Timeline for Restriction of a sheaf to an infinitely small neighbourhood of a closed submanifold: how to work with this ind-sheaf?
Current License: CC BY-SA 3.0
32 events
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| Jun 26, 2012 at 15:47 | vote | accept | Mikhail Bondarko | ||
| Oct 14, 2011 at 17:32 | comment | added | Alex | Actually, excluding finite covers is not unknown territory. The relationship between sheaves on the site that you can get by allowing only locally finite covers and ind-sheaves has already been studied. See for example the book of Kashiwara and Schapira, "Ind-sheaves" (Astérique 271), especially chapter 6. | |
| Oct 13, 2011 at 11:22 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
Some remarks (that do not help me, but clarify the situation somehow) are added.; Post Made Community Wiki
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| Oct 13, 2011 at 11:16 | comment | added | Mikhail Bondarko | Yes; this is exactly the case! I don't want to sheafify the limit presheaf. I will now add some observations to my post. | |
| Oct 13, 2011 at 10:35 | comment | added | algori | The only bad thing that can happen is that in general $\mathop{\mathrm{inj}}\lim_{a\in A} F_a(U)$ is not the same as the inductive limit of $F_a(U)$. | |
| Oct 13, 2011 at 10:34 | comment | added | algori |
Dear Mikhail -- now I'm confused by the statement that $S_Z$ does not have to be a sheaf. By definition, if $A$ is any directed set and $F=(\{F_a\}_{a\in A}, \{f_a^b:F_a\to F_b\mid a,b\in A, a<b\}$ is a directed system of sheaves on some topological space, then the inductive limit $\mathop{\mathrm{inj}}\lim_{a\in A} F_a$ is the sheaf generated by the presheaf $U\to \mathop{\mathrm{inj}}\lim_{a\in A} F_a(U)$. This sheaf has all the expected properties (e.g. it is the colimit of $F$ in the category of sheaves; its stalks are the inductive limits of the stalks of $F_a$'s etc.), apart from one..
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| Oct 13, 2011 at 6:14 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
Mistake corrected: the adjunctions I wrote about in the previous versions of the question seem to be wrong; yet I would like to understand W better.
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| Oct 13, 2011 at 3:56 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
It turned out that my $S_Z$ is only an injective limit of sheaves; it is not a sheaf since the topology is not noetherian. Yet I wonder whether such presheaves were studied previously.
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| Oct 13, 2011 at 3:17 | comment | added | Mikhail Bondarko | Yes, it seems that I do get something weird! I will write an update to my question. | |
| Oct 12, 2011 at 23:36 | answer | added | algori | timeline score: 3 | |
| Oct 12, 2011 at 22:29 | comment | added | Ben Wieland | If you exclude infinite covers, you are entering unknown territory and should be careful. It's not even obvious what that means. I think the three most likely possibilities are: that they come for free; that you don't get a topos; or that you end up with sheaves on a weird space. | |
| Oct 12, 2011 at 22:18 | comment | added | algori | Dear Mikhail -- I'm not sure I understand this remark. But if we have a sheaf $F$ on an arbitrary topological space all of whose stalks are zero, then for any open $U$ we have $F(U)=0$, so $F$ is obtained by sheafifying the zero presheaf. | |
| Oct 12, 2011 at 22:03 | comment | added | Mikhail Bondarko | Dear algori, it seems that one has to allow 'infinite' covers in order to prove that stalks of an non-zero sheaf (on a non-compact space) cannot be zero? | |
| Oct 12, 2011 at 21:52 | comment | added | algori | Mikhail -- if all stalks of a sheaf over an open set $U$ are zero, then this sheaf has no non-zero sections over any open subset of $U$. | |
| Oct 12, 2011 at 21:46 | comment | added | D.-C. Cisinski | Dear Mikhail, if $i_\ast i^\ast$ has a left adjoint say $W$, then $i^\ast$ has a left adjoint as well, as one can see, using the fact that $i_\ast$ is fully faithful, as follows: we have $Hom(F,i^\ast G)=Hom(i_\ast F, i_\ast i^\ast G)=Hom(W i_\ast F, G)$. In other words, $W i_\ast$ is then a left adjoint of $i^\ast$. Therefore, the obstructions against the existence of $W$ are the same as the ones against the existence of a left adjoint of $i^\ast$. | |
| Oct 12, 2011 at 21:46 | comment | added | Mikhail Bondarko | Yes, all these stalks are zero; yet the section $S_Z(X\setminus Z)\neq 0$. This does not seem to contradict those arguments on this subject (on stalks) that I know about. | |
| Oct 12, 2011 at 21:38 | comment | added | algori | Mikhail -- here is a different argument: if the stalk of $S_Z$ at some $x\in X\setminus Z$ were non-zero, then for any neighborhood $U$ of $x$ there would be a smaller neighborhood $V$ with $S_Z(V)\neq 0$. Now take a $U$ such that its closure does not intersect $Z$. | |
| Oct 12, 2011 at 21:30 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
added 383 characters in body
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| Oct 12, 2011 at 21:25 | comment | added | Mikhail Bondarko | $S_Z(U)=0$ if the closure of $U$ is disjoint from $Z$. This does not seem to imply that $S_Z(X\setminus Z)=0$. Maybe, I should have specified what does a 'sheaf' mean in this context; I do not allows coverings by infinite disjoint unions of manifolds. | |
| Oct 12, 2011 at 21:11 | comment | added | algori |
Mikhail -- I don't see why $S_Z$ does not vanish on $X\setminus Z$. If say $dist(x,Z)=a>0$, then the stalk of any $S_\epsilon, \epsilon<a$ at $x$ is 0, and hence so is the stalk of the limit. I think that $S_Z$ is just $i_* i^{-1}S$.
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| Oct 12, 2011 at 20:54 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
some explanations added
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| Oct 12, 2011 at 20:52 | comment | added | Mikhail Bondarko | Well, $S_Z$ is a sheaf on $X$, and it does not vanish on $X\setminus Z$. So, what is your formula for $S_Z$? Note that $Z_{\epsilon}\cap U$ is usually larger than $(Z\cap U)_ϵ$. | |
| Oct 12, 2011 at 20:40 | comment | added | algori | Mikhail -- yes, I see. But then it looks like $S_Z$ will be just the restriction of $S$ to $Z$, so the adjunction formula wouldn't work. | |
| Oct 12, 2011 at 20:34 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Oct 12, 2011 at 20:27 | comment | added | Mikhail Bondarko | So, $S_ϵ$ is exactly $i_{\epsilon \ast} i_{\epsilon}^\ast S$, whereas $i_{ϵ!}\neq i_{ϵ\ast}$. | |
| Oct 12, 2011 at 20:15 | comment | added | Mikhail Bondarko | No, I don't extend by $0$. Possibly the confusion is caused by my notation: $i_{\epsilon}$ is an open embedding and not a closed one. Now for a point $x$ lying on the distance exactly $=\epsilon$ from $Z$ (so, it does not belong to $Z_{\epsilon}$, but belongs to its closure) the intersection of any neighbourhood of $x$ with $Z_{\epsilon}$ is not empty; hence the residue of $S_{\epsilon}$ at $x$ is non-zero (for example, for a constant $S$). | |
| Oct 12, 2011 at 20:04 | comment | added | algori | Mikhail -- the way I understand what you say, you restrict the sheaf to $Z_{\epsilon}$ and then extend by 0. If this is correct, then when we apply this to a smaller neighborhood, we get something that maps to the result of the same procedure applied to a larger neighborhood. This corresponds to the fact that we can extend compactly supported cohomology classes from a neighborhood to the whole thing but we can't restrict them to a neighborhood. | |
| Oct 12, 2011 at 19:52 | comment | added | Mikhail Bondarko | Well, $Z_{\epsilon_1}\subset Z_{\epsilon_2}$ in this case; hence we have a map $S_{\epsilon_2}\to S_{\epsilon_1}$. This is a sort of a stalk at $Z$. | |
| Oct 12, 2011 at 19:29 | comment | added | algori |
Mikhail -- I'm not sure I understand your construction of $S_Z$. It looks to me that if $\epsilon_1<\epsilon 2$, then $S_{\epsilon_1}$ maps to $S_{\epsilon_2}$, not the other way around, so the limit should be projective, not injective. In general, I don't know if $i_*i^{-1}$ has a left adjoint, but it has a right adjoint, `$i_!i^!$.
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| Oct 12, 2011 at 18:03 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Oct 12, 2011 at 17:14 | history | edited | Mikhail Bondarko | CC BY-SA 3.0 |
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| Oct 12, 2011 at 16:26 | history | asked | Mikhail Bondarko | CC BY-SA 3.0 |