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Given an element in the (first) homology group of a surface, I would like to know if it can be represented as a simple closed curve. For orientable surfaces, this is well-known, but I wasn't able to find a reference for non-orientable surfaces.

For orientable surfaces, the sphere with $g$ handles has homology $\mathbb{Z}^{2g}$, and an element $(a_1, \dots, a_{2g}) \in \mathbb{Z}^{2g}$ can be represented by a simple closed curve if and only if $gcd(a_1, \dots, a_{2g})=1$. This is classical and actually not too hard to prove.

For non-orientable surfaces, the sphere with $k$ crosscaps has homology $\mathbb{Z}_2 \times \mathbb{Z}^{k-1}$. Let $(a, b_1, \dots, b_{k-1}) \in \mathbb{Z}_2 \times \mathbb{Z}^{k-1}$. If $k$ is odd and $gcd(b_1, \dots, b_{k-1})=1$, then we can represent this element as a simple curve, regardless of the value of $a$ by the orientable case. But there are other homology classes not of this form which can be represented by simple curves. For example, I think $(0,2,0) \in \mathbb{Z}_2 \times \mathbb{Z}^2$ can be represented by a simple curve on the sphere with 3 crosscaps. Incidentally, while I am at it, I'd also like to know why it is common to use $\mathbb{Z}_2$-homology when working with non-orientable surfaces, as opposed to $\mathbb{Z}$-homology (which is what I am using).

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I don't know the answer to your first question, but people often use $\mathbb{Z}/2$-homology when working with nonorientable surfaces so that Poincare duality will be available. In particular, one wants a reasonable notion of an "algebraic intersection" pairing. –  Andy Putman Nov 1 '10 at 16:28
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(continued) For nonorientable surfaces, this is false. In fact, there are three "types" of non-nullhomologous SCC's on a nonorientable surface. Namely 2-sided curves, 1-sided curves whose complement is orientable, and 1-sided curves whose complement is not orientable. This will translate into three families of conditions on homology classes. –  Andy Putman Nov 1 '10 at 16:40
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@Andy: what about for even genus, e.g. the Klein bottle? –  Tony Huynh Nov 1 '10 at 16:58
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Whoops! You're right. This is evidence that I never think about nonorientable surfaces... –  Andy Putman Nov 1 '10 at 17:07
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Tony - FWIW, my guess is that this might be the sort of thing that is easier to work out for yourself than to hunt down a reference for. –  HJRW Nov 1 '10 at 17:31

2 Answers 2

up vote 7 down vote accepted

The complete answer follows from the result of the paper referenced below (as the math review points out the result was also obtained slightly earlier by McCarthy and Pinkall).

@article {MR2161731, AUTHOR = {Gadgil, Siddhartha and Pancholi, Dishant}, TITLE = {Homeomorphisms and the homology of non-orientable surfaces}, JOURNAL = {Proc. Indian Acad. Sci. Math. Sci.}, FJOURNAL = {Indian Academy of Sciences. Proceedings. Mathematical Sciences}, VOLUME = {115}, YEAR = {2005}, NUMBER = {3}, PAGES = {251--257}, ISSN = {0253-4142}, MRCLASS = {57M60 (20F38 57N05)}, MRNUMBER = {2161731 (2006f:57019)}, MRREVIEWER = {Mustafa Korkmaz}, DOI = {10.1007/BF02829656}, URL = {http://dx.doi.org/10.1007/BF02829656}, }

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Igor, how in heck did you know about this? –  Deane Yang Dec 31 '10 at 4:37
    
Thanks a lot Igor! Sorry for the late acceptance, I didn't get a chance to look at the paper until now. –  Tony Huynh Apr 27 '11 at 17:07
    
@Tony: no problem... –  Igor Rivin Apr 27 '11 at 17:32

(I found this MO thread while searching for a reference similar to the one requested by the OP, so I'm adding the following reference in case it is helpful to other future visitors)

Representing codimension one homology classes on closed nonorientable manifolds by submanifolds: William H. Meeks, III. Illinois Journal of Mathematics, Volume 23, Number 2, June 1979

In dimension two the theorem in this paper implies that an integer homology class on a connected closed nonorientable surface can be represented by an embedded circle if and only if the class is primitive or twice a primitive class.

(A class in $H_1(M,Z)$ is called primitive if the induced class in $H_1(M, Z)/\text{Torsion}$ is the zero class or is not a nontrivial multiple of any other class.)

In a further paper (Representing homology classes by embedded circles on a compact surface) Meeks appears to be claiming that the same result also holds for nonorientable surfaces with boundary, but the referenced paper is listed as a preprint and does not appear to have been actually been published (it doesn't, for instance, appear on the list of publications in his CV)

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Thanks alot! Is it true that for non-orientable classes, primitive classes are represented by non-separating circles and twice primitive classes are represented by separating circles? –  Tony Huynh Apr 24 '12 at 5:34

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