Gauss' Linking Number III

The pullback computation: from cohomology to Gauss' double integral

\newcommand{\kl}{\widehat{K\text{-}L}}

In the previous post, we showed that the linking number can be computed as

$$\operatorname{Link}(K,L)=\int_{M\times N}F^*\omega,\qquad F(s,t)=K(s)-L(t),$$

where $\omega$ is any closed form on $\mathbb{R}^{m+n+1}\setminus\{0\}$ representing the normalized generator of the top de Rham cohomology. We constructed these generators explicitly by solving an ODE, and we tested the formula in $\mathbb{R}^2$, where it recovers the classical winding number.

Now we turn to the main event: two closed curves in $\mathbb{R}^3$. The pullback computation in this case is more involved—there are more differentials to track—but the reward is Gauss’ original double integral, with every piece of its kernel explained.

Setup

Specialize to $m=n=1$, so that $K,L\colon S^1\to\mathbb{R}^3$ are two disjoint closed curves. From the table in the previous post, the normalized generator of $H^2(\mathbb{R}^3\setminus\{0\})$ is

$$\omega=\frac{1}{4\pi r^3}\left(x\,dy\wedge dz+y\,dz\wedge dx+z\,dx\wedge dy\right),$$

where $r=\sqrt{x^2+y^2+z^2}$.

To match Gauss’ notation, write

$$K(s)=(x,y,z),\qquad L(t)=(x',y',z').$$

The difference map is $F(s,t)=(x-x',\;y-y',\;z-z')$. We need to compute $F^*\omega$ as a $2$-form on $S^1\times S^1$.

Pulling back the differentials

The pullback of the coordinate differentials to $S^1\times S^1$ is:

$$d(x-x')=x_s\,ds-x'_t\,dt,\qquad d(y-y')=y_s\,ds-y'_t\,dt,\qquad d(z-z')=z_s\,ds-z'_t\,dt,$$

where subscripts denote derivatives with respect to the curve parameters.

The structural simplification

Before plunging into the algebra, notice a key simplification. The domain $S^1\times S^1$ is two-dimensional, so any $2$-form on it is a multiple of $ds\wedge dt$. When we expand wedge products like $d(y-y')\wedge d(z-z')$, the terms $ds\wedge ds$ and $dt\wedge dt$ vanish identically. Only the mixed terms proportional to $ds\wedge dt$ survive.

This means the final answer must be bilinear in the tangent vectors $K'(s)$ and $L'(t)$—one factor from each curve. The algebra is bookkeeping; the structure is predetermined.

Computing the wedge products

We compute each of the three wedge products appearing in $\omega$:

$$\begin{aligned} d(y-y')\wedge d(z-z') &=(y_s\,ds-y'_t\,dt)\wedge(z_s\,ds-z'_t\,dt)\\ &=(-y_s z'_t+z_s y'_t)\,ds\wedge dt, \end{aligned}$$ $$\begin{aligned} d(z-z')\wedge d(x-x') &=(z_s\,ds-z'_t\,dt)\wedge(x_s\,ds-x'_t\,dt)\\ &=(-z_s x'_t+x_s z'_t)\,ds\wedge dt, \end{aligned}$$ $$\begin{aligned} d(x-x')\wedge d(y-y') &=(x_s\,ds-x'_t\,dt)\wedge(y_s\,ds-y'_t\,dt)\\ &=(-x_s y'_t+y_s x'_t)\,ds\wedge dt. \end{aligned}$$

Assembling the numerator

The numerator of $F^*\omega$ is

$$(x-x')\,d(y-y')\wedge d(z-z') +(y-y')\,d(z-z')\wedge d(x-x') +(z-z')\,d(x-x')\wedge d(y-y').$$

Substituting the wedge products from above and collecting, this becomes

$$\Big[ (x{-}x')(-y_s z'_t+z_s y'_t) +(y{-}y')(-z_s x'_t+x_s z'_t) +(z{-}z')(-x_s y'_t+y_s x'_t) \Big]\,ds\wedge dt.$$

We can rewrite this expression as a dot product:

$$\bigl(K(s)-L(t)\bigr)\cdot\mathbf{v}\,\;ds\wedge dt,$$

where $\mathbf{v}$ is the vector with components

$$(y_s z'_t-z_s y'_t,\quad z_s x'_t-x_s z'_t,\quad x_s y'_t-y_s x'_t).$$

This vector is exactly the cross product $K'(s)\times L'(t)$, so the numerator is the scalar triple product $(K(s)-L(t))\cdot(K'(s)\times L'(t))\,ds\wedge dt$.

Gauss’ integral

Putting everything together:

$$F^*\omega = \frac{1}{4\pi} \frac{(K(s)-L(t))\cdot(K'(s)\times L'(t))} {\|K(s)-L(t)\|^3}\,ds\wedge dt.$$

Integrating over $S^1\times S^1$:

$$\boxed{ \operatorname{Link}(K,L) = \frac{1}{4\pi} \int_{S^1\times S^1} \frac{(K(s)-L(t))\cdot(K'(s)\times L'(t))} {\|K(s)-L(t)\|^3}\,ds\,dt. }$$

This is Gauss’ original linking integral.

Reading the formula

Each piece of the integrand now has a clear origin:

• The displacement $K(s)-L(t)$ comes from the difference map $F$.
• The cross product $K'(s)\times L'(t)$ arises from the wedge product of pulled-back differentials.
• The dot product of these two comes from the contraction with the position coordinates in $\omega$.
• The $1/r^3$ denominator comes from the fact that $\omega$ is the Hodge dual of $d(-1/4\pi r)$, reflecting the decay of the Coulomb potential.
• The $1/(4\pi)$ normalization ensures that $\omega$ integrates to $1$ over $S^2$—it is the total solid angle.

Nothing is put in by hand.

A remark on the surface-curve case

For completeness: the same machine works in $\mathbb{R}^4$ to give a linking formula for a closed curve $K\colon S^1\to\mathbb{R}^4$ and a closed surface $L\colon \Sigma^2\to\mathbb{R}^4$. The generator of $H^3(\mathbb{R}^4\setminus\{0\})$ is the $d=4$ entry from the table in Post II, and the pullback computation produces a triple integral over $S^1\times\Sigma$ involving the $4$-dimensional analogue of the scalar triple product. We will not carry this out here, but the structure is identical.

What comes next

We have derived the formula. In the next post, we will put it to work: compute the linking number of the Hopf link by hand, verify it numerically, and prove that nontrivial links exist.