Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a manifold, $i: Z\to X$ is a closed embedding.

For a sheaf $S$ (of abelian groups) on a manifold $X$ and each $\varepsilon>0$ we denote by $Z_\varepsilon$ the set of points of $X$ that lie at distance $<\varepsilon$ from $X$. Consider the sheaf $S_\varepsilon:U\mapsto S(U\cap Z_\varepsilon)$, and also the section-wise limit $S_Z=\injlim_{\varepsilon\to 0} S_\varepsilon$.

I would like to understand the functor $W: S\to S_Z$ (I do not want to describe it section-wisely). Unfortunately, $S_Z$ is only an in-sheaf; it does not have to be a sheaf (in the 'usual' sense; note that the topological space $X$ is not noetherian). In particular, the stalk of $S_Z$ at any point $x\notin Z$ is easily seen to be $0$ (since $S_Z(U)=0$ for any $U$ such that its closure is disjoint from $Z$). On the other hand, the section $S_Z(X\setminus Z)$ does not seem to be $0$ (for example, if $S$ is constant); note that $X\setminus Z$ is not compact! See here the answer of algori to a previous version of this question (yesterday I believed that one can call $S_Z$ a sheaf).

Any hints (or references) for dealing with my $W$ would be very welcome! Note that $W$ is the limit of $i_{\varepsilon\ast}i_{\varepsilon}^\ast$. Unfortunately, the adjunctions that I wrote about previously probably do not hold (in any sense).

P.S. Some observations that do not seem to help me.

  1. My definition of $S_Z$ extends (without any changes) to presheaves.

  2. $W':P\mapsto P_Z$ sends ind-sheaves to ind-sheaves.

  3. It seems that ind-sheaves are (exactly) sheaves for the Grothendieck topology that admits only 'finite' covers.

So, perhaps one should use (somehow) the interplay between sheaves, ind-sheaves, and presheaves (and the corresponding topologies).

share|improve this question
1  
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, `$i_!i^!$. –  algori Oct 12 '11 at 19:29
1  
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 $i_* i^{-1}S$. –  algori Oct 12 '11 at 21:11
2  
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 $i^\ast$. –  Denis-Charles Cisinski Oct 12 '11 at 21:46
1  
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. –  Ben Wieland Oct 12 '11 at 22:29
1  
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.. –  algori Oct 13 '11 at 10:34
show 17 more comments

1 Answer 1

up vote 3 down vote accepted

Let me show that $i^{-1}$ can't have a left adjoint when $X$ is a connected topological space and $Z\neq X$ is a point. From the remark by Denis-Charles Cisinski it would follow that $i_* i^{-1}$ can't have a left adjoint either.

Suppose $Z=\{x\}$ and $i^{-1}$ had a left adjoint $J$. Then we would have $$Hom (JF,G)=Hom (F,i^{-1} G)$$ for any $F$ a sheaf on $Z$ and $G$ a sheaf on $X$. Take a non-zero sheaf $F\in Sh(Z)$, i.e. a non-zero abelian group. Note that the stalk $(JF)_x\neq 0$ (this can be seen by taking $G$ to be the constant sheaf with stalk $F$). Let us show that for any $F$ the sheaf $JF$ must be supported at $x$. Suppose there is a $y\neq x$ such that $(JF)_y\neq 0$. Then take $G=i'_* i'^{-1}JF$ where $i'$ is the inclusion $\{y\}\to X$. We have $i^{-1} G=0$, and so the right hand side of the above adjunction formula is zero. The left hand side part is non-zero since the canonical map $JF\to i'_* i'^{-1}JF$ is non-zero.

So $JF$ must be supported on $x$. Take $G$ to be the constant sheaf with stalk $F$. This time ii is the left hand side of the formula that is zero (here we use that $X$ is connected and $Z\neq X$, so $JF=i_*i^{-1} JF$ can't map into the constant sheaf $G=\underline{F}_X$ in a non-zero way) and the right hand side that isn't.

share|improve this answer
    
Thank you! See the update to my question. –  Mikhail Bondarko Oct 13 '11 at 4:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.