Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading "A Combinatorial Introduction to Topology" by Michael Henle and have come across something.

Suppose we have a plane diagram for a surface. I don't know how to include a picture here, but these are the diagrams where we label the edges of a polygon and such. For instance, the torus is a square with the face labeled $A$ top an bottom edges labeled $a$ and left and right edges labeled $b$ and all the corners are the same vertex $v$. For $k=0,1,2$, the group of $k$-chains is the set of all subsets of $k$-simplices. For our friend the torus, $C_0=\{\emptyset, v\}$, $C_1=\{\emptyset, a,b,\{a\cup b\}\}$, and $C_2=\{\emptyset,A\}$. We add two $k$-chains by taking the symmetric difference as sets.

Now, our boundary operator, $\partial$, is to map a $k$-chain to the sum of $k-1$-chains with which it is incident an odd number of times. For the torus, every boundary is the constant function $\partial x=\emptyset$, because $A$ has incidence $2$ with both $a$ and $b$. And $a$ and $b$ meet with $v$ in two spots.

Finally, the homology groups are the kernel of $\partial$ mod the image, just like usual.

For the torus, or any orientable, closed, surface, the homology works out exactly as we'd like. We get $\mathbb{Z}/2$ for $H_0$ and $H_2$, and the correct number of $\mathbb{Z}/2$'s for $H_1$. My issue is with non-orientable surfaces. For instance $\mathbb{R}P^2$ has a square diagram, just like the torus, only you twist the edges before you identify them. Henle's construction would not take the twist into consideration and we'd end up with the same groups as the torus.

My first question is, what is the functor that Henle has created? My second question is, can we modify the construction to make non-orientable surfaces distinct from orientable surfaces?

In order to answer the second question, I have tried to find a way to change the definition of incidence. Like, maybe a twist adds another point of incidence with the face. Suppose now that $v$, $a$, $b$, and $A$ are the vertex, edges, and face of $\mathbb{R}P^2$. With the new definition of incidence, $\partial A=a+b$. This will give us the correct $\mathbb{Z}/2$ for $H_1$. Does this correct the issue? Or is there some hidden problem here?

EDIT: Actually, there seems to be a problem. If I draw a diagonal edge, $c$, to divide $A$ into two triangles $A_1$ and $A_2$, then each will have incidence $1$ with each edge $a$, $b$, and $c$, since no face meets a twisted edge more than once.

share|improve this question
    
@Tom Bachmann: My main issue is that for $\mathbb{R}P^2$ it doesn't give homology with $\mathbb{Z}/2$ coefficients. There should be only one copy of $\mathbb{Z}/2$. But, as it stands, Henle's definition will give two copies. –  Joe Johnson 126 Dec 1 '11 at 18:25
1  
Despite looking almost identical, the diagram for the torus has only 1 vertex and the diagram for $\mathbb RP^2$ has 2 vertices. Hence different homology. –  Grigory M Dec 1 '11 at 18:33
    
(My comment used to erroneously claim that $\mathbb{Z}/2$-homology was just obtained by tensoring with $\mathbb{Z}/2$ which is wrong in the presence of 2-torsion. Therefore I deleted it.) –  Tom Bachmann Dec 1 '11 at 18:48
    
@Grigory M: Ah, I was thinking of the diagram for the Klein bottle. Head cold make brain slow. –  Joe Johnson 126 Dec 2 '11 at 2:20
add comment

1 Answer

up vote 3 down vote accepted

(Just not to leave this unanswered.)

This functor is $H_n(-;\mathbb Z/2)$ already.

And yes, this functor can't tell the torus from the Klein bottle. I don't think it can be fixed without changing coefficients to $\mathbb Z$ (and introducing signs — that's why the author starts with $\mathbb Z/2$-homology, I think).

share|improve this answer
    
Thank you for clearing up my confusion and answering. It is indeed the homology and should not be able to tell the Klein bottle from the torus. Let us close this silly business immediately and allow it to be buried. –  Joe Johnson 126 Dec 2 '11 at 12:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.