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I have a question /understanding problem with some Heegaard diagrams (see below).

As far as I understand the concept correctly then the Heegaard diagram explains uniquelly how two handlebodies $H_1, H_2$ with same genus $g$ (therefore they are especially isomorphically) are glued together along their boundaries.

With this aim in mind it suffice to say how the boundaries of the minimal system of discs $D_1, D_2, ..., D_g$ of $H_1$ (therefore closed paths $\gamma_i = \partial D_i$) are glued together with boundaries of the minimal system of discs $D_1', ..., D_g' $ of $H_2$.

Remark: Minimal systems of discs referes to the minimal number of discs inside a handlebody such that after cutting along them the handlebody becomes homeomorphic to a $3$-ball $D^3$.

Here an example which coincides with my understanding of Heegaard diagrams:

enter image description here

The green paths are pairwise glued with the red paths. This data suffice to define uniquelly (up to isomorphy) how the hendlebody $H_1 $ is glued with $H_2$ (here $g=3$) along the boundaries $\partial H_1, \partial H_2$, if we enumerate the tree red and green parts with $1,2,3$ and endow them with an orientation.

Now, here are the two diagrams which I dont understand:

(1)

enter image description here

Should the "gluing paths" be given pairwise? How to cope here with the green, glue and red paths? What kind of glueing occures here? I'm not sure if it is allowed to glue all three toghether (source: http://www.profmath.uqam.ca/~powell/HFHseminar.html)

(2)

enter image description here

Here the problem for me is that neither the red nor the blue paths are boundaries of discs which have property that if one cuts along them then the handlebody become a $3$-ball. I thought that by definition either the discs of $H_1$ or $H_2$ should have this property.

Could anybody explain to me the errors in my resonings considering these Heegaard diagrams and their general concept?

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Regarding your first example, until you label the red and green paths --- 1,2,3 for red and 1,2,3 for green so that you know which goes to which, and put arrows on all paths to indicate orientation so you know which orientation goes to which orientation --- the gluing is not uniquely specified.

Regarding your figure (1), I looked at that link and there is no explanation, it seems to be an illustration plucked out of context and put at the top of the web page because it looks cool. Without context, I have no idea what is intended.

Finally, regarding your figure (2), a Heegard diagram can be specified by drawing two curve systems on the surface, a red curve system and a blue curve system, as long as the surface minus the red curve system is a 2-sphere with $2g$ holes removed, and similarly for the blue curve system. The reason this works is because those two curve systems tell you how to abstractly glue in the discs --- red discs on one side of the surface (abstractly), and blue discs on the other side of the surface (abstractly) --- which is all that is needed in order to specify a Heegard decomposition of a 3-manifold.

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  • $\begingroup$ Hi, thank you for the answer. One point regarding your explanations about figure (2) stays unclear. You describe that the two curve systems determine how (abstractly) glue in the discs. Sorry for maybe penible question but it confuses me a bit when you mean "in the discs". Do you refer to the inner of the discs? I thought that the glueing process takes place only on the surface (so on $\partial D_1 \subset \partial H_1$ and resp $\partial D_2 \subset \partial H_2$) The inner points $\mathring{D}$ and $\mathring{D'} $ aren't identified in this construction, right? $\endgroup$
    – KarlPeter
    Jan 10 '19 at 21:31
  • $\begingroup$ This is simply a matter of the quotient topology. To express this business of "gluing in" more formally, starting from the surface $S$, with red curves $r_1,...,r_g \subset S$ and blue curve $b_1,..,b_g \subset S$, one takes red discs $R_1,...,R_g$ and blue discs $B_1,...,B_g$. Next one forms a quotient space: for each $i=1,...,g$, identify $r_i$ to the boundary of $R_i$ and identify $b_i$ to the boundary of $B_i$. The final thing one does to form the three manifold is (and now I revert to the informal language) one "glues in" a couple of 3-balls to form the final manifold. $\endgroup$
    – Lee Mosher
    Jan 11 '19 at 1:34
  • $\begingroup$ If this quotient topology language is unfamiliar to you, I would suggest reading up on the quotient topology, with as many examples as you can find, in topology textbooks. $\endgroup$
    – Lee Mosher
    Jan 11 '19 at 1:35
  • $\begingroup$ Ah so you attach discs to the surface along the curves and the "fill the cavities" to $3$-balls. Up to now I tried the Heegaard diagram construction in following way(without finding up to now concretely a paper explaining this in the same way). Could you say if it (intuitively I guess yes but I'm not sure) coinside finally with yours? $\endgroup$
    – KarlPeter
    Jan 12 '19 at 23:18
  • $\begingroup$ I take two handlebodies $H_1$ and $H_2$ of same genus $g$ and consider $g$ arbitrary curves $c_1, ..., c_g$ on $\partial H_1$ with property as you explaned in your main post: $\partial H_1 / \cup_i c_i$ is homeomorphic to $2$-sphere without $2g$ holes. On $\partial H_2$ I take another set of curves $d_1,...,d_g$ with same property. Then I glue the surfaces via homeomorphism $\phi: \partial H_1 \to H_2$ with following condition: $d_i = \phi(c_i)$ for all $i$. I think that this condition should determine up to homeomorphism $\phi$ and the resulting manifold $H_1 \cup_{\phi} H_2$ up to homeo. $\endgroup$
    – KarlPeter
    Jan 12 '19 at 23:18

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