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I am looking into the definition of the linking number. I've considered these two definitions. Consider a link $L$ with components $K_1$ and $K_2$, and respectively their embeddings $\gamma_1$ and $\gamma_2$.

Firstly, $Lk^{(1)}$ defined by counting the positive and negative intersections of a diagram of the link $L$ with +1 and -1 respectively, ultimately dividing this number by 2.

Secondly, one can define the linking number of $L$, by $Lk^{(2)}$, by taking the degree of the map

$\Psi: S^1\times S^1 \to S^2: (s,t)\mapsto \frac{\gamma_1(s) - \gamma_2(t)}{|\gamma_1(s) - \gamma_2(t)|}$

We now know the linking number is symmetric w.r.t the component knots. However $Lk^{(2)}(K_2,K_1) = deg(\mu\circ\Psi)$. with $\mu:S^2\to S^2$ that maps points to their antipodal points. However the degree of $\mu$ is -1, and thus

$Lk^{(2)}(K_2,K_1) = deg(\mu\circ\Psi) = -deg(\Psi) = -Lk^{(2)}(K_1,K_2) $.

And thus $Lk^{(2)}$ is not symmetric. Where did I go wrong?

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I would think that $Lk^{(2)}\ne deg(\mu\circ \Psi)$, since this refers to

$\Psi:S^1\to S^1: (s,t)\mapsto \frac{\gamma_2(t)-\gamma_1(s)}{||\gamma_2(t)-\gamma_1(s)||}$.

However $Lk^{(2)}$ refers to $\Psi:S^1\to S^1: (t,s)\mapsto \frac{\gamma_2(t)-\gamma_1(s)}{||\gamma_2(t)-\gamma_1(s)||}$

Here one sees that also the variables are switched and this adds an extra factor $-1$ to your degree. This should solve your sign problem.

Could someone confirm my idea is correct (comment or otherwise)? Since I am relatively new to these orientations and degrees and may have missed something.

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    $\begingroup$ Yes, I believe you are correct. The antipodal map will give the mirror image of the link and thus change every crossing's sign. Which is exactly the problem we see in the question. Knots are usually defined up to orientation preserving homeomorphisms, which $\mu$ is not. $\endgroup$ – N. Owad Jun 20 '15 at 12:51
  • $\begingroup$ Aha, indeed. I hadn't thought of that interpretation. $\endgroup$ – Tom Ultramelonman Jun 20 '15 at 13:39

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