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Definition Given disjoint manifolds $M$, $N\subset \Bbb{R}^{k+1}$, the linking map $\lambda: M\times N\to S^k$ is defined by $\lambda(p, q) = (p - q)/||p - q||$. If $M$ and $N$ are compact, oriented, and boundaryless, with total dimension $m + n = k$, then the degree of $\lambda$ is called the linking number $l(M, N)$.

Now let $K$ be the unit circle in the $xy$-plane with center (-1,0,0), oriented clockwise and let $M$ be the unit circle in the $xy$-plane with center $(2,0,0)$ oriented clockwise. How do I explicitly compute $l(K,M)$?

Since these circles are not "linked" the result must be $0$. First, for a regular value $z$ of $\lambda$ I need to determine the number of elements of $\lambda^{-1}(z)$. Then I should compute the Jacobian matrix of $\lambda$, BUT it is not a square matrix? So how do I obtain the sign of its determinant? I don't even know how many points are there in $\lambda^{-1}(z)$. Please help.

Lee Mosher
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jamal
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  • Probably, you should consider your circles lying in $\mathbb{R}^3$ instead of in $\mathbb{R}^2$. In the plane, two circles either intersect or are not linked in the 'intuitive' sense of the word. If they lie in $\mathbb{R}^3$, the condition on the dimensions from your definition would be satisfied, and you'll have a map from a $2$-dimensional torus to a $2$-dimensional sphere, so you should be able to compute the Jacobian. – ivanpenev Jun 01 '14 at 12:02
  • I'm so confused. Could you calculate the Jacobian for me? – jamal Jun 01 '14 at 12:06
  • I'll try, but I'll probably need some time. Do we agree to use the following circles as examples? $C = { (x_1-a)^2 + (x_2-b)^2 = r^2, x_3 = 0}$ and $\Gamma = {x_1 = 0, (x_2-\alpha)^2 + (x_3-\beta)^2 = \rho^2}$, where $a,b,r,\alpha,\beta,\rho \in \mathbb{R}$. I propose to use parameters rather than concrete numbers in order to more easily compute both linked and non-linked examples. – ivanpenev Jun 01 '14 at 12:17
  • @ivanpenev alright – jamal Jun 01 '14 at 12:20
  • @jamal: I fixed a "clash of variables". You were using $x$ and $y$ both for points in $\mathbb{R}^{k+1}$ and for the first two coordinates of $\mathbb{R}^3$. – Lee Mosher Jun 01 '14 at 12:26
  • On second thoughts, I would rather set $a = \alpha = \rho = 1$ and $b = \beta = 0$, so that there would be only one parameter (the radius of $C$) instead of six. This would be enough to produce a linked, as well as an unlinked example. – ivanpenev Jun 01 '14 at 12:26
  • @jamal, I'm really sorry. It seems that in order to compute the degree directly from the definition, I won't be able to avoid a very tedious calculation, which I won't have the time to do soon. I'm sorry! – ivanpenev Jun 01 '14 at 13:27

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In your example, where $K,M$ are disjoint circles in the $x,y$ plane of $\mathbb{R}^3$, the vector $\lambda(p,q)$ is parallel to the $x,y$ plane, and therefore $\lambda(p,q)$ is a point on the equator of $S^2$. So, $\lambda :K \times M \to S^2$ is a map from the 2-torus $T^2$ to $S^2$ whose image is contained in the equator. This map is not surjective. You may compute the degree using any point on the range $S^2$, so feel free to choose a point $z$ which is not in the image of the map. Since $\lambda^{-1}(z)$ is empty, $degree(\lambda)=l(K,M)$ is therefore equal to zero.

Lee Mosher
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