Gauss' Linking Number IV

Turning the machine on: evaluating Gauss' integral and proving links exist

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

We have spent three posts building a machine. In Post I, we defined the linking number as the degree of the normalized difference map. In Post II, we used de Rham cohomology to convert degree into an integral. In Post III, we carried out the pullback in $\mathbb{R}^3$ and derived Gauss’ double integral:

$$\operatorname{Link}(K,L) = \frac{1}{4\pi} \int_0^{2\pi}\int_0^{2\pi} \frac{(K(s)-L(t))\cdot(K'(s)\times L'(t))} {\|K(s)-L(t)\|^3}\,ds\,dt.$$

It is time to turn the machine on. We will write down three pairs of curves—one linked, one obviously unlinked, and a one-parameter family interpolating between them—and set up the Gauss integral explicitly for each. We will evaluate these integrals, see integers emerge from continuous integrands, and prove that the Hopf link cannot be pulled apart.

I. The Hopf link

Let

$$K(s) = (\cos s,\; \sin s,\; 0), \qquad L(t) = (1+\cos t,\; 0,\; \sin t),$$

for $s,t\in[0,2\pi]$. The curve $K$ is a unit circle in the $xy$-plane; the curve $L$ is a unit circle in the $xz$-plane, centered at $(1,0,0)$.

These curves are disjoint: $K$ lies in $z=0$, $L$ lies in $y=0$, and checking all four candidate intersection points ($\sin s=0$, $\sin t=0$) confirms no common point. Yet $L$ passes through the disk bounded by $K$—at $t=\pi$, the point $L(\pi)=(0,0,0)$ is the center of $K$.

Setting up the integrand

Tangent vectors:

$$K'(s) = (-\sin s,\;\cos s,\;0), \qquad L'(t)=(-\sin t,\;0,\;\cos t).$$

Cross product:

$$K'(s)\times L'(t) = (\cos s\cos t,\;\sin s\cos t,\;\cos s\sin t).$$

Displacement:

$$K(s)-L(t) = (\cos s-1-\cos t,\;\sin s,\;-\sin t).$$

Taking the dot product of the displacement with the cross product and simplifying (using $\cos^2+\sin^2=1$ throughout):

Numerator:

$$\mathcal{N}(s,t) = (K-L)\cdot(K'\times L') = \cos t - \cos s\,(1+\cos t).$$

Denominator (before the $3/2$ power):

$$\|K-L\|^2 = 3+2\cos t-2\cos s\,(1+\cos t).$$

The explicit integral

Gauss’ formula becomes

$$\operatorname{Link}(K,L) = \frac{1}{4\pi} \int_0^{2\pi}\!\!\int_0^{2\pi} \frac{\cos t-\cos s\,(1+\cos t)} {\bigl[3+2\cos t-2\cos s\,(1+\cos t)\bigr]^{3/2}} \,ds\,dt.$$

This is a completely explicit double integral of a smooth, bounded function on the torus $[0,2\pi]^2$.

A few observations before evaluating:

The integrand is smooth. Since the curves are disjoint, $\|K(s)-L(t)\|>0$ everywhere. In fact, the minimum distance is $1$, achieved at $(s,t)=(0,\pi/2)$.

The numerator changes sign. At $(s,t)=(\pi,0)$ the numerator is $1-(-1)(2)=3>0$. At $(s,t)=(0,\pi)$ it is $-1-(1)(0)=-1<0$. The integrand oscillates between positive and negative values across the torus.

The integral has no closed form. The inner integral (in either variable) cannot be evaluated in terms of elementary functions.

Numerical evaluation

Evaluating by composite Simpson’s rule on an $N\times N$ grid:

Grid size $N$	Computed value
$20$	$-0.99906$
$50$	$-0.999997$
$100$	$-1.00000000$
$200$	$-1.00000000$

The integral converges rapidly (the integrand is smooth and periodic) and stabilizes to

$$\operatorname{Link}(K,L) = -1.$$

The sign is negative—this reflects the handedness of the link (how the orientations of $K$ and $L$ relate to the threading direction). Reversing the orientation of either curve would flip the sign. What matters topologically is that the linking number is nonzero.

II. Parallel circles

For contrast, consider two parallel circles at different heights:

$$K(s) = (\cos s,\;\sin s,\;0), \qquad L(t)=(\cos t,\;\sin t,\;1).$$

These are congruent unit circles, one in $z=0$ and one in $z=1$, clearly not linked—each can be slid away from the other without crossing.

Setting up the integrand

Tangent vectors:

$$K'(s) = (-\sin s,\;\cos s,\;0), \qquad L'(t)=(-\sin t,\;\cos t,\;0).$$

Cross product:

$$K'(s)\times L'(t) = \det\begin{pmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-\sin s&\cos s&0\\-\sin t&\cos t&0\end{pmatrix} = \bigl(0,\;0,\;\sin(t-s)\bigr).$$

Since both tangent vectors are horizontal (zero $z$-component), their cross product is purely vertical. The sign is worth noting: when $t>s$, the cross product points upward.

Displacement:

$$K(s)-L(t) = (\cos s-\cos t,\;\sin s-\sin t,\;-1).$$

Numerator:

$$\mathcal{N}(s,t) = (K-L)\cdot(K'\times L') = (-1)\cdot\sin(t-s) = \sin(s-t).$$

Only the $z$-components contribute to the dot product, since the cross product has zero $x$- and $y$-components.

Squared distance:

$$\|K-L\|^2 = (\cos s-\cos t)^2+(\sin s-\sin t)^2+1 = 2-2\cos(s-t)+1=3-2\cos(s-t).$$

The explicit integral

$$\operatorname{Link}(K,L) = \frac{1}{4\pi} \int_0^{2\pi}\!\!\int_0^{2\pi} \frac{\sin(s-t)} {\bigl[3-2\cos(s-t)\bigr]^{3/2}} \,ds\,dt.$$

Something remarkable has happened: the integrand depends only on the difference $s-t$. This is a consequence of the rotational symmetry of the configuration—rotating both curves by the same angle leaves everything unchanged.

Analytic evaluation

Because the integrand depends only on $u=s-t$, the double integral factors. Fix any $t$ and integrate over $s$; by the substitution $u=s-t$ (with $2\pi$-periodicity):

$$\int_0^{2\pi}\frac{\sin(s-t)}{[3-2\cos(s-t)]^{3/2}}\,ds =\int_0^{2\pi}\frac{\sin u}{[3-2\cos u]^{3/2}}\,du.$$

This integral has a closed-form antiderivative. Setting $w=\cos u$, $dw=-\sin u\,du$:

$$\int\frac{\sin u}{[3-2\cos u]^{3/2}}\,du = \int\frac{-dw}{[3-2w]^{3/2}} = \left.\frac{-1}{\sqrt{3-2w}}\right| = \frac{-1}{\sqrt{3-2\cos u}}.$$

Evaluating from $0$ to $2\pi$:

$$\frac{-1}{\sqrt{3-2\cos(2\pi)}}-\left(\frac{-1}{\sqrt{3-2\cos 0}}\right) = \frac{-1}{\sqrt{1}}+\frac{1}{\sqrt{1}}=0.$$

The inner integral vanishes identically, independent of $t$. Therefore:

$$\operatorname{Link}(K,L) = 0.$$

No numerical quadrature needed. The rotational symmetry of the configuration forces the integrand into a form whose antiderivative is periodic, so the integral over a full period vanishes exactly.

III. A one-parameter family

We now connect the previous two examples by continuously deforming the linked pair into an unlinked pair.

Fix $K(s)=(\cos s,\sin s,0)$ as before, and let

$$L_c(t) = (c+\cos t,\;0,\;\sin t), \qquad c>0.$$

This is a circle in the $xz$-plane centered at $(c,0,0)$. When $c=1$, this is the Hopf link. As $c$ increases, the center of $L_c$ slides along the $x$-axis.

When do the curves intersect? Setting $K(s)=L_c(t)$ requires $\sin s=0$ and $\sin t=0$. Checking the four cases: the only positive solution is $s=0,\,t=\pi$, giving $K(0)=(1,0,0)$ and $L_c(\pi)=(c-1,0,0)$. These coincide when $c=2$.

So the curves are disjoint for $c\neq 2$, and they touch at the single point $(1,0,0)$ when $c=2$.

The general integrand

Since $L_c'=L'$ (the derivative is independent of $c$), the cross product $K'\times L_c'$ is the same as in the Hopf link computation. Only the displacement changes.

Numerator:

$$\mathcal{N}_c(s,t) = \cos t - \cos s\,(1+c\cos t).$$

Squared distance:

$$\|K-L_c\|^2 = 2+c^2+2c\cos t-2\cos s\,(c+\cos t).$$

The integral:

$$\operatorname{Link}(K,L_c) = \frac{1}{4\pi} \int_0^{2\pi}\!\!\int_0^{2\pi} \frac{\cos t-\cos s\,(1+c\cos t)} {\bigl[2+c^2+2c\cos t-2\cos s\,(c+\cos t)\bigr]^{3/2}} \,ds\,dt.$$

One can check that $c=1$ recovers the Hopf link integrand exactly (set $c=1$: denominator becomes $3+2\cos t-2\cos s(1+\cos t)$, as before).

What happens as $c$ varies

The integrand varies smoothly with $c$ for $c\neq 2$. But the linking number—being an integer—cannot change continuously. Something discontinuous must happen at $c=2$.

Evaluating numerically:

$c$	$\operatorname{Link}(K,L_c)$
$0.5$	$-1.000$
$1.0$	$-1.000$
$1.5$	$-1.000$
$1.9$	$-1.000$
$1.99$	$-1.000$
$2.01$	$\phantom{-}0.000$
$2.1$	$\phantom{-}0.000$
$3.0$	$\phantom{-}0.000$

For all $c<2$ the integral is $-1$; for all $c>2$ it is $0$. The transition is sharp: the integer jumps at the exact moment the curves touch.

This is the topological content made vivid. As $c$ approaches $2$ from below, the integrand develops a growing peak near the point $(s,t)=(0,\pi)$ where the curves nearly touch—the denominator $\|K-L_c\|^3$ shrinks toward zero there. But the integral stubbornly holds at $-1$, forced by topology to remain an integer. At $c=2$ the curves intersect, the integrand becomes singular, and the integral is undefined. For $c>2$ the singularity has passed, the integrand is smooth again, and the integral snaps to $0$.

There is no gradual transition. The linking number does not fade from $-1$ to $0$. It is $-1$, then undefined, then $0$. This is what it means for a topological invariant to be discrete.

What have we proved?

Theorem. The Hopf link is nontrivially linked: the curves $K$ and $L_1$ cannot be separated by any link-homotopy.

Proof. The linking number is invariant under link-homotopy (any deformation keeping the curves disjoint preserves the homotopy class of the normalized difference map, and hence its degree). We have shown:

• $\operatorname{Link}(K,L_1)=-1$, by direct evaluation of Gauss’ integral.
• Any pair of curves that can be separated has linking number $0$: a separating homotopy connects them to a distant configuration. The linking number is continuous in the homotopy and integer-valued, so once the Gauss integrand (which decays as $O(1/R^2)$ with distance $R$ between the curves) makes the integral smaller than $1/2$, the only possible value is $0$.

Since $-1\neq 0$, no link-homotopy can separate $K$ and $L_1$. $\square$

More generally, the one-parameter family shows that the linking number is the obstruction to separating the curves. As long as $L_c$ threads through the disk of $K$ (i.e., $c<2$), the linking number is nonzero and separation is impossible. The only way to unlink is to pass one curve through the other—which is precisely what happens at $c=2$.

Looking back

Across these four posts, we have followed a single idea through four stages:

1. Topology (Post I). Linking is a question about homotopy classes of maps into spheres. The normalized difference map $\kl$ encodes disjointness, and its degree is the linking number.

2. Cohomology (Post II). Degree can be computed as the integral of a pulled-back volume form. De Rham cohomology gives us the freedom to choose any convenient representative, and the Hodge star ansatz reduces the construction to an ODE for the Green’s function of the Laplacian.

3. Algebra (Post III). In $\mathbb{R}^3$, pulling back the cohomological generator through the difference map produces Gauss’ kernel: the scalar triple product divided by the cube of the distance, normalized by $1/(4\pi)$.

4. Evaluation (Post IV). Applied to specific curves, the formula yields exact integers—certifying that certain links are nontrivial and distinguishing them from unlinked curves.