14
$\begingroup$

Let $\mathcal{F}$ be a sheaf over some topological space. The stalks are $\mathcal{F}_x= \underset{{x\in U}}{ \underrightarrow{\lim}} \mathcal{F}(U)$. Is there a special name for a sheaf that satisfies $\mathcal{F}(U) = \underset{{x\in U}}{ \underleftarrow{\lim}} \mathcal{F}_x$?

Obviously this is a very restrictive property but here's a possible example:

Let $X=Spec A$ be an affine integral scheme with structure sheaf $\mathcal{O}_X$. We have:

$$\mathcal{O}_{X,x}= \underset{{x\in X_f}}{ \underrightarrow{\lim}} \mathcal{O}_X(X_f)=\underset{{f \notin \mathfrak{p}_x}}{\underrightarrow{\lim}} A_f = \bigcup_{f \notin \mathfrak{p}_x} A_f$$ But we also have (I hope):

$$\mathcal{O}_X(X_f)=A_f= \bigcap_{f \notin \mathfrak{p}_x \subset A} A_{\mathfrak{p}_x}=\underset{{f \notin \mathfrak{p}_x}}{\underleftarrow{\lim}} A_{\mathfrak{p}_x}=\underset{{x \in X_f}}{\underleftarrow{\lim}} \mathcal{O}_{X,x}$$

So we can recover the structure sheaf as a limit of the stalks. Does it still hold for non affine scheme? More generally:

When is a sheaf the inverse limit of its stalks?

Can I turn this into a technique for constructing sheaves?

Let $F: |X| \to Ab$ be a functor from the category of points of $X$ to Abelian groups. Now define:

$$\mathcal{F}(U) = \underset{{x\in U}}{\underleftarrow{\lim}} F(x)$$

If I take stalks and then do the above will I get back to the same sheaf? (Possibly after sheafication).

EDIT: Some details are missing. Whenever I'm taking limit of stalks, the category I'm taking the limit over is the poset of the points of the space. Where we have $x_0 \to x$ Iff $x$ is a generization of $x_0$ (i.e. if $x_0 \in \overline{\{x\}}$).

$\endgroup$
  • $\begingroup$ @Hoot The field of fractions $K(X)=A_{\eta}$ where $\eta$ is the generic point - 0. $\endgroup$ – Saal Hardali Dec 25 '15 at 13:20
  • 1
    $\begingroup$ I'm worried that your second display is going to fail if I cook up some non-Noetherian scheme where specialization/generalization doesn't tell the full topological story. Do you have a proof of this? $\endgroup$ – Hoot Dec 25 '15 at 14:47
  • 5
    $\begingroup$ What are the transition maps between the $\mathscr F_x$ for varying $x$ that you use to define the limit? $\endgroup$ – Remy Dec 25 '15 at 19:02
  • $\begingroup$ @Remy Added a more explicit construction $\endgroup$ – Saal Hardali Dec 25 '15 at 19:33
  • $\begingroup$ @SaalHardali have you gotten any kind of answer for this question? How about posting it on MO? $\endgroup$ – Arrow Jan 13 '16 at 14:33
7
$\begingroup$

First of all:

Let $X=Spec A$ be an affine integral scheme with structure sheaf $\mathcal{O}_X$.

We have: $$ \mathcal{O}_{X,x}= \underset{{x\in X_f}}{ \underrightarrow{\lim}} \mathcal{O}_X(X_f)=\underset{{f \notin \mathfrak{p}_x}}{\underrightarrow{\lim}} A_f = \bigcup_{f \notin \mathfrak{p}_x} A_f $$ But we also have (I hope): $$ \mathcal{O}_X(X_f)=A_f=\bigcap_{f\notin\mathfrak{p}_x\subset A} A_{\mathfrak{p}_x}=\underset{{f \notin \mathfrak{p}_x}}{\underleftarrow{\lim}} A_{\mathfrak{p}_x}=\underset{{x \in X_f}}{\underleftarrow{\lim}} \mathcal{O}_{X,x} $$

The first equality is false, because for any commutative ring $A$ with unit: \begin{equation} \forall\mathfrak{p}\in\operatorname{Spec}A,\,\lim_{\overrightarrow{f\notin\mathfrak{p}}}A_f=A_{\mathfrak{p}} \end{equation} and the second equality is partially true, that is: \begin{equation} \mathcal{O}_X(D(f))=A_f=\bigcap_{\stackrel{\mathfrak{p}\in\operatorname{Spec}A}{f\notin\mathfrak{p}}}A_{\mathfrak{p}} \end{equation} where I prefer the notation $D(f)$ for the open set \begin{equation} \{x\in\operatorname{Spec}A\mid f(x)\neq0\}. \end{equation} At most in general, the following lemma holds.

Lemma. Let $(X,\mathcal{T})$ be a topological space with topology $\mathcal{T}$ and $\mathfrak{B}$ a basis for $\mathcal{T}$; that is a system of open subsets of $X$ such that:

  1. $U,V\in\mathfrak{B}\Rightarrow U\cap V\in\mathfrak{B}$,
  2. every open subset of $X$ is a union of sets from $\mathfrak{B}$.

We can view $\mathfrak{B}$ as a category with objets its elements and the inclusions between the sets as morphisms.

Let $\mathcal{O}:\mathfrak{B}\to\mathbf{C}$ be a controvariant functor (or $\mathfrak{B}$-presheaf), where $\mathbf{C}$ is a category closed with respect to projective limits, such that $\mathcal{O}$ satisfies the sheaf conditions for coverings of type $\displaystyle U=\bigcup_{i\in I}U_i$, where $\forall i\in I,\,U,U_i\in\mathfrak{B}$.

Then $\mathcal{O}$ can be extended to a sheaf $\overline{\mathcal{O}}$ on $X$, where: \begin{equation} \forall U\in\mathcal{T},\,\overline{\mathcal{O}}(U)=\lim_{\overleftarrow{V\in\mathfrak{B}\,\text{with}\,V\subseteq U}}\mathcal{O}(V) \end{equation} and this extension is unique up to canonical isomorphism.

For a proof one can consult Bosch S. - Algebraic Geometry and Commutative Algebra, chapter 6, section 6, lemma 4.

Let $(X,\mathcal{T})$ be a topological space and let $\mathcal{F}$ be a sheaf on $X$ with values in a category $\mathbf{C}$ closed with respect to inductive and projective limits.

Let $x,y\in X$ such that $x\in\overline{\{y\}}$, or in other words: \begin{equation} \forall U\in\mathcal{T},\,x\in U\Rightarrow y\in U; \end{equation} then the following diagrams commute: \begin{equation} \require{AMScd} \forall V\subseteq U\in\mathcal{T},x,y\in V,x\in\overline{\{y\}},\, \begin{CD} \mathcal{F}(U) @>r_{U,x}>> \mathcal{F}_x\\ @V{r^U_V}VV @VV{=}V\\ \mathcal{F}(V) @>>\dot\exists r_{V,x}> \mathcal{F}_x \end{CD}, \begin{CD} \mathcal{F}(U) @>r_{U,y}>> \mathcal{F}_y\\ @V{r^U_V}VV @VV{=}V\\ \mathcal{F}(V) @>>\dot\exists r_{V,y}> \mathcal{F}_y \end{CD},\dot\exists r_{y,x}:\mathcal{F}_x\to\mathcal{F}_y \end{equation} and therefore $(\mathcal{F}_x,r_{y,x})_{x,y\in X}$ is a projective system in $\mathbf{C}$; where I get: \begin{gather} x\succcurlyeq y\iff x\in\overline{\{y\}}\,\text{or}\,x=y;\\ \forall U\in\mathcal{T},\,\mathcal{G}(U)=\lim_{\overleftarrow{x\in U}}\mathcal{F}_x. \end{gather} Let $\mathfrak{B}$ be a basis of $(X,\mathcal{T})$, let $V\subseteq U\in\mathfrak{B}$, let $x,y\in V$ such that $y\in\overline{\{x\}}\iff x\prec y$; then one can consider the diagram \begin{equation} \mathcal{G}(U)\stackrel{\displaystyle\left(r_y^U\right)^{\prime}}{\longrightarrow}\mathcal{F}_y\stackrel{\displaystyle r_{x,y}}{\longrightarrow}\mathcal{F}_x\stackrel{\displaystyle\left(r_x^V\right)^{\prime}}{\longleftarrow}\mathcal{G}(V) \end{equation} by the universal property of $\mathcal{G}(V)$: \begin{equation} \dot\exists r^U_V:\mathcal{G}(U)\to\mathcal{G}(V)\mid\left(r_x^V\right)^{\prime}\circ r^U_V=r_{x,y}\circ\left(r_y^U\right)^{\prime}, \end{equation} by definition $\mathcal{G}$ is a $\mathfrak{B}$-presheaf.

The $\mathfrak{B}$-sheaf axioms for $\mathcal{G}$ are equivalent to affirm that for any $U\in\mathfrak{B}$ and for any (open) covering $\{U_i\in\mathfrak{B}\}_{i\in I}$, $\mathcal{G}(U)$ is the equilizer of the diagram \begin{equation} \prod_{i\in I}\mathcal{G}(U_i)\rightrightarrows\prod_{i,j\in I}\mathcal{G}(U_{ij}) \end{equation} where I get $U_{ij}=U_i\cap U_j$ and the double arrows is the categorical product (in $\mathbf{C}$) of the morphism $r^{U_i}_{U_{ij}}$ and $r^{U_j}_{U_{ij}}$.

By definition of $\mathcal{G}$: $\mathcal{G}(U)$ equalizes the previous diagram; let $E$ be the equalizer of the previous diagram, then: \begin{equation} \forall i,j\in I,x_i\in U_i,y_{ij}\in U_{ij},\dot\exists\varphi_i:E\to\mathcal{F}_{x_i},\varphi_{ij}:E\to\mathcal{F}_{y_{ij}} \end{equation} such that the $\varphi_i$'s and $\varphi_{ij}$'s commute opportunely with the $r_{x_{ij},x_i}$'s; by definition $(E,\varphi_{ij})_{i,j\in I}$ is a cone (in $\mathbf{C}$) on the projective system $(\mathcal{F}_x,r_{y,x})_{x,y\in U}$, by the universal properties of $\mathcal{G}(U)$ and $E$: they are canonically isomorphic, that is $\mathcal{G}$ is a $\mathfrak{B}$-sheaf; by previous lemma $\mathcal{G}$ is extendible to a sheaf on $X$.

Whithout confusion, I can continue to write $\mathcal{G}$ for both sheaves!

For any $U\in\mathcal{T}$, by previous reasoning, one can consider $\mathcal{F}(U)$ as a cone over the projective system $(\mathcal{F}_x,r_{y,x})_{x,y\in U}$; then by universal property of $\mathcal{G}(U)$, there exists a unique morphism $\varphi_U:\mathcal{F}(U)\to\mathcal{G}(U)$ such that it (opportunely) commutes whit the $r_{y,x}$'s; in particular, the data of $\varphi_U$'s defines a morphism $\varphi:\mathcal{F}\to\mathcal{G}$ of sheaves. In this way, one can define the canonical morphism $\varphi_x:\mathcal{F}_x\to\mathcal{G}_x$, for any $x\in X$.

For any $x\in X$, $\left(\mathcal{F}_x,\left(r^U_x\right)^{\prime}\right)_{x\in U}$ and $\left(\mathcal{G}_x,r^U_x\right)_{x\in U}$ are cocones for the inductive system $\left(\mathcal{G}(U),r^U_V\right)_{x\in U,V}$; by the couniversal property of $\mathcal{G}_x$, there exists a unique morphism $\psi_x:\mathcal{G}_x\to\mathcal{F}_x$; in particular, the data of $\psi_x$'s defines a morphism $\psi:\mathcal{G}\to\mathcal{F}$ of sheaves.

Using the couniversal property of the stalks of a sheaf, one can prove that: \begin{equation} \forall x\in X,\,\psi_x\circ\varphi_x=Id_{\mathcal{F}_x},\varphi_x\circ\psi_x=Id_{\mathcal{G}_x}; \end{equation} in other words, $\mathcal{F}$ and $\mathcal{G}$ are canonical isomorphic sheaves on $X$ with values in $\mathbf{C}$. $\Box\,(Q.E.D.)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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