The Tetratorus and Other Multi-Layered Polyhedra
We usually regard an ordinary polyhedron as a 3-dimensional solid
with a single-layered 2-dimensional surface. On this basis we can
easily count the number of vertices, edges, and faces, and compute
the Euler characteristic of the surface
e(S) = V - E + F
It can be shown that any simply-connected closed 2-dimensional
surface (i.e., a surface that has the same topology as a sphere)
has an Euler characteristic of 2. This applies to all the most
familiar polyhedra, such as the Platonic and Archimedian solids,
and many others as well. (Apparently Descartes was the first person
to ever notice the simple numerical fact that V-E+F = 2 for all
these solid shapes, which is remarkable, considering how numero-
logically inclined the ancient Pythagoreans were.)
To be a bit more rigorous about the meaning of "simply-connected",
note that a surface S is defined to be "connected" if any pair of
its points can be joined by a continuous curve lying entirely within
the surface. Further, a surface is said to be "simply-connected"
if any CLOSED curve C on the surface divides the surface into TWO
distinct regions, each of which is internally "connected" in the
sense just described, and such that any continuous curve which joins
a point in one of those regions to a point in the other must cross
the closed curve C.
Another, somewhat less formal, way of characterizing a simply-
connected surface is to say that any closed curve on such a surface
can be "shrunk" continuously to a point, without ever departing the
surface. We mentioned previously that the surface of a sphere is
simply-connected, and obviously any circle drawn on a sphere can be
shrunk to a point. On the other hand, the surface of a torus (a
doughnut) is NOT simply connected, because we can draw a closed
curve around a torus shape in such a way that the curve cannot be
shrunk any further. Also, from our previous definition, we can see
that some closed curves C on the surface of a torus do NOT bisect
the surface into disconntected parts, as shown below.
In this sense, the torus is topologically distinct from the surface
of a sphere, and consequently it has a different Euler characteristic.
If we triangulate the surface of a torus (in any way we like) and
count the number of edges, faces, and vertices, we find that the
Euler characteristic is 0.
Incidentally, the Euler characteristic of a surface is directly related
to the total Gaussian curvature of that surface. Specifically, the
Gauss-Bonnet theorem states that for any compact oriented 2-dimensional
Riemannian manifold S, if we let K denote the Gaussian curvature as a
function of position on the manifold, then we have
/
| K dA = 2pi e(S)
/
S
In other words, the integral of the Gaussian curvature over the entire
area of the manifold equals 2pi times the Euler characteristic. For
example, the Gaussian curvature of a sphere of radius r is a
constant 1/r^2, so the integral of this over the entire surface of
the sphere is simple 1/r^2 times the total surface area A of the
sphere. Now, since e(S)=2 for a sphere, we must have
1
--- A = 2pi (2) = 4pi
r^2
and therefore the surface area of a sphere must be 4 pi r^2, which
of course it is. Compare this with the reasoning that Archimedes
originally used to deduce the surface area of a sphere (see the note
Archimedes on Spheres and Cylinders). The Gauss-Bonnet theorem
is a truly beautiful fact, relating a metrical property (curvature)
to a purely topological property (the Euler characteristic). Notice
that this theorem can be applied to polyhedra as well, if we agree to
assign the appropriate amount of "curvature" to the vertices (since
there is no curvature on any face or edge).
It follows from this theorem that the "net" Gaussian curvature of a
torus is zero, so it's possible to define a perfectly flat metric
globally over the entire surface of a torus. On the other hand, it's
obviously not possible to define a flat metric globally on a sphere,
or on any surface homeomorphic to a sphere. This is true even though
the Gaussian curvature on the surface of a polyhedron is zero at all
but a finite number of points (the vertices).
But this leads us to an interesting question, namely, whether it is
possible to "wrap" a completely flat closed surface (i.e., a torus)
onto the surface of a convex polyhedron if we allow multi-layered
surfaces. For example, if we create a surface out of squares, we
must have a certainly number of squares meeting at each vertex, and
each square represents a right angle. Now, for a single-layered flat
surface there must be 4 right angles meeting at each vertex, whereas
for a single-layered cube there are only 3 right angles meeting at
each vertex. However, our surface need not be connected after just
a single loop around a given vertex. Indeed we can imagine a "spiral
staircase" consisting of a surface that circles a vertex many times
before re-connecting with the original sheet. Of course, if this
2D surface is embedded in 3D space we won't be able to re-connect
without self-intersecting, but this is just a restriction related
to the embedding in 3D space, similar to the fact that a "Klein
bottle can't be realized in 3D space without self-intersecting,
whereas it CAN be realized in 4D space. By the same token, we can
imagine a sheet circling a vertex arbitrarily many times before
finally re-connecting to the original sheet.
The underlying "shape" of this multi-layered manifold will still
depend on the angular travel around the vertex that brings us back
to the same location in the manifold, but now we can allow multiples
of the basic angular requirements. If we must travel through some
multiple of 4 right angles to get back where we started, then we
can imagine that these sheets are molded in a flat shape, whereas
if we must travel through a multiple of 3 right angles to get back
where we started, we can imagine that that this surface is molded
in the shape of a corner of a cube. But what if the total angular
travel around a vertex is a multiple of BOTH 3 right angles AND
4 right angles, i.e., what if it is a multiple of 12 right angles?
In that case we have a multi-sheeted surface that can be interpreted
as either flat or as the vertex of a convex solid, i.e., as a point
with non-zero curvature.
Actually it isn't difficult to construct this surface out of paper.
Just cut three identical squares, and draw two perpendicular lines
though the center of each square so that each is divided into four
faces. Then cut along one of those lines from one edge to the center,
and tape the squares together in a spiral pattern, as indicated below.
The cuts in the three original squares of paper are along the thick
vertical lines, and the right-hand edge of "d" is connected to the
left-hand edge of "e" (which is underneath "a"). Similarly, the edge
of "h" is connected to the edge of "i". Lastly, after making three
complete circuits around the central vertex, the edge of "l" on the
bottom sheet is connected to the left-hand edge of "a" on the top
sheet. This last connection obviously can't be achieved in 3D
without the surface intersecting itself, but just as with the Klein
bottle we are free to imagine an embedding in a higher dimensional
space that allows us to adjoin all the edges as described. As an
alternative, we can neglect the embedding entirely, and simply
conceive of this as a 2D surface with its own intrinsic properties.
The interesting thing about this surface is that it "works" as a
3-layered flat plane, bit it also works as a 4-layered corner of
a cube. To see this, it's helpful to first crease the sheets along
the orthogonal lines, and then simply "wind up" the faces around
the vertex, so that three (rather than four) square faces meet at
the vertex. Notice that the intrinsic connections between all the
faces of this "wound up" version are identical to the connections
of the "flat" version, including the alignment of the adjoining
edges of the "l" and "a" faces. In effect, we began with a 3-layered
flat surface, but this can also be interpreted as a 4-layered
surface that follows the shape of a cubical vertex. If we consider
the surface "modulo the cube", we see that it has intrinsic curvature.
If we consider the surface modulo the "flat" arrangement we group
the faces into the four equivalence classes
{a,e,i}, {b,f,j}, {c,g,k}, {d,h,l}
whereas if we consider the surface modulo the curved arrangement we
group the faces into the three equivalence classes
{a,d,g,j}, {b,e,h,k}, {c,f,i,l}
Of course, neither of these interpretations yields a closed surface.
This raises the interesting question of whether it's possible to
devise a closed surface that can be interpreted equally well as flat
or curved, depending on which "shape modulus" we select. It turns
out that it IS possible. The simplest example is based on the
tetrahedron. To construct a model of this, cut out a piece of paper
in the diamond shape shown below, comprised of 8 equilateral
triangles
Now make a cut along the boundary between regions "f" and "g", and
wind up the surface around the central point, so that three triangles
meet at that point. This results in a 2-layered tetrahedron, and the
four faces correspond to the four colors on the above figure. Thus
the eight original triangles are grouped into the four equivalence
classes
{a,h}, {b,g}, {c,f}, {d,e}
The important point to notice is that this arrangement brings all
the edges into alignment in a way that is perfectly consistent with
a FLAT arrangement of the faces. Note that if we take many copies
of the above diamond figure and set them side-by-side, we have the
doubly-periodic tiling illustrated below
This shows how all the 8 regions of the original diamond must be
interconnected to form a torus, which then is given by identifying
the upper left boundary of the original diamond with the lower right,
and identifying the upper right with the lower left. In this flat
interpretation we have only a single layer, so we have the 8
equivalence classes
{a}, {b}, {c}, {d}, {e}, {f}, {g}, {i}
The connections between these eight triangles are identical, whether
we view them as comprising a single-layered torus or a 2-layered
tetrahedron. Of course, the curvature of the latter arises only
when we reduce the surface modulo the tetrahedron. By the way,
notice that we have the extrinsic arrangement of layers (i.e.,
"inner" and "outer") indicated below
{b,e,f,h} {a,c,d,g}
Of course, we can make either of these sets the "inner" surfaces
and the other the "outer" surfaces, depending on how we choose to
"wind up" the faces around the original central vertex.
Another way of constructing the "tetratorus" is to begin with the
same 8-part diamond shaped piece of paper as above, but this time
make a cut along the boundary between regions "d" and "g". This
results in the same instrinsic arrangement as before, but the
extrinsic knotting of the surfaces is different. With this
construction we get the "outer\inner" partition
{e,f,g,h} {a,b,c,d}
The actual extrinsic distinction between these two arrangements is
clarified by considering the left-most and middle graphs of the
tetrahedron shown below.
Each edge can be regarded as a permutation of two elements, since
we must decide for each edge how to map the two sheets (inner and
outer) of one face to the two sheets of the adjoining face. In the
diagrams above we have used the Greek letters "alpha" and "beta"
to denote the identity permutation [12] and the transposition [21]
respectively. Now, for each vertex to be "flat", it is necessary
that we must circle around it through 6 equilateral triangles to
get back where we started, which means that a single circuit around
the vertex (as seen modulo the tetrahedron) must give a net
transposition, so we must have an odd number of [21] edges meeting
at each vertex.
Indeed, for the first arrangement (on the left above) there are two
edges carrying the [21] (beta) permutation, and exactly one of these
meets each vertex. On the other hand, for the second arrangement
(in the center above) we have swapped the outer\inner positions of
the congruent faces "b" and "g", which implies that every "alpha"
on the edge of this face becomes a "beta", and vice versa. This
still works, as we now have 3 transpositions meeting at one vertex,
and 1 transposition at each of the remaining vertices, i.e., we
still have an odd number of "beta's" meeting at each vertex. From
an intrinsic standpoint these two surfaces are equivalent, but they
are knotted differently in the embedding space. To see this, observe
that it's possible in the first arrangement to make a complete loop
around all the outer (or all the inner) faces without ever crossing
the same edge twice and always remaining on the same level. In
contrast, this is not possible with the second arrangement.
An exploded view of the tetratorus is shown below
Depending on which of the embeddings we choose, we can make the
appropriate identifications of edges so that the surface of the
two-layered tetrahedron has the same connectivity as the flat
torroidal surface. For example, to give the embedding corresponding
to the middle graph in the preceding schematics we would make the
natural connections of the outer layers to the outer base (yellow)
triangle, and then make transposing identifications along the
remaining three edges. In other words, connect g,e, and h with
f, and then make the connections gd,be and ga,bh and ae,dh.
One other possible embedding arrangement now immediately presents
itself. If we swap the outer\inner positions of the congruent faces
"c" and "f" in the middle graph above, we produce the right-hand
graph, which must have transpositions for ALL 6 of its edges. In
this case we have 3 "beta" edges meeting at each vertex, and this
embedding is obviously distinct from each of the other two, because
we can't travel from any face to any other face without switching
layers. Nevertheless, all three of these multi-layered tetrahedral
surfaces are intrinsically equivalent, not only to each other, but
also to the single-layered torus.
Recall that in order to make a surface that corresponds to both a
multi-layered plane and a multi-layered CUBE we needed the flat
arrangement to have 3 layers while the cubical arrangement had 4
layers. This was necessary in order for a circuit around a vertex
to be a multiple both of 2pi and of (3/2)pi. In contrast, to make
a tetratorus we found that the flat surface was single-layered and
the tetrahedron was 2-layered. In general it's not difficult to
see what the requirements are for the various regular polyhedra.
We have the basic attributes
vertices edges faces
-------- ----- -----
tetrahedron 4 6 4
octahedron 6 12 8
cube 8 12 6
icosahedron 12 30 20
dodecahedron 20 30 12
and the Euler characteristic is V-(E-F). Also, the number of vertices
doesn't change when we transform the surface from flat to curved, so
the total number of vertices will be Vj if we have j layers in the
flat interpretation. On the other hand, the numbers of edges and
faces is proportional to the number of layers we have in the polyhedral
interpretation. Letting k denote this number, we see that the
intrinsic Euler characteristic is Vj - (E-F)k, and we need to make
this vanish. The minimal solutions for each type of solid are
j k
--- ---
tetrahedron 1 2
octahedron 2 3
cube 3 4
icosahedron 5 6
dodecahedron 9 10
For example, here's an image of "flat" vertex where pentagons meet.
It's easy to see that these are the numbers of layers necessary, in
each case, to arrange around a single vertex so that it can be
interpreted as a multiple of a flat surface AND as a multiple of
the vertex of the respective polyhedron. It would be interesting to
know if complete closed surfaces equivalent to multi-layered tori
can be constructed in cases beyond the tetrahedron. In each case
it would be a matter of identifying the arrangement(s) of permutations
of k elements that would yield the appropriate layer changes for
each edge, in such a way that the entire closed surface "works" as
both a flat surface and (modulo the polyhedron) a curved surface.
This involves considerations that are closely related to those
discussed in the note on Permutation Loops.
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