The Stacks project

20.37 Derived limits

Let $(X, \mathcal{O}_ X)$ be a ringed space. Since the triangulated category $D(\mathcal{O}_ X)$ has products (Injectives, Lemma 19.13.4) it follows that $D(\mathcal{O}_ X)$ has derived limits, see Derived Categories, Definition 13.34.1. If $(K_ n)$ is an inverse system in $D(\mathcal{O}_ X)$ then we denote $R\mathop{\mathrm{lim}}\nolimits K_ n$ the derived limit.

Lemma 20.37.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. For $U \subset X$ open the functor $R\Gamma (U, -)$ commutes with $R\mathop{\mathrm{lim}}\nolimits $. Moreover, there are short exact sequences

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) \to H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits H^ m(U, K_ n) \to 0 \]

for any inverse system $(K_ n)$ in $D(\mathcal{O}_ X)$ and any $m \in \mathbf{Z}$.

Proof. The first statement follows from Injectives, Lemma 19.13.6. Then we may apply More on Algebra, Remark 15.86.10 to $R\mathop{\mathrm{lim}}\nolimits R\Gamma (U, K_ n) = R\Gamma (U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ to get the short exact sequences. $\square$

Lemma 20.37.2. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. Then $Rf_*$ commutes with $R\mathop{\mathrm{lim}}\nolimits $, i.e., $Rf_*$ commutes with derived limits.

Proof. Let $(K_ n)$ be an inverse system in $D(\mathcal{O}_ X)$. Consider the defining distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits K_ n \to \prod K_ n \to \prod K_ n \]

in $D(\mathcal{O}_ X)$. Applying the exact functor $Rf_*$ we obtain the distinguished triangle

\[ Rf_*(R\mathop{\mathrm{lim}}\nolimits K_ n) \to Rf_*\left(\prod K_ n\right) \to Rf_*\left(\prod K_ n\right) \]

in $D(\mathcal{O}_ Y)$. Thus we see that it suffices to prove that $Rf_*$ commutes with products in the derived category (which are not just given by products of complexes, see Injectives, Lemma 19.13.4). However, since $Rf_*$ is a right adjoint by Lemma 20.28.1 this follows formally (see Categories, Lemma 4.24.5). Caution: Note that we cannot apply Categories, Lemma 4.24.5 directly as $R\mathop{\mathrm{lim}}\nolimits K_ n$ is not a limit in $D(\mathcal{O}_ X)$. $\square$

Remark 20.37.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)$ be an inverse system in $D(\mathcal{O}_ X)$. Set $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. For each $n$ and $m$ let $\mathcal{H}^ m_ n = H^ m(K_ n)$ be the $m$th cohomology sheaf of $K_ n$ and similarly set $\mathcal{H}^ m = H^ m(K)$. Let us denote $\underline{\mathcal{H}}^ m_ n$ the presheaf

\[ U \longmapsto \underline{\mathcal{H}}^ m_ n(U) = H^ m(U, K_ n) \]

Similarly we set $\underline{\mathcal{H}}^ m(U) = H^ m(U, K)$. By Lemma 20.32.3 we see that $\mathcal{H}^ m_ n$ is the sheafification of $\underline{\mathcal{H}}^ m_ n$ and $\mathcal{H}^ m$ is the sheafification of $\underline{\mathcal{H}}^ m$. Here is a diagram

\[ \xymatrix{ K \ar@{=}[d] & \underline{\mathcal{H}}^ m \ar[d] \ar[r] & \mathcal{H}^ m \ar[d] \\ R\mathop{\mathrm{lim}}\nolimits K_ n & \mathop{\mathrm{lim}}\nolimits \underline{\mathcal{H}}^ m_ n \ar[r] & \mathop{\mathrm{lim}}\nolimits \mathcal{H}^ m_ n } \]

In general it may not be the case that $\mathop{\mathrm{lim}}\nolimits \mathcal{H}^ m_ n$ is the sheafification of $\mathop{\mathrm{lim}}\nolimits \underline{\mathcal{H}}^ m_ n$. If $U \subset X$ is an open, then we have short exact sequences

20.37.3.1
\begin{equation} \label{cohomology-equation-ses-Rlim-over-U} 0 \to R^1\mathop{\mathrm{lim}}\nolimits \underline{\mathcal{H}}^{m - 1}_ n(U) \to \underline{\mathcal{H}}^ m(U) \to \mathop{\mathrm{lim}}\nolimits \underline{\mathcal{H}}^ m_ n(U) \to 0 \end{equation}

by Lemma 20.37.1.

The following lemma applies to an inverse system of quasi-coherent modules with surjective transition maps on a scheme.

Lemma 20.37.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(\mathcal{F}_ n)$ be an inverse system of $\mathcal{O}_ X$-modules. Let $\mathcal{B}$ be a set of opens of $X$. Assume

  1. every open of $X$ has a covering whose members are elements of $\mathcal{B}$,

  2. $H^ p(U, \mathcal{F}_ n) = 0$ for $p > 0$ and $U \in \mathcal{B}$,

  3. the inverse system $\mathcal{F}_ n(U)$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits $ for $U \in \mathcal{B}$.

Then $R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$ and we have $H^ p(U, \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n) = 0$ for $p > 0$ and $U \in \mathcal{B}$.

Proof. Set $K_ n = \mathcal{F}_ n$ and $K = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Using the notation of Remark 20.37.3 and assumption (2) we see that for $U \in \mathcal{B}$ we have $\underline{\mathcal{H}}_ n^ m(U) = 0$ when $m \not= 0$ and $\underline{\mathcal{H}}_ n^0(U) = \mathcal{F}_ n(U)$. From Equation (20.37.3.1) and assumption (3) we see that $\underline{\mathcal{H}}^ m(U) = 0$ when $m \not= 0$ and equal to $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$ when $m = 0$. Sheafifying using (1) we find that $\mathcal{H}^ m = 0$ when $m \not= 0$ and equal to $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$ when $m = 0$. Hence $K = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Since $H^ m(U, K) = \underline{\mathcal{H}}^ m(U) = 0$ for $m > 0$ (see above) we see that the second assertion holds. $\square$

Lemma 20.37.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)$ be an inverse system in $D(\mathcal{O}_ X)$. Let $x \in X$ and $m \in \mathbf{Z}$. Assume there exist an integer $n(x)$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that for $U \in \mathfrak {U}_ x$

  1. $R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) = 0$, and

  2. $H^ m(U, K_ n) \to H^ m(U, K_{n(x)})$ is injective for $n \geq n(x)$.

Then the map on stalks $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)_ x \to H^ m(K_{n(x)})_ x$ is injective.

Proof. Let $\gamma $ be an element of $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)_ x$ which maps to zero in $H^ m(K_{n(x)})_ x$. Since $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)$ is the sheafification of $U \mapsto H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ (by Lemma 20.32.3) we can choose $U \in \mathfrak {U}_ x$ and an element $\tilde\gamma \in H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ mapping to $\gamma $. Then $\tilde\gamma $ maps to $\tilde\gamma _{n(x)} \in H^ m(U, K_{n(x)})$. Using that $H^ m(K_{n(x)})$ is the sheafification of $U \mapsto H^ m(U, K_{n(x)})$ (by Lemma 20.32.3 again) we see that after shrinking $U$ we may assume that $\tilde\gamma _{n(x)} = 0$. For this $U$ we consider the short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{m - 1}(U, K_ n) \to H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits H^ m(U, K_ n) \to 0 \]

of Lemma 20.37.1. By assumption (1) the group on the left is zero and by assumption (2) the group on the right maps injectively into $H^ m(U, K_{n(x)})$. We conclude $\tilde\gamma = 0$ and hence $\gamma = 0$ as desired. $\square$

Lemma 20.37.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume that for every $x \in X$ there exist a function $p(x, -) : \mathbf{Z} \to \mathbf{Z}$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that

\[ H^ p(U, H^{m - p}(E)) = 0 \text{ for } U \in \mathfrak {U}_ x \text{ and } p > p(x, m) \]

Then the map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ of Derived Categories, Remark 13.34.4 is an isomorphism in $D(\mathcal{O}_ X)$.

Proof. Set $K_ n = \tau _{\geq -n}E$ and $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. The canonical map $E \to K$ comes from the canonical maps $E \to K_ n = \tau _{\geq -n}E$. We have to show that $E \to K$ induces an isomorphism $H^ m(E) \to H^ m(K)$ of cohomology sheaves. In the rest of the proof we fix $m$. If $n \geq -m$, then the map $E \to \tau _{\geq -n}E = K_ n$ induces an isomorphism $H^ m(E) \to H^ m(K_ n)$. To finish the proof it suffices to show that for every $x \in X$ there exists an integer $n(x) \geq -m$ such that the map $H^ m(K)_ x \to H^ m(K_{n(x)})_ x$ is injective. Namely, then the composition

\[ H^ m(E)_ x \to H^ m(K)_ x \to H^ m(K_{n(x)})_ x \]

is a bijection and the second arrow is injective, hence the first arrow is bijective. Set

\[ n(x) = 1 + \max \{ -m, p(x, m - 1) - m, -1 + p(x, m) - m, -2 + p(x, m + 1) - m\} . \]

so that in any case $n(x) \geq -m$. Claim: the maps

\[ H^{m - 1}(U, K_{n + 1}) \to H^{m - 1}(U, K_ n) \quad \text{and}\quad H^ m(U, K_{n + 1}) \to H^ m(U, K_ n) \]

are isomorphisms for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. The claim implies conditions (1) and (2) of Lemma 20.37.5 are satisfied and hence implies the desired injectivity. Recall (Derived Categories, Remark 13.12.4) that we have distinguished triangles

\[ H^{-n - 1}(E)[n + 1] \to K_{n + 1} \to K_ n \to H^{-n - 1}(E)[n + 2] \]

Looking at the associated long exact cohomology sequence the claim follows if

\[ H^{m + n}(U, H^{-n - 1}(E)),\quad H^{m + n + 1}(U, H^{-n - 1}(E)),\quad H^{m + n + 2}(U, H^{-n - 1}(E)) \]

are zero for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. This follows from our choice of $n(x)$ and the assumption in the lemma. $\square$

reference

Lemma 20.37.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume that for every $x \in X$ there exist an integer $d_ x \geq 0$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that

\[ H^ p(U, H^ q(E)) = 0 \text{ for } U \in \mathfrak {U}_ x,\ p > d_ x, \text{ and }q < 0 \]

Then the map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ of Derived Categories, Remark 13.34.4 is an isomorphism in $D(\mathcal{O}_ X)$.

Proof. This follows from Lemma 20.37.6 with $p(x, m) = d_ x + \max (0, m)$. $\square$

Lemma 20.37.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume there exist a function $p(-) : \mathbf{Z} \to \mathbf{Z}$ and a set $\mathcal{B}$ of opens of $X$ such that

  1. every open in $X$ has a covering whose members are elements of $\mathcal{B}$, and

  2. $H^ p(U, H^{m - p}(E)) = 0$ for $p > p(m)$ and $U \in \mathcal{B}$.

Then the map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ of Derived Categories, Remark 13.34.4 is an isomorphism in $D(\mathcal{O}_ X)$.

Proof. Apply Lemma 20.37.6 with $p(x, m) = p(m)$ and $\mathfrak {U}_ x = \{ U \in \mathcal{B} \mid x \in U\} $. $\square$

Lemma 20.37.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume there exist an integer $d \geq 0$ and a basis $\mathcal{B}$ for the topology of $X$ such that

\[ H^ p(U, H^ q(E)) = 0 \text{ for } U \in \mathcal{B},\ p > d, \text{ and }q < 0 \]

Then the map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ of Derived Categories, Remark 13.34.4 is an isomorphism in $D(\mathcal{O}_ X)$.

Proof. Apply Lemma 20.37.7 with $d_ x = d$ and $\mathfrak {U}_ x = \{ U \in \mathcal{B} \mid x \in U\} $. $\square$

The lemmas above can be used to compute cohomology in certain situations.

Lemma 20.37.10. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be an object of $D(\mathcal{O}_ X)$. Let $\mathcal{B}$ be a set of opens of $X$. Assume

  1. every open of $X$ has a covering whose members are elements of $\mathcal{B}$,

  2. $H^ p(U, H^ q(K)) = 0$ for all $p > 0$, $q \in \mathbf{Z}$, and $U \in \mathcal{B}$.

Then $H^ q(U, K) = H^0(U, H^ q(K))$ for $q \in \mathbf{Z}$ and $U \in \mathcal{B}$.

Proof. Observe that $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K$ by Lemma 20.37.9 with $d = 0$. Let $U \in \mathcal{B}$. By Equation (20.37.3.1) we get a short exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{q - 1}(U, \tau _{\geq -n}K) \to H^ q(U, K) \to \mathop{\mathrm{lim}}\nolimits H^ q(U, \tau _{\geq -n}K) \to 0 \]

Condition (2) implies $H^ q(U, \tau _{\geq -n} K) = H^0(U, H^ q(\tau _{\geq -n} K))$ for all $q$ by using the spectral sequence of Example 20.29.3. The spectral sequence converges because $\tau _{\geq -n}K$ is bounded below. If $n > -q$ then we have $H^ q(\tau _{\geq -n}K) = H^ q(K)$. Thus the systems on the left and the right of the displayed short exact sequence are eventually constant with values $H^0(U, H^{q - 1}(K))$ and $H^0(U, H^ q(K))$. The lemma follows. $\square$

Here is another case where we can describe the derived limit.

Lemma 20.37.11. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(K_ n)$ be an inverse system of objects of $D(\mathcal{O}_ X)$. Let $\mathcal{B}$ be a set of opens of $X$. Assume

  1. every open of $X$ has a covering whose members are elements of $\mathcal{B}$,

  2. for all $U \in \mathcal{B}$ and all $q \in \mathbf{Z}$ we have

    1. $H^ p(U, H^ q(K_ n)) = 0$ for $p > 0$,

    2. the inverse system $H^0(U, H^ q(K_ n))$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits $.

Then $H^ q(R\mathop{\mathrm{lim}}\nolimits K_ n) = \mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ for $q \in \mathbf{Z}$.

Proof. Set $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. We will use notation as in Remark 20.37.3. Let $U \in \mathcal{B}$. By Lemma 20.37.10 and (2)(a) we have $H^ q(U, K_ n) = H^0(U, H^ q(K_ n))$. Using that the functor $R\Gamma (U, -)$ commutes with derived limits we have

\[ H^ q(U, K) = H^ q(R\mathop{\mathrm{lim}}\nolimits R\Gamma (U, K_ n)) = \mathop{\mathrm{lim}}\nolimits H^0(U, H^ q(K_ n)) \]

where the final equality follows from More on Algebra, Remark 15.86.10 and assumption (2)(b). Thus $H^ q(U, K)$ is the inverse limit the sections of the sheaves $H^ q(K_ n)$ over $U$. Since $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ is a sheaf we find using assumption (1) that $H^ q(K)$, which is the sheafification of the presheaf $U \mapsto H^ q(U, K)$, is equal to $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$. This proves the lemma. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BKN. Beware of the difference between the letter 'O' and the digit '0'.