Mobius Strip and Homology Groups
I've been reading Croom, Basic concepts of Algebraic Topology.
I have a question.
What is the minimum number of vertices in a triangulation of Mobius strip?
Theorem 2.8 says:
$\alpha_0 \geq \frac{1}{2} (7+ \sqrt{49 – 24 \chi(M)})$
I know that Homology Groups of the strip is:
$H_2(M) = {0}$, $H_1(M) = \mathbb{Z}$, $H_0(M) = \mathbb{Z}$. So the sum for the 2-manifold is:
$\chi(M) = \sum_{p=0}^{p=2} R_p(M) = + 1 – 1 + 0 = 0$ where $R_p(M)$ is the Betti number which is defined as the rank of the free part in the decomposition of the homology group for a particular p.
Substituting back into 2.8 tell me I need at least 7 vertices. My confusion is, throughout the book it uses the following triangulation which has only 6 vertices:
So my conclusion is, either $\chi(M)$ is 1 not 0 in order to obey 2.8 or the above triangulation is false. Maybe 2.8 is not applicable to non-orientable n-manifolds?
Thanks in advance.
evrim.
Best Answer
The result you are referring to (Theorem 2.8) holds for $n-$pseudomanifolds, while the Mobius strip has a boundary (look at Exercise 24 for the definition of $n-$pseudomanifold with boundary). The point is that the 1-simplices of the boundary are a face of just one 2-simplex (in contrast with point (b) of the definition of $n-$pseudomanifold).
Moreover in exercise 7 there is an example of triangulation of the cylinder that is the analogue of the one you are using for the Mobius strip.