Curvature of Polyhedra | Nicolas James Marks Ford

This put up is customized from a brief discuss that I gave a pair years in the past at Mathcamp. It’s meant for a common mathematically curious viewers.

Introduction

Think about you’re standing on a sphere on the north pole. Stroll straight south till you get to the equator, then flip left. Stroll alongside the equator till you’ve traced out of longitude, after which flip left once more, so that you just’re heading north. If you get again to the north pole, the trail you’ve traced will appear to be Figure 1. Discover that this path you adopted kinds a triangle with three proper angles.

We are able to draw an identical image on the floor of a dice moderately than a sphere; it’s proven in Figure 2. You would possibly initially be skeptical of our use of the phrase “triangle” for this form. In spite of everything, it appears to have six edges moderately than three. However there’s a sense wherein the strains that wrap across the edges of the dice are straight: if we took simply the 2 faces of the dice that contact that edge, we might unfold them to type a flat rectangle, and after we did that our line would certainly be straight.

This may all the time be what we imply once we discuss polygons on the floor of a polyhedron. The strains that make up the perimeters — strains like those in Figure 2 that enter and exit every fringe of the polyhedron on the identical angle — are referred to as geodesics, and so they’re the closest analogue to straight strains we are able to discover on the floor of a polyhedron: like straight strains in a aircraft, they type the shortest paths between any two close by factors mendacity alongside them.

As soon as once more, we see that it has three proper angles, including as much as , greater than we might get from drawing the identical image in a flat aircraft. Furthermore, the additional appears to be particularly the fault of the nook: in the event you draw a triangle like in Figure 3, you’ll be able to see that the angles sum as much as the same old . In actual fact, in the event you tear off simply the 2 faces of the dice which have the triangle on them, you’ll be able to unfold them to make the triangle lie flat in a aircraft.

So there’s one thing happening on the corners of the dice that isn’t taking place anyplace else, one thing that pushes out on these triangles, rising the measures of their angles. We’ll say that the dice has curvature at its eight vertices, and is flat in all places else.

Moreover, there’s a way wherein the dice has “ price of curvature” sitting at every of its vertices: a triangle enclosing one in all these vertices appears to get added to the sum of its angles. If we used one other polyhedron as a substitute of a dice, we’d get a unique quantity: the sum of the angles within the triangle on the fitting in Figure 4 is greater than , however not a complete lot extra; the triangle on the left seems to be much more just like the one in our first instance, and the sum of its angles is clearly fairly a bit bigger.

Angle deficit

Placing all this collectively, we appear to have arrived at a method to quantify curvature: draw somewhat triangle round a vertex, add up its angles, subtract , and that’s how a lot curvature that vertex has. However this leaves open an enormous query: if we’re making an attempt to say that the quantity of curvature at a vertex is only a property of that vertex and never a property of the triangle we selected to measure it with, then it had higher be true that any triangle we use will give us the identical reply.

In actual fact, that is true. One method to see it’s to take a triangle that’s wrapped round a single vertex and lower it up into smaller triangles like in Figure 5: draw a line from the vertex to the triangle alongside every fringe of the polyhedron, and likewise join it to the vertices of the triangle.

Let’s suppose there are such triangles. (Within the image, .) We’d prefer to relate the sum of the angles of the triangle to one thing associated to simply the vertex itself. With this in thoughts, write for the sum of the angles of the triangle, and for the sum of the angles across the vertex. Then, on the one hand, the sum of the angles of all triangles is . Alternatively, now we have strains touching an fringe of the triangle and, because the edges of the triangle are geodesics, the angles at these factors every add as much as . So we get that

so

Now, we had been making an attempt to get an expression for a way a lot further angle now we have within the triangle, that’s, for . And we’re in luck: rearranging this final equation provides us that

That’s, the quantity that will get added to the angle sum of a triangle is identical because the distinction between and the sum of the angles on the vertex. This quantity, , is named the angle deficit on the vertex, and that is the quantity we need to connect to a vertex to measure its curvature. Word that if the vertex is flat, then the sum of the angles at that vertex shall be , making either side of the equation zero, as anticipated.

What if the sum of the angles round a vertex is greater than ? In that case, is damaging, and so we are saying that that vertex has damaging curvature. An instance of that is proven in Figure 6. The vertex that the triangle is drawn round has 5 proper angles at it, so the sum of its angles is , giving it a curvature of . So we must always count on the sum of the angles of the triangle so as to add as much as , and certainly the triangle within the image has three angles.

On this final part, we’re going to speak about one thing outstanding that occurs while you add up the curvatures for each vertex in a polyhedron. The quantity you get while you do that is referred to as the complete curvature. Now that now we have the angle deficit description of the curvature at a vertex, the overall curvature is pretty simple to compute. We are able to do it fairly shortly for the entire Platonic solids, pictured in Figure 7.

To show this end result, we’ll first have to triangulate our polyhedron, that’s, cowl it with triangles in order that any two triangles meet at an edge and any two edges meet at a vertex. For this proof, we’re additionally going to require that not one of the vertices of the triangles lies on prime of a vertex of the polyhedron. A doable triangulation of a dice is proven in Figure 8.

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Suppose our triangulation has vertices, edges, and triangles. (It’s essential to keep in mind that the vertices, edges, and triangles we’re speaking about now are not the vertices, edges, and faces of the polyhedron itself!) Write for the overall curvature of the polyhedron. On the one hand, once we add up all of the angles inside all of the triangles, we get , because the vertices of the triangles all lie at factors with no curvature. Alternatively, we all know that the sum of the angles of every triangle is plus the curvature inside it. All of the vertices of the polyhedron are inside some triangle, so altogether we’re going to get : for every triangle, plus all of the changes now we have to make to account for the curvature.

Placing these collectively, we see that

so

We must always pause right here to take inventory of what we simply proved. The left-hand facet of that method has nothing to do with the triangulation we selected; it’s simply the sum of the curvatures at every vertex of the polyhedron. However the right-hand facet is the alternative: it’s solely in regards to the triangulation, and it has nothing to do with the geometry of the polyhedron we drew it on. So it doesn’t matter what triangulation you utilize for a polyhedron, is all the time the identical quantity. And if we’re in a position to wrap the identical triangulation round two completely different polyhedra, then these two polyhedra will need to have the identical complete curvature. It’s this final end result that tells us why we noticed what we did for the Platonic solids: it’s simple to see that any triangulation that works for one in all them will work for any of the others. (For instance, you may put the 2 Platonic solids on prime of one another in order that they’re each centered on the identical level, and mission the vertices of the triangulation out from one in all them onto the opposite.)

The quantity is named the Euler attribute of the triangulation, and it’s normally written in a unique type, . In actual fact, since we solely used triangles, these two numbers are the identical. To see this, discover that you would be able to rely the perimeters of the triangulation by multiplying the variety of triangles by 3; every triangle has three edges, so each edge is accounted for. However this counts every edge twice, since every edge reveals up for precisely two of the triangles. So , that’s, , so

Does the Gauss-Bonnet Theorem imply that each polyhedron has the identical complete curvature? Think about the polyhedron in Figure 9.

Mathematicians prefer to say that the Euler attribute relies upon solely on the “topology” of the polyhedron and never on its “geometry.” The topology of the polyhedron, roughly talking, consists of these properties that don’t change once we repeatedly deform it, pushing vertices outward or inward, making dents or bubbles, or anything that doesn’t require us to tear or glue the floor. A triangulation is detached to those types of modifications: if a face of the polyhedron strikes round barely, we are able to nudge the strains in our triangles accordingly. This, after all, breaks down if now we have to tear a gap like in Figure 9: this would possibly lower an fringe of the triangulation, and there’s no apparent method to reattach it.

That is what makes Gauss-Bonnet so outstanding. It provides us a relationship between geometry — within the type of the curvatures at every of the vertices — and topology — within the type of the triangulation — two completely different facets of our floor that would appear to not have quite a bit to do with one another. That is the kind of end result that mathematicians crave; at any time when we see a connection between two seemingly distant fields, it’s an indication that there’s rather more depth there to discover.