Timeline for Discriminant locus in knot space
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2010 at 8:08 | vote | accept | Nikita Kalinin | ||
| Oct 26, 2010 at 2:33 | comment | added | Ryan Budney | I'll append further responses to your questions above to my answer below. | |
| Oct 26, 2010 at 2:13 | answer | added | Daniel Moskovich | timeline score: 2 | |
| Oct 25, 2010 at 22:59 | answer | added | Ryan Budney | timeline score: 2 | |
| Oct 25, 2010 at 21:26 | comment | added | Nikita Kalinin | >Ryan Does your bump-function formed a "discriminant tangent space" basis in some sense? | |
| Oct 25, 2010 at 21:23 | comment | added | Nikita Kalinin | >Ryan Yes, I see. So, my question: what is co-dimension mean in this case? Could you give me definition? I get tired with "naive" description of this situation, but in th oter hand I can't define codimension strictly. Therefore the question may be reformulated as "Does codimension in this context be present only as intuitive conception". What does "slice of the neighbourhood" mean? Does it be the embedding $\mathbb R^n$ image? | |
| Oct 25, 2010 at 21:18 | comment | added | Ryan Budney | The actual neighbourhood of $f$ consists of $f+\epsilon$ where $\epsilon : S^1 \to \mathbb R^3$ is smooth and "small". If you were using $C^k$-immersions, that would mean $\epsilon$ through $\epsilon^{(k)}$ are all uniformly small. If you were demanding $C^\infty$-smooth maps, then you would demand arbitrarily many but finitely many derivatives of $\epsilon$ to be uniformly small. Your $\mathbb R^n$ "slice" corresponds to decomposong $\epsilon$ into the sum of some bump-functions about the double points, together with complementary functions (which are zero at the double points). | |
| Oct 25, 2010 at 20:53 | comment | added | Ryan Budney | Oh, it sounds like you're looking at a slice of the neighbourhood, not the neighbourhood of $f$ itself. The parameters of the $\mathbb R^n$ space indicate how one chooses to resolve the $n$ double points. This isn't a neighbourhood of $f$, not at all. What you can do though is view it as part of a particular coordinate system on a neighbourhood of $f$ -- you're ignoring all but finitely-many coordinates because the other coordinates aren't very interesting. This is because the $n$ double point subspace has co-dimension $n$. | |
| Oct 25, 2010 at 20:49 | comment | added | Ryan Budney | Where did you get the $\mathbb R^n$ from? A neighbourhood of $f$ in $K$ looks like an infinite-dimensional Hilbert/Frechet space (the space of maps $S^1 \to \mathbb R^3$), depending on how differentiable your space of immersions is. | |
| Oct 25, 2010 at 20:27 | history | edited | Nikita Kalinin |
edited tags; edited tags
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| Oct 25, 2010 at 20:18 | history | asked | Nikita Kalinin | CC BY-SA 2.5 |