Timeline for Linking number and intersection number
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 13 at 21:10 | comment | added | Tom Goodwillie | @user429294: Yes, that is what I mean. For an example, take $p=1$ and $q\ge 2$. Let $D$ be a $2$-dimensional disk. Choose a map $f:D\to \mathbb R^2$ (for example, the $z\mapsto z^L$ map for complex numbers) such that only $0$ goes to $0$, and such that the winding number of the restricted map $\partial D\to \mathbb R^2\backslash 0$ is $L$. Choose an embedding $e:D\to \mathbb R^q$. Together these give a map $(f,e):D\to \mathbb R^2\times \mathbb R^q$ that is an embedding, and such that its restriction to the boundary of $D$ has linking number $L$ with $0\times \mathbb R^q$. | |
| Apr 13 at 20:38 | comment | added | user429294 | @TomGoodwillie: Do you mean the higher dimensional analog of the fact '$B$ must intersect $D$ at least $|link(A,B)|$ times' does not hold? If yes, can you please give any reference or idea about why this would fail? | |
| Apr 13 at 20:38 | comment | added | user429294 | Thank you @AndyPutman for a very nice explanation. | |
| Apr 13 at 20:10 | vote | accept | user429294 | ||
| Apr 13 at 20:05 | history | edited | Andy Putman | CC BY-SA 4.0 |
deleted 73 characters in body
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| Apr 13 at 20:03 | comment | added | Andy Putman | @TomGoodwillie: Your gentle criticism that I should have been less insistent about the One True Way is well-taken, and I’ll edit it to tone that down (in my defense, I am massively jet lagged right now!). Of course, the right thing to do is to know a bunch of different ways to think about things. | |
| Apr 13 at 19:52 | comment | added | Tom Goodwillie | Interesting. I would have said the right way to think about linking numbers is via Poincare (or Alexander) duality. But that would not have led me to your strong conclusion (unless I had used homology/cohomology with twisted coefficients). And the conclusion does not hold for a linked $p$-sphere and $q$-sphere in $(p+q+1)$-space if $p$ and $q$ are greater than $1$ (in which case covering spaces are not the right way to think). | |
| Apr 13 at 19:36 | history | answered | Andy Putman | CC BY-SA 4.0 |