Timeline for Formal completion of the normal bundle
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2014 at 12:15 | vote | accept | Sasha Pavlov | ||
| Mar 9, 2014 at 0:03 | answer | added | answer_bot | timeline score: 2 | |
| Mar 8, 2014 at 23:51 | answer | added | anonymous | timeline score: 7 | |
| Mar 8, 2014 at 22:05 | comment | added | Will Sawin | I mean the completion is locally isomorphic to a product of Z with the completion of a point inside affine space. | |
| Mar 8, 2014 at 21:02 | comment | added | Damian Rössler | @Will Savin: PS: The OP considers the formalisation of $Z$ inside $N_{Z/X}$, not the formalisation of $N_{Z/X}$ (unless I have been completely mislead). | |
| Mar 8, 2014 at 20:49 | comment | added | Damian Rössler | @Will Savin: what do you mean by 'the completion is a bundle' ? | |
| Mar 8, 2014 at 20:12 | comment | added | Will Sawin | It's true if $X$ is a bundle over $Z$. Then the completion is a bundle, and the automorphism group of the bundle is an extension of $GL_n$ by a unipotent group. The bundle is classified by cohomology of this group. $GL_n$ cohomology is just a vector bundle, the normal bundle. Unipotent cohomology on an affine variety is trivial, so it's just the completion of the normal bundle. | |
| Mar 8, 2014 at 18:32 | comment | added | Damian Rössler | PS The above is only valid in char. 0 because it uses the exponential map. | |
| Mar 8, 2014 at 18:18 | comment | added | Damian Rössler | Suppose for a moment that the ideal $Z$ is generated by an element, which is not a zero divisor and that $Z$ is smooth over the base field. Then Lemma 3.6 in 'Differential algebra - a scheme theory approach' by H. Gillet will prove that the answer is yes. Note that the isomorphism depends on the choice of a derivation. The general case (if $Z$ is smooth) should be provable along the same lines. For a general schematic regular immersion, I think this still works, if you can produce the relevant derivations (my guess is: if the normal bundle is trivial). | |
| Mar 8, 2014 at 14:00 | comment | added | Sasha Pavlov | Yes, in fact I'm ready to assume that $X$ and $Z$ are smooth. | |
| Mar 8, 2014 at 13:23 | comment | added | Allen Knutson | Non-example: $Z$ a cusp in a curve $X$. Then the normal bundle and its completion are nonreduced, but the completion in $X$ is reduced. (So whatever condition you use must rule that out.) | |
| Mar 8, 2014 at 12:32 | history | asked | Sasha Pavlov | CC BY-SA 3.0 |