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.

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    $\begingroup$ Marcel Berger gives an illuminating presentation of the Theorema Egregium and discusses the connection of the Gauss-Bonnet formula with parallel transport in ["A panoramic view of Riemannian geometry"][1]. [1]: books.google.co.uk/… $\endgroup$ Jul 27, 2010 at 13:46
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    $\begingroup$ I taught Gauss-Bonnet theorem in one class following the book of Rashevskii (in Russian; you can find a copy on the I-net). His geometric explanation using parallel transport is very good, but maybe it's more than one can reasonably hope to impart in a 5-day course. By GB I mean the version with a closed piecewise-smooth contour, $$\iint Kd\sigma +\int \kappa_g ds +\sum\Delta \psi = 2\pi.$$ $\endgroup$ Jul 27, 2010 at 21:51
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    $\begingroup$ I agree that it's an interesting question, but as a courtesy to fellow MOers, can you, please, remove [STILL OPEN] from the title? This is visual spam and not particularly accurate, as you explain in the opening paragraph. $\endgroup$ Aug 5, 2010 at 2:22
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    $\begingroup$ @Victor: What do you mean it's not accurate? The reason I put it there is that I don't want people to think that this question is answered even though it has an accepted answer, is colored accordingly in lists, etc. How else should I communicate this information? $\endgroup$ Aug 5, 2010 at 18:49
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    $\begingroup$ @Victor - I think the point is that some people who come to MO just to answer questions might see the dark green box next to this one and pass it by without a glance. I myself do not treat MO this way, but I guess some people might. For attracting such people the [Still Open] is needed. $\endgroup$ Mar 3, 2011 at 14:25

7 Answers 7


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.

  • $\begingroup$ By the way, it was basically this conceptual argument that led to the introduction of Chern classes by Chern. $\endgroup$ Aug 4, 2010 at 13:51
  • $\begingroup$ Hello Andrey, thanks for the idea. However, it is not at all clear to me: why is the exterior derivative of the canonical form the pullback of the curvature? $\endgroup$ Aug 4, 2010 at 15:29
  • $\begingroup$ Ah, for me this is due to some miraculous coincidence (and should be verified by a direct calculation). Have a look at Section 15.7.1 on page 736 of Berger's book, where he presents the proof in more detail. $\endgroup$ Aug 4, 2010 at 16:27
  • $\begingroup$ This is because the canonical form is the connection 1-form for the principal bundle, which is the orthonormal frame bundle, so its curvature is the curvature. $\endgroup$
    – Ben McKay
    Nov 13, 2018 at 11:17

A discrete version of curvature may help highschool students. Take a polyhedron in $R^3$ and define the curvature at a vertex $v$ by $2 \pi K(v) = 2 \pi - \theta_1 - \ldots - \theta_k$, where $\theta_j$ is the angle at $v$ of the j-th 2-face containing $v$. This gives a measure of how sharp that vertex is. For simplicity, assume there are only three faces meeting at $v$. Let the normal vectors to the faces be $n_1$, $n_2$, $n_3$. They are the corners of a geodesic triangle on the sphere, which is an analog of the region spanned by the Gauss map on a surface. By spherical geometry we have that the area of this triangle is $A = \beta_1 + \beta_2 + \beta_3 - \pi$, where $\beta_j$'s are the angles. One then shows that $\beta_j = \pi - \theta_j$, so we have the nice formula $ A = 2 \pi K(v)$. I think that similarily one can argue with parallel transport (see comment by pasquale below). I have not tried the computation, but it should work. Also the global Gauss-Bonnet theorem holds: if you sum the curvature of the vertices on a closed polyhedron the result is the Euler charcteristic (I think this was orginally proved by Descartes, but I'm not sure).

  • $\begingroup$ I would agree that some kind of discrete version of Gauss-Bonnet makes more sense for high school students than the standard continuous version. $\endgroup$
    – Deane Yang
    Mar 3, 2011 at 13:54
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    $\begingroup$ When defining (isometric) parallel transport on a polyhedron you should not use projections. Rather cut the polyhedron along some edges, lay it flat open on a plane, transport the standard way and then glue again. This gives another intuition for the $2 \pi - \theta_1 \dots $ formula. $\endgroup$ Mar 3, 2011 at 15:01
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    $\begingroup$ I guess one can further justify the geodesic triangle analogy as follows: the image of the Gauss map around a vertex with three faces consists of three points. Appropriately smoothing the polyhedron (for example, by letting the surface evolve for a short time as a wave, following Huygens' principle) you get the geodesic triangle: smoothing the edges of the polyhedron joins the three points on the sphere and smoothing the vertex fills the triangle. $\endgroup$ Mar 3, 2011 at 15:17
  • $\begingroup$ But I don't see an intuitive explanation of $A = \beta_1 + \beta_2 + \beta_3 - \pi = 2 \pi -\theta_1 -\theta_2 -\theta_3$ (which is probably a translation of the original question in the polyhedral framework). $\beta_j = \pi -\theta_j$ seems wrong to me, as $\theta_j$ is a property of one face at the vertex, while $\beta_j$ depends on the relative disposition of the other two faces at the vertex (i.e. it depends on $\theta_l, \theta_m \ l,m \neq j$). $\endgroup$ Mar 3, 2011 at 15:25
  • $\begingroup$ @pasquale zito. $\theta_1$ is the angle between $n_3 \wedge n_1$ and $n_1 \wedge n_2$. I think what I said is right. Assume $n_1$, $n_2$ and $n_3$ are positively oriented. Draw spherical geodesic arcs from $n_3$ to $n_1$ and from $n_2$ to $n_1$ and denote the tangent vectors to these arcs at $n_1$ by $v_3$ and $v_2$ respectively. $n_3 \wedge n_1$ is $v_3$ rotated 90 degrees in the direction of $v_2$ and $n_1 \wedge n_2$ is $v_2$ rotated 90 degrees in the direction of $v_3$. This gives the formula. $\endgroup$ Mar 3, 2011 at 15:54

For the high school students in question, I might suggest an infinitesimal approach, translated from the abstract language of Jacobi fields and first variation to concretely working with very small angles and areas.

A first example can come from taking a pair of equal length geodesic segments originating at a point, with a very small angle (e.g., $\epsilon$ with $\epsilon^2=0$) between them, and consider how the area of the thin geodesic triangle that is formed changes as you increase the length. To first order in $\epsilon$, you find that the second derivative of area is minus the curvature. This also holds for any "very nearby" geodesics that don't necessarily intersect anywhere, but define opposite sides of a very thin quadrilateral.

Similarly, you can take a degenerate geodesic triangle $ABC$, where $AB$ and $BC$ lie along $AC$, and push $B$ away from $AC$ infinitesimally (equivalently, let the angles at $A$ and $C$ become suitable infinitesimals, with the angle at $B$ being $\pi$ minus another infinitesimal). You end up with the angle excess formula, expressed as a suitably weighted integral of the curvature along $AC$, or equivalently, an integral of curvature over the area of the triangle.

Now you just need to assemble or deform some very thin triangles into a big one.

  • $\begingroup$ Hello Scott, thank you very much for your suggestion. It looks nice, but I don't understand how you find that the second derivative of area is minus the curvature in the second paragraph. (I assume you mean extrinsic curvature?) I guess this should come out of Jacobi field stuff, but is there some specific statement I could prove or at least state? $\endgroup$ Jul 30, 2010 at 2:52
  • $\begingroup$ I made a mistake in my assertion: it is the third derivative that is proportional to curvature. You can write this heuristically using discrete approximation: The derivative of area for a given length is the distance between the endpoints of the two geodesics. If you mark off lengths $\ell + \delta$, $\ell$, and $\ell - \delta$ on each geodesic, and draw three short geodesics between the corresponding distances, you can relate the second difference of length to the curvature. I'll try to explain a little better soon. $\endgroup$
    – S. Carnahan
    Jul 30, 2010 at 14:49

There is a physical "proof" of this fact which I learned from Mark Levy; it is in his book "THE MATHEMATICAL MECHANIC: Using Physical Reasoning to Solve Problems".

Imagine that you keep the axis of a bicycle wheel and move it in such a way that the bicycle wheel lies in the tangent plane to the surface. In the initial position the wheel stays still; you go along the loop in the surface and stop. After that your wheel rotates by some angle $\alpha$. If your loop was triangular this angle is its defect; this is (2) in your list.

The parallel motion does not rotates the wheel, so the same result will be the same if you only rotate the axis without moving the center of the wheel. This tells you that $\alpha$ depends only on the spherical image of the loop and from here it is easy to see that it is proportional to the algebraic area of the domain bounded by the spherical image of the loop. I.e., the area of the sphere that Gauss map traces out on a region; this is (1) in your list.


(Really wanted to make this just a comment. And, besides, you probably know all of this already)

I guess I don't see why showing the equivalence of the three statements would be helpful to the students, unless you want to show them how calculus can be used to define certain concepts and prove theorems about them. All of the concepts involved are for me rather hard to define in a purely geometric fashion but very easy using calculus and linear algebra.

For a more geometric (but non-rigorous) view, wouldn't it be better to do something like the following (more or less taken from Guillemin and Pollack):

  1. Build a model of the hyperbolic plane.
  2. Show how the sum of the angles of a geodesic triangle depends on the area for a sphere or hyperbolic plane (I'm not sure how to do this) and use this to motivate the notion of Gauss curvature.
  3. Or define the Gauss map $G$ and define Gauss curvature using $A(G(R))/A(R)$ for a region $R$.
  4. Using pictures of examples, show how the degree of the Gauss map is related to the genus of an orientable surface.
  5. Using c) and d), motivate the Gauss-Bonnet theorem

Another possibility is to work with a polygonal surface (maybe with only triangular faces). I don't have any constructive suggestions, but surely others do. Also, I stumbled onto this elementary proof of Gauss-Bonnet

  • $\begingroup$ That's a nice explanation. Polya's proof also looks very good. I wonder how many different proofs of the Gauss-Bonnet formula are known. $\endgroup$ Jul 27, 2010 at 16:19
  • $\begingroup$ Thank you for your suggestions. I'm also having trouble with step 2: how do you figure out the sum of angles of a triangle on a saddle-shaped surface? When I taught a hyperbolic geometry class, I did everything on the Poincare disk, and it worked fine. But I'd really like to relate the statements "hyperbolic plane has negative curvature" and "the saddle has negative curvature". (Now I think that perhaps it wasn't worth it, but I don't have enough time to invent a new course). $\endgroup$ Jul 27, 2010 at 23:31
  • $\begingroup$ My answer above was not really an answer but at best a speculative comment, because I don't know enough details. It seems difficult to work with an isometric embedding of the hyperbolic space. I was hoping that someone more knowledgeable would provide a better answer. Especially the stuff on polygonal surfaces. $\endgroup$
    – Deane Yang
    Jul 28, 2010 at 16:55

This is by no means a complete answer, but a key part of the correspondence is found in the area formulas for spherical and hyperbolic triangles in terms of angular excess/defect. Once you've convinced yourself (or your students) that a sphere is the model surface for constant positive curvature and that "wedges" on the circle have area directly proportional to their area and inversely proportional to the curvature, then there's a direct visual proof of the area formula in terms of angular excess: Proofs without words

In the case of negative curvature, once you have the Poincare disc as the model surface with constant negative curvature, you can take a similar tack. Replace the notion of a "wedge" of angle $\theta$ with the notion of a doubly asymptotic geodesic triangle whose finite vertex has angle $\theta$. Once it's established that this region has area directly proportional to $\pi - \theta$ and inversely proportional to the (absolute value of) curvature, then there's a similar direct visual proof of the area formula for a geodesic triangle in terms of angular defect: Proofs without words

  • $\begingroup$ Thanks for the advice. I thought of doing this, but the trouble is: how do you relate the geometry of the Poincare disk and the geometry of a saddle? $\endgroup$ Jul 27, 2010 at 23:34
  • $\begingroup$ Yes, I see your point. Perhaps the observation that in both cases the circumference of a circle is longer than it "should" be (from the Euclidean point of view), corresponding to the negative curvature, can serve as a didactic bridge. But of course that's far from a complete explanation. $\endgroup$ Jul 28, 2010 at 5:01

Let $K_e$ be the extrinsic Gauss curvature, defined as the product of the principal curvatures. Let $K_i$ be the intrinsic Gauss curvature, defined by how fast geodesics deviate from each other (which is a bit more intuitive to me than using parallel transport).

While it may be complicated to show that $K_e=K_i$, I claim that it's intuitive that if $K_e$ and $K_i$ are both nonzero, then they should have the same sign: When $K_e<0$, there are orthogonal tangent directions in which the surface bends in opposite directions away from the tangent plane. So you'd expect that the geodesics in those orthogonal directions are deviating faster than in Euclidean space. When $K_e>0$, the surface (nearby the point) stays on one side of the tangent plane. Since all of the geodesics are now bending in the same direction, you'd expect that they are deviating slower than in Euclidean space. Okay, that's not really a proof, but it makes sense. And it seems more than adequate for high school kids.

To address your actual question, I believe that at the end of the day, Gauss's Theorema Egregium has to boil down to a computation. I think the best we can do is use enough geometry to make the computation easy. A good example is the proof in Frank Morgan's Riemannian Geometry: A Beginner's Guide, page 23, which is short and sweet. However, this only shows that $K_e$ is intrinsic. One might still want to show that the intrinsic formula that you get matches some nicer geometric description (such as $K_i$ above), which will require some further computations.


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