Timeline for Diagonal map and "infinitesimal points"
Current License: CC BY-SA 2.5
16 events
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| Jan 9, 2021 at 0:28 | comment | added | Sándor Kovács | @Arrow, OK, I think you are right and I hadn't realized until now, but apparently I have been using the term "$n^\text{th}$ infinitesimal neighbourhood" incorrectly in the above comment. Then just ran with it... Anyway, I meant $I^n/I^{n+1}$, not $\mathscr O/I^{n+1}$. I don't know if that has a name... | |
| Jan 8, 2021 at 21:03 | comment | added | Arrow | Dear @SándorKovács, I think I agree the kernel of the map "omitting $n^\text{th}$ derivatives" is the $n^\text{th}$ symmetric power of the tangent bundle, since in coordinates the kernel consists of the degree $n$ homogeneous polynomials. However, I think the $n$-jets do correspond to the $n^\text{th}$ infinitesimal neighborhood. | |
| Jan 8, 2021 at 20:52 | comment | added | Sándor Kovács | @Arrow, OK, I think I see the issue. The n-jets don't correspond to the $n^\text{th}$ infinitesimal neighbourhood. The latter is the "difference" between n-jets and (n-1)-jets. More precisely the $n^\text{th}$ infinitesimal neighbourhood corresponds to the kernel of the restriction map from n-jets to (n-1)-jets. If you look at Goodwillie's previous comment, then he says exactly what I am saying (well, for n=2 at least). Does this resolve your worry? | |
| Jan 8, 2021 at 20:40 | comment | added | Arrow | Dear @SándorKovács, the phrase "globally polynomial" was poor chosen. I am asking about the seeming contradiction between your assertion and the comment by Tom Goodwillie asserting that jet bundles are not obtained by applying a functor fiberwise to the tangent bundle. | |
| Jan 8, 2021 at 20:05 | comment | added | Sándor Kovács | @Arrow, I'm not entirely sure what you are asking. The symmetric power of a bundle is a local construction, it does not guarantee that the global sections of the $n^\text{th}$ symmetric power is the $n^\text{th}$ symmetric power of the global sections of the original bundle. I also don't see how this relates to the linked comments. Can you try to explain what you mean? | |
| Jan 8, 2021 at 16:16 | comment | added | Arrow | Dear @SándorKovács, how is the $n^\text{th}$ infinitesimal neighborhood isomorphic to the $n^\text{th}$ symmetric power of the tangent bundle? This suggests $n$-jets are "globally polynomial" and seems to contradict this comment. What am I missing? | |
| Feb 8, 2011 at 5:46 | comment | added | Sándor Kovács | OK, let's say it "analytically": $X$ is a manifold, $Y$ is a point, so $X\times_YX$ is just $X\times X$. The order $n$ infinitesimal neighborhood of the diagonal in $X\times X$ (which is isomorphic to $X$) can be identified with the $n^{\rm th}$ symmetric power of the tangent bundle of $X$. | |
| Feb 8, 2011 at 1:21 | comment | added | Brian | Thanks! This is still a bit too "algebraic". What is the, say, "analytic" idea behind it, if there is any? Thanks! | |
| Feb 7, 2011 at 17:35 | comment | added | Sándor Kovács | Brian, if say $Y$ is an algebraically closed field and $X$ is smooth over $Y$, then $\mathscr I^n/\mathscr I^{n+1}$ is isomorphic to the $n^{\rm th}$-symmetric power of $\mathscr I/\mathscr I^2$, so you could identify them with pluriforms. | |
| Feb 7, 2011 at 14:49 | comment | added | Brian | Thanks a lot for your answer! How does this apply to higher infinitesimal? | |
| Feb 7, 2011 at 7:38 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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| Feb 7, 2011 at 5:43 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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| Feb 7, 2011 at 2:38 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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| Feb 7, 2011 at 2:33 | vote | accept | Brian | ||
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| Feb 7, 2011 at 2:33 | history | edited | Sándor Kovács | CC BY-SA 2.5 |
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| Feb 7, 2011 at 2:27 | history | answered | Sándor Kovács | CC BY-SA 2.5 |