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 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
Lemma 20.37.7.reference 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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