TITLE: Is there a nice way to "unravel" a great dodecahedron?
QUESTION [12 upvotes]: The great dodecahedron is a non-convex regular polyhedron bounded by 12 pentagonal faces, crossing each other, arranged in a star-shaped manner around each of its 12 vertices (see the Wikipedia page for a picture.) From a topological point of view, the abstract surface that you get if you glue the faces along the corresponding sides is a compact orientable surface; hence it is homeomorphic to a sphere with some number of handles. It is easy to check that it has genus 4. (You can also see it as a 3-sheeted Riemann surface with 12 branch-points.)
Thus it is possible to draw, on the surface of a 4-holed torus, a graph consisting of 12 vertices and 30 edges that subdivides the surface into 12 pentagonal regions, with 5 such regions meeting at each vertex. To put it more concisely: there exists a sort of 4-holed polyhedron which is regular in at least a topological (or, if you prefer, combinatorial) sense and whose Schläfli symbol would be {5 5}.
I was actually able to draw such a picture, but all I see is a tangle of lines twisting and wrapping around in all directions. My picture has no obvious symmetry. Is it possible to embed this object in ordinary 3-space without self-intersections, but with, let's say, at least an axis of 5-fold rotation? (Unfortunately, I suspect the answer is "no".) Or is it possible to at least make sure that there is a group of easily described transformations that coincides, at least partially, with the abstract symmetry group of the configuration (which has order 120)?

REPLY [9 votes]: No, it is not possible to embed the genus 4 surface in $\mathbb R^3$ with 5-fold axial symmetry.  The $2\pi/5$ rotation of the great dodecahedron has four fixed points.  Two of these fixed points are at the center of a pentagonal face, and the other two are at a vertex where five pentagons meet.  The combinatorial rotation angle at the center-of-face fixed points is, of course, $2\pi/5$.  But the combinatorial rotation angle at the vertex fixed points is $-4\pi/5$.  But there do not exist order-5 diffeomorphisms of $\mathbb R^3$ (or $S^3$) whose fixed points have two different rotation angles.  (cf. http://en.wikipedia.org/wiki/Smith_conjecture )
I think it is, however, possible to embed the genus 4 surface in the lens space $L(5,2)$ with 5-fold symmetry.

[Added later] Here's how to embed the surface in $L(5,2)$ with 5-fold symmetry.  (I apologize for the lack of pictures.)  $L(5,2)$ is the union of two solid tori.  Let $X$ and $X'$ be two meridian disks of the first solid torus, and let $Y$ and $Y'$ be two meridian disks of the second solid torus.  The boundaries of $X,X',Y,Y'$ divide the torus $T^2$ into 20 parallelograms.  Color these parallelograms black and white in an alternating, checkerboard fashion.  Now consider $$X \cup X' \cup Y \cup Y' \cup \mbox{the 10 black parallelograms} .$$  This is a (piecewise linear) surface of genus 4 (which of course can be smoothed if one so desires).  The surface has an obvious 5-fold symmetry, and this symmetry has four fixed points (the centers of $X,X',Y,Y'$) with rotation angles as described above.
All that remains is to divide this surface into 12 pentagons.  The first two pentagons are easy: they are $X$ and $X'$.  Now for the remaining ten pentagons.  $Y$ and $X'$ intersect in five points on the boundary of $Y$.  Draw lines from these five points to the center of $Y$.  This divides $Y$ into five pieces $Y_0, Y_1, Y_2, Y_3, Y_4$.  Similarly divide $Y'$ into five pieces $Y'_0,\ldots,Y'_4$ using intersections with $X$.  Each of the ten (black) parallelograms intersects each of $X,X',Y,Y'$ on one side of the parallelogram.  Cut each parallelogram into two triangles so that one triangle contains the $X$ and $Y$ sides of the parallelogram and the other triangle contains the $X'$ and $Y'$ sides of the parallelogram.  Each $Y_i$ is adjacent to two of these triangles; add those two triangles to $Y_i$.  Similarly add two triangles to each $Y'_i$.  We have now divided the surface into 12 pieces, and one can check that combinatorially each piece is a pentagon, and furthermore that the combinatorics match the great dodecahedron.
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