# Why isn't the Euler characteristic equal to 2 for this polyhedron?

I've managed to confuse myself on a topic I thought I understood.

To set up, if a surface $$M$$ is a compact Riemannian manifold with boundary $$\partial M$$, then the Gauss-Bonnet theorem states $$\int_{M}KdS+\int_{\partial M}kds=2\pi \chi(M)$$ where $$K$$ is the Gaussian curvature of $$M$$, $$k$$ is the geodesic curvature of $$\partial M$$, and $$\chi(M)$$ is the Euler characteristic of the surface. Now, if our surface was $$\textit{almost}$$ smooth everywhere, except at some finite set of points $$\left\{ x_{k}\right\}_{k\le n}\subset M$$ where there were vertex singularities with angular defects $$\left\{ \theta_{k}\right\}_{k\le n}$$ at these points, then the Gauss-Bonnet theorem generalizes to $$\int_{M}KdS+\int_{\partial M}kds+\sum_{k\le n}\theta_{k}=2\pi \chi(M).$$ This is intuitively very pleasing, because it's essentially the statement that a surface's curvature, when viewed as a measure, can either live on the subset of the surface that's Riemannian, the boundary, or at vertices. This generalizes both Gauss-Bonnet for smooth surfaces and Descarte's theorem for the total angular defect of polyhedra.

In particular, if $$M$$ is a polyhedron with $$n$$ vertices, then $$K=k=0$$, and the theorem reduces to Descarte's theorem $$\boxed{\sum_{k\le n}\theta_{k}=2\pi \chi(M)}$$ This formula works beautifully for convex or toroidal polyhedra. But for some nonconvex polyhedra, I have no idea what's going on and am completely lost. To illustrate, I'll focus my attention particularly on two examples.

1. First, the $$\textit{pentakis dodecahedron}$$ is simply a regular dodecahedral surface with (short) pentagonal pyramids attached to each face. From reading its Wiki page, its Euler characteristic is $$2$$. No surprises here, since it has no holes or handles.

2. Then we come to the $$\textit{small stellated dodecahedron}$$, which is very similar, but with pentagonal pyramids substantially taller. It's Wiki page, Wolfram page, and this source all state, quite strangely, that it has twelve pentagrams as faces with thirty edges and only twelve vertices (which I would never even consider being the case). Its Euler characteristic then drops to $$-6$$. The same peculiar counting method seems to be implemented for several other nonconvex polyhedra.

I tested this by counting the small stellated dodecahedron's total angular defect, as shown here. Since each triangular face is isosceles with one angle of $$36^{\circ}$$ at the top and two angles of $$72^{\circ}$$ at the base, this is straightforward. There are twelve vertices (the pyramidal vertices) with a positive defect of $$360^{\circ}-5\cdot36^{\circ}=180^{\circ}$$ and twenty vertices (the original dodecahedron's vertices) with a defect of $$360^{\circ}-6\cdot72^{\circ}={-72^{\circ}}.$$ The total angular defect then comes to $$12\cdot 180^{\circ}-20\cdot72^{\circ}=720^{\circ}.$$ This is exactly what we would expect given that the small stellated dodecahedron has no holes or handles. This would also suggest that it has the same topology as a sphere and has an Euler characteristic of $$2$$. Moreover, whenever I count the vertices, edges, and faces, I get $$V-E+F=32-90+60=2,$$ which I'd expect anyways.

My question is, how could this $$\textit{not}$$ be the correct answer? My background is in analysis and differential geometry, not topology. If I'm wrong, I have no idea how. I also doubt all these sources are simultaneously wrong. But nowhere in the Gauss-Bonnet/Descarte theorem is convexity assumed. Moreover, it sounds preposterous from an analysis perspective that increasing the heights of the pentagonal pyramids to a certain point suddenly causes the total curvature to discontinuously become negative.

The sources for the stellated dodecahedron all state that this polyhedron confused early topologists. If this is just a funny way of counting faces, edges, and vertices, then who decided this was the $$\text{"correct"}$$ way to count them, and which way is $$\text{"correct"}$$ for the Gauss-Bonnet/Descarte theorem?

• The phrase "twelve pentagrams as faces" in your quoted description of the small stellated dodecahedron tells me that these sources are defining the small stellated dodecahedron not as the non-convex surface you have in mind but as that surface plus the faces of the original dodecahedron. After all, those faces are part of the pentagrams mentioned there. So this surface not only fails to be convex (which, as you said, shouldn't affect the Euler characteristic) but is self-intersecting, which leads beyond my intuition at this time of night. Commented Nov 24, 2020 at 3:20
• The small stellated dodecahedron admits the "peculiar [face] counting method" in order gain access to the class of "regular polyhedra" which, among other things, must be "vertex transitive" (any vertex can be carried to any other by some isometry), a property that the mere spiky-dodecahedron interpretation lacks. Regularity an added benefit of the ambiguity MathWorld notes about the stellation process, making the peculiar interpretation popular; however, one isn't "wrong" to prefer the spiky interpretation in certain contexts.
– Blue
Commented Nov 24, 2020 at 3:29
• So does that imply that the definition of the Euler characteristic of a surface depends upon the context in which we're using it? And, is there an assumption in the Gauss-Bonnet/Descartes theorem that forces a particular choice of definition? Commented Nov 24, 2020 at 3:36
• FYI: For even-more-peculiar face counting shenanigans, take a look at Petrie polygons. These things allow us to interpret the Platonic tetrahedron as a having quadrilateral faces, and the dodecahedron as having decagonal faces, etc, yielding yet more Euler characteristics. It's just another way to look at the figures.
– Blue
Commented Nov 24, 2020 at 3:37
• V=12, E=30, F=12 (pentagrams) makes V-E+F=-6 so the genus = 4 (from $V-E+F=2-2g$) Commented Nov 24, 2020 at 16:04

Here is an anlogue from a lower dimension:

For polygons (aka. polygonal lines), the analogue of angular defect are the interior angles. For a closed non-self-intersecting polygonal line with $$n$$ corners, the sum of these angles is $$\pi(n-2)$$.

But the pentagram (like the small stellated dodecahedron) is self-intersecting, and so this rule does not apply. Note that for the pentagram these intersections are not vertices. If you consider them as vertices you actually study the star on the right.

Still, I find it strange to state that the Euler characteristic is negative, since topologically the pentagram is a circle (and the small stellated dodecahedron is a sphere).

You are treating the small stellated dodecahedron as a collection of triangles glued together along their edges -- just the exposed shapes that you can see in the figure. You are also treating each intersection of line segments as a vertex. As a result, you end up with something that is topologically equivalent to a sphere.

But when someone says the small stellated dodecahedron is made of $$12$$ pentagrams, the pentgram they refer to is the star polygon $$\left\{\frac52\right\}$$. Although this figure looks something like a pentagon with a triangle glued to each edge, it has only $$5$$ vertices and $$5$$ edges: each edge connects one of the "points" of the "star" to another "point", and in between these two vertices is crosses two other edges without creating any additional vertices. The pentagram is a self-intersecting polygon.

So we have edges that don't end when they intersect another edge, and faces that don't end when they intersect another face. This is getting weird already, even before we calculate the Euler characteristic.

In short, the small stellated dodecahedron is not just a pentakis dodecahedron with the pyramids stretched outward to make certain faces coplanar. It's a completely different construction.

It you try to project the pentagram faces of the small stellated dodecahedron onto a concentric sphere, they will overlap each other. I find it hard to see a spherical topology in this object--if it's there, it is strangely twisted around itself, just as the pentagram is a pentagon twisted around itself--so it should be no surprise really that it does not have the same Euler characteristic.

• @M.Winter I could not justify that statement, so I've removed it. It's an interesting question. I note that while it is possible to unwrap a pentagram into a convex pentagon by manipulating the sides within the plane, one of the edges gets "flipped" (in some way) in the process. The interior angles at some vertices go to zero, then suddenly you have very small exterior angles and very large interior angles. So one might also try to "unwrap" the small stellated dodecahedron and see what changes. Commented Nov 24, 2020 at 13:31

Who needs a dodecahedron? The same thing happens with an octahedron.

Say you have a regular octahedron and eight regular tetrahedra whose edges are congruent with those of the octahedron. If you affix a tetrahedron to the octahedron in such a way that one face of the tetrahedron eclipses a face of the octahedron, then the other faces of the tetrahedron become coplanar with the adjacent octahedral faces.

Now affix all eight tetrahedra to the octahedron in this way, covering the original octahrdron completely. If you count only the outer surfaces, you have 14 vertices (the original octahedral vertices plus the additional outer ones from the tetrahedra), 36 edges (which are counted as separate vertex-to-vettex connections even though they may be collinear), and the 24 exposed faces of the tetrahedra. That gives an Euler characteristic of $$+2$$, no problem.

But the stellated octahedron is actually defined as including the now hidden faces of the original octahedron, which merge with the coplanar outer surfaces. Now there are only the eight outer vertices (the original octahedral vertices are no longer counted after the merging), twelve edges (each group of three collinear edges from the "outer surface only" treatment has now been merged into a single line), and four faces that look like large triangles (merging the outer triangles of the tetrahedra with the inner triangles of the octahedron). Suddenly the Euler characteristic is $$0$$ instead of $$+2$$!

We can correlate this change in topology with the effects of cutting the polyhedron. If you just count the outer surfaces, matching the topology of a convex polyhedron or sphere and having an Euler characteristic of $$+2$$, the polyhedron falls apart by cutting just one loop around it just like cutting a loop in a sphere. But when you include the hidden surfaces to form the true stellated octahedron, that is no longer equivalent to a convex polyhedron or sphere; now you can cut a loop in the inner, octahedral faces without any pieces falling away. The outer faces of the tetrahedron keep the structure intact. You need a second loop cut to break the polyhedron either internally or externally. It thus behaves like a torus, which effectively has an inner layer surrounding the hole, instead of a sphere.

With the stellated dodecahedron, too, including the inner surfaces destroys the equivalence to a sphere and allows you to cut loops in those internal regions without breaking the polyhedron apart. Only in this case, the more complex internal structure allows multiple loops to be cut, which shows up as a bigger drop in the Euler characteristic.