Notes for AKT-140108/0:08:12: Difference between revisions
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\textbf{Jordan Curve Theorem} |
\textbf{Jordan Curve Theorem} |
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If $C$ is a simple closed curve in \mathbb{R}^2, then the complement R^2-J has two components |
If $C$ is a simple closed curve in \mathbb{R}^2, then the complement R^2-J has two components, the interior and the exterior, with $C$ the boundary of each. |
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The Jordan curve theorem implies that two distinct components in a diagram for 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 $\ |
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 afactor of $\frac{1}{2}$. |
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and hence an integral linking number. |
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Revision as of 13:25, 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.
\textbf{Jordan Curve Theorem} If $C$ is a simple closed curve in \mathbb{R}^2, then the complement R^2-J 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 afactor of $\frac{1}{2}$.
