Choosing a way to represent a 2-manifold

Suppose you want to portray a 2-manifold on a flat piece of paper – how can you find a representation with properties that might be suitable for your purpose? This page provides some suggestions.

These pages use two methods: a suitably-labelled polygon, and a "tunnelled" polyhedron. We will deal with these in turn.

• labelled polygons
• tunnelled polyhedra
• how to choose a suitable representation

An example of a labelled polygon is shown to the right. In choosing a polygon, there are three decisions to make:

• how many sides for the polygon
• how to label the edges
• how to arrange the arrows

Number of sides

This must be even (each side must be paired up with one other side). Any positive even number may have its uses.

How to arrange the labels

A symmetrical arrangement is likely to yield better results. Some possibilities are:

• two full cycles: abcde... abcde... .
• "commutator style": aabb ccdd eeff ...   
• A "skip-2 chain": axbacbdced ...         
• A "skip-4 chain": abcdeafcgehf ...       

How to arrange the arrows

There are three ways to do this:

• "parallel"
• "anti-parallel"
• mixed

"Anti-parallel" means that for each pair of edges with the same label, the two arrows are anti-parallel – they both run clockwise (or counterclockwise) around the outside of the polygon. This produces an non-orientable manifold. I recommend, for the sake of simplicity, having all the arrows run the same way.

Mixed involves some mix of parallel and anti-parallel. I recommend avoiding this, as it is hard to handle, and lacking in symmetry. However, it seems that sometimes there is no way to avoid using a "mixed" arrangement, if we want to draw the map on a polygon; see for example C⁴:{6;4}. I suggest a convention that in a mixed arrangement of arrows, the parallel arrowheads be drawn differently from the antiparallel ones.

What manifold does an arrangement give?

Once we have a labelled arrowed polygon, we will want to know what manifold its represents – or better, we can choose an appropriate labelled arrowed polygon for the manifold that we are interested in. We can find which manifold the polygon represents by triangulating it. This is most easily done by placing one vertex in its centre and connecting it by a half-edge to each side of the polygon, counting the resulting faces, and calculating its Euler number. This, with the fact of whether the manifold is orientable (it is orientable iff the arrows are all arranged parallel), is enough to establish what the manifold is.

To save you the trouble of triangulating polygons yourself, the table below lists some polygons. The black values are the genera of orientable manifolds, the red ones, of non-orientable manifolds.

No. of sides	2	4	6	8	10	12	14	16	18	20
two full cycles	S0 C1	S1 C1	S1 C1	S2 C1	S2 C1	S3 C1	S3 C1	S4 C1	S4 C1	S5 C1
commutator-style		S1 C1		S2 C2		S3 C3		S4 C4		S5 C5
skip-2 chain			S1 C1	S1 C4	S2 C5	S2 C4	S3 C7	S3 C8	S4 C7	S4 C10
skip-4 chain					S2 C1	S1 C6	S3 C7	S3 C8	S3 C9	S4 C6

The table below repeats the information in the table above, at greater length, and with the addition of a few irregular labellings.

An example of a tunnelled polyhedron is shown to the right. It is a cube with three tunnels.

If you want an orientable manifold of genus n, a simple way is to find a genus-0 orientable polyhedron (i.e. a Platonic solid) with at least 2n faces, and add n tunnels, each connecting a pair of faces, in a way that preserves most of the symmetry. However this only works for n=6, 8, 12, and 20.

More generally, if you want an orientable manifold of genus n, find a genus-m (m<n) orientable regular map with at least 2(n–m) faces, and add n–m tunnels. This can be done most symmetrically if the polyhedron has exactly 2(n–m) faces, in antipodal pairs – the tunnels can then connect the centres of antipodal faces. However, for polyhedra with more than ten faces, it may be better to sacrifice some symmetry and uses pairs of faces that are not antipodal.

Once we have drawn the regular map with its tunnels, we need to put labels around each tunnel-mouth. If the resulting manifold is to be orientable, the labels must go in opposite directions around the two ends of each tunnel. But if one tunnel-mouth surrounds the entire diagram (as in the example to the right and the two to the left) this means that the labels actually go in the same direction, because that tunnel-mouth is inside-out.

However, while we have no choice about which directions the labels go in around the tunnel mouths, we do have choices about exactly where the labels go: there is freedom in how much "twist" to put in each tunnel. I recommend not actually placing the labels and committing yourself to the amount of twist at this stage. Instead, draw your regular map, Cayley graph, or whatever on the diagram, and decide as late as possible how much twist to specify by the placement of the labels.

The two diagrams to the left differ only in that one has all its tunnels twisted through π relative to the other. They show different maps in the orientable genus-3 manifold. The upper one is a regular map, the lower one is not.

This must depend on what you want to do with it. All I can offer is a couple of hints.

• The group PSL(2,7) has S4 as a subgroup. The cube has S4 as its rotational symmetry group. So, the Cayley diagram for PSL(2,7) (which has genus 3) fits well on a manifold diagram formed from a cube with three tunnels.
• The rather dull regular map G3:{12,12} has one vertex and six edges. If we try to draw this in a regular-looking way, we find that each of the six edges leaves and re-enters the edge of the diagram, so it is appropriate to use a dodecagon. The arrows must be parallel for this orientable manifold. The table above shows that if we want a dodecagon of genus 3, "two full cycles" will work, and that is what I have used at S3:{12,12}.