change of linking number along Seifert surface is given by algebraic intersection number

Question

Let $J,K,L$ be knots in $S^3$ and $F$ be some surface such that $\partial F = K\cup J$. I want to prove that $$\text{lk}(L,K)=\text{lk}(L,J)+F\cdot L,$$ where lk is the linking number and $\cdot$ denotes the algebraic intersection number.
I know that we can express the linking number via the algebraic intersection form and a Seifert surface. This is: $$\text{lk}(L,K)=F_L\cdot K,$$ where $F_L$ is some Seifert-surface of $L$.
Further we can describe the algebraic intersection form in terms of the cup product. To do this one considers a thickening of $L$, this is $B^2\times L$, and then its complement $X:=S^3\setminus(B^2\times L)$. Denote by $PD_X$ the inverse map of the isomorphism given by Poincare duality, this is $H^k(X,\partial X)\overset{\cong}{\rightarrow}H_{3-k}(X)$ via capping with the fundamental class, then - if I understand the theorem I looked up correctly - we have $$F\cdot J = (PD_X([J])\cup PD_X([F']))\cap [X].$$ Here $[J]$ resp. $[F']$ denotes the image of the fundamental class under the canonical inclusion and $F':= F\cap X$. This makes the statement now obvious to me if $K$ and $J$ are homologous in the complement of $L$, but I can not prove the statement in general as I seem to have forgotten everything about cup products. What could be of use is that $$(PD_X([J])\cup PD_X([F']))\cap [X]=PD_X([F'])\cap(PD_X([J])\cap[X]).$$ Thanks in advance for any help!

Answer 1

When dealing with linking numbers, especially when there are more than two knots involved, one thing to be careful about is orientations. The formula give is correct with the following modification: the oriented boundary of $F$ is $K\cup -J$, where $-J$ denotes $J$ with reversed orientation.

The main fact underlying this is that linking number distributes over unions: $$ \operatorname{lk}(L,K\cup -J) = \operatorname{lk}(L,K) + \operatorname{lk}(L,-J) = \operatorname{lk}(L,K) - \operatorname{lk}(L,J). $$ Then if $F$ is the surface described above (such that it intersects $L$ transversely) then $\operatorname{lk}(L,K\cup -J)=F\cdot L$. Thus, $$ \operatorname{lk}(L,K) = \operatorname{lk}(L,J) + F\cdot L. $$

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If you want a cup product version of this, then something worth knowing is that a Seifert surface of a link $L$ is Poincaré dual to the Alexander dual of the fundamental class of $L$. That is, if $[L]\in H_1(L)$ is the fundamental class (which is determined by the orientation of each component of $L$), then the Alexander dual of $[L]$ is an element $\alpha_L\in H^1(S^3-B^2\times L)$, whose Poincaré dual in $H_2(S^3-B^2\times L,S^1\times L)$ is represented by a connected surface $F_L\subseteq S^3-B^2\times L$. What $\alpha_L$ defines is a homomorphism $H_1(S^3-B^2\times L)\to\mathbb{Z}$ that computes linking numbers of homology classes with $L$. That is to say $\operatorname{lk}(L,K)=\langle\alpha_L,[K]\rangle$, where $\langle-,-\rangle$ is the natural pairing between cohomology and homology (you might write $\alpha_L([K])$ for this, or perhaps even $\int_K\alpha_L$ if you want to think about it as integration along $K$ of a differential form).

The way the Alexander dual works is that there is an isomorphism $H^1(S^3-L)\to H_1(L)$ defined by $\beta\mapsto \partial([S^3-L]\frown \beta)$, where by $S^3-L$ I mean $S^3-B^2\times L$ (warning/note: with everything I say, I want a manifold-with-boundary, so interpret accordingly). For sake of recall: $[S^3-L]\in H^3(S^3-B^2\times L,S^1\times L)$ is the fundamental class, and then we have $[S^3-L]\frown \beta\in H_2(S^3-B^2\times L,S^1\times L)$, whose boundary is in $H_1(B^2\times L) = H_1(L)$. Hence, $\alpha_L$ satisfies the equation $\partial([S^3-L]\frown \alpha_L)=[L]$. Hidden in here is Poincaré duality: $[S^3-L]\frown\alpha_L$ is $[F_L]$.

Hence, at the level of (co)chains (i.e., I'm not being careful saying where each of the intermediate results live): \begin{align*} \operatorname{lk}(L,K) &= \langle \alpha_L,[K]\rangle \\ &= \langle \alpha_L, \partial([S^3-K]\frown \alpha_K) \rangle \\ &= \langle \delta\alpha_L, [S^3-K]\frown\alpha_K \rangle \\ &= \langle \alpha_K \smile \delta\alpha_L, [S^3-K] \rangle \\ &= \langle PD(F_K) \smile \delta PD(F_L), [S^3-K] \rangle \\ &= \langle PD(F_K) \smile PD(\partial F_L), [S^3-K] \rangle \\ &= \langle PD(F_K) \smile PD(L), [S^3-K] \rangle \\ &= F_K \cdot L. \end{align*}

If $F$ is a surface whose boundary is $K\cup -J$, then in $X=S^3-(K\cup J)$ we have $PD(F)=\alpha_K-\alpha_J$, where these are cohomology classes restricted to $X$. This is because $\partial [F]$ is $[K]-[J]$.

Then we can see, for example, that $$ \operatorname{lk}(L,K\cup -J) = \langle (\alpha_K-\alpha_J)\smile \delta\alpha_L,[S^3-(K\cup J)]\rangle $$ and then making use of linearity of cup products and the bilinear form, and also using excision, we can write that as a difference of linking numbers.

Answer 2

First of all let me refer to Poincare dual of the Alexander dual of the fundamental class of a knot is given by a Seifert surface - proof, where I wrote down a proof of your statement, that the Poincare dual of the Alexander dual of the fundamental class is represented by a Seifert surface of the knot. More precisely we have the following statement:
Let $J\subset S^3$ be a knot and let $F_J$ be some Seifert surface for $J$. Further let $T:\overline{B^2}\times J\rightarrow S^3$ be a thickening with self linking number $0$. We will just write $B^2\times J$ resp. $\overline{B^2}\times J$ for $T(B^2\times J)$ resp. $T(\overline{B^2}\times J)$, if it is clear from the context what we mean. We consider the map $$A_J:\tilde{H}_1(J;\mathbb{Z})\rightarrow \tilde{H}^1(S^3\setminus (B^2\times J);\mathbb{Z})=H^1(S^3\setminus (B^2\times J);\mathbb{Z})$$ and $$\text{PD}_{S^3\setminus(B^2\times J)}:H_2(S^3\setminus(B^2\times J),S^1\times J;\mathbb{Z})\rightarrow H^1(S^3\setminus(B^2\times J);\mathbb{Z}).$$ Then we have $$\underbrace{A_J(\left[ J\right] )}_{\in H^1(S^3\setminus (B^2\times J);\mathbb{Z})}\frown\underbrace{\left[ S^3\setminus (B^2\times J)\right]}_{\in H_3(S^3\setminus(B^2\times J),S^1\times J;\mathbb{Z})} = \left[ F_J'\right]\in H_{2}(S^3\setminus (B^2\times J),S^1\times J;\mathbb{Z}),$$ or in other words $$\text{PD}_{S^3\setminus(B^2\times J)}(\left[ F^\prime_J\right] ) = A_J(\left[ J\right] )=\left[\mu_J^\star\right]\in H^1(S^3\setminus(B^2\times J);\mathbb{Z}),$$ where $F_J':=F_J\cap (S^3\setminus (B^2\times J))$ and $\left[ S^3\setminus (B^2\times J)\right]\in H_3(S^3\setminus(B^2\times J),S^1\times J;\mathbb{Z})$ denotes the fundamental class.
Now let $K,J$ be knots in $S^3$ that do not intersect. Further let $F_K$ be some Seifert-surface of $K$ that is transverse to $J$.
Since $K$ and $J$ are compact and thus some minimal distance apart from each other we can pick some tubular neighborhood $\overline{B^2}\times K$ such that $J\cap(\overline{B^2}\times K)=\emptyset$ and $S^1\times K$ intersects $F_K$ transversally in a single curve.
In particular $F^\prime_K:=F_K\cap(S^3\setminus(B^2\times K))$ is a compact connected codimension $0$ submanifold of $F_K$. Thus we can calculate:
$$\begin{array}{ccl} F_K\cdot J = J\cdot F_K \overset{J\subset X_K}{=} J\cdot F_K^\prime & = & \left.\left(\text{PD}_{S^3\setminus(B^2\times K)}\left(\left[J\right]\right)\smile\text{PD}_{S^3\setminus(B^2\times K)}\left(\left[F^\prime_K\right]\right)\right)\right.\frown\left[S^3\setminus(B^2\times K)\right]\\ & = & \text{PD}_{S^3\setminus(B^2\times K)}\left(\left[F^\prime_K\right]\right)\frown\left.\left(\text{PD}_{S^3\setminus(B^2\times K)}\left(\left[J\right]\right)\frown\left[S^3\setminus(B^2\times K)\right]\right)\right.\\ & = & \left[\mu_K^\star\right]\frown\left[J\right] = \langle\left[\mu_K^\star\right],\left[J\right]\rangle = \text{lk}(K,J) \end{array}$$ Further the statement and proof I refered to above easily translates to the following statement:
Let $K,J\subset S^3$ be two knots that do not intersect and let $F$ be some surface with $\partial F=K\sqcup(-J)$. Further let $T_1:\overline{B^2}\times K\rightarrow S^3$ and $T_2:\overline{B^2}\times J\rightarrow S^3$ be two thickenings with self linking number $0$ and such that $T_1(B^2\times K)\cap T_2(B^2\times J)=\emptyset$. Then we have $$(\left[\mu_K^\star\right]-\left[\mu_J^\star\right])\frown\left[S^3\setminus(B^2\times(K\sqcup J))\right]=\left[F^\prime\right]\in H_2(S^3\setminus(B^2\times (K\sqcup J)),S^1\times(K\sqcup J);\mathbb{Z})$$ or in other words $$\text{PD}_{S^3\setminus(B^2\times(K\sqcup J))}(\left[F^\prime\right])=\left[\mu_K^\star\right]-\left[\mu_J^\star\right]\in H^1(S^3\setminus(B^2\times (K\sqcup J));\mathbb{Z}),$$ where $F^\prime:=F\cap(S^3\setminus(B^2\times(K\sqcup J)))$ and $$\left[S^3\setminus(B^2\times(K\sqcup J))\right]\in H_3(S^3\setminus(B^2\times (K\sqcup J)),S^1\times(K\sqcup J);\mathbb{Z})$$ denotes the fundamental class.
Now let $K,J,M$ be knots in $S^3$ that do not intersect.
Since $K,J$ and $M$ are compact and thus some minimal distance apart from each other we can by the tubular neighborhood theorem pick thickenings $\overline{B^2}\times K$ and $\overline{B^2}\times J$ with $$\text{1. }M\cap\left(\overline{B^2}\times K\right)=\emptyset$$ $$\text{2. }M\cap\left(\overline{B^2}\times J\right)=\emptyset$$ $$\text{3. }\left(\overline{B^2}\times J\right)\cap\left(\overline{B^2}\times K\right)=\emptyset$$ $$\text{4. }S^1\times K\text{ and }S^1\times J\text{ intersect }F\text{ transversally in a single curve.}$$ We denote by $X_L:=S^3\setminus\left(B^2\times K\sqcup B^2\times J\right)$ the link-complement and make the identification $F^\prime:=F\cap X_L$. With this we can now calculate:
$$\begin{array}{rcl} F\cdot M = M\cdot F \underset{M\subset X_L}{=} M\cdot F^\prime & = & \left(\text{PD}_{X_L}\left(\left[M\right]\right)\smile\text{PD}_{X_L}\left(\left[F^\prime\right]\right)\right)\frown\left[X_L\right]\\ & = & \left(\text{PD}_{X_L}\left(\left[F^\prime\right]\right)\right)\frown\left(\text{PD}_{X_L}\left(\left[M\right]\right)\frown\left[X_L\right]\right)\\ & = & \left(\left[\mu_K^\star\right]-\left[\mu_L^\star\right]\right)\frown\left[M\right]\\ & = & \left[\mu_K^\star\right]\frown\left[M\right] - \left[\mu_L^\star\right]\frown\left[M\right]\\ & = & \text{lk}(K,M) - \text{lk}(J,M)\\ & = & \text{lk}(M,K) - \text{lk}(M,J) \end{array}$$ yielding the desired equality.
The problem in your calculation above is that different Poincare duality maps get muddled.