The Euler characteristic (triangulations 2)

I want to talk about some things I saw when I was looking at triangulations. This also continues my reading in Bloch.

The major thing I saw was the Euler characteristic. We usually define it for polyhedra, as the alternating sum / difference of the number of vertices, edges, and faces (0-, 1- and 2- simplices)…

\chi = V - E + F

Then we would define it for a surface by taking the Euler characteristic of any polyhedron which is homeomorphic to that surface.

For that to be well-defined, of course, requires that all polyhedra which are homeomorphic to a surface S have the same Euler characteristic (as they do).

By the same token, we could define the Euler characteristic of a surface from any triangulation of the surface, after we show that all triangulations of a surface have the same Euler characteristic. Oh, and we’d better actually prove that every topological surface (every topological 2-manifold) can be triangulated.

Yes, they can be. Not true for topological 4-manifolds, and I think it’s still wide open for higher dimensions. In contrast, I believe that every differentiable n-manifold supports a unique piecewise linear (PL) structure (which is the generalization, they tell me, of triangulations).

Details.

So we have the Euler characteristic in one hand. With the other hand, let’s grab the classification theorem. One form said that every compact connected surface is a connected sum of tori T or of projective planes P:

gT, g = 0,1,2,3…

or

gP, g = 1,2,3,…

(where the sphere is 0T, the “torus” with no holes; 2T = T#T, etc., and the integer g is called the genus of the surface.)

Is there any chance we can easily compute the Euler characteristic of the connected sum A#B from the Euler characteristics of A and B? If we could, then we could compute the Euler characteristic of nT and nP, hence of all connected compact surfaces.

Of course we can, and here’s the equation:

\chi \left( A \#B \right) = \chi \left( A \right) + \chi \left( B \right) - 2

Oh, now go get the Euler characteristics of triangulations of the sphere, torus, and projective plane.

Ok, we can do that.

But it would be simpler to get the Euler characteristic from the polygonal disks from which we constructed the basic surfaces. We would have to know that counting the v, e, f (after identifications) of the polygonal disk gives us the same answer, but it does.

Yes, having specifically studied Bloch because he distinguished topological surfaces from simplicial surfaces, I am now effectively going to compute the Euler characteristics of topological surfaces instead of simplicial surfaces.

I never said I didn’t want to mix it all together; I just wanted to see the separate ingredients that went into the stew.

In principal we only need 3 surfaces: sphere, torus, and projective plane. Nevertheless, since the Klein bottle has an equally simple polygonal disk, I’ll show it too. One set of representations is


Edges which are to be identified (“glued”) have the same symbol; an arrow shows the directions which are used. Points which are to be identified have the same color. I considered using colors for the edges, but we can actually do algebra with the symbols: the sphere can be called a\ b\ b^{-1}\ a^{-1}\ , the torus a\ b\ a^{-1}\ b^{-1}\ , etc.) If it helps, the colors of the vertices are a consequence of the identifications of the edges.

From them, we compute that the Euler characteristics are

Note that the largest Euler characteristic is 2, and it corresponds to a sphere. Note that all four are non-negative. Note that an Euler characteristic of 1 corresponds to the (non-orientable) projective plane. Then for Euler characteristic of 0, we have two surfaces, one orientable, one not.

Then we compute that the Euler characteristic of gT is 2 – 2g, i.e. the set of even numbers less than or equal to 2. And the Euler characteristic of gP is 2 – g, i.e. the set of integers less than or equal to 1. (Maybe I should restrict g > 0 for nP, but I’m not sure it’s worth worrying about.) So if we have an odd negative Euler characteristic, the surface is unique and non-orientable; if the Euler characteristic is even and negative, there are two surfaces, one orientable, one not.

Note that all the new surfaces we construct have Euler characteristic less than zero.

Note that the connected sum of a sphere and any surface A has the same Euler characteristic as A. in fact, the connected sum is homeomorphic to A; gluing spheres to surfaces does not change them.

We know, however, another form of the classification theorem. The non-orientable surfaces gP can be written as a connected sum of tori plus either one projective plane or one Klein bottle:

mT#P

or

mT#K

for some m (and I am paying no attention to the relationships between these m’s and the genus g). To prove that, we need to know that

P # P = K

and

P # P # P = P # T.

We can now compute that the Euler characteristic of mT#P is (2 – 2m) + 1 – 2 = 1 – 2m, i.e. the set of odd integers less than or equal to 1.

The Euler characteristic of mT#K is (2-2m) + 0 – 2 = -2m, i.e. the set of even integers less than or equal to 0.

So, in fact, if the Euler characteristic is an odd negative, there is a unique surface of the form mT#P; if the Euler characteristic is even negative, there is an orientable surface of the form nT and a non-orientable surface of the form mT#K.

References for this material are

  • Bloch, “A First Course in Geometric Topology & Differential Geometry”.
  • Lee, “Introduction to Topological Manifolds”.
  • Firby & Gardiner, “Surface Topology”.
  • Massey, “Algebraic Topology: An Introduction” or “A Basic Course in Algbraic Topology”

See the bibliography.

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3 Responses to “The Euler characteristic (triangulations 2)”

  1. cutelyaware Says:

    I’d not found much before connecting orientability and Euler characteristics, so this is beautiful. Thank you!

    Now my question is, given a triangulation of a surface, how can you determine whether that surface is orientable?

  2. Math Shirt | Truncated Icosahedron | The Nerdiest Shirts Says:

    […] our Truncated Icosahedron Math Shirt, we have chosen to show the Euler characteristic equation chi=vertices-edges+faces=2. The surface of any convex polyhedron is a sphere, hence will […]


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