Notes for AKT-140108/0:08:12: Difference between revisions
From Drorbn
Jump to navigationJump to search
No edit summary |
No edit summary |
||
Line 3: | Line 3: | ||
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. |
||
'''Jordan Curve Theorem.''' If $C$ is a simple closed curve in $\mathbb{R}^2$, then the complement $R^2 |
'''Jordan Curve Theorem.''' If $C$ is a simple closed curve in $\mathbb{R}^2$, then the complement ${\mathbb R}^2\setminus C$ 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 |
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 |
Revision as of 12:31, 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 ${\mathbb R}^2\setminus C$ 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 $\frac12$.