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Let $\Sigma$ be a compact oriented surface with boundary. Assume that the genus of $\Sigma$ is positive. We say that an element $h \in H_1(\Sigma)$ can be realized by a simple closed curve if there exists an oriented simple closed curve $\gamma$ on $\Sigma$ such that $[\gamma] = h$.

If $\Sigma$ has $0$ or $1$ boundary components, then $h \in H_1(\Sigma)$ can be realized by a simple closed curve if and only if $h$ is primitive, that is, if we cannot write $h = n \cdot h'$ for some $n \in \mathbb{Z}$ and $h' \in H_1(\Sigma)$ with $n > 1$. This is a standard fact; for instance, it is contained in Farb and Margalit's Primer on Mapping Class Groups.

This brings me to my question : if $\Sigma$ has more than $1$ boundary component, then what elements of $H_1(\Sigma)$ can be realized by simple closed curves?

One might guess that the answer is still the primitive elements. However, this guess is wrong. Indeed, assume that $\Sigma$ has at least $2$ boundary components. Let $\delta$ be an oriented simple closed nonseparating curve in the interior of $\Sigma$ and let $b$ be one of the boundary components of $\Sigma$. Observe that $[b] \neq 0$, and hence that $2[\delta]+[b]$ is a primitive element of $H_1(\Sigma)$. Assume that $\gamma$ is an oriented simple closed curve in $\Sigma$ such that $[\gamma] = 2[\delta]+[b]$. Let $S$ be the surface obtained by gluing discs to all the boundary components of $\Sigma$. There is then an inclusion map $i : \Sigma \hookrightarrow S$, and we have $$[i(\gamma)] = 2[i(\delta)] + [i(b)] = 2[i(\delta)],$$ a contradiction.

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    $\begingroup$ Someone downvoted this. I assume they did so because they skimmed Ryan's answer and assumed that the question is trivial. However, if they read my comments they would learn that his argument is (at best) incomplete : all it shows is that if $h \in H_1(\Sigma)$ is such that $h$ maps to a primitive element in the capped-off surface, then there is a simple closed curve $\gamma$ on $\Sigma$ such that $[\gamma] =h+b$ for some $b$ in the subspace of $H_1(\Sigma)$ spanned by the boundary components. The heart of the question is which $b$ can occur, and Ryan's argument says nothing about this. $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 20:25
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    $\begingroup$ IMO it is the general tone of the interactions of the OP here that are quite downvotable :-( $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 3, 2014 at 0:10
  • $\begingroup$ @MarianoSuárez-Alvarez it is bad enough that the OP is asking other people to do his research, but being obnoxious about it seems really beyond the pale. $\endgroup$
    – Igor Rivin
    Commented Jan 3, 2014 at 0:43
  • $\begingroup$ I can but commend you two for your gentlemanship. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jan 3, 2014 at 0:44

2 Answers 2

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This is related to a nontrivial question, address in this paper of Chas and Krongold (there are other related papers of Moira Chas with Fabiana Krongold and Dennis Sullivan, which a google search will bring up).

The original question, however, is trivial, since if we take some curve (think of it as a hyperbolic geodesic) realizing the homology class, we can perform a surgery on each crossing, which removes it, and possibly disconnects the curve, so eventually we will have a multicurve realizing the homology class. Some components of this multicurve will be boundary-parallel. from this multicurve it is pretty easy to see when the class is realizable (unless I am confused, which is quite possible, you need to be realizable in the cupped-off surface, plus something that is not a multiple of a boundary component).

EDIT Firstly, the OP is apparently trying to win friends and influence people for downvoting my answer and Ryan's. Not cool at all.

Secondly, if you want a different answer, knock yourself out (notice that he is solving a more general, thus harder, problem):

MR2335737 (2008e:57015) Reviewed Granda, Larry M.(1-STL) Representing homology classes of a surface by disjoint curves. (English summary) Houston J. Math. 33 (2007), no. 3, 807–813. 57M50 (57M20 57N05)

A more extensive discussion of the same problem solved in Granda's paper (without, however, a complete answer) is given by Allen Edmonds in:

Edmonds, Allan L.(1-IN) Systems of curves on a closed orientable surface. Enseign. Math. (2) 42 (1996), no. 3-4, 311–339.

Another edit

The best reference is W. Meeks and J. Patrusky, where Theorem 1 is: enter link description here

For the link-challenged, it says that a class can be represented by simple closed curve if and only if it is primitive in the capped-off surface OR it is a sum of (some of the) boundary curves, which is what Ryan and I have been saying.

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    $\begingroup$ @RyanBudney : How? Let $\Sigma$ be a surface with multiple boundary components, let $S$ be the capped-off surface, and let $\pi : H_1(\Sigma) \rightarrow H_1(S)$ be the map induced by the inclusion $\Sigma \hookrightarrow S$. Please tell me how to write down a map $H_1(S) \rightarrow H_1(\Sigma)$ splitting $\pi$ that does not depend on some choices. Note that this mystery map does not come from a map $S \rightarrow \Sigma$ splitting the inclusion (maybe up to homotopy or something) since no such map exists... $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 21:51
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    $\begingroup$ I'm using the word natural in the usual mathematical sense : independent of arbitrary choices (in particular, in this case I mean independent of a choice of basis or something equivalent to a basis). What, pray tell, do you think it should mean? $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 22:03
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    $\begingroup$ What sort of answer do you expect? You are asking questions about a surface with boundary, and you expect the answer to not mention the boundary??? $\endgroup$
    – Igor Rivin
    Commented Jan 2, 2014 at 22:08
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    $\begingroup$ @IgorRivin : There is essentially a natural basis for the boundary subspace $B$ (given by the boundary components; this isn't quite true because they sum to $0$, but that is not all that important). Certainly the answer will have something to do with that. But there is no natural choice of splitting $H_1(\Sigma) = B \oplus H_1(S)$. The thing that is natural is the short exact sequence $0 \rightarrow B \rightarrow H_1(\Sigma) \rightarrow H_1(S) \rightarrow 0$. It splits only because $H_1(S)$ is a free abelian group; there is no canonical choice of such a splitting. Since "realizable by $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 22:13
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    $\begingroup$ @Roger see the edit, but being a jerk and downvoting people will not get you any other answers. $\endgroup$
    – Igor Rivin
    Commented Jan 2, 2014 at 22:57
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The case with boundary quickly reduces to the case without boundary. The idea is to cap off your surface with discs to create a closed surface. The map on homology is not injective. But a homology class is realizable whenever it is realizable in the capped-off surface. So you get a fairly simple answer:

A homology class in $\Sigma$ is realizable if and only if either 1) it is realizable in the capped-off surface or 2) it is a sum of boundary classes (coherently oriented) together with perhaps a realizable class in the capped-off surface.

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  • $\begingroup$ I'm not sure I follow your answer. Assume that $S$ be the result of capping off all of the boundary components of $\Sigma$ and let $i : \Sigma \hookrightarrow S$ be the inclusion map. Are you claiming that an element $h \in H_1(\Sigma)$ can be realized by a simple closed curve if and only if $i_{\ast}(h)$ can be so realized? I don't see at all how to do this; how do you "add" arbitrary boundary component homology classes to a curve? Or are you claiming that we can write $H_1(\Sigma) = H_1(S) \oplus B$ for $B$ the boundary classes and then giving a condition with respect to this $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 19:58
  • $\begingroup$ (continued) decomposition? This also sound really fishy to me since while we can write $H_1(\Sigma) = H_1(S) \oplus B$, there is no natural such decomposition. Any description of which homology classes can be realized would have to be independent of the choice of decomposition, but your description seems to depend on it. $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 19:59
  • $\begingroup$ Yes, a curve is realizable in $\Sigma$ if and only if it is realizable in the capped-off surface. This is transversality -- you push the curve off the discs you use to cap-off the surface. $\endgroup$
    – Ryan Budney
    Commented Jan 2, 2014 at 20:02
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    $\begingroup$ That can't be right since it says nothing about the contribution to homology from the boundary. It only proves that if $i_{\ast}(h)$ is primitive, then we can find a simple closed curve $\gamma$ with $[\gamma] = h + b$ for some $b$ in the subspace of homology spanned by the boundary components. $\endgroup$
    – Roger
    Commented Jan 2, 2014 at 20:06
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    $\begingroup$ I second this answer. A nontrivial class in the capped-off surface is a meridian of a handle; the rest (handling the holes) is an easy exercise: bring them to the meridian, one by one, "from the right side". A trivial class, if realizable, cuts the original surface, hence is a sum of boundary classes. Speaking about refs, I doubt a thing like this can be published as a "big theorem" somewhere. $\endgroup$
    – Alex Degtyarev
    Commented Jan 2, 2014 at 23:22

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