3.3. 下降

The idea of descents should be that, they are like sheaf axiom for fibered categories.

例 3.3.0.1. Let $X$ be a scheme and $C$ be the category $\mathsf{Op}(X)$. Then consider $\mathcal{F}=\mathsf{Vect}\to \mathcal{C}$ where $\mathcal{F}(U)$ be the category of vector bundles on $U$. Then, if $U={\bigcup }_i{U}_i$, a vector bundle on $U$ is not equivalent to ${\mathcal{E}}_i$ on $U_i$ with double intersections isomorphic (i.e. ${\sigma }_{ij}:{\mathcal{E}}_i{|}_{U_{ij}}\stackrel{\sim }{\to }{\mathcal{E}}_j{|}_{U_{ij}}$).

In this case, the naive sheaf axioms fit into the picture, i.e. we get$$\mathcal{F}(U)\to \prod _i\mathcal{F}(U_i)\Rightarrow \prod _{i,j}\mathcal{F}(U_{ij})$$and this diagram is not exact.

We are missing the “cocycle” condition (to make the above diagram exact/equalizer). This means that, on $U_{ijk}=U_i\cap U_j\cap U_k$, we get diagram

In other word, the right “exact” diagram we need will be something like

Therefore, we want to formalize this triple arrow thing in fibered categories.

Let $p:\mathcal{F}\to \mathcal{C}$ be a fibered category. Given $f:X\to Y$ in $\mathcal{C}$, let $\mathcal{F}(X\stackrel{f}{\to }Y)$ be the category defined as follows (this is called category of descent data). The object should be $(E,\sigma )$ with $E\in \mathcal{F}(X)$ andwhere the triple arrows are $p_{12},p_{13},p_{23}$ and the double arrows are $p_1,p_2$, and $\sigma :p_1^{\ast }E\to p_2^{\ast }E$ is an isomorphism in $\mathcal{F}(X{\times }_{Y}X)$ such that we get the following commutative diagramwhere $=$ means canonical isomorphism. This is called the cocycle condition.

注 3.3.0.2. Here is just a brife recall of what all the above notations (i.e. $p_i^{\ast }E$, $p_{ij}$, etc) means.

First, recall that $p_i^{\ast }E$ are defined as the pullback of the following diagramSimilarly, $p_{jk}^{\ast }{p}_i{E}^{\ast }$ are defined as the pullback of the following diagram

Second, we note $p_{ij}$ are projections come from the universal property of (fibered) products. In other word, note we would define $X_1{\times }_S{X}_2{\times }_S{X}_3$ as the unique object satisfies the following diagram

Finally, a word on the isomorphisms $p_{jk}^{\ast }{p}_i^{\ast }E$. Continue with the above diagram (where now we let $X_1=X_2=X_3=X$), we get the followingwhere the two pullbacks along $p_{13}$ and $p_{12}$ must be the same object living over $X_1\times X_2\times X_3$ , hence the canonical isomorphisms between $p_{13}^{\ast }{p}_1^{\ast }E\cong p_{12}^{\ast }{p}_1^{\ast }E$. The others are similar.

The idea is that, if we have $\mathcal{F}(Y)\stackrel{ϵ}{\to }\mathcal{F}(X\stackrel{f}{\to }Y)$, then stack would be $\mathcal{F}$ fibered in groupoids where $ϵ$ is an equivalence¹.

Last time, we let $p:\mathcal{F}\to \mathcal{C}$ be a fibered category (not necessarily fibered in groupoids). Given $X\stackrel{f}{\to }Y$ in $\mathcal{C}$, we defined a category $\mathcal{F}(X\stackrel{f}{\to }Y)$ be the category of descent data.

The objects of this category is $(E,\sigma )$ where $E\in \mathcal{F}(X)$ and $\sigma :p_1^{\ast }E\to p_2^{\ast }E$ is an isomorhism between $p_1^{\ast }E\stackrel{\sim }{\to }{p}_2^{\ast }E$. Here we have a diagramand hence the two pullbacks of $E$ should be isomorphic, and we just choose a particular $\sigma$. However, this $\sigma$ cannot be arbitrary, as we need one additional condition called the cocycle condition.

The cocycle condition says that, all the ways to pullback $E$ should all be commuting (we have three arrows $p_{ij}$ from $X\times X\times X$ to $X\times X$, and two arrows from $X\times X$ to $X$, then we want $p_{23}^{\ast }\sigma \circ p_{12}^{\ast }\sigma =p_{13}^{\ast }\sigma$).

The point is that, we then get $\mathcal{F}(Y)\stackrel{ϵ}{\to }\mathcal{F}(X\stackrel{f}{\to }Y)$ by$$F\mapsto (f^{\ast }F,{\sigma }_{\mathrm{can}})$$This is because, $p_1^{\ast }{f}^{\ast }F$ and $p_2^{\ast }{f}^{\ast }F$ are pullbacks of $F$ along $g:X{\times }_{Y}X\to Y$. For example, we get $p_1^{\ast }{f}^{\ast }F$ by the following diagram (where $g$ is equal both $f{p}_1$ and $f{p}_2$ at the same time)

Therefore we get a canonical map ${\sigma }_{\mathrm{can}}:p_1^{\ast }{f}^{\ast }F\stackrel{\sim }{\to }{p}_2^{\ast }{f}^{\ast }F$.

In this case, we say $f$ is an effective descent morphism if $ϵ$ is equivalence of categories. In this case we also say $\mathcal{F}$ satisfies descent for $f$.

Before we say explain what this means, we recall the morphisms of $\mathcal{F}(X\stackrel{f}{\to }Y)$ is the following. From $(E^{\prime },{\sigma }^{\prime })$ to $(E,\sigma )$, a morphism is $\alpha :E^{\prime }\to E$ in $\mathcal{F}(X)$ such that we get the following commuting diagram

More generally, we can define $\mathcal{F}(\{X_i\stackrel{f_i}{\to }Y{\}}_{i\in I})$ as $(\{E_i{\}}_{i\in I},\{{\sigma }_{ij}{\}}_{i,j})$ such that $E_i\in \mathcal{F}(X_i)$ and ${\sigma }_{ij}:p_i^{\ast }{E}_i\stackrel{\sim }{\to }{p}_j^{\ast }{E}_j$ is an isomorhism, whereWe also need the cocycle condition ${\sigma }_{jk}\circ {\sigma }_{ij}={\sigma }_{ik}$.

Normally, we do not need to think about this more general case because of the following lemma.

引理 3.3.0.3 (Olsson, Lemma 4.2.7). Assume coproducts exist in $\mathcal{C}$ and coproducts commute with fiber products when they exists. Assume for all sets of objects $\{X_i{\}}_{i\in I}$ in $\mathcal{C}$ the natural map $\mathcal{F}(\coprod X_i)\to \prod \mathcal{F}(X_i)$ is an equivalence. Then if $\{X_i\to Y\}$ are morphisms in $\mathcal{C}$ and $Q={\coprod }_i{X}_i$, then $\mathcal{F}(Y)\to \mathcal{F}(\{X_i\to Y\})$ is equivalence (of categories) if and only if $\mathcal{F}(Y)\to \mathcal{F}(Q\to Y)$ is equivalence (of categories).

Note this is coproduct, not co-fiber product. Co-fiber product may not exists in schemes. On the other hand, coproducts exists in schemes, as they are just disjoint union of schemes.

Thus, the point is that, we can always think $\mathcal{F}(\{X_i\to Y\})$ as $\mathcal{F}(\coprod X_i\to Y)$, which means we back to the first case.