The Stacks project

Lemma 20.34.10. 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 20.34.9 with $d = 0$. Let $U \in \mathcal{B}$. By Equation (20.34.3.1) 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. $\square$


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Your email address will not be published.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 0BKT. Beware of the difference between the letter 'O' and the digit '0'.