Notes for AKT-140108/0:08:12: Difference between revisions
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18S-AKT Question: Why is this sum divisible by 2? Why the $\frac{1}{2}$? |
'''18S-AKT Question:''' Why is this sum divisible by 2? Why the $\frac{1}{2}$? |
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The factor $\frac{1}{2}$ I think is as a result of the projection of the link unto the plane making each sign appear twice. |
The factor $\frac{1}{2}$ I think is as a result of the projection of the link unto the plane making each sign appear twice. |
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\textbf{Jordan Curve Theorem} |
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The Jordan curve theorem implies that two distinct components in a diagram for a link $L$ intersect an even number of times. Hence we add up an even |
The Jordan curve theorem implies that two distinct components in a diagram for a link $L$ intersect an even number of times. Hence we add up an even |
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number of $\pm1$’s in the computation of lk(L), which yields an even number. It is always an integer. This is why we have |
number of $\pm1$’s in the computation of $lk(L)$, which yields an even number. It is always an integer. This is why we have a factor of $\frac12$. |
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Revision as of 13:30, 23 May 2018
18S-AKT Question: Why is this sum divisible by 2? Why the $\frac{1}{2}$?
The factor $\frac{1}{2}$ I think is as a result of the projection of the link unto the plane making each sign appear twice.
Jordan Curve Theorem. If $C$ is a simple closed curve in $\mathbb{R}^2$, then the complement $R^2-\setminus C$ has two components, the interior and the exterior, with $C$ the boundary of each.
The Jordan curve theorem implies that two distinct components in a diagram for a link $L$ intersect an even number of times. Hence we add up an even number of $\pm1$’s in the computation of $lk(L)$, which yields an even number. It is always an integer. This is why we have a factor of $\frac12$.
