Gauss' Linking Number III

The pullback computation: from cohomology to Gauss' double integral

Third in a four-part series deriving Gauss’ linking integral from first principles. The setup is in the first post.

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

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

The three coefficients here are the components of a single familiar vector, up to an overall sign. Setting

$$\mathbf{v}=(y_s z'_t-z_s y'_t,\quad z_s x'_t-x_s z'_t,\quad x_s y'_t-y_s x'_t),$$

which is exactly the cross product $K'(s)\times L'(t)$, each bracket above is $-\mathbf{v}$ in the corresponding slot. So the numerator is

$$-\bigl(K(s)-L(t)\bigr)\cdot\bigl(K'(s)\times L'(t)\bigr)\,ds\wedge dt = \bigl(L(t)-K(s)\bigr)\cdot\bigl(K'(s)\times L'(t)\bigr)\,ds\wedge dt.$$

That minus sign is worth pausing on rather than absorbing, because it is easy to lose and it decides the answer. Its source is that the second curve enters $F$ with a minus sign: $F(s,t)=K(s)-L(t)$, so

$$\partial_s F = K'(s), \qquad \partial_t F = -L'(t),$$

and the frame the difference map actually carries onto the sphere is $(K', -L')$, whose cross product is $-\,K'\times L'$. The cross product of the two tangent vectors is the natural thing to write down, but it is not the oriented frame of the map—it is that frame with one axis flipped. Keeping track of the flip is what makes the next line agree with Gauss.

Gauss’ integral

Putting everything together:

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

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

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

This is Gauss’ original linking integral—and now literally so. Writing $K=(x,y,z)$ and $L=(x',y',z')$ and expanding the triple product back out, the numerator is

$$(x'-x)(dy\,dz'-dz\,dy')+(y'-y)(dz\,dx'-dx\,dz')+(z'-z)(dx\,dy'-dy\,dx'),$$

which is the expression in Gauss’ notebook, primes and all. The displacement really does run from the first curve to the second, exactly as he wrote it.

Reading the formula

Each piece of the integrand now has a clear origin:

• The displacement $L(t)-K(s)$ comes from the difference map $F$, with the sign the wedge products handed us.
• 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.

Below, every one of those pieces is on the curves at once. Move either point: $K(s)$ and its tangent $K'(s)$ in blue, $L(t)$ and $L'(t)$ in orange, the displacement between them in black, and their cross product $K'(s)\times L'(t)$ in violet. The displacement is drawn from $K$ to $L$—the direction Gauss wrote, and the one the minus sign above earned. Watch the violet arrow swing from one side of the displacement to the other as the pair moves: that is the numerator changing sign.

A remark on the surface-curve case

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 analog of the scalar triple product.

This is not just a formal exercise. Here $m=1$ and $n=2$, so $m+n+1=4$ and we are in the Goldilocks dimension of Post I—which is to say a circle and a sphere in $\mathbb{R}^4$ can genuinely be linked, and neither can be pulled free of the other. That deserves its own post, and it will get one.

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.