## A Rickety Bridge between Tilings and Homotopy

If you map a circle into a torus it takes only two integers to identify
which homotopy equivalence class the map belongs to—how many times the long way and how many times the short way.
If you map
the circle into a two holed torus then it takes much more information.
If the two tori are of similar size then a circle whose image is “long
enough to go around 10 times”, is long enough to belong to any of about
10^{2} homotopy sets when mapped into a torus, but into any of about
e^{10} sets when mapped into a TH (two holed) torus.
This is because
the circle image can thread the first hole alternating with the second
hole in many combinations of ways that are not homotopically equivalent.
We can cut the ordinary torus with two cuts both ends of each of which
meet at a point.
We can then flatten the torus onto a square in the plane.
A neighborhood of the point where the cuts met is cut into four parts which
end up in the corners of the square.
We can extend the pattern in the plane
from the square with more squares and define a map from the plane back
into the torus which includes the inverse of out flattening map and maps
each new square back onto the torus.
Squares will meet four at a time at
their corners.
This inverse map restores the continuity that was lost when
the torus was cut.

The TH torus can be cut and unrolled onto an octagonal
figure in the (hyperbolic) plane.
Each of the four cuts terminate at
both end at one point on the TH torus.
A neighborhood of this point is
thus split into eight regions, each destined to a different corner of the
octagon.
A similar extension can be made in the hyperbolic plane with new
eight sided figures patched together as were the squares above.
This is
hard to imagine but this is tiling {8, 8} as described here
and drawn here.

More commentary & Images

Now imagine a circle mapped into a TH torus.
Unroll that TH torus and
follow the image of the circle.
It will, in general cross the cuts into
adjacent octagonal patches and eventually end up at a similar point in
a different patch.
If we take a different circle image that is homotopic
to the first and do a similar map we have an arc on the plane connecting
two other points in the hyperbolic plane.
The collection of all homotopic
circle images thus constitutes a mapping of the hyperbolic plane onto itself.
This mapping is a congruence if we have made the octagonal patches congruent.

There are two groups now, the homotopy group and the translation group
in the hyperbolic plane.
Are they related?

Relying on the simple torus case as an example, I think that the following
is the case:

Consider the group of motions of the tiled plane that takes tiles to
tiles.
There is a subgroup that is homomorphic to the (circle) homotopy
group of the corresponding torus, if any.
There is also a subgroup which
is homomorphic to the homotopies of ...

I suspend that thought while I follow another lead.
How many ways can a simple torus T map into itself? I think that if I know the image of two generating circles on the torus then I know the T to T map within a homotopy.