2
$\begingroup$

Suppose that $F$ is an orientable, compact, connected surface of genus $g$ with $n$ boundary components, i.e. $F$ can be drawn like this (see Lickorish's 'An introduction to knot theory'): [link to picture].

One can via induction on $n$ show that we have $H_1(F)=\bigoplus_{2g+n-1}\mathbb{Z}$ for the first (singular) homology group.

I will give now the proposition 6.3 from Lickorish's 'An introduction to knot theory' and its proof, my question later refers to the proof.

Proposition. Suppose that $F$ is like above with non-empty boundary, piecewise lineary contained in $\mathbb{S}^3$. Then the homology groups $H_1(\mathbb{S}^3\setminus F)$ and $H_1(F)$ are isomoprhic, and there is a unique non-singular bilinear form $$\beta:H_1(\mathbb{S}^3\setminus F)\times H_1(F)\rightarrow\mathbb{Z}$$ such that $\beta([c],[d])=\text{lk}(c,d)$ for any oriented simple closed curves.

Proof (only first part). $F$ is now embedded in $\mathbb{S}^3$. $H_1(F)=\bigoplus_{2g+n-1}\mathbb{Z}$ generated by $\{[f_i]\}$ (see picture).

Let$V$ be a regular neighbourhood of $F$ in $\mathbb{S}^3$, so that $V$ is just a $3$-ball with $(2g+n-1)$ $1$-handles attached. The inclusion of $F$ in $V$ is a homotopy equivalence, and $H_1(\partial V)=\bigoplus_{2g+n-1}\mathbb{Z}\oplus\bigoplus_{2g+n-1}\mathbb{Z}$. For this, generators $\{[f_i']:1\leq i\leq 2g+n-1\}$ and $\{[e_i]:1\leq i\leq 2g+n-1\}$ can be chosen so that each $e_i$ is the boundary of a small disc in $V$ that meets $f_i$ at one point and the inclusion $\partial V\subset V$ induces on homology a map sending $[f_i']$ to $[f_i]$ and $[e_i]$ to zero.

Questions. 1) What changes if we embed $F$ in $\mathbb{S}^3$ (and no more in $\mathbb{R}^3)$?

2) What is a regular neighbourhood of $F$ in $\mathbb{S}^3$ and why is this just a $3$-ball with $(2g+n-1)$ handles?

3) Why is $\partial V\subset V$ an inclusion? A boundary of a space needn't to be contained in the space.

$\endgroup$
  • $\begingroup$ Please cite that this proposition and image are from Lickorish, "An Introduction to Knot Theory." (Proposition 6.3) $\endgroup$ – Kyle Miller May 11 '17 at 20:02
  • $\begingroup$ Done! Thanks for the hint. $\endgroup$ – user444847 May 11 '17 at 20:08
2
$\begingroup$

Question 1): this is easy to answer with a Mayer-Vietoris sequence, showing that removal of a single point from a 3-dimensional manifold does not alter the 1st homology.

Question 2): Suppose you are given a manifold $M$ in your case $\mathbb{S}^3$ (more generally you might be given some kind of cell complex, but I'll stick to manifolds for $M$). Suppose also that you are given a finite subcomplex $\Sigma \subset M$. A regular neighborhood $N(\Sigma) \subset M$ of $\Sigma$ is a generalization of a tubular neighborhood of a compact submanifold (with empty boundary). Intuitively the concept is similar, and the key feature from the perspective of homotopy theory is that $N(\Sigma)$ is a compact submanifold-with-boundary that deformation retracts to $\Sigma$.

In your situation, you know that $\Sigma = F$ itself deformation retracts to a graph $\Gamma$ having one vertex and $2g+n-1$ edges. From this you can conclude that $N(\Sigma)$ is not only a regular neighborhood of $F$ but it is also a regular neighborhood of $\Gamma$ (one is applying the uniqueness theorem for regular neighborhoods at this point, similar to the uniqueness theorem for tubular neighborhoods of submanifolds). Finally, by a direct construction you can show that a regular neighborhood of a graph in $\mathbb{S}^3$ having one vertex and $2g+n-1$ edges is a 3-ball with $2g+n-1$ handles.

For question 3), you might be confusing topological boundary with manifold boundary? In this setting $V$ is a compact manifold-with-boundary, and $\partial V$ denotes its manifold boundary which is a subset of $V$ by definition of manifold boundary.

$\endgroup$
  • 1
    $\begingroup$ Regarding 1), the statement applies to any 3-manifold, and particular to the 3-manifold $\mathbb{S}^3 \setminus F$. Since $\mathbb{R}^3 \setminus F = (\mathbb{S}^3 - \{\infty\}) \setminus F = (\mathbb{S}^3 \setminus F) - \{\infty\}$, it follows that $\mathbb{R}^3 \setminus F$ has the same first homology as $(\mathbb{S}^3 \setminus F) - \{\infty\}$ which has the same first homology as $\mathbb{S}^2 \setminus F$. $\endgroup$ – Lee Mosher May 11 '17 at 21:47
  • 1
    $\begingroup$ Regarding 2), deformation retraction is a special case of homotopy equivalence, so yes, every deformation retract of a space $X$ has the same homology groups as $X$. $\endgroup$ – Lee Mosher May 11 '17 at 21:49
  • 1
    $\begingroup$ As for your other question regarding 2), perhaps a picture might help alexsisto.files.wordpress.com/2012/01/handlebody.png $\endgroup$ – Lee Mosher May 11 '17 at 21:51
  • 1
    $\begingroup$ That's a big question, too much for a comment, but here's a few specific reasons. One is that $\mathbb{R}^3$ is not compact but its one point compactification $\mathbb{S}^3$ is compact. Another is that Alexander duality holds in $\mathbb{S}^3$ but not in $\mathbb{R}^3$. Another is the complement of a compact knot or surface in $\mathbb{S}^3$ has contractible universal cover but the complement in $\mathbb{R}^3$ does not. Perhaps what these add up to is that studying objects in $\mathbb{S}^3$ is often simpler than studying objects in $\mathbb{R}^3$. $\endgroup$ – Lee Mosher May 12 '17 at 13:41
  • 1
    $\begingroup$ The choice of generators is again best explained with pictures, which are not hard to draw. For example, take the standard picture for generators of $H_1(\partial V)$ in the case of genus 2 researchgate.net/profile/Philip_Argyres/publication/228540074/…, and superimpose it on the previous picture linked in my earlier comment. $\endgroup$ – Lee Mosher May 12 '17 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy