## 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 $D(\mathcal{O}_ X)$.
$\square$

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

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

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

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 $U \in \mathfrak {U}_ x$

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

$H^ m(U, K_ n) \to H^ m(U, K_{n(x)})$ 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 asssociated 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

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

$H^ p(U, H^{m - p}(E)) = 0$ 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

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

$H^ p(U, H^ q(K)) = 0$ 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

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

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

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

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.
$\square$

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