An Alexander polynomial for MOY graphs

Abstract

We introduce an Alexander polynomial for MOY graphs. For a framed trivalent MOY graph ${\mathbb {G}}$, we refine the construction and obtain a framed ambient isotopy invariant $\Delta _{({\mathbb {G}},c)}(t)$. The invariant $\Delta _{({\mathbb {G}}, c)}(t)$ satisfies a series of relations, which we call MOY-type relations, and conversely these relations determine $\Delta _{({\mathbb {G}}, c)}(t)$. Using them we provide a graphical definition of the Alexander polynomial of a link. Finally, we discuss some properties and applications of our invariants.

Appendix A: Replacing \({{\mathrm {sign}}}(s)\cdot m(s)\) with M(s)

The goal of this appendix is to show Proposition 2.16. To this end, it suffices to prove the following Proposition. Let $(D, \delta )$ be a decorated MOY graph diagram. For any given total orders on ${\mathrm {Cr(D)}}$ and ${\mathrm {Re(D)}}$, recall that ${\mathrm {sign}}(s)$ is defined to be the sign of the state $s\in S(D, \delta )$ as a permutation with respect to the given orders.

Proposition A.1

Let $(D, \delta )$ be a decorated MOY graph diagram, we have

$$\begin{aligned} \frac{{\mathrm {sign}}(s_{1})}{{\mathrm {sign}}(s_{2})} =\frac{M(s_{1})m(s_{1})}{M(s_{2})m(s_{2})}, \end{aligned}$$

(10)

for any two states $s_{1}, s_{2} \in S(D, \delta )$. Namely we have ${\mathrm {sign}}(s)=\prod _{p=1}^{N}{\mathrm {sign}}_{C_p}^{s(C_{p})}$ up to an overall sign change, where ${\mathrm {sign}}_{C_p}^{s(C_{p})}$ is the local contribution defined in Fig. 17.

Proof

It is easy to see that any diagram D can be transformed into a trivalent graph diagram by surgery ($\Lambda I$) or ($I\Lambda$) and surgery (I), which are defined in Figs. 22 and 21. Therefore the proof is a combination of Lemmas A.3, A.4 and A.5. $\square$

Fig. 17

The local contribution of ${\mathrm {sign}}_{C_p}^{\Delta }$

Lemma A.2

Let D be a diagram of a singular link. Namely at each vertex, there are two edges pointing inwards and two outwards. Then the statement of Theorem A.1 holds.

The Heegaard Floer homology and the Alexander polynomial of a singular link are studied in [12]. Using statements there we can get Lemma A.2. However to keep the combinatorial flavor of the paper, we give a proof by recalling some facts in [6].

A universeU is a connected diagram of a link in ${\mathbb {R}}^{2}$ where the data of which strand is over and which is under at each crossing is suppressed. Choose a base point $\delta$ on U and consider the set of states $S(U, \delta )$. The Clock Theorem in [6] says that any two states in $S(U, \delta )$ are connected by a sequence of clockwise or counterclockwise transpositions, which are defined in Fig. 18.

Fig. 18

Transpositions in the Clock Theorem

Two states that are connected by a transposition satisfy (10). By the Clock Theorem, any two states in $S(U, \delta )$ satisfy (10). Therefore the sign of a state for a universe is given by the local contribution in Fig. 17.

Proof of Lemma A.2

Since D is a connected diagram of a singular link, we call the underlying universe $U_D$. The difference of ${\mathrm {Cr}}(D)$ and ${\mathrm {Cr}}(U_D)$ occurs around the singular crossings, and the same is true for ${\mathrm {Re}}(D)$ and ${\mathrm {Re}}(U_D)$.

Suppose $C_i$ is a crossing of $U_D$ which corresponds to a singular crossing of D. There are two crossings in ${\mathrm {Cr}}(D)$ and one circle region in ${\mathrm {Re}}(D)$ around $C_i$, which we call $C_i, C_{i.5}$ and $R_{i.5}$, as in Fig. 19. We choose orders on ${\mathrm {Cr}}(D)$ and ${\mathrm {Re}} (D)$ so that

$$\begin{aligned} C_{i}< C_{i.5}<C_{i+1} \text { and } R_i<R_{i.5}<R_{i+1}. \end{aligned}$$

Then by forgetting $C_{i.5}$ and $R_{i.5}$ we get orders on ${\mathrm {Cr}} (U_D)$ and ${\mathrm {Re}} (U_D)$.

We see that any state s in $S(D, \delta )$ corresponds to a state ${\tilde{s}}$ in $S(U_D, \delta )$ by ignoring the circle regions:

$$\begin{aligned} {\tilde{s}}(C_i)={\left\{ \begin{array}{ll} \text {souther corner} ,&{} \,\,s(C_i)=\text {east corner} \vee s(C_{i.5})=\text {west corner} \\ \text {west corner}, &{} \,\, s(C_i)=\text {west corner} \\ \text {east corner}, &{} \,\, s(C_i)=\text {east corner} \end{array}\right. } \end{aligned}$$

Considering the orders above, we have

$$\begin{aligned} {\mathrm {sign}}(s)={\left\{ \begin{array}{ll} {\mathrm {sign}}({\tilde{s}}), &{}\text { if }s(C_{i.5})=R_{i.5} \\ -{\mathrm {sign}}({\tilde{s}}), &{}\text { if }s(C_{i})=R_{i.5}. \end{array}\right. } \end{aligned}$$

Up to an overall sign change, it is easy to check that the signs defined from the local contributions in Fig. 19 satisfies the relation above. $\square$

Fig. 19

The local contribution of sign for a universe $U_D$ (left) and a graph D (right)

Lemma A.3

When D is a diagram of an oriented trivalent graph without sinks or sources, the statement of Theorem A.1 holds.

To prove Lemma A.3, we separate the vertices of a trivalent graph into two groups. We call a vertex with indegree two and outdegree one an even vertex, and a vertex with indegree one and outdegree two an odd vertex. It is easy to see that the number of even vertices equals that of odd vertices. We use an oriented simple arc to connect an even vertex to an odd vertex, and call it surgery (X), which transforms two trivalent vertices into singular crossings.

Fig. 20

The surgery (X) transforms a trivalent graph into a singular link

Proof of Lemma A.3

Let $D_{1}$ and $D_{2}$ be the graph diagrams before and after applying a surgery (X). We show that if the statement of Theorem A.1 holds for $D_{2}$, it also holds for $D_{1}$.

As shown in Fig. 20, the interior of the newly added arc intersects $D_{1}$ at several points. We assume that the regions that the arc goes across, which we call $R_1, R_2, \ldots , R_k$, are distinct regions, otherwise we can replace the arc by an arc with less intersection points with $D_{1}$.

The arc separates each region $R_q$ for $1\le q \le k$ into two regions since $D_{1}$ is a connected diagram. We label the regions in $D_{2}$ around the arc by $R_1, R_{1.5}, R_2, R_{2.5}, \ldots , R_k, R_{k.5}$ as shown in Fig. 20, and label the newly created crossings by $C_1, C_2, \ldots , C_k$. Note that ${\mathrm {Cr}}(D_2)={\mathrm {Cr}}(D_1)\cup \{C_1, C_2, \ldots , C_k\}$. Consider an order on ${\mathrm {Cr}}(D_1)$ and extend it to an order on ${\mathrm {Cr}}(D_2)$ by requiring that

$$\begin{aligned} C_1< C_2<\cdots<C_k < \text { any other crossing in } {\mathrm {Cr}}(D_1). \end{aligned}$$

Consider an order on ${\mathrm {Re}}(D_2)$ so that

$$\begin{aligned} R_1<R_{1.5}<R_{2}<R_{2.5}<\cdots<R_k<R_{k.5}. \end{aligned}$$

Then we get an order on ${\mathrm {Re}}(D_1)$ by forgetting $R_{i.5}$’s.

Surgery (X) naturally induces an injective map $\phi : S(D_{1}, \delta ) \rightarrow S(D_{2}, \delta )$ described as below. Given $s\in S(D_{1}, \delta )$, $\phi (s)$ sends the crossing $C_p$ to the region $R_p$ (resp. $R_{p.5}$) if s does not occupy any corner in $R_p$ (resp. $R_{p.5}$), for $1\le p \le k$.

By considering the orders above, we can check that

$$\begin{aligned} \displaystyle \frac{{\mathrm {sign}} (s_{1})}{{\mathrm {sign}} (s_{2})}=\frac{{\mathrm {sign}} (\phi (s_{1})) \prod _{p=1}^{k} {\mathrm {sign}}_{C_p}^{\phi (s_{1})(C_{p})}}{\mathrm {sign} (\phi (s_{2}))\prod _{p=1}^{k} \mathrm {sign}_{C_p}^{\phi (s_{2})(C_{p})}}=\frac{\prod _{C_p>C_k} {\mathrm {sign}}_{C_p}^{\phi (s_{1})(C_{p})}}{\prod _{C_p>C_k} {\mathrm {sign}}_{C_p}^{\phi (s_{1})(C_{p})}} \end{aligned}$$

for any two states $s_{1}, s_{2} \in S(D_1, \delta )$. The first equality comes from a direct calculation of the sign using the orders defined above, and the second equality follows from the assumption that the statement of Theorem A.1 holds for $D_{2}$. Note that the crossings of $D_2$ that are greater than $C_k$ are exactly the crossings of $D_1$. Therefore

$$\begin{aligned} \displaystyle \frac{{\mathrm {sign}} (s_{1})}{{\mathrm {sign}} (s_{2})}=\frac{\prod _{C_{p}\in {\mathrm {Cr}}(D_{1})} {\mathrm {sign}}_{C_p}^{s_{1}(C_{p})}}{\prod _{C_{p}\in {\mathrm {Cr}}(D_{1})} {\mathrm {sign}}_{C_p}^{s_{1}(C_{p})}}. \end{aligned}$$

This completes the proof of the lemma. $\square$

Lemma A.4

If the statement of Theorem A.1 holds for the diagram after a surgery (I), then it also holds for the one before the surgery (I) (see Fig. 21).

Proof

Suppose the diagrams before and after a surgery (I) are $D_1$ and $D_2$. It is easy to see that the surgery (I) induces a one-one map $\phi : s(D_{1}, \delta )\rightarrow s(D_{2}, \delta )$ since $C_1$ must be mapped to $R_1$. Therefore

$$\begin{aligned} \frac{{\mathrm {sign}} (s_{1})}{{\mathrm {sign}} (s_{2})}=\frac{{\mathrm {sign}} (\phi (s_{1}))}{{\mathrm {sign}} (\phi (s_{2}))}, \end{aligned}$$

the right-hand side of which, by assumption, is defined by the local contribution of sign. This completes the proof. $\square$

Fig. 21

Surgery (I)

Fig. 22

Surgery (${\mathrm {I}}\Lambda$) and ($\Lambda {\mathrm {I}}$)

Lemma A.5

If the statement of Theorem A.1 holds for the diagram after a surgery ($\Lambda I$) or ($I\Lambda$), then it also holds for the one before the surgery ($\Lambda I$) or ($I\Lambda$) (see Fig. 22).

Proof

We prove the lemma for surgery ($\Lambda {\mathrm {I}}$), and the case of surgery (${\mathrm {I}}\Lambda$) can be proved similarly. Suppose the diagrams before and after surgery ($\Lambda {\mathrm {I}}$)) are $D_1$ and $D_2$. Consider an order on ${\mathrm {Cr}}(D_2)$, which induces an ordering in ${\mathrm {Cr}}(D_1)$ by forgetting $C_1$. Since ${\mathrm {Re}}(D_2)={\mathrm {Re}}(D_1)\cup \{R_1\}$, an order on ${\mathrm {Re}}(D_2)$ also induces an order on ${\mathrm {Re}}(D_1)$ by forgetting $R_1$.

The surgery ($\Lambda {\mathrm {I}}$)) induces an injective map $\phi : s(D_{1}, \delta )\rightarrow s(D_{2}, \delta )$ as below. Given $s\in s(D_1, \delta )$, if $s(C_{k})\ne R_2$, let $\phi (s)(C_{1})=R_1$ and $\phi (s)$ maps the other crossings the same way as s. If $s(C_{k})= R_2$ and $s(C_{k-1})$ is the east corner of $C_{k-1}$, let $\phi (s)(C_k)=R_1$, $\phi (s)(C_{k-1})=R_2$ and $\phi (s)(C_1)$ be the east corner. If $s(C_{k})= R_2$ and $s(C_{k-1})$ is the west corner, then $s(C_2)$ must be its west corner. In this case, let $\phi (s)(C_k)=R_1$, $\phi (s)(C_2)=R_2$ and $\phi (s)(C_1)$ the west corner.

In each case, we can check that ${\mathrm {sign}}(s)={\mathrm {sign}}(\phi (s))$ for any $s\in s(D_1, \delta )$. The right-hand side of the equality, by assumption, is defined by the local contribution of sign. Since $\phi$ is an injective map, the local definition of sign works as well for $D_1$. $\square$