## 30.8 Cohomology of projective space

In this section we compute the cohomology of the twists of the structure sheaf on $\mathbf{P}^ n_ S$ over a scheme $S$. Recall that $\mathbf{P}^ n_ S$ was defined as the fibre product $\mathbf{P}^ n_ S = S \times _{\mathop{\mathrm{Spec}}(\mathbf{Z})} \mathbf{P}^ n_{\mathbf{Z}}$ in Constructions, Definition 27.13.2. It was shown to be equal to

$\mathbf{P}^ n_ S = \underline{\text{Proj}}_ S(\mathcal{O}_ S[T_0, \ldots , T_ n])$

in Constructions, Lemma 27.21.5. In particular, projective space is a particular case of a projective bundle. If $S = \mathop{\mathrm{Spec}}(R)$ is affine then we have

$\mathbf{P}^ n_ S = \mathbf{P}^ n_ R = \text{Proj}(R[T_0, \ldots , T_ n]).$

All these identifications are compatible and compatible with the constructions of the twisted structure sheaves $\mathcal{O}_{\mathbf{P}^ n_ S}(d)$.

Before we state the result we need some notation. Let $R$ be a ring. Recall that $R[T_0, \ldots , T_ n]$ is a graded $R$-algebra where each $T_ i$ is homogeneous of degree $1$. Denote $(R[T_0, \ldots , T_ n])_ d$ the degree $d$ summand. It is a finite free $R$-module of rank $\binom {n + d}{d}$ when $d \geq 0$ and zero else. It has a basis consisting of monomials $T_0^{e_0} \ldots T_ n^{e_ n}$ with $\sum e_ i = d$. We will also use the following notation: $R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]$ denotes the $\mathbf{Z}$-graded ring with $\frac{1}{T_ i}$ in degree $-1$. In particular the $\mathbf{Z}$-graded $R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]$ module

$\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]$

which shows up in the statement below is zero in degrees $\geq -n$, is free on the generator $\frac{1}{T_0 \ldots T_ n}$ in degree $-n - 1$ and is free of rank $(-1)^ n\binom {n + d}{d}$ for $d \leq -n - 1$.

Lemma 30.8.1. Let $R$ be a ring. Let $n \geq 0$ be an integer. We have

$H^ q(\mathbf{P}^ n, \mathcal{O}_{\mathbf{P}^ n_ R}(d)) = \left\{ \begin{matrix} (R[T_0, \ldots , T_ n])_ d & \text{if} & q = 0 \\ 0 & \text{if} & q \not= 0, n \\ \left(\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]\right)_ d & \text{if} & q = n \end{matrix} \right.$

as $R$-modules.

Proof. We will use the standard affine open covering

$\mathcal{U} : \mathbf{P}^ n_ R = \bigcup \nolimits _{i = 0}^ n D_{+}(T_ i)$

to compute the cohomology using the Čech complex. This is permissible by Lemma 30.2.6 since any intersection of finitely many affine $D_{+}(T_ i)$ is also a standard affine open (see Constructions, Section 27.8). In fact, we can use the alternating or ordered Čech complex according to Cohomology, Lemmas 20.23.3 and 20.23.6.

The ordering we will use on $\{ 0, \ldots , n\}$ is the usual one. Hence the complex we are looking at has terms

$\check{\mathcal{C}}_{ord}^ p(\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{i_0 < \ldots < i_ p} (R[T_0, \ldots , T_ n, \frac{1}{T_{i_0} \ldots T_{i_ p}}])_ d$

Moreover, the maps are given by the usual formula

$d(s)_{i_0 \ldots i_{p + 1}} = \sum \nolimits _{j = 0}^{p + 1} (-1)^ j s_{i_0 \ldots \hat i_ j \ldots i_{p + 1}}$

see Cohomology, Section 20.23. Note that each term of this complex has a natural $\mathbf{Z}^{n + 1}$-grading. Namely, we get this by declaring a monomial $T_0^{e_0} \ldots T_ n^{e_ n}$ to be homogeneous with weight $(e_0, \ldots , e_ n) \in \mathbf{Z}^{n + 1}$. It is clear that the differential given above respects the grading. In a formula we have

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}} \check{\mathcal{C}}^\bullet (\vec{e})$

where not all summands on the right hand side occur (see below). Hence in order to compute the cohomology modules of the complex it suffices to compute the cohomology of the graded pieces and take the direct sum at the end.

Fix $\vec{e} = (e_0, \ldots , e_ n) \in \mathbf{Z}^{n + 1}$. In order for this weight to occur in the complex above we need to assume $e_0 + \ldots + e_ n = d$ (if not then it occurs for a different twist of the structure sheaf of course). Assuming this, set

$NEG(\vec{e}) = \{ i \in \{ 0, \ldots , n\} \mid e_ i < 0\} .$

With this notation the weight $\vec{e}$ summand $\check{\mathcal{C}}^\bullet (\vec{e})$ of the Čech complex above has the following terms

$\check{\mathcal{C}}^ p(\vec{e}) = \bigoplus \nolimits _{i_0 < \ldots < i_ p, \ NEG(\vec{e}) \subset \{ i_0, \ldots , i_ p\} } R \cdot T_0^{e_0} \ldots T_ n^{e_ n}$

In other words, the terms corresponding to $i_0 < \ldots < i_ p$ such that $NEG(\vec{e})$ is not contained in $\{ i_0 \ldots i_ p\}$ are zero. The differential of the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ is still given by the exact same formula as above.

Suppose that $NEG(\vec{e}) = \{ 0, \ldots , n\}$, i.e., that all exponents $e_ i$ are negative. In this case the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ has only one term, namely $\check{\mathcal{C}}^ n(\vec{e}) = R \cdot \frac{1}{T_0^{-e_0} \ldots T_ n^{-e_ n}}$. Hence in this case

$H^ q(\check{\mathcal{C}}^\bullet (\vec{e})) = \left\{ \begin{matrix} R \cdot \frac{1}{T_0^{-e_0} \ldots T_ n^{-e_ n}} & \text{if} & q = n \\ 0 & \text{if} & \text{else} \end{matrix} \right.$

The direct sum of all of these terms clearly gives the value

$\left(\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]\right)_ d$

in degree $n$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma).

Assume $NEG(\vec{e}) = \emptyset$. In this case the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ has a summand $R$ corresponding to all $i_0 < \ldots < i_ p$. Let us compare the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ to another complex. Namely, consider the affine open covering

$\mathcal{V} : \mathop{\mathrm{Spec}}(R) = \bigcup \nolimits _{i \in \{ 0, \ldots , n\} } V_ i$

where $V_ i = \mathop{\mathrm{Spec}}(R)$ for all $i$. Consider the alternating Čech complex

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$

By the same reasoning as above this computes the cohomology of the structure sheaf on $\mathop{\mathrm{Spec}}(R)$. Hence we see that $H^ p( \check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)}) ) = R$ if $p = 0$ and is $0$ whenever $p > 0$. For these facts, see Lemma 30.2.1 and its proof. Note that also $\check{\mathcal{C}}_{ord}^\bullet (\mathcal{V}, \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ has a summand $R$ for every $i_0 < \ldots < i_ p$ and has exactly the same differential as $\check{\mathcal{C}}^\bullet (\vec{e})$. In other words these complexes are isomorphic complexes and hence have the same cohomology. We conclude that

$H^ q(\check{\mathcal{C}}^\bullet (\vec{e})) = \left\{ \begin{matrix} R \cdot T_0^{e_0} \ldots T_ n^{e_ n} & \text{if} & q = 0 \\ 0 & \text{if} & \text{else} \end{matrix} \right.$

in the case that $NEG(\vec{e}) = \emptyset$. The direct sum of all of these terms clearly gives the value

$(R[T_0, \ldots , T_ n])_ d$

in degree $0$ as stated in the lemma. Moreover these terms do not contribute to cohomology in other degrees (also in accordance with the statement of the lemma).

To finish the proof of the lemma we have to show that the complexes $\check{\mathcal{C}}^\bullet (\vec{e})$ are acyclic when $NEG(\vec{e})$ is neither empty nor equal to $\{ 0, \ldots , n\}$. Pick an index $i_{\text{fix}} \not\in NEG(\vec{e})$ (such an index exists). Consider the map

$h : \check{\mathcal{C}}^{p + 1}(\vec{e}) \to \check{\mathcal{C}}^ p(\vec{e})$

given by the rule

$h(s)_{i_0 \ldots i_ p} = s_{i_{\text{fix}} i_0 \ldots i_ p}$

(compare with the proof of Lemma 30.2.1). It is clear that this is well defined since

$NEG(\vec{e}) \subset \{ i_0, \ldots , i_ p\} \Leftrightarrow NEG(\vec{e}) \subset \{ i_{\text{fix}}, i_0, \ldots , i_ p\}$

Also $\check{\mathcal{C}}^0(\vec{e}) = 0$ so that this formula does work for all $p$ including $p = - 1$. The exact same (combinatorial) computation as in the proof of Lemma 30.2.1 shows that

$(hd + dh)(s)_{i_0 \ldots i_ p} = s_{i_0 \ldots i_ p}$

Hence we see that the identity map of the complex $\check{\mathcal{C}}^\bullet (\vec{e})$ is homotopic to zero which implies that it is acyclic. $\square$

In the following lemma we are going to use the pairing of free $R$-modules

$R[T_0, \ldots , T_ n] \times \frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}] \longrightarrow R$

which is defined by the rule

$(f, g) \longmapsto \text{coefficient of } \frac{1}{T_0 \ldots T_ n} \text{ in }fg.$

In other words, the basis element $T_0^{e_0} \ldots T_ n^{e_ n}$ pairs with the basis element $T_0^{d_0} \ldots T_ n^{d_ n}$ to give $1$ if and only if $e_ i + d_ i = -1$ for all $i$, and pairs to zero in all other cases. Using this pairing we get an identification

$\left(\frac{1}{T_0 \ldots T_ n} R[\frac{1}{T_0}, \ldots , \frac{1}{T_ n}]\right)_ d = \mathop{\mathrm{Hom}}\nolimits _ R((R[T_0, \ldots , T_ n])_{-n - 1 - d}, R)$

Thus we can reformulate the result of Lemma 30.8.1 as saying that

30.8.1.1
$$\label{coherent-equation-identify} H^ q(\mathbf{P}^ n, \mathcal{O}_{\mathbf{P}^ n_ R}(d)) = \left\{ \begin{matrix} (R[T_0, \ldots , T_ n])_ d & \text{if} & q = 0 \\ 0 & \text{if} & q \not= 0, n \\ \mathop{\mathrm{Hom}}\nolimits _ R((R[T_0, \ldots , T_ n])_{-n - 1 - d}, R) & \text{if} & q = n \end{matrix} \right.$$

Lemma 30.8.2. The identifications of Equation (30.8.1.1) are compatible with base change w.r.t. ring maps $R \to R'$. Moreover, for any $f \in R[T_0, \ldots , T_ n]$ homogeneous of degree $m$ the map multiplication by $f$

$\mathcal{O}_{\mathbf{P}^ n_ R}(d) \longrightarrow \mathcal{O}_{\mathbf{P}^ n_ R}(d + m)$

induces the map on the cohomology group via the identifications of Equation (30.8.1.1) which is multiplication by $f$ for $H^0$ and the contragredient of multiplication by $f$

$(R[T_0, \ldots , T_ n])_{-n - 1 - (d + m)} \longrightarrow (R[T_0, \ldots , T_ n])_{-n - 1 - d}$

on $H^ n$.

Proof. Suppose that $R \to R'$ is a ring map. Let $\mathcal{U}$ be the standard affine open covering of $\mathbf{P}^ n_ R$, and let $\mathcal{U}'$ be the standard affine open covering of $\mathbf{P}^ n_{R'}$. Note that $\mathcal{U}'$ is the pullback of the covering $\mathcal{U}$ under the canonical morphism $\mathbf{P}^ n_{R'} \to \mathbf{P}^ n_ R$. Hence there is a map of Čech complexes

$\gamma : \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}', \mathcal{O}_{\mathbf{P}_{R'}}(d))$

which is compatible with the map on cohomology by Cohomology, Lemma 20.15.1. It is clear from the computations in the proof of Lemma 30.8.1 that this map of Čech complexes is compatible with the identifications of the cohomology groups in question. (Namely the basis elements for the Čech complex over $R$ simply map to the corresponding basis elements for the Čech complex over $R'$.) Whence the first statement of the lemma.

Now fix the ring $R$ and consider two homogeneous polynomials $f, g \in R[T_0, \ldots , T_ n]$ both of the same degree $m$. Since cohomology is an additive functor, it is clear that the map induced by multiplication by $f + g$ is the same as the sum of the maps induced by multiplication by $f$ and the map induced by multiplication by $g$. Moreover, since cohomology is a functor, a similar result holds for multiplication by a product $fg$ where $f, g$ are both homogeneous (but not necessarily of the same degree). Hence to verify the second statement of the lemma it suffices to prove this when $f = x \in R$ or when $f = T_ i$. In the case of multiplication by an element $x \in R$ the result follows since every cohomology groups or complex in sight has the structure of an $R$-module or complex of $R$-modules. Finally, we consider the case of multiplication by $T_ i$ as a $\mathcal{O}_{\mathbf{P}^ n_ R}$-linear map

$\mathcal{O}_{\mathbf{P}^ n_ R}(d) \longrightarrow \mathcal{O}_{\mathbf{P}^ n_ R}(d + 1)$

The statement on $H^0$ is clear. For the statement on $H^ n$ consider multiplication by $T_ i$ as a map on Čech complexes

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) \longrightarrow \check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d + 1))$

We are going to use the notation introduced in the proof of Lemma 30.8.1. We consider the effect of multiplication by $T_ i$ in terms of the decompositions

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_ i = d} \check{\mathcal{C}}^\bullet (\vec{e})$

and

$\check{\mathcal{C}}_{ord}^\bullet (\mathcal{U}, \mathcal{O}_{\mathbf{P}_ R}(d + 1)) = \bigoplus \nolimits _{\vec{e} \in \mathbf{Z}^{n + 1}, \ \sum e_ i = d + 1} \check{\mathcal{C}}^\bullet (\vec{e})$

It is clear that it maps the subcomplex $\check{\mathcal{C}}^\bullet (\vec{e})$ to the subcomplex $\check{\mathcal{C}}^\bullet (\vec{e} + \vec{b}_ i)$ where $\vec{b}_ i = (0, \ldots , 0, 1, 0, \ldots , 0))$ the $i$th basis vector. In other words, it maps the summand of $H^ n$ corresponding to $\vec{e}$ with $e_ i < 0$ and $\sum e_ i = d$ to the summand of $H^ n$ corresponding to $\vec{e} + \vec{b}_ i$ (which is zero if $e_ i + b_ i \geq 0$). It is easy to see that this corresponds exactly to the action of the contragredient of multiplication by $T_ i$ as a map

$(R[T_0, \ldots , T_ n])_{-n - 1 - (d + 1)} \longrightarrow (R[T_0, \ldots , T_ n])_{-n - 1 - d}$

This proves the lemma. $\square$

Before we state the relative version we need some notation. Namely, recall that $\mathcal{O}_ S[T_0, \ldots , T_ n]$ is a graded $\mathcal{O}_ S$-module where each $T_ i$ is homogeneous of degree $1$. Denote $(\mathcal{O}_ S[T_0, \ldots , T_ n])_ d$ the degree $d$ summand. It is a finite locally free sheaf of rank $\binom {n + d}{d}$ on $S$.

Lemma 30.8.3. Let $S$ be a scheme. Let $n \geq 0$ be an integer. Consider the structure morphism

$f : \mathbf{P}^ n_ S \longrightarrow S.$

We have

$R^ qf_*(\mathcal{O}_{\mathbf{P}^ n_ S}(d)) = \left\{ \begin{matrix} (\mathcal{O}_ S[T_0, \ldots , T_ n])_ d & \text{if} & q = 0 \\ 0 & \text{if} & q \not= 0, n \\ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}( (\mathcal{O}_ S[T_0, \ldots , T_ n])_{- n - 1 - d}, \mathcal{O}_ S) & \text{if} & q = n \end{matrix} \right.$

Proof. Omitted. Hint: This follows since the identifications in (30.8.1.1) are compatible with affine base change by Lemma 30.8.2. $\square$

Next we state the version for projective bundles associated to finite locally free sheaves. Let $S$ be a scheme. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ S$-module of constant rank $n + 1$, see Modules, Section 17.14. In this case we think of $\text{Sym}(\mathcal{E})$ as a graded $\mathcal{O}_ S$-module where $\mathcal{E}$ is the graded part of degree $1$. And $\text{Sym}^ d(\mathcal{E})$ is the degree $d$ summand. It is a finite locally free sheaf of rank $\binom {n + d}{d}$ on $S$. Recall that our normalization is that

$\pi : \mathbf{P}(\mathcal{E}) = \underline{\text{Proj}}_ S(\text{Sym}(\mathcal{E})) \longrightarrow S$

and that there are natural maps $\text{Sym}^ d(\mathcal{E}) \to \pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)$.

Lemma 30.8.4. Let $S$ be a scheme. Let $n \geq 1$. Let $\mathcal{E}$ be a finite locally free $\mathcal{O}_ S$-module of constant rank $n + 1$. Consider the structure morphism

$\pi : \mathbf{P}(\mathcal{E}) \longrightarrow S.$

We have

$R^ q\pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)) = \left\{ \begin{matrix} \text{Sym}^ d(\mathcal{E}) & \text{if} & q = 0 \\ 0 & \text{if} & q \not= 0, n \\ \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}( \text{Sym}^{- n - 1 - d}(\mathcal{E}) \otimes _{\mathcal{O}_ S} \wedge ^{n + 1}\mathcal{E}, \mathcal{O}_ S) & \text{if} & q = n \end{matrix} \right.$

These identifications are compatible with base change and isomorphism between locally free sheaves.

Proof. Consider the canonical map

$\pi ^*\mathcal{E} \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)$

and twist down by $1$ to get

$\pi ^*(\mathcal{E})(-1) \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})}$

This is a surjective map from a locally free rank $n + 1$ sheaf onto the structure sheaf. Hence the corresponding Koszul complex is exact (More on Algebra, Lemma 15.28.5). In other words there is an exact complex

$0 \to \pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1) \to \ldots \to \pi ^*(\wedge ^ i\mathcal{E})(-i) \to \ldots \to \pi ^*\mathcal{E}(-1) \to \mathcal{O}_{\mathbf{P}(\mathcal{E})} \to 0$

We will think of the term $\pi ^*(\wedge ^ i\mathcal{E})(-i)$ as being in degree $-i$. We are going to compute the higher direct images of this acyclic complex using the first spectral sequence of Derived Categories, Lemma 13.21.3. Namely, we see that there is a spectral sequence with terms

$E_1^{p, q} = R^ q\pi _*\left(\pi ^*(\wedge ^{-p}\mathcal{E})(p)\right)$

converging to zero! By the projection formula (Cohomology, Lemma 20.50.2) we have

$E_1^{p, q} = \wedge ^{-p} \mathcal{E} \otimes _{\mathcal{O}_ S} R^ q\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(p)\right).$

Note that locally on $S$ the sheaf $\mathcal{E}$ is trivial, i.e., isomorphic to $\mathcal{O}_ S^{\oplus n + 1}$, hence locally on $S$ the morphism $\mathbf{P}(\mathcal{E}) \to S$ can be identified with $\mathbf{P}^ n_ S \to S$. Hence locally on $S$ we can use the result of Lemmas 30.8.1, 30.8.2, or 30.8.3. It follows that $E_1^{p, q} = 0$ unless $(p, q)$ is $(0, 0)$ or $(-n - 1, n)$. The nonzero terms are

\begin{align*} E_1^{0, 0} & = \pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})} = \mathcal{O}_ S \\ E_1^{-n - 1, n} & = R^ n\pi _*\left(\pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1)\right) = \wedge ^{n + 1}\mathcal{E} \otimes _{\mathcal{O}_ S} R^ n\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)\right) \end{align*}

Hence there can only be one nonzero differential in the spectral sequence namely the map $d_{n + 1}^{-n - 1, n} : E_{n + 1}^{-n - 1, n} \to E_{n + 1}^{0, 0}$ which has to be an isomorphism (because the spectral sequence converges to the $0$ sheaf). Thus $E_1^{p, q} = E_{n + 1}^{p, q}$ and we obtain a canonical isomorphism

$\wedge ^{n + 1}\mathcal{E} \otimes _{\mathcal{O}_ S} R^ n\pi _*\left(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)\right) = R^ n\pi _*\left(\pi ^*(\wedge ^{n + 1}\mathcal{E})(-n - 1)\right) \xrightarrow {d_{n + 1}^{-n - 1, n}} \mathcal{O}_ S$

Since $\wedge ^{n + 1}\mathcal{E}$ is an invertible sheaf, this implies that $R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)$ is invertible as well and canonically isomorphic to the inverse of $\wedge ^{n + 1}\mathcal{E}$. In other words we have proved the case $d = - n - 1$ of the lemma.

Working locally on $S$ we see immediately from the computation of cohomology in Lemmas 30.8.1, 30.8.2, or 30.8.3 the statements on vanishing of the lemma. Moreover the result on $R^0\pi _*$ is clear as well, since there are canonical maps $\text{Sym}^ d(\mathcal{E}) \to \pi _* \mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)$ for all $d$. It remains to show that the description of $R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)$ is correct for $d < -n - 1$. In order to do this we consider the map

$\pi ^*(\text{Sym}^{-d - n - 1}(\mathcal{E})) \otimes _{\mathcal{O}_{\mathbf{P}(\mathcal{E})}} \mathcal{O}_{\mathbf{P}(\mathcal{E})}(d) \longrightarrow \mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1)$

Applying $R^ n\pi _*$ and the projection formula (see above) we get a map

$\text{Sym}^{-d - n - 1}(\mathcal{E}) \otimes _{\mathcal{O}_ S} R^ n\pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d)) \longrightarrow R^ n\pi _*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(-n - 1) = (\wedge ^{n + 1}\mathcal{E})^{\otimes -1}$

(the last equality we have shown above). Again by the local calculations of Lemmas 30.8.1, 30.8.2, or 30.8.3 it follows that this map induces a perfect pairing between $R^ n\pi _*(\mathcal{O}_{\mathbf{P}(\mathcal{E})}(d))$ and $\text{Sym}^{-d - n - 1}(\mathcal{E}) \otimes \wedge ^{n + 1}(\mathcal{E})$ as desired. $\square$

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