\label{sec:checking_movie_moves}

Here we will prove Theorem \ref{thm:1}, which asserts functoriality for the theory. This requires showing that chain maps in $\Kob{3}$, induced by link cobordisms, are well-defined, i.e., they are invariant under changing a cobordism presentation by a movie move. The overall strategy will be very similar to the one used in \cite{ClarMW:Fix}.

\subsection{Duality and homotopy isolation}
\label{ssec:duality}
We begin by stating a result about duality with respect to Hom-sets. For tangles $P$ and $Q$, denote any gluing of them by $P \bullet Q$, and let $\overline{Q}$ denote the reflection of $Q$. The following proposition was first presented in \cite{ClarMW:Fix}:

\begin{prop}
\label{prop:duality}
Given oriented tangles $P$, $Q$ and $R$, there is an isomorphism
between the spaces of chain maps up to homotopy
$$F : \Hom{Kh}{\Kh{P \bullet Q}}{\Kh{R}} \IsoTo \Hom{Kh}{\Kh{P}}{\Kh{R \bullet \overline{Q}}}.$$
\end{prop}

While this result was originally proven in the context of Khovanov's $\su{2}$ theory, it clearly holds for the $\su{3}$ case without any changes to the statement or proof (for whose details we refer the interested reader to \cite{ClarMW:Fix}). It's important to note, however, that the proof assumes the theory is already invariant under MM9, the ninth movie move. As such, invariance under MM9, shown in Section \ref{sssec:MM9}, must take place with complete independence of the material in this section.

From this we get an easy corollary, also given in \cite{ClarMW:Fix}, which will be highly useful during our movie move checks. In particular, note that it applies to (the first and last frames of) every movie move.

\begin{cor}
\label{cor:post-duality}
Let $T_1$ and $T_2$ be tangles with $k$ endpoints such that
$\overline{T_1} \bullet T_2$ is an unlink with $m$ components. Then the space of
chain maps modulo chain homotopy from $\Kh{T_1}$ to $\Kh{T_2}$ in
grading $m-k$ is 1-dimensional, and all chain maps of grading higher
than $m-k$ are chain homotopic to zero.
\end{cor}

The second component of machinery we'll need is the ``homotopy isolation" idea from \cite{ClarMW:Fix}.

\begin{defn}
Let $C^{\bullet}$ and $D^{\bullet}$ be complexes in a graded additive category, with $A$ a direct summand in some $C^i$. We'll say $A$ is $C$-$D$ \emph{homotopically isolated} if, for any grading zero homotopy $h:C^{\bullet} \rightarrow D^{\bullet - 1}$, the restriction of $dh + hd$ to $A$ is zero.
\end{defn}

\begin{lemma}
\label{lem:hom-iso1}
Let $f,g:C^\bullet \rightarrow D^\bullet$ be chain maps, and say $f \htpy \alpha g$ are homotopic for some scalar $\alpha$. If $f$ and $g$ agree and are nonzero on a $C$-$D$ homotopically isolated object $A$ in $C^\bullet$, then we have that $f \htpy g$ are homotopic.
\end{lemma}

By Corollary \ref{cor:post-duality}, we know that any movie move (except for MM9) changes the induced map in $\Kob{3}$ by at most a scalar. We'll show this scalar is always 1 by computing with homotopically isolated objects, which have a convenient description in the $\su{3}$ web case.

\begin{lemma}
\label{lem:hom-iso2}
Let $\Kh{T_1}$ and $\Kh{T_2}$ be the complexes for two tangle diagrams, and let $D_1$ be a web appearing as a direct summand somewhere in $\Kh{T_1}$. Then
\begin{enumerate}
\item $A$ is $\Kh{T_1}$-$\Kh{T_1}$ homotopically isolated if $D_1$ contains no cycles (as a graph) and is not connected by differentials to webs containing cycles;
\item $A$ is $\Kh{T_1}$-$\Kh{T_2}$ homotopically isolated if $\Kh{T_1}$ and $\Kh{T_2}$ contain only acyclic webs.
\end{enumerate}
\end{lemma}

To prove this, we'll first need a more general result about Hom-sets of foams.

\begin{defn}
The \emph{bare grading} of a morphism $C$ between webs $D_1$ and $D_2$ is given by $$\deg'(C) = 2\chi(C) - B + \frac{V}{2},$$
where $B$ is the number of boundary points on $h$ and $V$ is the total number trivalent vertices in $D_1$ and $D_2$.\footnote{Here we've simply neglected the contributions from formal shifts of boundary webs present in the usual $\Kob{3}$ grading.}
\end{defn}

\begin{prop}
\label{prop:no-positive-grading}
If $D_1$ and $D_2$ are acyclic then there are no nonzero morphisms with positive bare grading between them.
\end{prop}

\begin{proof}
Let $C:D_1 \rightarrow D_2$. Our first task is to remove the closed seams from $C$, producing a new foam called $\widetilde{C}$. Begin by performing neck cutting on each sheet incident on a closed seam. (If there are $k$ closed seams to begin with, there will be $3k$ operations). This will produce $3^{3k}$ terms, in which all the original closed seams are sequestered in closed foams. We can evaluate each one of these closed foams to a scalar by the remark in Section \ref{ssec:categorifying}, leaving us with a new presentation of $C = \sum C_i$, a degree-homogeneous linear combination. Here the $C_i$ may still have closed seams, but only of the variety appearing in the neck cutting relation: locally, they will all look like ``choking handles" $\left(\mathfig{0.1}{cobordisms/handle_disc_bdy_left}\right)$. At this point, we can perform neck cutting once again (to remove unwanted tubes connecting sheets) so that each $C_i$ has the following pieces:
\begin{itemize}
\item 0-cells given by trivalent vertices and boundary points.
\item 1-cells given by seams, boundary lines, and edges in $D_i$.
\item 2-cells given either by discs in choking handles, or by sheets that intersect $D_1 \cup D_2$ nontrivially and that may have handles or choking handles.
\end{itemize}

Now pick any $C_i \neq 0$ and consider the foam $\widetilde{C}$ obtained from $C_i$ by removing all handles and choking handles. Since these pieces have bare grading $-1$ and $0$, respectively, we have that $\deg'(C) = \deg'(C_i) \leq \deg'(\widetilde{C})$. Also note that $\widetilde{C}$ has no closed seams and no handles, and can thus be decomposed as follows:
\begin{itemize}
\item 0-cells given by trivalent vertices and boundary points.
\item 1-cells given by seams, boundary lines, and edges in $D_i$.
\item 2-cells given by genus zero sheets that intersect $D_1 \cup D_2$ nontrivially.
\end{itemize}
It suffices to prove the result for this much simpler foam $\widetilde{C}$.

Let $F$ be the number of seams in $\widetilde{C}$, and let $S$ be the number of sheets. We now claim that $\chi(\widetilde{C}) = S-2F$, which we can see as follows. Imagine building $\widetilde{C}$ out of its $S$ disjoint sheets. We'll then add seams, joining together three sheets at each seam. Each of these operations will reduce the Euler characteristic by $2$, giving the formula. Thus, since $F=\frac{V}{2}$, $\deg'(\widetilde{C}) = 2S - 3F - B$.

Next we'll show that, if $D_1$ and $D_1$ are acyclic, this formula can be modified to $\deg'(\widetilde{C}) = 2S -4F - N$, where $N = \pi_0 (D_1) + \pi_0 (D_2)$, i.e., the total number of connected components of the boundary webs. Assuming $D_1$ and $D_2$ are acyclic just means that each of their components is a tree. This says that $B = V_i + 2n_i$, where $n_i$ is the number of connected components and $V_i$ is the number of trivalent vertices in $D_i$. Adding these two equations we get
\begin{align*}
    2B &= V_1 + V_2 + 2(n_1 + n_2)\\
    &= V + 2N\\
    \Rightarrow B &= \frac{V}{2} + N = F + N.
\end{align*}
Thus we have that $\deg'(\widetilde{C}) = 2S - 3F - B = 2S - 4F -N$.

Remember, our goal is to show that $\deg'(\widetilde{C}) \leq 0$, which is now equivalent to proving
\begin{align}
\label{eq:seams}
S \leq 2F + \frac{N}{2}.
\end{align}
We'll get there by considering the boundary 1-cells of the sheets in $\widetilde{C}$, which, as mentioned before, consist of seams, boundary lines, and edges in $D_i$. It's an easy observation that the total number of edges, $E$, in $D_1$ and $D_2$ is just $E = 2V + N$. These segments, as well as the boundary lines, can serve as part of the boundary for a single sheet. Each seam, however, will serve as a boundary component for three distinct sheets (from our acyclicity assumption).

For an example, let's say that every sheet in $\widetilde{C}$ were a $d$-gon. Then we would have that $S d = E + B + 3F$: each side of each sheet corresponds to a graph edge, a boundary line, or one third of a seam. Let's make a more general assumption: that every sheet in $\widetilde{C}$ has at least four sides. If this is the case, then
\begin{align*}
    4S &\leq E + B + 3F\\
    &\leq (2V + N) + (F + N) + 3F\\
    &\leq 8F + 2N.
\end{align*}

This would give us Equation \ref{eq:seams}. Note that there cannot be three-sided sheets (or in fact any odd-sided sheets, by acyclicity); unfortunately, there can be bigons. However, we observe that bigons in $\widetilde{C}$ must have one edge in one of the $D_i$ and the other edge a seam intersecting the same $D_i$ twice. In other words, this bigon must be part of a zip or unzip morphism. Thus we can factor $\widetilde{C}$ into a stack of zips, unzips, and a foam in which each sheet has at least four sides. Since the zip and unzip morphisms have bare degree $-1$, we have our result.
\end{proof}


\begin{proof}[Proof of Lemma \ref{lem:hom-iso2}]
A degree zero homotopy is a morphism $$h: q^{m}D_1 \rightarrow q^{m-1}D_2.$$ (Here, we could have $D_2$ in either $\Kh{T_1}$ or $\Kh{T_2}$, depending on which part of the lemma we're trying to prove.) Thus, by Equation \ref{eq:grading}, $\deg'(h) = 1$. And by Proposition \ref{prop:no-positive-grading}, $h$ must be the zero map.
\end{proof}

The first part of this lemma is well-suited for the reversible movie moves (MM6-10) and the second part for the those involving Morse moves. It's an easy observation that every web in the initial (and final) complex $C$ of movie moves 6, 7, and 8 is $C$-$C$ homotopically isolated, and every web in the initial and final complexes $C$ and $D$ of 11, 13, and 15 is $C$-$D$ homotopically isolated. This means we can compare induced chain maps simply by applying them to a single object of our choice. Movie moves 12 and 14, unfortunately, do not contain homotopically isolated objects, so we'll need to compute the induced maps on all objects; luckily the complexes are small, and this is not a great burden. We'll handle movie moves 9 and 10 with different techniques: the former because of the paragraph after Proposition \ref{prop:duality}, and the latter just to illustrate something fun.

Keep in mind that, as always, all crossings in these moves must be ordered, and they may need to be reordered to be consistent with the conventions we've defined for the Reidemeister maps. (Recall the discussion about planar compositions of complexes in Section \ref{ssec:link-homology} and Appendix \ref{ssec:sign-conventions}) However, the chain maps induced by crossing reorderings are trivial in every movie move except MM6 and MM9. The signs appearing in MM6 are particularly nasty, but we will show some sample calculations.
{"config": "arxiv", "file": "0806.0601/sections/movie_moves1.tex"}