http://www.georgehart.com/virtual-polyhedra/vrml/tetrahemihexahedron.wrl (this is a 3D model viewer)
I don’t get this shape at all. Can you build a paper model of it? It seems impossible. Here’s the main page, with more weird polyhedra:
http://www.georgehart.com/virtual-polyhedra/uniform-info.html
I can’t see the polyhedron you mean, but the site has this to say about it:
(italics mine) which seems to indicate that, like a mobius strip, if you start on one side of one of the polygons, and start drawing a line, you could make a continuous line to the other side of the polygon you started on.
Aha! I was able to replicate this with paper. Draw a square with four equilateral triangles on each edge, and cut this out. Fold two opposite triangles up so the points touch, and fold the other two down. The triangles make a regular octohedron, with half the faces missing. You should be able to see where the other two squares can go: both intersect the first square along a diagonal and touch all four triangles along an edge.
The tetrahemihexahedron is a representation of the projective plane, which is indeed a one-sided surface like the Moebius strip. However, to draw the continous line like ZenBeam describes you have to pretend that the three squares don’t really intersect. In other words if your line is travelling across one of the squares you have to let it pass through the other two squares, instead of jumping from one square to another like it would at a regular edge.
I think part of my confusion was caused by that model viewer not being very good. If you rotate the image around, some of the faces appear and disappear. Here is a better site. It still seems weird to me to allow faces to intersect each other, or I guess it’s pass through each other without intersecting.
Does anyone know if the figures on the page The Kepler-Poinsot Polyhedra, (e.g. the one on the lower right, which is easiest for me, at least, to see what’s going on) are topologically equivalent to the sphere?
The great icosahedron is topologically equivalent to a sphere, but the great dodecahedron is not. You can tell from the Euler characteristic. The great icosahedron has 20 faces, 30 edges (3 per face, but each edge is shared by 2 faces), and 12 vertices; 20-30+12=2, so the great icosahedron is topologically a sphere. The great dodecahedron has 12 faces, 30 edges, and 12 vertices, and 12-30+12=-6, so the great dodecahedron is topologically equivalent to a “genus four surface”, or if you prefer a sphere with four handles. (The genus is 1-E/2, where E is the Euler characteristic.)
As for the other two…well, a pentagram has 10 vertices and 10 edges. The great stellated dodecahedron then has 12 faces and 1210/2=60 edges, plus 125/3=20 vertices at the points (where each vertex is shared by 3 faces) and 12*5/5=12 vertices in the “valleys” (where each vertex is shared by 5 faces). And 12-60+32=-16. If I’ve counted right, the great stellated dodecahedron is a genus 9 surface.
Similarly, the small stellated dodecahedron has 12 faces, 60 edges, 12 vertices at the points and 20 vertices in the valleys. So the small stellated dodecahedron is topologically a genus 9 surface too.
Again, all this assumes that you ignore intersections that occur in the interior of a face, just like with the tetrahemihexahedron.
I figured out the deal with the model viewer. I originally thought it was a Java applet or something similar, but it’s actually a browser plugin that reads Virtual Reality Modeling Language (VRML) code. Internet Explorer 5.5 comes with a VRML viewer, but it isn’t very good. Netscape 6.2 doesn’t come with one. I’m still looking for a good VRML plugin, I’ll post a link here if I find one.
Someone told me when I was younger that stellated polyhedra were formed by extending the edges of a convex polyhedron so that they met at new vertices. But the proper definition is to extend the faces while preserving symmetry, and this is why a great stellated dodecahedron isn’t called a stellated icosahedron, right? Apparently you can even make polyhedra with discontinuous surfaces, such as stellated icosahedron #33!
Oh and according to both sites, a pentagram doesn’t have 10 faces and vertices, it has 5. So E for the small stellated dodecahedron is 12 - 30 + 12 = -6, making it genus 4, and the great stellated dodecahedron has E = 12 - 30 + 20 = 2, making it genus 0.
Hmm…thinking about this some more, I think we’re both right depending on how you define the surface whose Euler characteristic we’re trying to calculate.
The problem is that a pentagram can be viewed in two different ways. In one view, it’s a region in the plane bounded by 10 line segments. In another, it’s the image of a pentagon under a map that wraps the boundary of the pentagon twice around a point, similar to the map which takes z to z[sup]2[/sup] in the complex plane. The first view leads to my values for the Euler characteristic, the second view leads to yours.
In other words, the great stellated dodecahedron (for example) is both the image of a genus 9 surface and the image of a sphere, under two different maps. The map from a genus 9 surface to the great stellated dodecahedron is one-to-one except at those points where the faces pass through one another, while the map from the sphere to the g. s. dodec. is one-to-one at some points (in the “tips” of each pentagram) and two-to-one at others (inside the pentagon in the “core” of each pentragram).
I’d contend that the great stellated dodecahedron is closer to being “topologically equivalent” to a genus 9 surface than a sphere, but there does seem to be room for argument…especially since we’re stretching the definition of “topologically equivalent” in the first place by allowing the faces to intersect.
Isn’t whether surfaces intersect irrelevent in topology?
No, it’s not quite irrelevant. For example, consider a Klein bottle. The most common mental picture that people have of a Klein bottle is of a tube which intersects itself. But that picture is not topologically equivalent to a Klein bottle because a Klein bottle doesn’t intersect itself. To make that picture topologically equivalent to a Klein bottle we have to make the intersection “go away”, for example by embedding the picture in four spatial dimensions instead of three. Alternatively we can leave the points of intersection there and settle for something less than topological equivalence; for example we can say the picture of a tube intersecting itself is an “immersed image” of a Klein bottle.
The exact same thing is happening with the great icosahedron, for example. It’s not, strictly speaking, topologically equivalent to a sphere. But it’s the immersed image of a sphere, which is so close that only nitpicky grad students in topology like myself really care about the difference.
I’ve always understood that in the 3D representation of the Klein bottle, the intersection is in the representation, but not in the Klein bottle itself. Of course, I’m not a nit-picky topology grad student. 
I found a VRML browser which works as a plugin for IE and NetScape (but not Mozilla 
). This site has a VRML showing eversion of a sphere which is pretty cool. It has a slider, so you can step slowly through the process, and you can look at cutaway views. I still have problems visualizing one part of the process, however. Here’s a webpage with some views from the VRML model, but get the plugin. For those of you who remember the steps shown in Scientific American thirty six years ago ("Turning a surface inside out’’, May 1966, by Anthony Phillips), this is nothing like that.