The Stacks project

20.20 Vanishing on Noetherian topological spaces

The aim is to prove a theorem of Grothendieck namely Proposition 20.20.7. See [Tohoku].

Lemma 20.20.1. Let $i : Z \to X$ be a closed immersion of topological spaces. For any abelian sheaf $\mathcal{F}$ on $Z$ we have $H^ p(Z, \mathcal{F}) = H^ p(X, i_*\mathcal{F})$.

Proof. This is true because $i_*$ is exact (see Modules, Lemma 17.6.1), and hence $R^ pi_* = 0$ as a functor (Derived Categories, Lemma 13.16.9). Thus we may apply Lemma 20.13.6. $\square$

Lemma 20.20.2. Let $X$ be an irreducible topological space. Then $H^ p(X, \underline{A}) = 0$ for all $p > 0$ and any abelian group $A$.

Proof. Recall that $\underline{A}$ is the constant sheaf as defined in Sheaves, Definition 6.7.4. It is clear that for any nonempty open $U \subset X$ we have $\underline{A}(U) = A$ as $X$ is irreducible (and hence $U$ is connected). We will show that the higher Čech cohomology groups $\check{H}^ p(\mathcal{U}, \underline{A})$ are zero for any open covering $\mathcal{U} : U = \bigcup _{i\in I} U_ i$ of an open $U \subset X$. Then the lemma will follow from Lemma 20.11.8.

Recall that the value of an abelian sheaf on the empty open set is $0$. Hence we may clearly assume $U_ i \not= \emptyset $ for all $i \in I$. In this case we see that $U_ i \cap U_{i'} \not= \emptyset $ for all $i, i' \in I$. Hence we see that the Čech complex is simply the complex

\[ \prod _{i_0 \in I} A \to \prod _{(i_0, i_1) \in I^2} A \to \prod _{(i_0, i_1, i_2) \in I^3} A \to \ldots \]

We have to see this has trivial higher cohomology groups. We can see this for example because this is the Čech complex for the covering of a $1$-point space and Čech cohomology agrees with cohomology on such a space. (You can also directly verify it by writing an explicit homotopy.) $\square$


Lemma 20.20.3. Let $X$ be a topological space such that the intersection of any two quasi-compact opens is quasi-compact. Let $\mathcal{F} \subset \underline{\mathbf{Z}}$ be a subsheaf generated by finitely many sections over quasi-compact opens. Then there exists a finite filtration

\[ (0) = \mathcal{F}_0 \subset \mathcal{F}_1 \subset \ldots \subset \mathcal{F}_ n = \mathcal{F} \]

by abelian subsheaves such that for each $0 < i \leq n$ there exists a short exact sequence

\[ 0 \to j'_!\underline{\mathbf{Z}}_ V \to j_!\underline{\mathbf{Z}}_ U \to \mathcal{F}_ i/\mathcal{F}_{i - 1} \to 0 \]

with $j : U \to X$ and $j' : V \to X$ the inclusion of quasi-compact opens into $X$.

Proof. Say $\mathcal{F}$ is generated by the sections $s_1, \ldots , s_ t$ over the quasi-compact opens $U_1, \ldots , U_ t$. Since $U_ i$ is quasi-compact and $s_ i$ a locally constant function to $\mathbf{Z}$ we may assume, after possibly replacing $U_ i$ by the parts of a finite decomposition into open and closed subsets, that $s_ i$ is a constant section. Say $s_ i = n_ i$ with $n_ i \in \mathbf{Z}$. Of course we can remove $(U_ i, n_ i)$ from the list if $n_ i = 0$. Flipping signs if necessary we may also assume $n_ i > 0$. Next, for any subset $I \subset \{ 1, \ldots , t\} $ we may add $\bigcap _{i \in I} U_ i$ and $\gcd (n_ i, i \in I)$ to the list. After doing this we see that our list $(U_1, n_1), \ldots , (U_ t, n_ t)$ satisfies the following property: For $x \in X$ set $I_ x = \{ i \in \{ 1, \ldots , t\} \mid x \in U_ i\} $. Then $\gcd (n_ i, i \in I_ x)$ is attained by $n_ i$ for some $i \in I_ x$.

As our filtration we take $\mathcal{F}_0 = (0)$ and $\mathcal{F}_ n$ generated by the sections $n_ i$ over $U_ i$ for those $i$ such that $n_ i \leq n$. It is clear that $\mathcal{F}_ n = \mathcal{F}$ for $n \gg 0$. Moreover, the quotient $\mathcal{F}_ n/\mathcal{F}_{n - 1}$ is generated by the section $n$ over $U = \bigcup _{n_ i \leq n} U_ i$ and the kernel of the map $j_!\underline{\mathbf{Z}}_ U \to \mathcal{F}_ n/\mathcal{F}_{n - 1}$ is generated by the section $n$ over $V = \bigcup _{n_ i \leq n - 1} U_ i$. Thus a short exact sequence as in the statement of the lemma. $\square$


Lemma 20.20.4. Let $X$ be a topological space. Let $d \geq 0$ be an integer. Assume

  1. $X$ is quasi-compact,

  2. the quasi-compact opens form a basis for $X$, and

  3. the intersection of two quasi-compact opens is quasi-compact.

  4. $H^ p(X, j_!\underline{\mathbf{Z}}_ U) = 0$ for all $p > d$ and any quasi-compact open $j : U \to X$.

Then $H^ p(X, \mathcal{F}) = 0$ for all $p > d$ and any abelian sheaf $\mathcal{F}$ on $X$.

Proof. Let $S = \coprod _{U \subset X} \mathcal{F}(U)$ where $U$ runs over the quasi-compact opens of $X$. For any finite subset $A = \{ s_1, \ldots , s_ n\} \subset S$, let $\mathcal{F}_ A$ be the subsheaf of $\mathcal{F}$ generated by all $s_ i$ (see Modules, Definition 17.4.5). Note that if $A \subset A'$, then $\mathcal{F}_ A \subset \mathcal{F}_{A'}$. Hence $\{ \mathcal{F}_ A\} $ forms a system over the directed partially ordered set of finite subsets of $S$. By Modules, Lemma 17.4.6 it is clear that

\[ \mathop{\mathrm{colim}}\nolimits _ A \mathcal{F}_ A = \mathcal{F} \]

by looking at stalks. By Lemma 20.19.1 we have

\[ H^ p(X, \mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _ A H^ p(X, \mathcal{F}_ A) \]

Hence it suffices to prove the vanishing for the abelian sheaves $\mathcal{F}_ A$. In other words, it suffices to prove the result when $\mathcal{F}$ is generated by finitely many local sections over quasi-compact opens of $X$.

Suppose that $\mathcal{F}$ is generated by the local sections $s_1, \ldots , s_ n$. Let $\mathcal{F}' \subset \mathcal{F}$ be the subsheaf generated by $s_1, \ldots , s_{n - 1}$. Then we have a short exact sequence

\[ 0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}/\mathcal{F}' \to 0 \]

From the long exact sequence of cohomology we see that it suffices to prove the vanishing for the abelian sheaves $\mathcal{F}'$ and $\mathcal{F}/\mathcal{F}'$ which are generated by fewer than $n$ local sections. Hence it suffices to prove the vanishing for sheaves generated by at most one local section. These sheaves are exactly the quotients of the sheaves $j_!\underline{\mathbf{Z}}_ U$ where $U$ is a quasi-compact open of $X$.

Assume now that we have a short exact sequence

\[ 0 \to \mathcal{K} \to j_!\underline{\mathbf{Z}}_ U \to \mathcal{F} \to 0 \]

with $U$ quasi-compact open in $X$. It suffices to show that $H^ q(X, \mathcal{K})$ is zero for $q \geq d + 1$. As above we can write $\mathcal{K}$ as the filtered colimit of subsheaves $\mathcal{K}'$ generated by finitely many sections over quasi-compact opens. Then $\mathcal{F}$ is the filtered colimit of the sheaves $j_!\underline{\mathbf{Z}}_ U/\mathcal{K}'$. In this way we reduce to the case that $\mathcal{K}$ is generated by finitely many sections over quasi-compact opens. Note that $\mathcal{K}$ is a subsheaf of $\underline{\mathbf{Z}}_ X$. Thus by Lemma 20.20.3 there exists a finite filtration of $\mathcal{K}$ whose successive quotients $\mathcal{Q}$ fit into a short exact sequence

\[ 0 \to j''_!\underline{\mathbf{Z}}_ W \to j'_!\underline{\mathbf{Z}}_ V \to \mathcal{Q} \to 0 \]

with $j'' : W \to X$ and $j' : V \to X$ the inclusions of quasi-compact opens. Hence the vanishing of $H^ p(X, \mathcal{Q})$ for $p > d$ follows from our assumption (in the lemma) on the vanishing of the cohomology groups of $j''_!\underline{\mathbf{Z}}_ W$ and $j'_!\underline{\mathbf{Z}}_ V$. Returning to $\mathcal{K}$ this, via an induction argument using the long exact cohomology sequence, implies the desired vanishing for it as well. $\square$

Example 20.20.5. Let $X = \mathbf{N}$ endowed with the topology whose opens are $\emptyset $, $X$, and $U_ n = \{ i \mid i \leq n\} $ for $n \geq 1$. An abelian sheaf $\mathcal{F}$ on $X$ is the same as an inverse system of abelian groups $A_ n = \mathcal{F}(U_ n)$ and $\Gamma (X, \mathcal{F}) = \mathop{\mathrm{lim}}\nolimits A_ n$. Since the inverse limit functor is not an exact functor on the category of inverse systems, we see that there is an abelian sheaf with nonzero $H^1$. Finally, the reader can check that $H^ p(X, j_!\mathbf{Z}_ U) = 0$, $p \geq 1$ if $j : U = U_ n \to X$ is the inclusion. Thus we see that $X$ is an example of a space satisfying conditions (2), (3), and (4) of Lemma 20.20.4 for $d = 0$ but not the conclusion.

Lemma 20.20.6. Let $X$ be an irreducible topological space. Let $\mathcal{H} \subset \underline{\mathbf{Z}}$ be an abelian subsheaf of the constant sheaf. Then there exists a nonempty open $U \subset X$ such that $\mathcal{H}|_ U = \underline{d\mathbf{Z}}_ U$ for some $d \in \mathbf{Z}$.

Proof. Recall that $\underline{\mathbf{Z}}(V) = \mathbf{Z}$ for any nonempty open $V$ of $X$ (see proof of Lemma 20.20.2). If $\mathcal{H} = 0$, then the lemma holds with $d = 0$. If $\mathcal{H} \not= 0$, then there exists a nonempty open $U \subset X$ such that $\mathcal{H}(U) \not= 0$. Say $\mathcal{H}(U) = n\mathbf{Z}$ for some $n \geq 1$. Hence we see that $\underline{n\mathbf{Z}}_ U \subset \mathcal{H}|_ U \subset \underline{\mathbf{Z}}_ U$. If the first inclusion is strict we can find a nonempty $U' \subset U$ and an integer $1 \leq n' < n$ such that $\underline{n'\mathbf{Z}}_{U'} \subset \mathcal{H}|_{U'} \subset \underline{\mathbf{Z}}_{U'}$. This process has to stop after a finite number of steps, and hence we get the lemma. $\square$

Proof. We prove this lemma by induction on $d$. So fix $d$ and assume the lemma holds for all Noetherian topological spaces of dimension $< d$.

Let $\mathcal{F}$ be an abelian sheaf on $X$. Suppose $U \subset X$ is an open. Let $Z \subset X$ denote the closed complement. Denote $j : U \to X$ and $i : Z \to X$ the inclusion maps. Then there is a short exact sequence

\[ 0 \to j_{!}j^*\mathcal{F} \to \mathcal{F} \to i_*i^*\mathcal{F} \to 0 \]

see Modules, Lemma 17.7.1. Note that $j_!j^*\mathcal{F}$ is supported on the topological closure $Z'$ of $U$, i.e., it is of the form $i'_*\mathcal{F}'$ for some abelian sheaf $\mathcal{F}'$ on $Z'$, where $i' : Z' \to X$ is the inclusion.

We can use this to reduce to the case where $X$ is irreducible. Namely, according to Topology, Lemma 5.9.2 $X$ has finitely many irreducible components. If $X$ has more than one irreducible component, then let $Z \subset X$ be an irreducible component of $X$ and set $U = X \setminus Z$. By the above, and the long exact sequence of cohomology, it suffices to prove the vanishing of $H^ p(X, i_*i^*\mathcal{F})$ and $H^ p(X, i'_*\mathcal{F}')$ for $p > d$. By Lemma 20.20.1 it suffices to prove $H^ p(Z, i^*\mathcal{F})$ and $H^ p(Z', \mathcal{F}')$ vanish for $p > d$. Since $Z'$ and $Z$ have fewer irreducible components we indeed reduce to the case of an irreducible $X$.

If $d = 0$ and $X$ is irreducible, then $X$ is the only nonempty open subset of $X$. Hence every sheaf is constant and higher cohomology groups vanish (for example by Lemma 20.20.2).

Suppose $X$ is irreducible of dimension $d > 0$. By Lemma 20.20.4 we reduce to the case where $\mathcal{F} = j_!\underline{\mathbf{Z}}_ U$ for some open $U \subset X$. In this case we look at the short exact sequence

\[ 0 \to j_!(\underline{\mathbf{Z}}_ U) \to \underline{\mathbf{Z}}_ X \to i_*\underline{\mathbf{Z}}_ Z \to 0 \]

where $Z = X \setminus U$. By Lemma 20.20.2 we have the vanishing of $H^ p(X, \underline{\mathbf{Z}}_ X)$ for all $p \geq 1$. By induction we have $H^ p(X, i_*\underline{\mathbf{Z}}_ Z) = H^ p(Z, \underline{\mathbf{Z}}_ Z) = 0$ for $p \geq d$. Hence we win by the long exact cohomology sequence. $\square$

Comments (2)

Comment #5966 by Joseph Lipman on

For the case d=0 of Proposition 02UZ, is X={*} supposed to mean that X consists of a single point (as in Grothendieck's original argument)? In fact any space X is an irreducible 0-dimensional noetherian space iff X itself is its only nonempty closed subset. Would it be better to say that if d=0 then U is empty, so j_!(\bf Z)_U=0?

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