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 20.52.2) 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 caculation or by using that $\Omega _{L/S, k} \subset \Omega _{L^\star /S, k}$ (see references above). $\square$

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