I am confused by a certain point in Scharlemann's paper "Sutured Manifolds and Generalized Thurston Norms", which seems important enough to not just skip it. I mean the "2-surgery on disks" in the first paragraph on page 7 (or 563). I will try to explain what is going on for those who don't have this paper, but first the questions I have:
- What is this 2-surgery geometrically? Is there a picture somewhere?
- Why does it preserve homology classes?
- Why does it not increase the generalized Thurston norm?
So the situation is the following. We have two properly embedded surfaces in a 3-manifold that minimize the generalised Thurston norm. If there is a curve on one of the surfaces that bounds a disk in the surface without points of intersection of the $\beta$-graph in the disk (or the curve can be boundary parallel, but still without points of intersection with graph in the disk), we want to do this 2-surgery on this disk to eventually arrive at two surfaces that have no such curves of intersection. But we also want to not increase the generalised Thurston norm and preserve the homology classes. Can anybody help me to understand what is going on here? Thanks!
Intuition / Motivation
Letting $S$ and $T$ be our surfaces, the point of the 2-surgery operation is to eliminate intersection curves $\gamma \subset S \cap T$ that bound disks in $S$ or $T$ in such a way that the resulting surface $\Sigma$ represents the same homology class as $S\cup T$. This is a step in showing that the generalized Thurston (semi-) norm satisfies the triangle inequality $$\lVert a+b \rVert_{\beta}\leq \lVert a \rVert_{\beta} + \lVert b \rVert_{\beta}$$
How do we show that $[S\cup T]$ has an embedded representative? We first use the "innermost disk" argument sketched in Thurston, William P., A norm for the homology of 3-manifolds, Mem. Am. Math. Soc. 339, 99-130 (1986). ZBL0585.57006.
Per question 3 in your list (on why the generalized Thurston norm doesn't change under 2-surgery on innermost disks): remember the Thurston norm of an immersed surface $\Sigma$ is defined in terms of its homology class $[\Sigma]$; any operation $\tau$ on our surface that doesn't change this underlying homology class will not change the Thurston norm, i.e. $$[\Sigma]=[\tau(\Sigma)] \Rightarrow \lVert \Sigma \rVert_{\beta} = \lVert \tau(\Sigma) \rVert_{\beta}. $$
Our goal is to find a compact, embedded surface that represents the sum (union) of two compact embedded surfaces $S$ and $T$ at the level of homology. 2-surgery on innermost disks and the double-curve sum are somewhat violent operations, but homology classes are pretty robust and the resulting surface will represent the same homology class while also being embedded.
I'll try to give an account for why the innermost disk procedure preserves the underlying homology class of a surface. This (and much more) appears in some very nice notes by Daniel Alvarez-Gavela.
Innermost Disk Argument
Let $\Sigma := S \cup T$ where $S$ and $T$ are embedded, closed surfaces (compact and without boundary). If $\Sigma$ is embedded, we're done. Otherwise, $S$ and $T$ generically intersect transversely in a compact, 1-dimensional, closed submanifold, so a union of circles. The goal of 2-surgery is to replace $\Sigma$ with a new surface $\Sigma'$ which carries the same homology class but differs from $\Sigma$ in the sense that all circles in $S\cap T$ which are nullhomologous in either $S$ or $T$ have been eliminated.
Suppose I have a closed curve $\gamma\subset S\cap T$ which bounds a disk $D$ in either $S$ or $T$, say, $D\subset S$. This disk could contain another component of $S\cap T$ in its interior, so what we do is we go to the "innermost disk", by looking for a $\gamma' \subset (S\cap T)\cap D^{\circ}$ bounding a disk $D'$ such that $D'^{\mathrm{o}}\cap S\cap T = \emptyset$ (that such a disk exists follows from compactness of $S$ and $T$).
2-surgery on the curve $\gamma'$ means (in Thurston's words) that we modify $T$ by attaching a copy of $D'$ on "either side" of $S$ so that the resulting surface, $T'$, no longer has the component $\gamma'$ in its intesection with $S$.
Explicitly the tubular neighborhood theorem tells us there is a bicollar $D'\times [-\epsilon, \epsilon]$ of $D'$. Let $A(\gamma'):=\gamma' \times (-\epsilon,\epsilon)$ be a bicollar of $\gamma$ in $T$.
For $\alpha\in \{ \pm \epsilon\}$, let $\phi_{\alpha}$ be an attaching map taking $D'\times \{\alpha\}$ to $\gamma' \times \{\alpha\}$ we can construct $$T':= \Big(T\setminus A(\gamma) \Big) \sqcup_{\phi_{\epsilon} \sqcup \phi_{-\epsilon}} \Big(D'\times \{-\epsilon\} \sqcup D'\times \{\epsilon\}\Big)$$
Here's the picture I have in mind:
Once we've eradicated the $\gamma'$ bounding the innermost disk, we do it again on the intersection curve bounding the next-to-innermost disk and so on until there are no more curves in $S\cap T$ bounding a disk in either $S$ or $T$.
In the end, all homologically trivial components of $S\cap T$ will be gone. Whatever remains of $S\cap T$ is then dealt with by taking the "double-curve sum" to produce an embedded surface.
Invariance of homology class
Note that outside of a neighborhood $N$ of $\gamma'$, $S\cup T$ and $S\cup T'$ are identical, so if we can show that no cycles or boundaries were introduced in the interior of $N$, then we're done. Neither of the disks we attached contribute a cycle: they are bounded by $\gamma' \times \{-\epsilon\}$ and $\gamma' \times \{\epsilon\}$ respectively. Likewise, no 3-cell boundaries have been introduced.