Timeline for Power series expansions and limits of knot invariants
Current License: CC BY-SA 4.0
18 events
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| May 21 at 23:50 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 13:14 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 13:00 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 12:41 | comment | added | Eric Ley | @Andy Putman Pointwise convergence is not that strong. If we require uniform convergence, then yes, everything is false here. | |
| May 21 at 12:39 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 12:37 | comment | added | Andy Putman | Be that as it may, I’m extremely skeptical that anything is true in this level of generality. | |
| May 21 at 12:32 | comment | added | Eric Ley | Ah, I see, and it looks like both (2) and (3) are false from this viewpoint, but I cannot see a easy rigorous proof. But on the other hand, let's consider a similar case: the functions on $\mathbb Z$, and although they can be very wild and arbitrary, we can approximate them by polynomials pointwise. FTI are anologue of polynomials on set of knots. | |
| May 21 at 12:27 | comment | added | Andy Putman | I’m trying to give you a sense of how wild and arbitrarily a knot invariant can be. There is no reason to expect that they have any relation to the combinatorics of a knot diagram. Here’s another invariant: fix an enumeration of knots, and assign to the nth knot the nth digit of pi. | |
| May 21 at 12:25 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 12:23 | comment | added | Eric Ley | Yes, I do mean an invariant takes values in a field like reals, what does "there are uncountably many integer-valued invariants" imply? | |
| May 21 at 12:19 | comment | added | Andy Putman | By an invariant do you literally just mean a real valued function defined on knots? There is no way to say anything about things that are that general. For instance, the set of knots is countable, so there is an integer-valued “invariant” that records a knot’s position in an enumeration of all knots. This already gives uncountably many integer-valued “invariants”. | |
| May 21 at 12:15 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 12:10 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 11:48 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 9:08 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 9:03 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 8:53 | history | edited | Eric Ley | CC BY-SA 4.0 |
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| May 21 at 5:17 | history | asked | Eric Ley | CC BY-SA 4.0 |