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Topic: Triangulation torus problem
Replies: 5   Last Post: May 13, 2005 2:23 PM

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Posts: 806
Registered: 12/6/04
Re: Triangulation torus problem
Posted: May 7, 2005 4:17 AM
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Carl R. wrote:
> Hi, thanks to Robin for his reply, which I didn't
> understand quite well, so here's my question again:
> I'm trying to understand the concept of triangulation
> of a topological space.
> So far I've seen how to triangulate a torus with 10
> vertices, I think I understand this procedure, you
> just identity the opposite edges of the square in the
> same direction.
> What I _dont_ understand at all is how to find
> a triangulation of the torus with _seven_ vertices.
> I know how the picture looks like since the author has
> already solved this problem, but why I don't understand
> is how to construct it, i.e how did he came up with such
> picture.

First of all, you start off with a "brick wall" of bricks (of size 2x6)
which is 7-colored in a regular manner. (You may want to print this
out, and you should make sure it's in a fixed-width font like Courier.)


Note that the colors appear "regularly": The colors below a brick that
has been colored 7 are 4 and 5, for instance.

Now fill in the 1 in the upper left-hand corner of each "brick" with a
*, so that you get:


Now comes the part I can't do in plain text 8-). Connect the *'s with
straight lines to get a parallelogram. Remove everything from the
diagram outside of that parallelogram, and draw lines between the
different bricks; i.e., whereever you see


draw a line around this region.

You should now have a parallelogram consisting of regions, each colored
with a number between 1 and 7. If you identify the opposite sides of
the parallelogram (just like you would do for a square) to get a torus
(verify this!), the pieces of the bricks on opposite sides of the
original parallelogram will merge together to form bricks which are 2x6
in size.

Each of these bricks is one of 7 regions on the torus, and every region
is adjacent to every other region. (Verify this on your own!)

To get a triangulation of the torus, choose a vertex in each region,
and connect every dot with every other one, so that the edge between
regions A and B only crosses one edge, the one immediately between the
two regions; this is the usual "dual" construction, done in the manner
of planar graphs.

((BTW, the final graph proves that you will possibly need 7 colors to
color a map (or a graph) on the torus. This is sufficient because if
you have a graph embedded on the torus, there will always be a vertex
of degree <= 6, which you can "color last"; this latter fact follows
from the fact that the Euler characteristic of the torus is 0, and the
v - e + f = 0.))

> I have some questions:
> 1) How do you know from where to where connect the lines
> from the distinct vertices??
> 2) What determines to join them in such a way?

Given above.

> 3) Once you outlined the construction of the simplex how do you prove
> that the underlying space is actually homeomorphic to the torus?

If you do the construction which takes a square to the torus, and do
the same thing to a parallelogram, you get the same shape

> For instance in the case of 10 vertices, you construct a natural
> triangulation of the unit square I^2 and then extend linearly each
> vertex to the vertices of the torus, which is clear.
> 4) But for this example (that is the triangulation of the torus
> with seven vertices) how do you know that such triangulation does in
> fact works?

Proof by construction.

> 5) What do you have to check to know that it is in fact a
> triangulation??

Notice in the original coloring of the plane (as "bricks"), you will
always have 3 bricks meeting at a point. When you find the dual graph,
this fact will turn into a facial triangle.

> The book says "each simplex is uniquely determined
> by its vertices" I don't really understand what does
> this means. What are the steps to check that your
> triangulation really works?

In the plane, if you have 3 points P1, P2, P3, and pairwise connect
them, you will get the same shape (in a topological sense), no matter
how you draw the lines. That is, you will have edges for P1P2, P1P3,
P2P3, and the inside of this curve will represent the "triangle"

> Finally what is the usual approach to construct a triangulation for a
> given space?

There's a way to get the double torus by identifying certain sides of
an octogon P1P2P3P4P5P6P7P8 in the plane. If you start off with a
triangulation T of this octogon (i.e., you have the edges P1P2, P2P3,
P3P4, etc. in T, and every face of T inside of these paths is a
triangle), and let the edges follow along with the identification of
the edges, it will induce a triangulation of the double torus.

This idea can be used along with the "usual constructions" of the
2-manifolds, whether they're oriented or not.

--- Christopher Heckman

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