I am looking for a practical description of the first homology group of $S_{g,b}$, the connected compact surface of genus $g$ with $b\geq 1$ boundary components. I think of $S_{g,b}$ as the $g$-holed torus with $b$ discs removed.

I know that $H_1(S_{g,b},\mathbb{Z})$ is of rank $2g+b-1$ and I'm trying to find out good generators.

We first get the $2$ cycles around each hole : the longitudinal and latitudinal loops. Then, I need to choose $b-1$ other cycles.

If I fix a closed loop around each removed disc and then I choose $b-1$ of them, does it complete my list of generators ?

  • $\begingroup$ If I recall correctly, you need all $b$. If you label the holes $H_1,\ldots,H_n$, you should choose a loop round each $H_i$, and a loop passing through $H_i$ and $H_{i+1}$ for $i=1,\ldots,n-1$, which gives you $2g-1$ loops. This is a comment rather than an answer because I don't remember at all how to prove it, but I'll make it an answer if people think I should. $\endgroup$ – mdp Aug 23 '12 at 15:45
  • $\begingroup$ Oops - in my last comment, $n=g$. $\endgroup$ – mdp Aug 23 '12 at 15:59
  • $\begingroup$ I think this is equivalent to my description. $\endgroup$ – Bebop Aug 23 '12 at 16:02
  • $\begingroup$ Sorry, I miss clicked... Correct me if I am wrong, if I denote by $a$ (resp. $b$) the loop around $H_i$ (resp. $H_{i+1}$) and by $c$ the one passing through $H_i$ and $H_{i+1}$, isn't there a relation between $a,b$ and $c$ ? $\endgroup$ – Bebop Aug 23 '12 at 16:12
  • $\begingroup$ I don't think so, although I'm by no means expert in this area. I'm remembering a picture in the case $b=0$ (although I'm now wondering if that was $\pi_1$ rather than $H_1$). $\endgroup$ – mdp Aug 23 '12 at 16:14

The short answer is your generators are correct. I'm not quite sure how to draw images on this site yet (my first post!), so I'll try to explain in just words. Perhaps you could draw your own pictures. My apologies if this is long-winded, but I am trying to provide you with your sought-after practical description.

First, let's think about a genus $g$-surface, which we imagine as a $2g$-gon with edges cyclically labelled $a_1, b_1, a_1^{-1}, b_1^{-1},\dots, a_g, b_g, a_g^{-1}, b_g^{-1}$. All this means is that we can use this labeling to glue the edges of the $2g$-gon to get our genus $g$-surface. I'll assume that you know how to get the homology of this closed surface, and know that it is the rank $2g$ free abelian group generated by $a_1,b_1,\dots,a_g,b_g$.

So what happens to this picture if we add a single boundary component? To do this, we can remove a small open disk $D_1$ from the interior of our labelled $2g$-gon to get a surface $\Sigma_{g,1},$ where $\partial\Sigma_{g,1}=\partial D_1=:B_1\simeq S^1$. As it turns out and as you are well aware, in this case for $b=1$, the first homology doesn't change; that is, $H_1(\Sigma_{g,0})=H_1(\Sigma_{g,1})$, and their generators are the same. Let's see why.

Note that our $2g$-gon with a disc removed deformation retracts to the labeled edges of the polygon. Since nothing weird is going on at the edges (i.e. the deformation retract is the identity here), we can push this deformation retract forward to the glued quotient space. Thus, we get a deformation retract $r$ from $\Sigma_{g,1}$ to a wedge of $2g$ circles $\bigvee_{i=1}^{2g}S^1$. Since a deformation retract is a homotopy equivalence we have another map $$i:\bigvee_{i=1}^{2g}S^1\rightarrow\Sigma_{g,1}$$ such that $$i\circ r\simeq\mathbb{1}_{\Sigma_{g,1}} \text{and} r\circ i\simeq\mathbb{1}_{\bigvee_{i=1}^{2g}S^1}.$$ In fact, if we identify $\bigvee_{i=1}^{2g}S^1$ with the image of our surface after the deformation retract, we can take $i$ to be the inclusion map, in which case we know that the generators of $H_1(\bigvee_{i=1}^{2g}S^1)$ are precisely the generators of $H_1(\Sigma_{g,1}),$ namely $a_1,\dots,b_g.$

Trying to generalize this deformation retract approach to $b>1$ is a tad difficult, so I'll explain it intuitively first. Start with our labelled $2g$-gon and draw $b$ circles inside. Start blowing up one of the circles like a balloon, without letting it cross itself. We see that the farthest the balloon can go is to completely cover the edges of the polygon and the $b-1$ other boundary components, in the process introducing some new edges between the boundary components and the edges of the polygon. Of course, as before, nothing weird is going on at the edges of the polygon, so we're really doing all of this on the glued-up surface. The resulting image of the balloon--or (dropping the analogy) the image of our surface after the deformation retract--is now a graph with edges given by $a_1,b_1,\dots,a_g,b_g$ and the $b-1$ boundary components we didn't move, as well as the new edges we introduced in blowing up our balloon. However, with some imagination we can convince ourselves that the new edges we introduced add no new cycles because we didn't allow our balloon to pass through itself. Thus we can deformation retract all of the new edges we introduced to get a wedge of $2g+b-1$ circles.

This can all be made rigorous by showing that we can consider our surface with boundary as being obtained by gluing a disk with a hole in it to a graph with $2g+b-1$, essentially working backwards from the intuition we built from the previous paragraph.

For the sake of a complete picture (without actually providing one!) here's how I imagine doing this. Start, as always, with the labelled $2g$-gon and draw $b-1$ circles $c_1,\dots,c_{b-1}$ inside of it, lining them up nicely from left to right in a nicely ordered fashion. Now draw an edge from $c_i$ to $c_{i+1}$ for $1\leq i\leq b-2$, making sure not to cross any of the edges (that would just be silly!). Draw one final edge from $c_{b-1}$ to the nearest vertex of the polygon, again making sure not to add any crossings. We now see before us a connected graph, whose complement in the pre-gluing polygon is a disk. After gluing according to the edge labeling, we get a surface with boundary with a graph drawn on it such that the complement of the graph is a disk. Thus, if we introduce a $b^{\rm th}$ boundary component, we have to remove a smaller disk from the interior of this disk, allowing us to construct a deformation retract from our surface with $b$ boundary components to the graph we drew before (i.e. the graph that tells us how to glue in the punctured disk). We can now use the same argument for the $b=1$ case to show that your generators are correct.

The lesson is that while we can generate $H_1$ of our surface by all of the $2g$ latitude/longitude generators $a_1,b_1,\dots,a_g,b_g$ and all of the $b$ boundary curves $B_1,\dots,B_b$, the deformation retract shows that (assuming appropriate orientations or working over $\mathbb{Z}/2\mathbb{Z}$) we can always identify, say $B_b$ with a sum of all of the other boundary components.

  • $\begingroup$ Thank you for such a fully detailled answer! About the last part of your explanation, when you want to retract the $g$-surface with $b$ boundary components to the graph : you mean that you extend the retraction we know from the disk into the circle by the identity outside of this disk in order to obtain a map $j:\Sigma_{g}\bigvee (\vee_{i=1}^{b-1}S^1)\rightarrow\Sigma_{g,b}$ that gives a equivalence homotopy between this spaces ? $\endgroup$ – Bebop Aug 27 '12 at 16:43
  • $\begingroup$ If by "the retraction we know from the disk into the circle" you mean the retraction from the disk with the new $b^{\rm th}$ boundary component, that is more-or-less what I meant. We have our graph, let's call it $G$, and we're gluing on a disk with a hole $H$, i.e. a cylinder. The gluing map sends one of the boundary components of $H$ to $G$. We know how to deformation retract $H$ onto this boundary component. Thus we can push this deformation retract forward into the actual surface. You might want to note that deformation retract is actually stronger than homotopy equivalence. $\endgroup$ – Eric S. Aug 31 '12 at 1:43
  • $\begingroup$ Ok, this is clear now. I really thank you for your help! $\endgroup$ – Bebop Aug 31 '12 at 9:00
  • $\begingroup$ Glad to help :-D $\endgroup$ – Eric S. Aug 31 '12 at 14:46

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