Tuesday, June 13, 2017

Constructible sheaves

Let $X$ be a topological space with an open cover $\mathcal U = \{U_i\}$, and category $Op(X)$ of open sets of $X$. The goal is to define constructible sheaves and consider some applications. Thanks to Joe Berner for helpful pointers in this area.

Definition: Constructible subsets of $X$ are the smallest family $F$ of subsets of $X$ such that
  • $Op(X)\subset F$,
  • $F$ is closed under finite intersections, and
  • $F$ is closed under complements.
This idea can be applied to sheaves. Recall that a locally closed subset of $X$ is the intersection of an open set and a closed set.

Definition: A sheaf $\mathcal F$ over $X$ is constructible if there exists, equivalently,
  • a filtration $\emptyset=U_0\subset \cdots \subset U_n=X$ of $X$ by opens such that $\mathcal F|_{U_{i+1}\setminus U_i}$ is constant for all $i$, or
  • a cover $\{V_i\}$ of locally closed subsets of $X$ such that $\mathcal F|_{V_i}$ is constant for all $i$.
Since the category of abelian sheaves over a topological space has enough injectives, we may consider an injective resolution of a sheaf $\mathcal F$ rather than the sheaf itself. The resolution may be considered as living inside the derived category of sheaves on $X$.

Definition: Let $A$ be an abelian category.
  • $C(A)$ is the category of cochain complexes of $A$, 
  • $K(A) = C(A)$ modulo cochain homotopy, and
  • $D(A) = K(A)$ modulo $F\in K(A)$ such that $H^n(F)=0$ for all $n$, called the derived category of $A$.
Next we consider an example. Recall the Ran space $\Ran(M) = \{X\subset M\ :\ 0<|X|<\infty\}$ of non-empty finite subsets of a manifold $M$ and the Čech complex of radius $t>0$ of $P\in \Ran(M)$, a simplicial complex with $n$-cells for every $P'\subset P$ of size $n+1$ such that $d(P'_1,P'_2)<t$ for all $P'_1,P'_2\in P'$.

Example: Consider the subset $\Ran^{\leqslant 2}(M) = \{X\subset M\ :\ 1\leqslant |X|\leqslant 2\}$ of the Ran space. Decompose $X=\Ran^{\leqslant 2}(M)\times \R_+$ into disjoint sets $U_\alpha\cup U_\beta$, where
\[
U_\alpha = \underbrace{\left(\Ran^1(M)\times \R_+\right)}_{U_{\alpha,1}} \cup \underbrace{\bigcup_{P\in \Ran^2(M)}\{P\}\times (d_M(P_1,P_2),\infty)}_{U_{\alpha,2}},
\hspace{1cm}
U_\beta = \bigcup_{P\in \Ran^2(M)} \{P\} \times (0,d_M(P_1,P_2)],
\]
with $d_M$ the distance on the manifold $M$. The idea is that for every $(P,t)\in U_\alpha$, the Čech complex of radius $t$ on $P$ has the homotopy type of a point, whereas on $U_\beta$ has the homotopy type of two points. With this in mind, define a constructible sheaf $F\in\text{Shv}(\Ran^{\leqslant 2}(M)\times \R_+)$ valued in simplicial complexes, with $F|_{U_\alpha}$ and $F|_{U_\beta}$ constant sheaves. Set
\[
F_{(P,t)\in U_\alpha} = F(U_\alpha) = \left(0\to \{*\} \to 0\right),
\hspace{1cm}
F_{(P,t)\in U_\beta} = F(U_\beta) = \left(0\to \{*,*\}\to 0\right).
\]
Note that the chain complex $F(U_\alpha)$ is chain homotopic to $0\to \{-\}\to \{*,*\}\to 0$, where $-$ is a single 1-cell with endpoints $*,*$. To show that this is a constructible sheaf, we need to filter $\Ran^{\leqslant 2}(M)\times \R_+$ into an increasing sequence of opens. For this we use a distance on $\Ran^{\leqslant 2}(M)\times \R_+$, given by $d((P,t),(P',t'))=d_{\Ran(M)}(P,P')+d_\R(t,t'),$ where $d_\R(t,t')=|t-t'|$ and
\[
d_{\Ran(M)}(P,P')=\max_{p\in P}\left\{\min_{p'\in P'}\left\{d_M(p,p')\right\}\right\} + \max_{p'\in P'}\left\{\min_{p\in P}\left\{d_M(p,p')\right\}\right\}.
\]
Note that $U_\alpha$ is open. Indeed, for $(P,t)\in U_{\alpha,1}$, every other $P'\in \Ran^1(M)$ close to $P$ is also in $U_{\alpha,1}$, and if $P'\in \Ran^2(M)$ is close to $P$, then the non-zero component $t\in\R_+$ still guarantees the same homotopy type. The set $U_{\alpha,2}$ is open as well, so $U_\alpha$ is open. The whole space is open, so a filtration $\emptyset\subset U_\alpha\subset X$ works for us.

References: Hartshorne (Algebraic geometry, Section II.3), Hartshorne (Residues and Duality, Chapter IV.1), Kashiwara and Schapira (Sheaves on manifolds, Chapters 2 and 8), Lurie (Higher algebra, Section 5.5.1)

No comments:

Post a Comment