Presenting 3-manifolds by planar graphs

Question

From a planar graph $\Gamma$, equipped with an integer-valued weight function $d:E(\Gamma) \sqcup V(\Gamma) \to \mathbb{Z}$, one can build a $3$-manifold $M_{\Gamma}$ as follows. For each vertex $v$, draw a small planar unknot centered at $v$. For each edge $e$ connecting vertices $v$ and $w$, add a series of $d(e)$ clasps between the corresponding unknots (with positive weights represented by right-handed clasps and negative weights represented by left-handed clasps). The result is a link with unknotted components - call it $L_{\Gamma}$. To get the $3$-manifold $M_{\Gamma}$, perform Dehn surgery on each component of $L_{\Gamma}$, with framings $$ f(v) = d(v) + \sum_{e \ni v} d(e). $$

In fact, every 3-manifold $M$ is diffeomorphic to a manifold of the form $M_{\Gamma}$. This can be proved using ideas from a paper of Matveev and Polyak (A Geometrical Presentation of the Surface Mapping Class Group and Surgery), as follows. Choose a Heegaard splitting for $M$, and write the gluing map as a composition of Lickorish twists. Then use the graphical calculus in sections 3 and 4 of that paper (omitting the twists denoted $\epsilon_i$) to produce a tangle whose plat closure is a framed link of the form $L_{\Gamma}$, such that the result of surgery on this link is $M$. This argument is given by Polyak in slides available on his website.[1][2]

There is another proof as well, which involves taking an arbitrary link surgery presentation and repeatedly simplifying it. There is a natural way to measure the complexity of a link diagram, such that links of minimal complexity are of the form $L_{\Gamma}$. It is always possible to reduce this complexity by adding cancelling unknotted components and doing handleslides.

One can take the idea further and show that there is a finite set of local moves which suffice to relate any two weighted planar graphs representing the same 3-manifold. Indeed, this follows abstractly from the fact that the mapping class group is finitely presented. However, the moves produced by Wajnryb's presentation (for example) are rather complicated and nasty-looking. It is therefore natural to ask if one can find a more appealing set of moves.

One must certainly include the following moves (please excuse the lack of pictures):

• Self-loops and edges with weight $0$ can be eliminated.
• Any two parallel edges can be combined, at the expense of adding their weights.
• Suppose that $v$ is a vertex incident to exactly one edge $e$, with $d(e) = \pm f(v) = \pm 1$. Then $v$ and $e$ can be deleted, at the expensing of changing the framing on the other endpoint of $e$, call it $w$. If $f(v) = 0$, then $w$ (and all of its incident edges) can be removed from the graph .
• Suppose that $v$ is a vertex incident to exactly two edges $e_1$ and $e_2$, with $d(e_1) = d(e_2) = - f(v) = \pm 1$, then the vertex $v$ can be replaced with a single edge joining the opposite endpoints of $e_1$ and $e_2$, call them $w_1$ and $w_2$. If $d(e_1) = - d(e_2)$ and $f(v) = 0$, then $e_1$ and $e_2$ can be contracted, with the resulting vertex having weight $d(w_1) + d(w_2)$.
• Suppose that $v$ is a vertex incident to exactly three edges, and suppose that $d(e_1) = d(e_2) = -d(e_3) = f(v)$. Then $v$, together with all $e_i$, can be eliminated at the expense of adding a triangle connecting the opposite endpoints of the edges $e_i$.

Let's call any of the above moves a "blowdown", and let's call their inverses "blowups". All of them can be easily deduced using Kirby calculus, or from relations in the mapping class group.

Question 1: Are blowups and blowdowns sufficient to relate any two planar graph presentations of a given $3$-manifold?

If the answer to this question is no, then there are additional non-local moves to consider:

• If two edges $e_1$ and $e_2$ connect the same pair of vertices (but are not necessarily parallel), then they can be combined, at the expense of adding their weights.
• If there is a vertex $v$ which divides $\Gamma$ into multiple components, those components can be "permuted around $v$". Any of these components which is connected to $v$ by a single edge $e$ with weight $\pm 1$ can also be "flipped over", at the expense of changing the weight on $e$.
• If there are two vertices $v$ and $w$ which separate the graph into multiple components, then any component which is joined to both $v$ and $w$ by a single pair of edges with opposite weights in $\{\pm 1\}$ can be "flipped over", at the expense of changing the weights on the edges.

Let's call any of the above moves a ``mutation''.

Question 2: Are blowups, blowdowns, and mutations sufficient to relate any two planar graph presentations of a given $3$-manifold?

If the answer to this question is also no, then it would be good to have a nice answer to the following (admittedly vague) question:

Question 3: What is the "simplest possible" set of moves which are sufficient to relate any two planar graph presentations of a given $3$-manifold?

One argument in favor of a positive answer to Questions 1 or 2, or at least a very nice answer to Question 3, is that Kirby calculus itself admits a finite set of simple local moves. One approach might be to find a canonical way of simplifying an arbitrary link surgery presentation to a planar graph presentation, and trace the effects of a Kirby move through the simplification process to see which graph moves are required to implement it.

There is also a relationship with double branched covers, which might be relevant. If all vertex weights $d(v) = 0$, then $M_{\Gamma}$ can be identified (after doing surgery on an essential 2-sphere) with the double cover branched over a link $Z \subset S^3$, whose "checkerboard graph" is $\Gamma$. In this picture, Reidemeister moves on $Z$ can be realized by blowdowns and blowups, and Conway mutations can be realized by graph mutations. Note that these moves do not suffice to relate links with diffeomorphic double branched covers - this might be viewed as evidence in favor of a negative answer to Questions 1 and 2.

Answer 1

I'm pretty sure that the answers to Question 1 and part of Question 2 are no, and this does come from looking at double branched covers.

There are examples due to Watson (1) of pairs of knots with diffeomorphic double branched covers but different Khovanov homology. So to show that these moves do not suffice, we just need to write down a chain complex for each of these graphs $\Gamma$ that computes Khovanov homology of an appropriate link when all of the $d(v)$ are $0$, and is invariant under these moves.

To show that (odd, or $\mathbb{F}_2$ coefficient) Khovanov homology is mutation invariant, Bloom constructs a complex computing Khovanov homology from some basically linear-algebraic data, and a slight adaptation of this gives the needed chain complex here.

More specifically, for a graph $\Gamma$ as in the question, let $\hat{\Gamma}$ be a graph obtained by adding a vertex $v_*$ to $\Gamma$, adding an edge labelled $d(v)$ from $v$ to $v_*$ for each vertex $v \in \Gamma$, replacing every edge $e$ with $d(e) > 0$ with $d(e)$ edges labelled $1$, and replacing every edge with $d(e) < 0$ with $-d(e)$ edges labelled $-1$.

Then let $E$ and $V$ be the $\mathbb{F}_2$ vector spaces spanned by the edges and vertices of $\hat{\Gamma}$, and let $I: E \to V$ be the map sending each edge to the sum of its endpoints. For each subset $S$ of the edges of $\hat{\Gamma}$, let $E_S \subset E$ be spanned by the edges in $S$, and let $I_S$ be the restriction of $I$ to $E_S$. Finally, let $C_0(S) = \operatorname{coker} I_S \oplus (\ker I_S)^*$, and let $C(S) = \Lambda^*C_0(s)$. (Apologies for all the notation)

Using the snake lemma, one can show that if $S$ and $S'$ differ only in a single edge $e$, then either there is a canonical element $\phi_e \in C_0(S)$ with $C_0(S') \cong C_0(S)/\langle \phi_e \rangle$, or the reverse: a canonical element $\phi_e \in C_0(S')$ with $C_0(S) \cong C_0(S')/\langle \phi_e \rangle$.

With this data, we can build a complex just like the odd Khovanov complex: call an assignment of $\pm 1$ to each edge of $\hat{\Gamma}$ a state, and for a state $s$ let $S(s)$ be the set of edges where the number assigned to the set by the state agrees with the label of the state. If $s'$ differs from $s$ by changing the $-1$ on $e$ to a $+1$, let $$\psi_{s,s'}: C(S(s)) \to C(S(s'))$$ be the map induced by the projection $C_0(S(s)) \to C_0(S(s'))$ if $C_0(S(s')) \cong C_0(S(s)) / \langle \phi_e \rangle$, and if $C_0(S(s)) \cong C_0(S(s')) / \langle \phi_e \rangle$ let $$ \psi_{s,s'}(x) = \phi_e \wedge x. $$ The chain complex then has the direct sum of all $C(S(s))$ where $s$ has $k$ ones in degree $k$, and the differential is the sum of all $\psi_{s,s'}$.

It's not hard to see that this complex computes the direct sum of two copies of the Khovanov homology of $L$ if $\Gamma$ was originally the checkerboard graph of $L$ (the two copies come from the extra vertex added). Any of the blowup or blowdown moves change $\hat{\Gamma}$ locally in exactly the way that a Redemeister move changes the checkerboard graph of a link, so the proofs that odd Khovanov homology is a knot invariant should work just as well for showing that this is invariant under blowups (this is the part of this argument that I'm least sure of). Similarly, the types of mutations that don't involve flipping an edge over and changing it's sign don't change $\hat{\Gamma}$ at all. Flipping an edge over seems more complicated to me, but I wouldn't be surprised if this still remains invariant.

With this set-up, I'd be really interested in other local moves on these graphs in order to better understand how Khovanov homology fails to be an invariant of the double branched cover.