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The classification theorem for surfaces says that the complete set of homeomorphism classes of surfaces is

{ $S_g : g \geq 0$ } $ \cup$ { $N_k : k \geq 1$ },

where $S_g$ is a sphere with $g$ handles, and $N_k$ is a sphere with $k$ crosscaps. The first homology groups are easy to compute. They are $H_1 (S_g) = \mathbb{Z}^{2g}$, and $H_1 (N_k)=\mathbb{Z}^{k-1} \times \mathbb{Z} / 2\mathbb{Z}$. My question concerns how the homology groups change once we start cutting holes in our surface.

In the orientable case, it is easy to see what happens. The first hole that we cut out does not change the homology. Every additional hole then introduces another factor of $\mathbb{Z}$.

In the non-orientable case something peculiar happens. Consider the projective plane with homology group $\mathbb{Z} / 2 \mathbb{Z}$. If I cut out a hole, then I get the Mobius strip, which has homology group $\mathbb{Z}$ (it is homotopic to a circle). In general, if I cut a hole out of $N_k$, then in the homology group I lose a factor of $\mathbb{Z} / 2 \mathbb{Z}$, and introduce a factor of $\mathbb{Z}$. Each additional hole will then just introduce another factor of $\mathbb{Z}$.

My question: In the non-orientable case what happened to the factor of $\mathbb{Z} / 2 \mathbb{Z}$? Is there a nice geometric explanation of why it went away? I'm slightly disturbed because I had the intuition that torsion was supposed to record non-orientability, but I guess this doesn't work for surfaces with holes.

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3 Answers 3

up vote 5 down vote accepted

Richard's answer captures the essence why there is a $\mathbb{Z}_2$ is in the first homology of a closed nonorientable surface.

If you remove a disk from a closed surface, the resulting object has a $1$-dimensional $CW$-complex as a strong deformation retract, so that the homology of the resulting object has no torsion.

A closed nonorientable surface is always the result of the connect sum of an orientable surface with a projective plane or two projective planes. That is you are gluing one Moebius band, or two Moebius bands into the boundary of an orientable surface. The core or cores of the Moebius bands, oriented appropriately represent the generator of the $2$-torsion in the first homology of the surface.

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Thanks Charlie. Your second paragraph was basically the answer I was looking for. –  Tony Huynh Mar 20 '10 at 17:27

Think of it in the other direction:

When you attach a disk to the boundary of the Mobius band, you get a map from $\pi_1$ of the Mobius band to $\pi_1$ of the projective plane, which is just the quotient map $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$.

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In RP2, you can literally draw a line on the surface twice around and upon sliding parts of this line across the surface, the line can shrink to a dot - thus, [2] = [0].

However, if you introduce a hole in the right place before sliding, then it's possible for the line to get 'stuck' and thus form a loop around this hole, so [2] now becomes its own equivalence class. Of course it might still be possible to slide the line so that it can shrink to a dot, granted the hole is elsewhere, but this then is simply [0] and still different from [2].

In other words, the hole can cause multiple loops like any hole in any surface, regardless of what the rest of that surface looks like.

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