The Stacks project

Lemma 50.16.1. For $a \geq 0$ we have

  1. the map $\Omega ^ a_{X/S} \to b_*\Omega ^ a_{L/S}$ is an isomorphism,

  2. the map $\Omega ^ a_{Z/S} \to p_*\Omega ^ a_{P/S}$ is an isomorphism, and

  3. the map $Rb_*\Omega ^ a_{L/S} \to i_*Rp_*\Omega ^ a_{P/S}$ is an isomorphism on cohomology sheaves in degree $\geq 1$.

Proof. Let us first prove part (2). Since $P = Z \times _ S \mathbf{P}^{n - 1}_ S$ we see that

\[ \Omega ^ a_{P/S} = \bigoplus \nolimits _{a = r + s} \text{pr}_1^*\Omega ^ r_{Z/S} \otimes \text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S} \]

Recalling that $p = \text{pr}_1$ by the projection formula (Cohomology, Lemma 20.54.2) we obtain

\[ p_*\Omega ^ a_{P/S} = \bigoplus \nolimits _{a = r + s} \Omega ^ r_{Z/S} \otimes \text{pr}_{1, *}\text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S} \]

By the calculations in Section 50.11 and in particular in the proof of Lemma 50.11.3 we have $\text{pr}_{1, *}\text{pr}_2^*\Omega ^ s_{\mathbf{P}^{n - 1}_ S/S} = 0$ except if $s = 0$ in which case we get $\text{pr}_{1, *}\mathcal{O}_ P = \mathcal{O}_ Z$. This proves (2).

By the material in Section 50.10 and in particular Lemma 50.10.4 we have $\pi _*\Omega ^ a_{L/S} = \Omega ^ a_{P/S} \oplus \bigoplus _{k \geq 1} \Omega ^ a_{L/S, k}$. Since the composition $\pi \circ 0$ in the diagram above is the identity morphism on $P$ to prove part (3) it suffices to show that $\Omega ^ a_{L/S, k}$ has vanishing higher cohomology for $k > 0$. By Lemmas 50.10.2 and 50.10.4 there are short exact sequences

\[ 0 \to \Omega ^ a_{P/S} \otimes \mathcal{O}_ P(k) \to \Omega ^ a_{L/S, k} \to \Omega ^{a - 1}_{P/S} \otimes \mathcal{O}_ P(k) \to 0 \]

where $\Omega ^{a - 1}_{P/S} = 0$ if $a = 0$. Since $P = Z \times _ S \mathbf{P}^{n - 1}_ S$ we have

\[ \Omega ^ a_{P/S} = \bigoplus \nolimits _{i + j = a} \Omega ^ i_{Z/S} \boxtimes \Omega ^ j_{\mathbf{P}^{n - 1}_ S/S} \]

by Lemma 50.8.1. Since $\Omega ^ i_{Z/S}$ is free of finite rank we see that it suffices to show that the higher cohomology of $\mathcal{O}_ Z \boxtimes \Omega ^ j_{\mathbf{P}^{n - 1}_ S/S}(k)$ is zero for $k > 0$. This follows from Lemma 50.11.2 applied to $P = Z \times _ S \mathbf{P}^{n - 1}_ S = \mathbf{P}^{n - 1}_ Z$ and the proof of (3) is complete.

We still have to prove (1). If $n = 1$, then we are blowing up an effective Cartier divisor and $b$ is an isomorphism and we have (1). If $n > 1$, then the composition

\[ \Gamma (X, \Omega ^ a_{X/S}) \to \Gamma (L, \Omega ^ a_{L/S}) \to \Gamma (L \setminus E, \Omega ^ a_{L/S}) = \Gamma (X \setminus Z, \Omega ^ a_{X/S}) \]

is an isomorphism as $\Omega ^ a_{X/S}$ is finite free (small detail omitted). Thus the only way (1) can fail is if there are nonzero elements of $\Gamma (L, \Omega ^ a_{L/S})$ which vanish outside of $E = 0(P)$. Since $L$ is a line bundle over $P$ with zero section $0 : P \to L$, it suffices to show that on a line bundle there are no nonzero sections of a sheaf of differentials which vanish identically outside the zero section. The reader sees this is true either (preferably) by a local calculation or by using that $\Omega _{L/S, k} \subset \Omega _{L^\star /S, k}$ (see references above). $\square$

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 0FUR. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 50.16.1 as pdf