As I read from the wikipedia page of the linking number, it says that the linking number of two curves $\gamma_1$ and $\gamma_2$ in space can be found using the integral $$\,\frac{1}{4\pi} \oint_{\gamma_1}\oint_{\gamma_2} \frac{\mathbf{r}_1 - \mathbf{r}_2}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \cdot (d\mathbf{r}_1 \times d\mathbf{r}_2)$$ It says the integrand is the Jacobian of the Gaussian map $$\Gamma(s,t) = \frac{\gamma_1(s) - \gamma_2(t)}{|\gamma_1(s) - \gamma_2(t)|}$$ Following Ted's comment I am able to show that

$$\oint_{\gamma_1}\oint_{\gamma_2} \left\|\frac{\partial\Gamma}{\partial s}\times \frac{\partial\Gamma}{\partial t}\right\|\,dsdt = \oint_{\gamma_1}\oint_{\gamma_2} \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|^3} \left| (\mathbf{r}_1 - \mathbf{r}_2)\cdot \left (\frac{d\mathbf{r}_1}{ds} \times \frac{d\mathbf{r}_2}{dt}\right)\right| ds dt$$

  • $\begingroup$ I am very unfamiliar with the pullback... could you explain with some college multivariable calculus what that means? $\endgroup$
    – wilsonw
    Jun 9, 2013 at 4:25

1 Answer 1


This is (up to sign) the integral of the pullback of the area $2$-form on the sphere by $\Gamma$. We get the same thing by pulling back the $2$-form $$\omega = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{\|\mathbf x\|^3}$$ by the unnormalized chord map $\tilde\Gamma(s,t)=\gamma_1(s)-\gamma_2(t)$.

Note that when we evaluate the pullback of the numerator, we get the determinant of the matrix with columns $\mathbf r_1-\mathbf r_2$, $\mathbf r_1'$, $-\mathbf r_2'$, which is $-(\mathbf r_1-\mathbf r_2)\cdot(\mathbf r_1'\times\mathbf r_2')$.

Indeed, I believe Wikipedia has the sign wrong — Gauss's original paper writes what is equivalent to $$\frac1{4\pi}\oint_{C_1}\oint_{C_2}\frac{\mathbf r_2-\mathbf r_1}{\|\mathbf r_2-\mathbf r_1\|^3}\cdot (d\mathbf r_1\times d\mathbf r_2)\,.$$ As often happens, the formula in Wikipedia appears in various sources. I think a lot of people, when they check the signs in their heads with a typical $+1$ knot, must forget the negative sign when they compute $\partial\Gamma/\partial t$.

More classically, to compute the Jacobian of the map, you need to calculate $$\left\|\frac{\partial\Gamma}{\partial s}\times \frac{\partial\Gamma}{\partial t}\right\|\,.$$ More precisely, we want the signed area of the map $\Gamma$ (keeping track of how the chord map covers the sphere, with signed orientation), so we erase the absolute value signs you have at the end of your calculation. (For example, given two complicated curves that are not linked, the chord map covers portions of the sphere—perhaps all of it—numerous times, but then hits the same regions with opposite orientation.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .