Curvature of Polyhedra (2021) | Hacker News Reader

Curvature of Polyhedra (2021)

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74 pointscgadski18 comments

In my experience, the discrete version of the Gauss-Bonnet theorem is a really great conversation starter to talk about the kinds of things mathematicians like. Everyone knows that the exterior angles in a polygon has to sum up to 2 pi, but did you know that you can define "exterior angles" on the vertices of a polyhedron that have to sum up to 4 pi? Can you think of a single line of reasoning that explains both facts? (Remember that the circumference of the unit circle is 2 pi and the area of the unit sphere is 4 pi...)

> Can you think of a single line of reasoning that explains both facts?

It's definitely not immediately obvious that this number should be an invariant for any shape topologically equivalent to a sphere.

However, it is easy to see why it should hold for a dihedron.

For a general convex shape (to start with) I think I would try to show people how the planar face orientations change at the corners, and point out that these must sweep out the whole sphere in the 3-dimensional space of planar orientations. This is similar to the way in which the linear orientations of the sides of a polygon must sweep out a whole circle (exterior angles sum to 2π). Non-convex corners can "subtract" some of the areas we already swept out, which can be a little tricky to reason about, but could probably be made clearish with a nice interactive computer diagram.

We just need to look at the corners because on the face there is no orientation change, and at the edges the orientation change is 1 dimensional and therefore has no "area" relative to the 2-dimensional quantity we are measuring. This might be easier if you look at the linear orientations normal to the faces.

It might be easier still to just start with tetrahedra.

* * *

I have a much harder problem for everyone: come up with some way of calculating the sum of angle deficit at the corners of a polyhedron constructed of great-spherical faces on a 3-sphere and relating it to the size of the polyhedron. Does this generalize the pattern for spherical polygons on the 2-sphere? (This is a genuine question: I'm not sure quite what is known about this.)

(This is the author of the article.) I agree! One of my favorite things about this topic is that you can prove the polyhedron version using very simple tools, so it's possible to get a feel for the concept of curvature (at least for surfaces) with a lot less machinery than you'd need to study the corresponding concept for smooth surfaces.

Draw a circle around a polygon, and call its radius "1 radian".

Drop perpendiculars from the center of the circle through each side. Pairs of adjacent perpendiculars each subtend an arc of the circle, of length equal to the "exterior angle" between the corresponding sides. Those arcs sum to 2π radians. The intersection points with the circle from a sort of "dual" polygon to the original, exchanging sides with vertices, and exchanging central angles with exterior angles. (But this dual construction is not uniquely reversible back to a non cyclic polygon or not using the same center)

Now take a polyhedron and choose a center and draw a sphere around it. Drop perpendiculars to faces, intersecting with the sphere to create new "dual" vertices from faces. This time, draw edges between adjacent-face perpendiculars , so each original edge generates a new dual (perpendicular!) edge. This now creates a new face for each vertex of the original shape. Now we have a dual polyhedron inscribed in a sphere, where each face subtends the 2D "exterior angle" of an vertex of the original polygon.

Note that an exterior (hyper!)angle of a hyperpolyhedron in N-dimensional space is an N-1 dimensional measure (face-dimensional dimensional measure) of a vertex (with its cluster of incident faces). In 2D, there is an isomorphism between pairs-of-edges and ordinary 1D angles. (and edges are 2D hyper-faces). But in higher dimensions we need use N-1 dimensional dual hyperfaces to measure N-1 dimensional exterior angles

I'm not actually sure this really works in n>3 dimensions.

Thinking about HN favorite Geometric Algebra is a little helpful for the intuition here.

For concave shapes, some of these angles are negative, as the relevant side/face is oriented the opposite way from its neighbors, with respect to the enclosing circle/sphere/hypersphere.

These N-1 dimensional angles in N dimensional are called "solid" angles. The unit is the "steradian", which is usually considered dimensionless, but cool cats know that its dimension is really the dimension of normalized hypersurface hyperarea. (Similar to how a https://en.m.wikipedia.org/wiki/Solid_angle

Here a visual illustration:

https://www.polydron.co.uk/resources/descartes-angle-defect-...

Watching my son when he was 8 write python code to draw regular polygons made the exterior angles summing to 2π really clear and I realized it made for an exceptionally nice way to prove that the sum of the interior angles of a polygon would be π(s-2) and I really wanted to teach geometry again because it was such a beautiful result.

It's also a fun exercise to take one of several images that show a football made out of hexagons and use the Euler characteristic to show that it is impossible to make such a football (no matter how much you bend or stretch your hexagons).

It's possible if you attach 6 pairs of your hexagons along two adjacent sides. https://archive.bridgesmathart.org/2017/bridges2017-237.html

For example, you can cover an octahedron with 4 regular hexagons, with 2 of the hexagons coming together at each octahedron vertex. Or you can divide these hexagons each into any Löschian number (integers of the form a² + b² + ab).

You might protest that these "hexagons" are now not connected like a honeycomb anymore, and now you have 12 "pentagons" mixed in, and that's true. But this idea turns out to still be very practically useful for making grids on a sphere, etc.

How do you attach a pair along 2 adjacent sides? By folding it?

You can cover e.g. an octahedron, a triangular prism, or a hexagonal dihedron, each of which has 6 corners where there is only 240° of surface meeting at the corner, exactly the amount covered by 2 corners from adjacent regular hexagons. So you end up attaching those hexagons along each of their two adjacent sides meeting at that corner. This makes them metrically hexagonal, but topologically pentagonal (if you consider the two distinct edges between adjacent faces to merge together).

Oh so it's like taking two hexagons attached at a common side and then stretching and pinching together another pair of sides.

In a computation sense, ignoring rigid geometry, you have 2 hexagons with sides 12 sides A1 to A6 and B1 to B6, and you are identifying A1 = B1 to make an edge AB1 and A2 = B2 to make a edge AB2, and then fusing AB1 + AB2 = AB12.

What's not obvious to me is why it's useful to fuse edges like that. I guess that it's only possible when the vertex between the edges has only 2 edges, with the same faces, and in discrete geometry that pre-fusion vertex isn't doing any work. But then why not fuse all possible edges, leaving you with a unigonal dihedron (double-sided circular disc)?

But it _is_ possible to make a ball out of hexagons if you use infinitely many of them!

A bit late but I think you're going to have points with infinite hexagons in any neighbourhood of them. Otherwise the set of hexagons would be finite because of compactness.

Which I think means those points can't really lie on the edge of or interior of a hexagon. But I can't rule out that they might lie on a corner, it's hard to reason about.

A polygon is half a dihedron, with one of faces ignored.

So the polygon has 4pi total exterior angle (vertex angle defect) is you consider it as a degenerate 3D object.

Ah- that's a much better definition of exterior angle than the old one I had in my head. Thanks for the puzzle!

Without opening the article I swore I thought they were going to talk about the concave curved surfaces of spiked, icosahedron-shaped "Artifact" of Baldur's Gate 3.

I can’t remember the last time I saw someone talk about the Gauss-Bonnet theorem without talking about soccer balls

Applications of (generalizations of the Chern-) Gauss-Bonnet theorem came up in the context of alternatives to General Relativity around fifteen to twenty years ago. You should probably stop reading here. :-)

https://en.wikipedia.org/wiki/Chern%E2%80%93Gauss%E2%80%93Bo... (the History section is pretty interesting)

https://en.wikipedia.org/wiki/Gauss%E2%80%93Bonnet_gravity

As briefly touched on there, Einstein-Gauss-Bonnet (EGB) gravity turned out to require higher dimensions (the start of this abstract is worth the click :-) <https://link.springer.com/article/10.1140/epjc/s10052-020-82...> ) so the idea that EGB might sweep up higher order curvature in the self-interaction of physical gravitation is mostly dead.

Mostly. There's still a trickle in the literature. For example, <https://doi.org/10.1103/PhysRevD.104.044026> ("General relativity from Einstein-Gauss-Bonnet gravity", 2021, open access; ref [12] therein corresponds to <https://arxiv.org/abs/0805.3575> (no singularity in EGB black holes -- lots of Ricci calculus there, nothing like soccer balls as far as I can see)). The references in <https://arxiv.org/abs/1904.00260v1> (no singularity in Ef(G)B cosmology, applying a correcting function on the Gauss-Bonnet term) can takes an interested reader to years of discussions about the accelerated expansion of the universe. As always the first two digits of an arXiv identifier are the submission year.

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