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Suppose you have two closed curves in $\mathbb{R}^3$, and allow them to continuously deform and possibly pass through themselves, but not each other. The linking number is an invariant of such deformations and in fact each equivalence class of pairs of curves has a unique linking number.

For a given pair of curves, the linking number can be computed using the Gauss integral formula, as described here: http://en.wikipedia.org/wiki/Linking_number#Gauss.27s_integral_definition

I'm interested in pairs of closed curves in SO(3) that again can pass through themselves, but not each other. Is there an analogue to the linking number in this setting, and is there a way to compute it for a given pair of curves? Does SO(3)'s non-trivial fundamental group add any wrinkles?

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I'm far from an expert, but linking numbers are often defined in the following way. Given two curves $C_1$ and $C_2$, $C_1$ bounds a disc since $\mathbb{R}^3$ is contractible. By isotoping $C_2$ an epsilon, one can arrange that the intersection of $C_2$ and the disc to be transverse. Now, one defines the linking number as the number of intersection points, counting orientation. On $SO(3)$ this fails precisely because of the nontrivial fundamental group: If $C_1$ is a curve representing the unique nontrivial element of $\pi_1(SO(3))$, then it does not bound a disc. (continued) –  Jason DeVito Jun 26 '13 at 1:14
    
I'm not sure if there are other methods which give the same answer in $\mathbb{R}^3$, but can still be applied in the $SO(3)$ setting, which is why I'm refraining from posting any kind of answer. –  Jason DeVito Jun 26 '13 at 1:15
    
I like this question a lot, but I don't know how to answer it. I have posted it on MathOverflow with the hope that it might receive some attention there. –  Jim Belk Jun 27 '13 at 6:49

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