the question.
One idea I had is to show the example of a sphere and the hyperbolic plane, and then explain that on very small scales the curvature is constant. However, then I would need a nice proof that the embeddings of the hyperbolic plane in $\mathbb R^3$ have curvature -1.
Thank you very much!
P.S. This question is related, but not quite the same (I hope), to this question:
Equivalent definitions of Gaussian curvature
P.P.S. Thank you to whomever recommended to Berger's "Panoramic View of Riemannian Geometry". it was quite useful to me. I do not know why you deleted your answer.
That books claims there is no conceptual proof. However, I'd still be very happy with a nice illustration of why one should believe this, especially for negative curvature.
Best Answer
There is a short conceptual proof of Gauß-Bonnet due to Chern (see also "A panoramic view of Riemannian geometry" by Berger). The argument assumes a basic familiarity with differential forms though.
Assume that the surface $S$ is oriented so that its canonical measure $dm$ is a 2-form. Let's consider the set of all unit vectors tangent to $S$, i.e. the unitary fiber $US$ of $S$. The canonical projection $p:US\to S$ associates with each unit vector a point on $S$ where it is tangent. Now, $US$ is a 3-manifold which posesses a canonical differential form $\zeta$. The exterior derivative $d\zeta$ is the form lifted by $p$ onto $US$ of the 2-form of the curvature $K(m)dm$. If the domain $D\subset S$ is simply connected, one can define a continuous field $\xi$ of unit vectors on $D$; therefore, $D$ can be lifted into $US$.
The Gauß-Bonnet formula follows directly from the Stokes formula applied to $\xi(D)$ since the canonical 1-form $\zeta$ is the geodesic curvature. This is actually more than you ask for because the boundary of $D$ does not have to consist of geodesics.