Gauss' Linking Number II

From degree to integral: de Rham cohomology and the winding number

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

In the previous post, we showed that the linking number of two disjointly embedded manifolds $M^m$ and $N^n$ in $\mathbb{R}^{m+n+1}$ can be expressed as the degree of the normalized difference map

$$\kl \colon M \times N \to S^{m+n}.$$

This description is conceptually satisfying, but it is not yet computational. The degree is defined by counting preimages with signs, which requires finding all solutions to a nonlinear equation—awkward at best.

The purpose of this post is to explain how to convert this topological invariant into an integral. The key tool is de Rham cohomology, and the key computation is a short ODE. At the end, we will have an explicit integral formula for the linking number in all dimensions—and we will test it on the simplest case: a point and a curve in $\mathbb{R}^2$.

The big idea: degree as an integral

The bridge between topology and analysis is the following remarkable fact.

Theorem. Let $f\colon X^k\to Y^k$ be a smooth map between closed oriented $k$-manifolds, and let $\alpha$ be a volume form on $Y$ normalized so that $\int_Y\alpha=1$. Then $\deg(f)=\int_X f^*\alpha.$

This theorem deserves a moment of reflection. Why should it be true?

Fix a regular value $q\in Y$, and think of $\alpha$ as a “bump” form concentrated near $q$. The pullback $f^*\alpha$ is then concentrated near the preimages $f^{-1}(q)$, and the integral picks up a contribution of $\pm 1$ at each preimage according to whether $f$ preserves or reverses orientation there. Summing these contributions gives the degree. The content of the theorem is that this sum is the same no matter where on $Y$ we concentrate the bump—which is ultimately a consequence of $\alpha$ being closed.

Applying this to the normalized difference map gives our first integral formula:

$$\operatorname{Link}(K,L)=\int_{M\times N}\kl^*\alpha.$$

This is already a formula for the linking number. But it is not yet practical: the map $\kl$ lands on the sphere $S^{m+n}$, which requires multiple coordinate charts. Everything that follows is about turning this abstract formula into something we can write down in a single integral.

Moving to Euclidean space

Factor the normalized difference map as

$$\kl = \pi\circ F, \qquad F(s,t)=K(s)-L(t), \qquad \pi(x)=\frac{x}{\|x\|}.$$

Then

$$\operatorname{Link}(K,L)=\int_{M\times N}F^*(\pi^*\alpha).$$

The map $F$ is just a subtraction—completely explicit. The map $\pi$ is the radial projection onto the sphere. Since $\mathbb{R}^{m+n+1}\setminus\{0\}$ deformation retracts onto $S^{m+n}$ via $\pi$, the pullback $\pi^*\alpha$ is a closed form on punctured Euclidean space, and we can work entirely in Euclidean coordinates.

But what is this form? To answer that, we need a way to translate between topology and differential forms.

De Rham cohomology

The de Rham cohomology of a smooth manifold $X$ is defined as

$$H^k_{\mathrm{dR}}(X)=\frac{\{\text{closed }k\text{-forms on }X\}}{\{\text{exact }k\text{-forms on }X\}}.$$

It measures the “topological complexity” of $X$ through the lens of calculus. Two closed forms represent the same cohomology class precisely when they differ by an exact form—and therefore give the same value when integrated over any closed submanifold. The foundational result (de Rham’s theorem) is that this construction recovers the same invariants as singular cohomology: the topology of $X$ determines which closed forms are exact and which are not.

For our purposes, the essential computation is:

$$H^{m+n}_{\mathrm{dR}}(\mathbb{R}^{m+n+1}\setminus\{0\})\cong\mathbb{R},$$

generated by $[\pi^*\alpha]$. Removing the origin from $\mathbb{R}^{m+n+1}$ creates exactly one “hole,” and this one-dimensional cohomology group detects it.

Why does this help? Because it gives us freedom. The form $\pi^*\alpha$ is one representative of the generator, but any closed $(m+n)$-form on $\mathbb{R}^{m+n+1}\setminus\{0\}$ whose integral over a sphere enclosing the origin is $1$ represents the same cohomology class, and therefore gives the same linking number when pulled back and integrated. We can choose whichever representative is easiest to work with:

General formula. Let $\omega$ be any closed $(m+n)$-form on $\mathbb{R}^{m+n+1}\setminus\{0\}$ with $\int_{S^{m+n}}\omega=1$. Then $\operatorname{Link}(K,L)=\int_{M\times N}F^*\omega, \qquad F(s,t)=K(s)-L(t).$

The entire computation has been reduced to a single task: find an explicit $\omega$.

Constructing the generator

We seek a closed $(m+n)$-form $\omega$ on $\mathbb{R}^{m+n+1}\setminus\{0\}$, normalized so its integral over the unit sphere is $1$. Several considerations guide the construction.

Rotational invariance. Since no direction in $\mathbb{R}^{m+n+1}$ is preferred, we look for a form that is invariant under rotations. This is both aesthetically natural and practically essential: it constrains the ansatz enough to make the problem tractable.

The Hodge star route. How do we systematically build closed forms on punctured Euclidean space? A clean approach: given any smooth function $\rho$ on $\mathbb{R}^{m+n+1}\setminus\{0\}$, the $1$-form $d\rho$ is exact, and its Hodge dual

$$\omega = *\,d\rho$$

is an $(m+n)$-form. When is $\omega$ closed? A calculation shows

$$d\omega = d*d\rho = (\Delta\rho)\,\mathrm{vol},$$

so $\omega$ is closed if and only if $\rho$ is harmonic: $\Delta\rho = 0$.

The key point is that for the ansatz $\omega = *\,d\rho$, closedness reduces to Laplace’s equation on $\rho$ alone.

Radial symmetry. Rotational invariance forces $\rho$ to depend only on $r=\|x\|$. So we need a radially symmetric harmonic function on punctured space: $\Delta\rho(r)=0$.

The ODE

In $\mathbb{R}^d$ (with $d=m+n+1$), the Laplacian of a radial function $\rho(r)$ is

$$\Delta\rho=\rho''+\frac{d-1}{r}\rho'=\frac{1}{r^{d-1}}\frac{d}{dr}\bigl(r^{d-1}\rho'\bigr).$$

Setting this to zero gives $r^{d-1}\rho'=C$, which integrates to:

$$\rho(r)= \begin{cases} a+b\ln r & \text{if } d=2,\\[6pt] a+\dfrac{b}{r^{d-2}} & \text{if } d\geq 3. \end{cases}$$

The constant $a$ is killed by $d\rho$, so only $b$ matters. We fix $b$ by the normalization $\int_{S^{d-1}}\omega=1$.

A computation (Appendix A) shows that on the unit sphere, $*\,d\rho$ restricts to $b\cdot(d-2)\,\mathrm{vol}_{S^{d-1}}$ for $d\geq 3$ and $b\,\mathrm{vol}_{S^1}$ for $d=2$. Setting the integral to $1$ determines $b$, and we also choose the sign of $\rho$ so that the resulting form has the correct orientation (compatible with $S^{d-1}$ as the boundary of a ball). The results:

Ambient dimension $d$	$\rho(r)$	Normalized generator $\omega = *\,d\rho$
$2$	$\dfrac{1}{2\pi}\ln r$	$\dfrac{1}{2\pi r^2}(x\,dy - y\,dx)$
$3$	$\dfrac{-1}{4\pi r}$	$\dfrac{1}{4\pi r^3}(x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy)$
general $d$	$\dfrac{-1}{(d{-}2)\,\mathrm{Vol}(S^{d-1})r^{d-2}}$	$\dfrac{1}{\mathrm{Vol}(S^{d-1})r^d}\displaystyle\sum_{i=1}^d(-1)^{i-1}x^i\,dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^d$

The reader may recognize the entries in the $\rho$ column. The function $-1/(4\pi r)$ is the fundamental solution of the Laplacian in $\mathbb{R}^3$—the Coulomb potential, or the gravitational potential of a point mass. The function $(1/2\pi)\ln r$ is its counterpart in $\mathbb{R}^2$. This is not a coincidence: the cohomological generator on $\mathbb{R}^d\setminus\{0\}$ is always the Hodge dual of the gradient of the Green’s function for the Laplacian. The linking number is, in a precise sense, measuring the “electrostatic flux” of one manifold as seen from the other.

Warm-up: the winding number in $\mathbb{R}^2$

Before tackling Gauss’ integral in $\mathbb{R}^3$, let us test the machine on the simplest possible case.

Take $m=0$, $n=1$, and $d=m+n+1=2$. The manifold $M^0$ is a point and $N^1=S^1$ is a closed curve. The “linking number” of a point $K\in\mathbb{R}^2$ with a disjoint closed curve $L\colon S^1\to\mathbb{R}^2$ is the classical winding number of $L$ around $K$.

The general formula says

$$\operatorname{Link}(K,L)=\int_{S^1}F^*\omega,\qquad F(t)=K-L(t),$$

where $\omega$ is the $d=2$ entry from our table:

$$\omega=\frac{1}{2\pi r^2}(x\,dy-y\,dx).$$

Write $K=(a,b)$ and $L(t)=(x(t),y(t))$. Then $F(t)=(a-x(t),\,b-y(t))$. Pulling back the differentials:

$$d(a-x) = -x'\,dt, \qquad d(b-y)=-y'\,dt.$$

Substituting into $\omega$:

$$F^*\omega = \frac{1}{2\pi}\frac{(a-x)(-y')-(b-y)(-x')}{(a-x)^2+(b-y)^2}\,dt.$$

Integrating:

$$\operatorname{Link}(K,L)=\frac{1}{2\pi}\int_0^{2\pi}\frac{(b-y)\,x'-(a-x)\,y'}{(a-x)^2+(b-y)^2}\,dt.$$

If we place $K$ at the origin, this simplifies to

$$\frac{1}{2\pi}\int_0^{2\pi}\frac{x\,y'-y\,x'}{x^2+y^2}\,dt,$$

which those who have taken complex analysis will recognize as $\frac{1}{2\pi}\oint d(\arg z)$—the total angle swept out by the curve around the origin, divided by $2\pi$.

And the formula was produced mechanically by the same procedure that will yield Gauss’ double integral.

What comes next

We have built the general machine:

1. The linking number is a degree.
2. Degree can be computed as the integral of a pullback form.
3. De Rham cohomology tells us that any normalized closed form will do.
4. The Hodge star ansatz reduces the construction to a harmonic function.
5. Radial symmetry reduces the PDE to an ODE, which we solve explicitly.

We have tested it in $\mathbb{R}^2$ and recovered the winding number. In the next post, we specialize to $\mathbb{R}^3$ and carry out the pullback computation that produces Gauss’ linking integral—the scalar triple product kernel, the $1/r^3$ denominator, and all.

---

Appendix A. Normalization of the generator

We verify the normalization integrals.

Case $d\geq 3$. Take $\rho(r) = b/r^{d-2}$. Then

$$d\rho = -b(d-2)\frac{1}{r^{d-1}}\,dr = -b(d-2)\frac{1}{r^d}\sum_i x^i\,dx^i.$$

On the unit sphere $S^{d-1}$ (where $r=1$), the Hodge dual of the radial $1$-form $dr$ is the volume form $\mathrm{vol}_{S^{d-1}}$. More precisely, the Hodge star maps the inward-pointing $1$-form $-dr/r^d$ to the positively oriented volume form on $S^{d-1}$ (with the convention that $S^{d-1}$ is oriented as the boundary of the ball). Thus

$$*\,d\rho\big|_{S^{d-1}}=b(d-2)\,\mathrm{vol}_{S^{d-1}},$$

and $\int_{S^{d-1}}*\,d\rho = b(d-2)\,\mathrm{Vol}(S^{d-1})$. Setting this to $1$ gives

$$b=\frac{1}{(d-2)\,\mathrm{Vol}(S^{d-1})}.$$

Since we need $d\rho$ to point inward (so that $*\,d\rho$ has the correct outward orientation on spheres), we take $\rho = -b/r^{d-2}$ with $b>0$.

Case $d=2$. Take $\rho(r) = b\ln r$. Then $d\rho = (b/r)\,dr$, and

$$*\,d\rho\big|_{S^1}=b\,\mathrm{vol}_{S^1},$$

so $\int_{S^1}*\,d\rho = 2\pi b$. Setting this to $1$ gives $b=1/(2\pi)$.