A Proof of the Invariant-Based Formula for the Linking Number and Its Asymptotic Behaviour

Abstract

In 1833 Gauss defined the linking number of two disjoint curves in 3-space. For open curves this double integral over the parameterised curves is real-valued and invariant modulo rigid motions or isometries that preserve distances between points, and has been recently used in the elucidation of molecular structures. In 1976 Banchoff geometrically interpreted the linking number between two line segments. An explicit analytic formula based on this interpretation was given in 2000 without proof in terms of 6 isometry invariants: the distance and angle between the segments and 4 coordinates specifying their relative positions. We give a detailed proof of this formula and describe its asymptotic behaviour that wasn’t previously studied.

Appendices

Appendix 1: Proofs of Lemmas about Isometry Invariants

Appendices 1, 2, and 3 provide extra details that are not included into the main paper.

Proof (Lemma 1 )

Any line segments $L_1,L_2\subset \mathbb {R}^3$ that are not in the same plane are contained in distinct parallel planes. For i = 1, 2, the plane Π_i is spanned by L _i and the line parallel to L _3−i and passing through an endpoint of L _i. Let $L^{\prime }_i$ be the orthogonal projection of the line segment L _i to the plane Π_3−i. The non-parallel lines through the segments L _i and $L^{\prime }_{3-i}$ in the plane Π_i intersect at a point, say O _i. Then the line segment O ₁ O ₂ is orthogonal to both planes Π_i, hence to both L _i for i = 1, 2.

By Theorem 1, to compute lk(L ₁, L ₂), one can apply a rigid motion to move the mid-point of the line segment O ₁ O ₂ to the origin $O=(0,0,0)\in \mathbb {R}^2$ and make O ₁ O ₂ vertical, i.e., lying within the z-axis. The signed distance d can be defined as the difference between the coordinates of O ₂ = Π₂ ∩ (z-axis) and $O_1=\Pi _1\cap {\left (z\text{-axis}\right )}$ along the z-axis. Then L _i lies in the horizontal plane $\Pi _i={\left \{z=(-1)^i d/2\right \}}$, i = 1, 2.

An extra rotation around the z-axis guarantees that the x-axis in the horizontal plane Π = {z = 0} is the bisector of the angle α ∈ [0, π] from $\mathrm {pr}_{xy}(\bar {L}_1)$ to $\mathrm {pr}_{xy}(\bar {L}_2)$, where $\mathrm {pr}_{xy} \colon \mathbb {R}^3\to \Pi$ is the orthogonal projection. Then the infinite lines $\bar L_i$ through L _i have the parametric form $(x,y,z)={\left (t\cos {}(\alpha /2),(-1)^i t\sin {}(\alpha /2),(-1)^i d/2\right )}$ with $s\in \mathbb {R}$.

The point O _i can be considered as the origin of the oriented infinite line $\bar L_i$. Let the line segment L _i have a length l _i > 0 and its initial point have the coordinate $a_i\in \mathbb {R}$ in the oriented line $\bar L_i$. Then the final endpoint of L _i has the coordinate b _i = a _i + l _i. To cover only the segment L _i, the parameter t should be replaced by a _i + l _i t, $t\in {\left [0,1\right ]}$. □

Proof (Lemma 2 )

The vectors along the segments are L _i = v _i −u _i, hence the lengths are , i = 1, 2. The angle α ∈ [0, π] between L ₁, L ₂ can be found from the scalar product as $\alpha =\arccos ((\mathbf {L_1}\cdot \mathbf {L_2})/(l_1l_2))$, because the function $\arccos x\colon [-1,1]\to [0,\pi ]$ is bijective.

Since L ₁ and L ₂ are not proportional, the normalised vector product ${\mathbf {e}}_3={\left (\mathbf {L_1}\times \mathbf {L_2}\right )}/{\left (|\mathbf {L_1}\times \mathbf {L_2}|\right )}$ is well-defined and orthogonal to both L ₁, L ₂. Then e ₁ = L ₁∕|L ₁|, e ₂ = L ₂∕|L ₂| and e ₃ have lengths 1 and form a linear basis of $\mathbb {R}^3$, where the last vector is orthogonal to the first two.

Let O be any fixed point of $\mathbb {R}^3$, which can be assume to be the origin (0, 0, 0) in the coordinates of Lemma 1, though its position relative to is not yet determined. First we express the points $O_i=(0,0,(-1)^i d/2)\in \bar L_i$ from Fig. 1 in terms of given vectors . If the initial endpoint A _i has a coordinate a _i in the line $\bar L_i$ through L _i, then and

By Definition 2, is orthogonal to the line $\bar L_i$ going through the vector e _i = L _i∕|L _i| for i = 1, 2. Then the product equals |e ₁ ×e ₂|d, where is in the z-axis, the signed distance d is the z-coordinate of O ₂ minus the z-coordinates O ₁.

The product doesn’t depend on a ₁, a ₂, because e ₁ ×e ₂ is orthogonal to both e ₁, e ₂. Hence the signed distance is

which can be positive or negative, see Fig. 1.

It remains to find the coordinate a _i of the initial endpoint of L _i relative to the origin $O_i\in \bar L_i$, i = 1, 2. The vector is orthogonal to both e _i if and only if the scalar products vanish: . Due to |e ₁| = 1 = |e ₂| and ${\mathbf {e}}_1\cdot {\mathbf {e}}_2=\cos \alpha$, we get

The determinant of the 2 × 2 matrix is $\cos ^2\alpha -1=-\sin ^2\alpha \neq 0$, because L ₁, L ₂ are not parallel. Then

and

The coordinates of the final endpoints are obtained as b _i = a _i + l _i, i = 1, 2. □

Proof (Lemma 3 )

Under the central symmetry CS, in the notation of Lemma 2 the vectors change their signs. Then the formulae for α, a ₁, b ₁, a ₂, b ₂ gives the same expression, but the triple product and d change their signs.

Since the central symmetry CS is an orthogonal map M with $\det M=-1$, the new linking number changes its sign as follows:

where we also make use of the invariance of lk under exchange of the segments from Theorem 1 (f). □

Appendix 2: Proofs of Lemmas for the lk Formula in Theorem 2

Proof (Lemma 4 )

The following computations assume that a ₁, a ₂, l ₁, l ₂, α are given and t, s ∈ [0, 1].

To simplify the last integral, introduce the variables p = (a ₁ + l ₁ t)∕d and q = (a ₂ + l ₂ s)∕d. In the new variables p, q the expression under the power 3∕2 in the denominator becomes

The old variables are expressed as t = (pd − a ₁)∕l ₁, ts = (pd − a ₂)∕l ₂ and have the differentials t = d∕l ₁ p, s = d∕l ₂ q. Since t, s ∈ [0, 1], the new variables p, q have the ranges [a ₁∕d, b ₁∕d] and [a ₂∕d, b ₂∕d], respectively. Then the linking number has the required expression in the lemma:

$$\displaystyle \begin{aligned} \mathrm{lk}(L_1,L_2) & =-\frac{dl_1 l_2\sin\alpha}{4\pi} \int\limits_{a_1/d}^{b_1/d} \int\limits_{a_2/d}^{b_2/d} \frac{d^2}{l_1l_2}\frac{{\mathrm{d}d}{\mathrm{d}q}}{d^3(1+p^2+q^2-2pq\cos\alpha)^{3/2}} \\ & =-\frac{1}{4\pi} \int\limits_{a_1/d}^{b_1/d} \int\limits_{a_2/d}^{b_2/d} \frac{\sin\alpha {\mathrm{d}p} {\mathrm{d}q}}{(1+p^2+q^2-2pq\cos\alpha)^{3/2}} . \end{aligned} $$

Due to Lemma 3, the above computations assume that the signed distance d > 0. □

Proof (Lemma 5 )

Complete the square in the expression under power 3∕2 in Lemma 4:

$$\displaystyle \begin{aligned} 1+p^2+q^2-2pq\cos\alpha=1+p^2\sin^2\alpha+(q-p\cos\alpha)^2 . \end{aligned}$$

The substitution $(q-p\cos \alpha )=(1+p^2\sin ^2\alpha )\tan ^2\psi$ for the new variable ψ simplifies the sum of squares to $1+\tan ^2\psi =1/\cos ^2\psi$. Since q varies within [a ₂∕d, b ₂∕d], for any fixed p ∈ [a ₁∕d, b ₁ d], the range [ψ ₀, ψ ₁] of ψ satisfies

Since we treat p, ψ as independent variables, the Jacobian of the substitution (p, q)↦(p, ψ) equals

$$\displaystyle \begin{aligned} \frac{\partial q}{\partial \psi}=\frac{\partial \psi}{\partial \left(p\cos\alpha+\tan\psi\sqrt{1+p^2\sin^2\alpha}\right)} =\frac{\sqrt{1+p^2\sin^2\alpha}}{\cos^2\psi} . \end{aligned}$$

In the variables p, ψ the expression under the double integral of Lemma 4 becomes

$$\displaystyle \begin{aligned} \frac{\sin\alpha{\mathrm{d}p}{\mathrm{d}q}}{(1+p^2+q^2-2pq\cos\alpha)^{3/2}} & =\frac{\sin\alpha{\mathrm{d}q}}{((1+p^2\sin^2\alpha)+(1+p^2\sin^2\alpha)\tan^2\psi)^{3/2}} \frac{\partial q}{\partial \psi} {\mathrm{d}\psi} \\ & =\frac{\sin\alpha{\mathrm{d}p}}{(1+p^2\sin^2\alpha)^{3/2}(1+\tan^2\psi)^{3/2}} \frac{{\mathrm{d}\psi}\sqrt{1+p^2\sin^2\alpha}}{\cos^2\psi} \\ & =\frac{\sin\alpha{\mathrm{d}p}\cos\psi {\mathrm{d}\psi}}{1+p^2\sin^2\alpha} . \end{aligned} $$

$$\displaystyle \begin{aligned} \mathrm{lk}(L_1,L_2) & =-\frac{1}{4\pi} \int\limits_{a_1/d}^{b_1/d} \frac{\sin\alpha{\mathrm{d}p} }{1+p^2\sin^2\alpha} \int\limits_{\psi_0}^{\psi_1} \cos\psi {\mathrm{d}\psi} \\ & =\frac{1}{4\pi} \int\limits_{a_1/d}^{b_1/d} \frac{\sin\alpha{\mathrm{d}p} }{1+p^2\sin^2\alpha}(\sin\psi_0-\sin\psi_1) . \end{aligned} $$

Express the sin functions for the bounds ψ ₀, ψ ₁ in terms of tan as $\sin \psi _0=\tan \psi _0/\sqrt {1+\tan ^2\psi _0}$. Using obtained above, we get

$$\displaystyle \begin{aligned} \sqrt{1+\tan^2\psi_0} & =\sqrt{\frac{(1+p^2\sin^2\alpha)+(\frac{a_2}{d}-p\cos\alpha)^2}{1+p^2\sin^2\alpha}} \\ & =\sqrt{\frac{1+p^2+(\frac{a_2}{d})^2-2\frac{a_2}{d}p\cos\alpha}{1+p^2\sin^2\alpha}} . \\ \sin\psi_0 & =\frac{\frac{a_2}{d}-p\cos\alpha}{\sqrt{1+p^2\sin^2\alpha}} \sqrt{\frac{1+p^2\sin^2\alpha}{1+p^2+(\frac{a_2}{d})^2-2\frac{a_2}{d}p\cos\alpha}} \\ & =\frac{\frac{a_2}{d}-p\cos\alpha}{\sqrt{1+p^2+(\frac{a_2}{d})^2-2\frac{a_2}{d}p\cos\alpha}} . \end{aligned} $$

Then $\sin \psi _1$ has the same expression with a ₂ replaced by b ₂. After substituting these expressions in the previous formula for the linking number, we get

$$\displaystyle \begin{aligned} \mathrm{lk}(L_1,L_2) = & \frac{1}{4\pi}\int\limits_{a_1/d}^{b_1/d} \frac{\sin\alpha{\mathrm{d}p}}{1+p^2\sin^2\alpha} \left( \frac{\frac{a_2}{d}-p\cos\alpha}{\sqrt{1+p^2+(\frac{a_2}{d})^2-2\frac{a_2}{d}p\cos\alpha}} \right. \\ & \quad -\left.\frac{\frac{b_2}{d}-p\cos\alpha}{\sqrt{1+p^2+(\frac{b_2}{d})^2-2\frac{b_2}{d}p\cos\alpha}}\right) \\ = & \frac{S(a_2/d)-S(b_2/d)}{4\pi} , \end{aligned} $$

where

$$\displaystyle \begin{aligned} I(r)=\int\limits_{a_1/d}^{b_1/d} \frac{\sin\alpha(r-p\cos\alpha){\mathrm{d}p}}{(1+p^2\sin^2\alpha)\sqrt{1+p^2+r^2-2pr\cos\alpha}} . \end{aligned}$$

□

Proof (Lemma 6 )

The easiest way is to differentiate the function arctanω for

$$\displaystyle \begin{aligned} \omega=\frac{pr\sin^2\alpha+\cos\alpha}{\sin\alpha\sqrt{1+p^2+r^2-2pr\cos\alpha}} \end{aligned}$$

with respect to the variable p remembering that r, α are fixed parameters. For notational clarity, we use an auxiliary symbol for the expression under the square root: $R=1+p^2+r^2-2pr\cos \alpha$. Then

$$\displaystyle \begin{aligned} \omega=\frac{pr\sin^2\alpha+\cos\alpha}{\sin\alpha\sqrt{R}} \end{aligned}$$

and

Since we got the required expression under the integral I(r), Lemma 6 is proved. □

Appendix 3: Proofs of Corollaries of Main Theorem 2

Proof of Corollary 1

By definition any simple orthogonal segments L ₁, L ₂ have α = π∕2 and initial endpoints a ₁ = 0 = a ₂, hence b ₁ = l ₁, b ₂ = l ₂. Substituting the values above into (3) gives

Then

□

Proof (Corollary 2 )

By Theorem 2 lk(L ₁, L ₂) is a sum of 4 arctan functions divided by 4π. Since each arctan is strictly between ± π∕2, the linking number is between ± 1∕2. □

Proof (Corollary 3 )

If α = 0 or α = π, then $\cot \alpha$ is undefined, so Theorem 2 sets . Then $\mathrm {lk}(L_1,L_2)= \operatorname {\mathrm {sign}}(d)\pi /2(1+1-1-1)=0$.

Theorem 2 also specifies that lk(L ₁, L ₂) = 0 for d = 0. If d ≠ 0 and α → 0 within [0, π] while all other parameters remain fixed, then . Hence each of the 4 arctan functions in Theorem 2 approaches π∕2, so lk(L ₁, L ₂) → 0. The same conclusion similarly follows in the case α → π when .

If L ₁, L ₂ are not parallel, the angle α between them belongs to (0, π). In d > 0, Lemma 4 says that

$$\displaystyle \begin{aligned} \mathrm{lk}(L_1,L_2)=-\frac{1}{4\pi} \int\limits_{a_1/d}^{b_1/d} \int\limits_{a_2/d}^{b_2/d} \frac{\sin\alpha{\mathrm{d}p}{\mathrm{d}q}}{(1+p^2+q^2-2pq\cos\alpha)^{3/2}} . \end{aligned}$$

Since the function under the integral is strictly positive, lk(L ₁, L ₂) < 0. By Lemma 3 both lk(L ₁, L ₂) simultaneously change their signs under a central symmetry. Hence, the formula $\operatorname {\mathrm {sign}}(\mathrm {lk}(L_1,L_2))=- \operatorname {\mathrm {sign}}(d)$ holds for all d including d = 0 above. □

Proof (Corollary 4 )

Recall that . By Corollary 3 assume that α ≠ 0, α ≠ π, so α ∈ (0, π). Then $\sin \alpha >0$, $a^2+b^2-2ab\cos \alpha >(a-b)^2\geq 0$ and

so Theorem 2 gives

$$\displaystyle \begin{aligned} \lim\limits_{d\to 0}\mathrm{lk}(L_1,L_2)=\frac{\operatorname{\mathrm{sign}}(d)}{8} {\left(\operatorname{\mathrm{sign}}(a_1)-\operatorname{\mathrm{sign}}(b_1)\right)}{\left(\operatorname{\mathrm{sign}}(b_2)-\operatorname{\mathrm{sign}}(a_2)\right)} . \end{aligned}$$

In the limit case d = 0, the line segments $L_1,L_2\subset {\left \{z=0\right \}}$ remain disjoint in the same plane if and only if both endpoint coordinates a _i, b _i have the same sign for at least one of i = 1, 2, which is equivalent to $\operatorname {\mathrm {sign}}(a_i)- \operatorname {\mathrm {sign}}(b_i)=0$, i.e., $\lim \limits _{d\to 0}\mathrm {lk}(L_1,L_2)=0$ from the product above. Hence formula (3) is continuous under d → 0 for any non-crossing segments. Any segments that intersect in the plane ${\left \{z=0\right \}}$ when d = 0 have endpoint coordinates a _i < 0 < b _i for both i = 1, 2 and have the limit

$$\displaystyle \begin{aligned} \lim\limits_{d\to 0}\mathrm{lk}(L_1,L_2)=\frac{\operatorname{\mathrm{sign}}(d)}{8}(-1-1)(1-(-1))=-\frac{\operatorname{\mathrm{sign}}(d)}{2} \end{aligned}$$

as required. □

Proof (Corollary 5 )

If , while all other parameters of L ₁, L ₂ remain fixed, then the function

from Theorem Theorem 2 has the limit ${\mathrm {arctan}}( \operatorname {\mathrm {sign}}(d)\cot \alpha )= \operatorname {\mathrm {sign}}(d)\left (\pi /2-\alpha \right )$. Since the four AT functions in Theorem 2 include the same d, α, their limits cancel, so lk(L ₁, L ₂) → 0. □

Proof (Corollary 6 )

If , then , i = 1, 2. If , then , i = 1, 2. Consider the former case , the latter is similar.

Since d, α are fixed,

$$\displaystyle \begin{aligned} a^2+b^2-2ab\cos\alpha+d^2\leq (a+b)^2+d^2\leq 5b^2 \end{aligned}$$

for large enough b. Since arctan(x) increases,

as . Since the four AT functions in Theorem 2 have the same limit when their first two arguments tend to , these 4 limits cancel and we get lk(L ₁, L ₂) → 0. □

Proof (Corollary 7 )

lk(L ₁, L ₂) = 0 for d = 0. It’s enough to consider the case d ≠ 0. Then

from Theorem 2 is continuous. Let (say for i = 1) a ₁ → b ₁, the case b ₁ → a ₁ is similar. The continuity of implies that and . In the limit all terms in Theorem 2 cancel, hence lk(L ₁, L ₂) → 0. □