What is Curvature?
Questions for small group investigation or individual study using Hyperbolic Games
1. Holonomy and Curvature

Start up the Hyperbolic Games and choose menu options as follows:

Game Sudoku
Options : Repeating Images Draw Full
Options : Geodesics Show No Geodesics
Topology Triple Torus
Click on the window to get a hand cursor, and then move the hand cursor to a corner of the central triangle. Carefully slide the hand cursor along the perimeter of the central triangle (shown here in green).

Figure 1.1

A. After you've slid the hand cursor once around the central triangle, do its fingers still point straight up? If the hand has come back rotated, say whether it’s rotated clockwise or counterclockwise, and estimate the angle of rotation (in radians, not degrees).

B. Does your answer to Question A depend on whether the hand travels clockwise or counterclockwise around the triangle’s perimeter?

C. Double-click to get an arrow cursor, and then click back into the window to get a pristine hand cursor, with its fingers pointing straight up. Move the hand cursor to the same starting position as before, at a corner of the central triangle. How many times must you circumnavigate the central triangle before the hand cursor returns with its fingers again pointing straight up? Technical Tip: Keep your physical mouse aligned with the hand cursor at all times, as explained in Hyperbolic Games Help : Using the Mouse : The solution. Don’t rush — for accurate results you must follow the triangle’s perimeter as precisely as possible.

D. Use your answer from Question C to decide how much the hand cursor gets rotated when it makes one trip around the central triangle’s perimeter.

Definition. The angle by which the hand cursor comes back rotated is called the holonomy angle. The word holonomy alone sometimes refers to the holonomy angle and sometimes refers to the holonomy effect in general.

Now switch from the hyperbolic plane to the sphere. Keeping other settings the same, select

Topology Sphere
Projection d = 1 — Stereographic Projection
The “d = 1” projection distance moves you to a point on the sphere itself, so you now see the sphere’s inner surface. Dismiss the warning that “Sudoku runs only on Triple Torus”. You won’t get a Sudoku puzzle but that’s OK — you’ll get an empty triangular grid which is all you’ll need for the following exercises.

As before, click on the window to get a hand cursor, move the hand cursor to a corner of the central triangle, and carefully slide the hand cursor along the perimeter of the central triangle.

Figure 1.2

E. Repeat Question A for the sphere. In what fundamental way does holonomy on the sphere differ from holonomy on the hyperbolic plane?

F. Repeat Question B for the sphere.

G. Repeat Question C for the sphere.

H. Repeat Question D for the sphere.

Now switch from the sphere to the Euclidean plane. Keeping other settings the same, select

Topology Torus
Projection d = 1
Again dismiss the warning that “Sudoku runs only on Triple Torus”.

Click on the window to get a hand cursor, move the hand cursor to a corner of a triangle, and carefully slide the hand cursor along that triangle’s perimeter.

Figure 1.3

I. Repeat Question A for the Euclidean plane. In what fundamental way does holonomy on the Euclidean plane differ from holonomy on the hyperbolic plane or sphere?

Now consider an arbitrary triangle, which may lie on a hyperbolic plane, a flat plane or a sphere. This is a good moment to decide whether clockwise or counterclockwise motion constitutes a positive rotation.

Definition. If we slide the hand cursor counterclockwise (resp. clockwise) around a triangle’s perimeter — or around any other path — then positive holonomy angles are measured counterclockwise (resp. clockwise) and negative holonomy angles are measured clockwise (resp. counterclockwise).

In other words, the holonomy angle is positive (resp. negative) when the hand cursor ends up rotated in the same (resp. different) direction that it circumnavigated the triangle.

J. Find a formula for the expected holonomy angle (for one trip around a triangle’s perimeter) as a function of the triangle’s angles. Prove that your formula is correct. (That is, prove the formula in general, as opposed to merely verifying that it works in the three examples considered above.) Hint: As the hand cursor slides along an edge of the triangle, its fingers keep a constant angle with that edge. Now analyze what happens at the corners.

K. When you apply your holonomy formula from Question J to a specific triangle on a specific surface, you may get an answer that is somewhat greater than -2π, equal to -2π, or somewhat less than -2π. What is the geometrical significance in each case?

At this point you might be wondering, where the heck did the -2π come from? Let's sit back for a moment and contemplate the -2π’s significance before we move on. Start with the following experiment:

Take a flat sheet of paper (zero curvature!) and draw a closed loop. Take an imaginary hand cursor and slide it around your loop. Because there's no curvature, the hand comes back to its starting place with its fingers pointing in the same direction as when it started. Relative to your sheet of paper, the hand's orientation remained constant (here I use the word orientation in the everyday sense of the word, as a synonym of heading or attitude, not in the mathematician’s sense as a synonym of chirality). Slide your imaginary hand cursor around the loop again, but this time watch its orientation relative to its direction of travel. Initially the fingers point in the direction of travel. But as hand begins its journey, the path starts bending around counterclockwise. For example, at Point A (see figure above), the hand direction and the travel direction differ by π/4 radians. Treating counterclockwise rotations as positive (recall the definition above), the travel direction has rotated +π/4 radians relative to the hand direction. Now here’s the crucial point, so please read this sentence three times out loud: relative to the direction of travel, the hand has rotated -π/4 radians. Relative to the page, the hand didn’t rotate at all; but relative to the travel direction, it rotated π/4 radians in the negative sense. Why the negative sense? Because while travelling counterclockwise around the loop, the hand rotates clockwise relative to the travel direction.

Continuing around the loop, the hand finds itself rotated by -π/2, then -3π/4, and so on, until finally it has rotated a full -2π relative to the travel direction (not relative to the paper!) when it returns to its starting point. Relative to the paper, of course, it’s the travel direction (the so-called “direction vector”) that rotates once clockwise while the hand direction stays constant, but that’s not how we want to think about it, because that way of thinking applies only to flat surfaces and doesn’t generalize to curved surfaces. Our reason for looking at how the hand direction rotates relative to the travel direction is that this approach applies equally well to curved surfaces.

In the absence of curvature — for example on the flat sheet of paper — the hand rotates -2π radians relative to the travel direction as it wends its way once around the loop. By contrast, in the presence of curvature the deviation from -2π will tell us how much curvature the path encloses. The important thing is that the -2π itself has nothing to do with the surface’s curvature, i.e. whether the surface is hyperbolic, flat or spherical. The difference between the observed holonomy and the default -2π value is what reveals the curvature. To better focus on the curvature, we adopt the following convention:

Convention. Add 2π to all holonomy angles. This guarantees that the holonomy angle will be exactly zero in the flat (Euclidean) case.

In light of this convention, Question K becomes

K'. When you apply your (freshly modified!) holonomy formula to a specific triangle on a specific surface, you may get an answer that is positive, zero or negative. What is the geometrical significance in each case?

L. Find a formula for the holonomy angle you’d expect when you slide the hand cursor around a quadrilateral. Apply your reasoning from Question J to prove that your formula is correct. Remember to include the +2π convention.

M. Generalize your result from Question L to find a formula for the holonomy angle you’d expect when you slide the hand cursor around an arbitrary n‑gon. Prove that your formula is correct. As always, include the +2π convention.

N. True or false: Your formula from Question M says that the holonomy angle equals the difference between the sum of the given n‑gon’s angles and the sum of the angles of a Euclidean n‑gon.

O. Proof or counterexample: If you cut an n‑gon into two pieces, the holonomy angle around the perimeter of the whole n‑gon is the sum of the holonomy angles around each of the pieces.

Question O illustrates the fact that holonomy is proportional to area. The proportionality constant is the curvature.

Definition. The curvature is the holonomy around a path divided by the area that the path encloses.

The sphere, the Euclidean plane and the hyperbolic plane each has constant curvature, in the sense that no matter what path you consider, the ratio of the holonomy around that path divided by the enclosed area always comes out to the same value. If instead you were to study an irregular blob’s surface, you’d find that its curvature varies: positive in some places, negative in other places, of varying magnitude. On an irregular blob’s surface, the local curvature at a point is the limiting value you get for the ratio holonomy/area as you consider smaller and smaller paths surrounding the point.

P. A unit sphere has area 4π.

Q. The coordinate planes slice a standard unit sphere into eight equilateral right triangles.

R. What is the curvature of a sphere of radius 2? Radius 3? Radius r?

S. What is the curvature of a sphere of radius 5cm ? What are the proper units for curvature?

T. In the hyperbolic plane (Figure 1.1) if the central triangle has area π/7 cm², what is the curvature of that hyperbolic plane?

If a surface has constant curvature, then the total curvature of the surface is defined to be its curvature times its area.

U. What is the total curvature of a sphere of radius 2? Radius 3? Radius r?

V. What is the total curvature of a sphere of radius 5cm ? What are the proper units for total curvature?

Consider a surface tiled by polygons, like the examples we’ve been studying. The surface’s total area A is the sum of the areas of the polygons comprising it: A = ∑Ai. For each polygon, the curvature k is the holonomy angle φi divided by the area Ai, that is, k = φi/Ai.

W. Use the facts in the preceding paragraph to prove that for a surface of area A and constant curvature k, the total curvature kA may be computed as the sum of the holonomies around the polygons in a tiling. In symbols, prove kA = ∑φi.

X. Return to the spherical tiling shown in Figure 1.2. How many triangles does that tiling contain? What is the holonomy around each triangle? What value do you get when you compute the sphere’s total curvature as ∑φi? Did you use the sphere’s radius in your computation? Does your answer agree with your answer from Question U?

Now return to the hyperbolic plane. Within the Hyperbolic Games, keeping other settings the same, select

Topology Double Torus
Options : Repeating Images Draw None
You’ll see a so-called fundamental domain for the double torus, with each triangle appearing exactly once. If you are unfamiliar with the concept of a fundamental domain, choose Game : Pool and shoot a little Pool, scrolling the board often to get a feel for what’s going on, and then choose Game : Sudoku to return.

Y. How many triangles does that tiling of the double torus contain? What is the holonomy around each triangle? What value do you get when you compute the double torus’s total curvature as ∑φi?
Within the Hyperbolic Games, choose Topology : Triple Torus. How many triangles does that tiling of the triple torus contain? What is the holonomy around each triangle? What value do you get when you compute the triple torus’s total curvature as ∑φi?

The beautiful thing about the formula ∑φi is that it doesn’t rely on the surface having constant curvature. This is our key to defining the total curvature of any surface, even the surface of an irregular blob.

Definition. The total curvature of a surface is the sum of the holonomies around each of the polygons in a tiling. In symbols, the total curvature is ∑φi.

Conceptually the holonomy around a single polygon tells the total curvature within that polygon, so really the above definition is just saying that the total curvature of the surface is the sum of the total curvatures contained within each of the polygons in the tiling.

Your proof from Question W shows that this formal definition of total curvature agrees with our earlier provisional definition of total curvature as kA. The advantage of the new ∑φi definition is that it applies to all surfaces, whereas the old kA definition applied only to surfaces of constant curvature.

For logical completeness we still need to prove that the total curvature formula ∑φi always gives the same answer, no matter what tiling we use (regular or irregular!). Fortunately it’s fairly easy to (informally) convince yourself that this is true, by considering a common subdivision of any two given tilings.

We conclude with the Gauss-Bonnet Theorem, one of the most beautiful theorems in all of mathematics.

Gauss-Bonnet Theorem. The total curvature ∑φi of a surface equals 2πχ, where the Euler number χ = v ‑ e + f is the alternating sum of the number v of vertices, the number e of edges and the number f of faces in a given tiling.

In effect, the Gauss-Bonnet Theorem is a Law of Conservation of Curvature for surfaces. Imagine the surface of an irregular blob. As you deform the blob (changing its shape) you create positive curvature in some parts of the surface and negative curvature in other parts. The Gauss-Bonnet Theorem says that the newly created positive curvature must exactly balance the newly created negative curvature, leaving the total curvature unchanged. No matter how you deform the surface, the total curvature remains equal to 2π times an integer χ. Pretty cool, eh?

Z. Compute the Euler number χ = v ‑ e + f for the following tilings.

Proof of the Gauss-Bonnet Theorem:

In Questions J, L and M you saw that at a vertex of angle θ, the direction of travel rotates by π ‑ θ relative to the direction of the hand. Equivalently, the direction of the hand rotates by θ ‑ π relative to the direction of travel. At each vertex, write down this contribution to the holonomy:

The angle θ is of course different at each vertex, so to be formally correct we’d need to use a different symbol at each vertex, but I didn’t want to clutter the picture with lots of ugly subscripts, so I used the same symbol θ for each angle, confident that you’d understand the intended meaning. Also keep in mind that each θ ‑ π is that vertex’s contribution to the holonomy; the angle itself is θ.

The +2π at the polygon’s center represents the +2π we added in the Convention following Question K (see above). All the θ ‑ π terms together with the +2π record the polygon’s full contribution to the total holonomy. So far so good?

Now here’s the crucial step: take the θ ‑ π that we just wrote down at each vertex, and slide the ‑π over a bit, to sit next to an adjacent edge:

Numerically nothing has changed: the sum of all those numbers still records the polygon’s full contribution to the total holonomy. The only thing that changed is where each number gets written within the polygon. What do we gain from this seemingly trivial change? Well, take a look at the tiling as a whole:

Each edge see two ‑π’s, one sitting on either side of it; combine those two ‑π’s into a single ‑2π and write it directly on top of the edge itself. Each vertex sees some number of θ’s surrounding it; those θ’s sum to 2π (because the angles surround a point) so replace the θ’s with a +2π and write it directly on top of the vertex itself. Now we see:

It’s now easy to add up the total holonomy for the whole tiling: each of the v vertices contributes +2π (shown in green), each of the e edges contributes ‑2π (shown in red) and each of the f faces contributes +2π (shown in blue). The total holonomy — which by definition gives the total curvature — is therefore 2πv ‑ 2πe + 2πf = 2π(v ‑ e + f) = 2πχ.

End of proof.

Just for the record: The Gauss-Bonnet Theorem works even for tilings with curvy edges. Introducing curviness in an edge increases the holonomy around the polygon on one side, while decreasing the holonomy around the polygon on the other side by exactly the same amount, so the total remains unchanged.

2. Geodesic Convergence/Divergence

To appear in some future version of this document

3. Models of the Hyperbolic Plane

To appear in some future version of this document

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© 2010 by Jeff Weeks