Handle decompositions using only 1-handles

Question

Let $\Sigma$ be an oriented, compact, connected 2-manifold with boundary. Assume that its boundary is equipped with a disjoint union decomposition into two non-empty parts: $$\partial\Sigma=\partial_{in}\Sigma\cup\partial_{out}\Sigma$$ (both $\partial_{in}\Sigma$ and $\partial_{out}\Sigma$ are disjoint unions of circles).

The following should be well known, and yet...

$\bullet\quad$ $\Sigma$ admits a relative handle decomposition that only uses $1$-handles (no need to use $0$-handles or $2$-handles). [Ok, that's easy]

$\bullet\quad$ One can go from any handle decomposition of the above type to any other one by using: (1) isotopies (2) handle slides (no need to ever introduce $0$-handles or $2$-handles). [That's the harder question]

If someone could provide a short and elegant proof of the above result, that would make me very happy.

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Here, by a relative handle decomposition that only uses $1$-handles, I mean an identification between $\Sigma$ and something of the form $$ \Sigma':=\text{pushout}\left((\partial_{in}\Sigma\times I) \stackrel\varphi\longleftarrow \bigsqcup_{i=1}^n(\partial I\times I) \hookrightarrow \bigsqcup_{i=1}^n(I\times I)\right), $$ where $\varphi$ is an embedding $\bigsqcup_{i=1}^n(\partial I\times I)\stackrel\varphi\to \partial_{in}\Sigma\cong \partial_{in}\Sigma\times\{1\}\hookrightarrow \partial_{in}\Sigma\times[0,1]$. Moreover, the identification $\Sigma\cong\Sigma'$ should commute with the obvious embeddings of $\partial_{in}\Sigma$ into $\Sigma$ and into $\Sigma'$.

Answer 1

The second statement ought to be in the literature somewhere but I don't know a reference so I'll give an argument.

The result can be rephrased in terms of graphs. Let $S$ be a compact connected surface with non-empty boundary and let $P$ be a non-empty finite set of points in the interior of $S$. Consider finite connected graphs $X$ in $S$ with $P$ as vertex set and such that the components of $S-X$ are half-open annuli each having its boundary circle a component of $\partial S$. From $X$ we obtain a handle decomposition of $S$ with no 2-handles, where the 0-handles are disks about the vertices of $X$ and the 1-handles are neighborhoods of the edges of $X$ minus these disks, the rest of $S$ being just a collar on $\partial S$.

Let $\cal G$ be the set of all isotopy classes of such graphs $X$, where isotopies fix $P$. The assertion to be proved is that any two graphs in $\cal G$ are related by a finite sequence of handle slides, moving one end of an edge across an adjacent edge.

To show this we first enlarge the set $\cal G$ by allowing extra edges to be added to graphs in $\cal G$, keeping the same vertex set. This creates new complementary disks, and we require that none of these is a monogon or bigon. (This restriction may not be necessary for the argument but is usually made to avoid getting graphs that are unnecessarily large.) The resulting enlargement $\overline{\cal G}$ of $\cal G$ is a poset under inclusion, whose geometric realization we also call $\overline{\cal G}$.

From a graph $X$ in $\overline{\cal G}-{\cal G}$ we can obtain a graph $X'$ in $\cal G$ by deleting some of its edges. There may be different ways to do this, but we claim that the resulting graphs $X'$ are all related by handle slides (and isotopy). To see this, consider the dual graph $X^*$ whose vertices correspond to the components of $S-X$ and whose edges correspond to the edges of $X$. Then a choice of edges of $X$ to delete to obtain $X'$ corresponds to a choice of maximal tree $T$ in the quotient graph $X^*_{\partial}$ of $X^*$ obtained by identifying all vertices corresponding to components of $S-X$ meeting $\partial S$. It is easy to see that any two choices of maximal tree in a finite connected graph are related by a sequence of elementary moves in which one edge is omitted from a maximal tree and replaced by another edge not in the tree.
Changing a maximal tree $T$ in $X^*_{\partial}$ by an elementary move corresponds to adding one edge to $X'$ in an annulus component of $S-X'$ and deleting another edge of $X'$. It is evident that this operation can also be achieved by a sequence of handle slides. This verifies the claim that different graphs $X'$ obtained from $X$ by deleting edges are all related by handle slides.

It is a well-known fact that $\overline{\cal G}$ is connected, and in fact contractible. A simple proof of connectedness will be sketched below. Assuming this, let us finish the argument. Given two graphs defining vertices of $\cal G$, there is a sequence of vertices $X_1, X_2,\cdots, X_n$ of $\overline{\cal G}$ with $X_1$ the first given graph and $X_n$ the second one, such that each $X_{i+1}$ is obtained from $X_i$ by adding or deleting a set of edges. (We could refine this sequence so that only a single edge is added or deleted at a time.) By what we have shown in the preceding paragraph, any two graphs $X'_i$ and $X'_{i+1}$ in $\cal G$ obtained from $X_i$ and $X_{i+1}$ by deleting edges are then related by handle slides since if $X_i$, say, is obtained from $X_{i+1}$ by deleting edges, we may choose $X'_{i+1}=X'_i$. Thus the two given graphs $X_1=X'_1$ and $X_n=X'_n$ are related by a finite sequence of handle slides and we are done.

We can show $\overline{\cal G}$ is connected by a surgery argument. Let $X$ and $Y$ be two graphs in $\overline{\cal G}$. We can isotope one of them to be transverse to the other and intersect it in the minimum number of points within its isotopy class. Let $x$ be a point of $(X\cap Y)-P$ closest to an endpoint of the edge of $X$ containing it, so $x$ cuts off an arc $\alpha$ in this edge whose interior is disjoint from $Y$, with the endpoints of $\alpha$ being $x$ and a point $p\in P$. We can then cut the edge of $Y$ containing $x$ into two arcs with $x$ as their common endpoint and drag the ends of these arcs at $x$ along $\alpha$ to $p$. After a small isotopy these two arcs can be made disjoint from $Y$ except at their endpoints. We can then enlarge $Y$ by adding these two arcs to get a new graph $Y'$ in $\overline{\cal G}$. If one of the two arcs happens to be isotopic to an existing edge of $Y$, so that a component of $S-Y'$ is a bigon, we can just delete this arc to avoid the duplication. The hypothesis that $X$ and $Y$ intersect minimally guarantees that no complementary monogons are created. There is then an edge in $\overline{\cal G}$ from $Y$ to $Y'$. If we delete from $Y'$ the edge of $Y$ that we cut, we obtain a new graph $Y''$ in $\overline{\cal G}$ joined to $Y'$ by an edge of $\overline{\cal G}$. Since $Y''$ meets $X$ in fewer non-vertex points than $Y$ did, this surgery process can be iterated until we obtain a new graph $Y$ meeting $X$ only in its vertices. Then there are edges of $\overline{\cal G}$ joining $X$ to $X\cup Y$ and $X\cup Y$ to $X$ (after deleting parallel edges of $X\cup Y$ as before).

Answer 2

I doubt you're going to find an exact citation for either statement. The first statement is easy to prove, and for now I think that's all I will do, except that I'll close with some comments on the second statement.

Let me start with some observations and definitions. First, the relative core graph of the handle decomposition, which is the union of $\partial_{in}\Sigma$ with the cores of the 1-handles, is a spine of $\Sigma$, meaning a finite graph embedded in $\Sigma$ to which $\Sigma$ deformation retracts ($\Sigma$ is obtained from this graph by attaching a collection of annuli along one boundary component per annulus). Next, this is a rather special spine, being the union of $\partial_{in}\Sigma$ with a finite, pairwise disjoint collection of properly embedded arcs whose endpoints lie on $\partial_{in}\Sigma$; I'll call refer to this as a $\partial_{in}$ arc spine. Finally, any $\partial_{in}$ arc spine is the relative core graph of some relative 1-handle decomposition.

Your first question is proved by explicit construction of a $\partial_{in}$ arc spine; I would say this is a standard argument applying the triangulation theorem, but I don't know a citation. First, start with a triangulation of $S$. Next, remove all edges in $\partial_{out}\Sigma$. Next, inductively remove edges not in $\partial_{in}\Sigma$ and not bounding on both sides regions incident to $\partial_{out}\Sigma$; the result is a spine containing $\partial_{in}\Sigma$. Next, this spine may touch $\partial_{out}\Sigma$ at isolated points; push the spine off $\partial_{out}\Sigma$ at those points. Next, inductively collapse any edge which does not have both endpoints on $\partial_{in}\Sigma$; the result is a spine in which every edge is either in $\partial_{in}\Sigma$ or is a properly embedded arc with both endpoints in $\partial_{in}\Sigma$. Finally, perturb the attaching maps of these arcs to make all the endpoints disjoint on $\partial_{in}\Sigma$. The result is a $\partial_{in}$ arc spine.

For your second question, I do know how to reduce it to a statement about spines of surfaces which is closely related to citable statements, but that "closely related" thing is probably is not in itself sufficient to prove this exact statement, so there's still something left to be proved. I could write a proof using Stallings fold methods, one of the primary analytic tools for studying maps between graphs. But I doubt I could write a proof so as to be "short and elegant", especially for folks who might now know Stallings fold methods, so I'll hold off writing such a proof for now.