$\operatorname{Lk}(C_1,C_2)= \operatorname{Lk}(C_2,C_1)$

Question

Let $D$ be an oriented link diagram with two components $C_1,C_2$. Then $\operatorname{Lk}(C_1,C_2)= \operatorname{Lk}(C_2,C_1)$.

By definition of Linking number is the following:

The linking number of $\operatorname{Lk}(C_1,C_2)$ is defined to be $[C_2] = \operatorname{Lk}(C_1,C_2)\cdot\mu_1\in H_1(S^3\setminus C_1)\simeq\Bbb Z=\langle\mu_1\rangle$.

I also have a pictorial definition (maybe same as wikipedia says). From that pictorial definition, the above statement is obvious. But from the definition using homology, it's not very obvious to me. How can I show they are the same numbers? * I know the concept of meridian

Note. I'm not very familiar with knot theory. I only know homology theory (Hatcher AT Chapter 2 level). That linking number definition is the first nontrivial definition I met.

Answer 1

The function $\operatorname{Lk(C_1,-)}$ that takes oriented curves in the complement of $C_1$ to the linking number can thought of as being comprised of two steps: (1) take the image of the second curve in $H_1(S^3\setminus C_1)$ and then (2) apply the isomorphism $H_1(S^3\setminus C_1)\to\mathbb{Z}$ given by the meridian $\mu_1$ of $C_1$.

The universal coefficient theorem lets us think of this isomorphism as being an element $\mu_1^*$ of $H^1(S^3\setminus C_1)$, hence we can think of the linking number in terms of the canonical pairing between $H^1(S^3\setminus C_1)$ and $H_1(S^3\setminus C_1)$, with $\operatorname{Lk}(C_1,C_2) = \langle \mu_1^*,[C_2]\rangle$. My goal is to transform this until it is obviously symmetric, though we will also see it is symmetric along the way to this destination. I should say that I don't feel like I succeeded in making it completely obvious, but by the end we more or less see why it calculates the Gauss linking integral.

Where does this $\mu_1^*$ element come from, though? It's Alexander duality: $H_1(C_1) \cong H^1(S^3\setminus C_1)$, and $\mu_1^*$ is the image of the fundamental class of $C_1$ (the one that gives $C_1$ its orientation). Let's recall how this isomorphism works. Let $\nu C_1\subseteq S^3$ denote an open tubular neighborhood of $C_1$, and let $S^3-C_1$ denote the closed manifold $S^3\setminus\nu C_1$. First of all, there is a relative cap product $$\frown:H_3(S^3-C_1,\partial(S^3-C_1))\otimes H^1(S^3-C_1) \to H_2(S^3-C_1,\partial(S^3-C_1))$$ and second of all there is a long exact sequence $$\cdots \to H_2(S^3) \to H_2(S^3,C_1) \xrightarrow{\partial} H_1(C_1) \to H_1(S^3) \to \cdots$$ Recall that $H_2(S^3,C_1)\cong H_2(S^3-C_1,\partial(S^3-C_1))$ by excision along with the fact that the closure of $\nu C_1$ is a collar neighborhood of $C_1$. Let $[S^3]\in H_3(S^3)$ be the fundamental class. The image of this in $H_3(S^3,C_1)\cong H_3(S^3-C_1,\partial(S^3-C_1))$ is the fundamental class of $S^3-C_1$, so we will identify them. Then Alexander duality is the map \begin{align*} D: H^1(S^3\setminus C_1) &\to H_1(C_1) \\ \alpha &\mapsto \partial([S^3] \frown \alpha) \end{align*} Let's work through the inverse to identify $\mu_1^*$. Given the fundamental class $[C_1]\in H_1(C_1)$, then since $H_1(S^3)=H_2(S^3)=0$, there is a unique class $[\Sigma_1]\in H_2(S^3,C_1)$ whose boundary is $[C_1]$. (By the way, we can represent the 2-cycle $\Sigma_1$ by an oriented embedded surface -- this is a Seifert surface for $C_1$. One reason Seifert surfaces are important is that they demarcate when you've gone around $C_1$.) We now want to solve the equation $[S^3]\frown\alpha=[\Sigma_1]$ for some $1$-cocycle $\alpha$. Recall that when $\beta\in H^2(S^3-C_1,\partial(S^3-C_1))$, $$\langle \beta,[\Sigma_1]\rangle = \langle \beta, [S^3]\frown\alpha\rangle = \langle \alpha\smile\beta, [S^3]\rangle$$ Since this holds for all $\beta$, this is saying that $\alpha$ must be the Poincaré dual of $[\Sigma_1]$. We can give a concrete description of the Poincaré dual: $\mu_1^*$ is the class such that $\langle \mu_1^*,[C_2]\rangle=\Sigma_1\cdot C_2$ for all oriented curves $C_2$ in the complement of $C_1$, where $\Sigma_1\cdot C_2$ is the algebraic intersection number, which we can define by $$\Sigma_1\cdot C_2 = \langle PD(\Sigma_1)\smile PD(C_2),[S^3]\rangle$$ where $PD$ denotes the Poincaré dual. Geometrically, it's the count of the signed intersections between $\Sigma_1$ and $C_2$ after they've been perturbed to lie transverse to one another. Incidentally, by taking $\Sigma_1$ to be the canonical Seifert surface given a diagram of $C_1$, you can check that this calculates the linking number defined by a signed count of crossings between $C_1$ and $C_2$ in the diagram. Given this, we might say we're done because that signed count is obviously symmetric.

Let $[\Sigma_2]\in H_2(S^3,C_2)$ represent the Alexander dual of $[C_2]\in H_1(C_2)$. We now have \begin{align*} \operatorname{Lk}(C_1,C_2) &= \Sigma_1 \cdot \partial \Sigma_2 \\ &= \langle PD(\Sigma_1) \smile PD(\partial\Sigma_2), [S^3]\rangle \\ &= \langle PD(\Sigma_1) \smile \delta PD(\Sigma_2), [S^3]\rangle \\ &= \langle \mu_1^* \smile \delta \mu_2^*, [S^3]\rangle \end{align*} where the last line is using the definition $\mu_i^*=PD(\Sigma_i)$.

The cup product $\mu_1^*\smile \mu_2^*$ is an element of $H^2(S^3,C_1\cup C_2)$, and with it we can perform the following calculation: \begin{align*} \operatorname{Lk}(C_1,C_2) - \operatorname{Lk}(C_2,C_1) &= \langle \mu_1^* \smile \delta \mu_2^*, [S^3]\rangle - \langle \mu_2^* \smile \delta \mu_1^*, [S^3]\rangle \\ &= \langle \mu_1^* \smile \delta \mu_2^*, [S^3]\rangle - \langle \delta\mu_1^* \smile \mu_2^*, [S^3]\rangle \\ &= \langle \delta(\mu_1^* \smile \mu_2^*), [S^3]\rangle \\ &= \langle \mu_1^* \smile \mu_2^*, \partial[S^3]\rangle \\ &= \langle \mu_1^* \smile \mu_2^*, 0\rangle \\ &= 0 \end{align*} Therefore the linking number is symmetric!

Let's take a moment to consider what $\delta(\mu_1^* \smile \mu_2^*)$ is. Applying Poincaré duals, this is $PD(\partial(\Sigma_1\cdot\Sigma_2))$. The algebraic intersection of $\Sigma_1$ and $\Sigma_2$ is calculated by putting the surfaces into general position and taking the oriented intersection, which is a bunch of arcs and circles. The boundary of a circle is $0$, but the boundary of an arc is the difference of the endpoints, which is $0$ in $H_0(S^3\setminus C_1\cup C_2)$ since the space is connected.

I've still only showed that the linking number is symmetric, but I promised putting it into a form where it is obviously symmetric. Let $\Delta\subseteq S^3\times S^3$. Inside $H^3(S^3\times S^3,S^3\times S^3\setminus\Delta)$ is a unique class $u$ (essentially the Thom class) that has to do with local orientation of $S^3$, which we can define in a relatively concrete way. Since $S^3$ is parallelizable, we can parameterize a closed tubular neighborhood of $\Delta$ as $S^3\times B^3$, and thus by applying excision, $$H^3(S^3\times S^3,S^3\times S^3\setminus\Delta) \cong H^3(S^3\times S^3,S^3\times S^3-\Delta) \cong H^3(S^3\times B^3, S^3\times S^2)$$ This is isomorphic to $H_3(S^3\times B^3)$ by Poincaré duality, and since $S^3\times B^3$ deformation retracts onto $S^3$, we see the group is isomorphic to $\mathbb{Z}$. Using $[S^3]\in H_3(S^3)$, then let $u$ be the corresponding class in $H^3(S^3\times S^3,S^3\times S^3\setminus\Delta)$. ​On relative $3$-cycles $\gamma$, its defining formula is $$\langle u, \gamma\rangle = \Delta_*([S^3])\cdot\gamma$$ That is, it's the algebraic intersection number with the diagonal. Related to this is the useful formula $$\langle u \smile (\alpha\times\beta),[S^3\times S^3]\rangle = \langle \alpha\smile \beta, [S^3]\rangle$$ So, for example, $$\langle u, \Sigma_1\times C_2\rangle = \Sigma_1\cdot C_2$$

For this next part, I'm going to use a trick to simplify the explanation. If we remove a point from $S^3$, then linking number calculations essentially don't change. Let $u'\in H^3(\mathbb{R}^3\times \mathbb{R}^3,\mathbb{R}^3\times \mathbb{R}^3-\Delta)$ be the restriction of $u$. Consider the long exact sequence $$\cdots \to H^2(\mathbb{R}^3\times\mathbb{R}^3) \to H^2(\mathbb{R}^3\times\mathbb{R}^3-\Delta) \xrightarrow{\delta} H^3(\mathbb{R}^3\times \mathbb{R}^3,\mathbb{R}^3\times \mathbb{R}^3-\Delta) \to H^3(\mathbb{R}^3\times\mathbb{R}^3) \to \cdots$$ Since the outer two cohomology groups are trivial, $\delta$ is an isomorphism, so we can pull the class we found back to a unique class $\alpha\in H^2(\mathbb{R}^3\times\mathbb{R}^3-\Delta)$ with $\delta\alpha=u'$. Thus, $$\langle u', \Sigma_1\times C_2\rangle = \langle \delta\alpha, \Sigma_1\times C_2\rangle = \langle \alpha, \partial(\Sigma_1 \times C_2)\rangle = \langle \alpha, C_1\times C_2\rangle$$ (This is also, equivalently, $\langle \alpha\smile (\mu_1^*\times\mu_2^*),[S^3\times S^3] \rangle$.)

Therefore, there is this $\alpha\in H^2(\mathbb{R}^3\times\mathbb{R}^3-\Delta)$ that does not depend on $C_1$ or $C_2$ such that $$\operatorname{Lk}(C_1,C_2) = \langle\alpha, C_1\times C_2\rangle$$ I'm not sure the best way to see this is symmetric directly, but one way is to go back to 1833 and take a page from Gauss's die Anzahl der Umschlingungen in Nachlass zur Electrodynamic where he gives the integral $$ \frac{1}{4\pi} \int_{S^1\times S^1} \frac{\gamma_1(s)-\gamma_2(t)}{\lvert \gamma_1(s)-\gamma_2(t)\lvert^3} \cdot (\gamma_1'(s)\times\gamma_2'(t))\,ds\,dt $$ where $\gamma_1$ and $\gamma_2$ are differentiable parameterizations of $C_1$ and $C_2$. Using $\mathbb{R}$ coefficients (and the isomorphism to de Rham cohomology), we can represent $\alpha$ by the closed $2$-form $\omega$ on $\mathbb{R}^3\times\mathbb{R}^3-\Delta$ given by $$\omega(v,w) = \frac{1}{4\pi\lvert x-y \rvert^3}\det(x-y,v,w)$$ where $x,y\in\mathbb{R}^3$ parameterize $\mathbb{R}^3\times\mathbb{R}^3-\Delta$. We can check that $\omega$ is closed and that it calculates the linking number of a Hopf link, so it must represent $\alpha$ since $H^2(\mathbb{R}^3\times\mathbb{R}^3-\Delta;\mathbb{R})$ is $1$-dimensional. Hence, $$\operatorname{Lk}(C_1,C_2) = \langle\alpha, C_1\times C_2\rangle = \int_{C_1\times C_2}\omega$$ We can see that $\omega_{y,x}(w,v)=\omega_{x,y}(v,w)$ since a double negative cancels out, so $$\operatorname{Lk}(C_2,C_1) = \int_{C_2\times C_1}\omega = \int_{C_1\times C_2}\omega = \operatorname{Lk}(C_1,C_2)$$ and we see the linking number is symmetric.

Answer 2

In my other answer, I use (co)homology and dualities to show this definition of linking number is symmetric - it's something I had worked out before and this was a good opportunity to write it down.

However, the question really only was "why is the homological definition the same as the diagram-based definition." Here's how I've thought about it. We're going to take the image of $C_2$ in $H_1(S^3\setminus C_1)$.

First, take a diagram of $C_1$ and $C_2$. Second, "lift" the $C_2$ part above the plane of the diagram, where when there's a part of $C_2$ that lies under $C_1$, add in line segments to detour under $C_1$ (this is illustrated below). The lifted curve is homologous to the original curve because it's homotopic, but it's also easy to work out what boundaries of 2-chains you're adding in to make this happen. Third, at the places where $C_2$ still goes under $C_1$, add in the boundary of a 2-chain to "bubble off" a loop around $C_1$.

This shows the 2-chain whose boundary, when added in, bubbles off the loop:

After bubbling off the loops, the remaining part of $C_2$ above the diagram lies in a subspace homeomorphic to an open ball $B$. Recall that $H_1(B)=0$, so this means this part of $C_2$ is nullhomologous in $B$ (i.e., it is the boundary of a 2-chain in $B$), so it is the boundary of a 2-chain in $S^3\setminus C_1$ and can be removed.

All that's left are these rings on $C_1$. If you are familiar with those "bead maze" toys for babies, the principle is now very similar: take all of the loops ("beads") and drag them along $C_1$ until they are co-located. Unlike beads in a bead maze, loops have an orientation. If two of the loops have opposite orientation, they cancel out. What's left after cancellation is some multiple of $\mu_1$, and if you think through what we did, the coefficient in front of $\mu_1$ is the number of positive crossings where $C_2$ goes under $C_1$ minus the number of negative crossings where $C_2$ goes under $C_1$. That matches Wikipedia's formula $\operatorname{Lk}(C_1,C_2)=n_1-n_4$ for linking number!

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Side note: I'll take a moment to illustrate what the relation is between this and Alexander duality in the other answer. If you lift up $C_2$ (letting it pass through $C_1$!) up above the plane of the diagram, then it is nullhomologous using a 2-chain in the ball $B$ that contains it. Adding this to the 2-chains involved in the lift, we get a 2-chain $\Sigma_2$ in $S^3$ whose boundary is $C_2$. Notice that with this setup, $C_1$ intersects $\Sigma_2$ exactly where we'd be bubbling off loops:

Pretty much the entire idea is that to account for the bubbled off loops, you can instead calculate the algebraic intersection number between $C_1$ and $\Sigma_2$. Algebraic intersection number in this exemplary situation is from taking an oriented basis of tangent vectors of $\Sigma_2$ at the intersection along with the tangent vector of $C_1$ at the intersection, then taking the sign of the determinant of the $3\times 3$ matrix of these vectors. This number tells you whether it's a $\mu_1$ or a $-\mu_1$ that bubbles off.