Cohomology of projective space

Written by GPT-6.1 Sol (OpenAI), in Codex, at Ultra effort, October 2026. Self-checked by the writing AI, GPT-6.1 Sol, at Ultra effort. Public domain (CC0).

Projective space has a finite affine cover, so its cohomology can be calculated from localizations of a polynomial ring. The calculation separates according to individual Laurent monomials. Their negative exponents determine which intersections can carry them, and this elementary combinatorics explains both the vanishing of intermediate cohomology and the duality between ordinary and reciprocal monomials.

We assume Cohomology of affine schemes and Serre's criterion. The constructions of Proj, twisting sheaves and relative projective space are provided by the planned lessons Proj of a graded ring and Projective space, relative Proj and maps to projective space in Sheaves and schemes. Their precise local descriptions are recalled below; the corresponding open reference is [Stacks, Tag 01M3]. No calculation here requires a field, a reduced ring, or a characteristic-zero hypothesis.

1. The monomial section complex

Fix a ring \(A\), an integer \(n\ge0\), and the standard graded ring \[ S=A[T_0,\ldots,T_n],\qquad\deg T_i=1. \] Write \(\mathbf P_A^n=\operatorname{Proj}S\). The twist \(\mathcal O(d)\) is the sheaf of the graded module \(S(d)\), with \(S(d)_m=S_{d+m}\). On a nonempty intersection of standard charts, \[ \Gamma(D_+(T_{i_0})\cap\cdots\cap D_+(T_{i_p}),\mathcal O(d)) =S[T_{i_0}^{-1},\ldots,T_{i_p}^{-1}]_d. \] The subscript means homogeneous degree \(d\). Inverting the product of these variables is equivalent to inverting each of them. These intersections are affine; for example, the chart indexed by \(i\) has coordinates \(T_j/T_i\). Thus affine-cover comparison identifies cohomology with the ordered Čech complex \[ C^p(d)=\bigoplus_{i_0<\cdots<i_p} S[T_{i_0}^{-1},\ldots,T_{i_p}^{-1}]_d. \] It stops in degree \(n\). The differential is the alternating sum of restrictions, with increasing indices and the sign of the deleted index's position.

Each term is a free \(A\)-module on Laurent monomials. This remains true when \(A\) has torsion or zero divisors: the monomials are formal basis vectors of the polynomial localization. A weight is an integer vector \(e=(e_0,\ldots,e_n)\), with total \(\sum e_i=d\). The monomial \(T^e\) occurs in the factor indexed by a subset \(I\) exactly when \[ N(e)=\{i:e_i<0\}\subset I. \] The differential preserves weights. The whole complex is therefore a direct sum of complexes \(C^\bullet(e)\), each having a copy of \(A\) for the nonempty subsets \(I\) containing \(N(e)\). Its differential uses only zero maps and signed identity maps between these copies.

These are direct sums rather than products over the weights: every Laurent polynomial has finite support, and only finitely many chart subsets occur in a fixed degree. Direct sums are exact, so cohomology can be computed separately at each weight.

2. Negative exponents determine cohomology

Lemma 2.1 (the three weight cases). For a weight \(e\), the cohomology of \(C^\bullet(e)\) is:

Negative indices Surviving cohomology
None One copy of \(A T^e\) in degree zero
All \(n+1\) indices One copy of \(A T^e\) in degree \(n\)
A nonempty proper subset Zero in every degree

Proof. If every exponent is negative, only the subset \(I=\{0,\ldots,n\}\) is allowed. Its copy of \(A\) is the whole complex, in degree \(n\).

If no exponent is negative, every nonempty subset is allowed. This is the ordinary simplex cochain complex, with \(A\) as its degree-zero cohomology. More explicitly, augment it by a copy of \(A\) in degree \(-1\), whose map is the diagonal. Inserting a fixed index gives a contracting homotopy of this augmented complex. Thus the original complex has the stated cohomology.

Now suppose \(N(e)\) is nonempty and proper. Choose \(j\notin N(e)\). Regard an ordered cochain as an alternating function of distinct indices, with value zero on any tuple whose set does not contain \(N(e)\), and zero on repeated indices. Define \[ (hc)_{i_0\ldots i_{p-1}}=c_{j i_0\ldots i_{p-1}}. \] Insertion is followed by reordering with its sign. This map is well-defined: adjoining \(j\) does not change whether a subset contains \(N(e)\). The identity \(dh+hd=1\) follows by pairing each deletion other than the inserted \(j\) with its opposite-sign counterpart; deletion of \(j\) contributes the original cochain. In degree zero the would-be augmented term is absent because \(N(e)\ne\varnothing\), and the same calculation still works: the value at the singleton \(\{j\}\) is zero. This is a contraction of the unaugmented complex, so all its cohomology is zero. \(\square\)

For a concrete example with three variables, take \(N(e)=\{0\}\). The weight complex is \[ A\longrightarrow A^2\longrightarrow A, \qquad a\longmapsto(-a,-a),\qquad(b,c)\longmapsto b-c. \] The two middle entries correspond to \(\{0,1\}\) and \(\{0,2\}\). The last map is surjective and its kernel is the diagonal, exactly the preceding image. With \(N(e)=\{0,1\}\), the only terms are \(A\xrightarrow{1}A\) in degrees one and two. With all three exponents negative, only the last \(A\) survives. These small complexes illustrate the general cancellation without dividing by an integer.

Put \(r=n+1\). For \(n\ge1\), let \[ L_d=\bigoplus_{\substack{e_i<0\text{ for all }i\\\sum e_i=d}} A T^e. \] It is zero unless \(d\le-r\). It can also be written as the homogeneous degree-\(d\) part of \[ \frac1{T_0\cdots T_n}A[T_0^{-1},\ldots,T_n^{-1}]. \] The expression is a module of reciprocal monomials, not a ring.

Theorem 2.2 (cohomology of every twist). If \(n\ge1\), then \[ H^q(\mathbf P_A^n,\mathcal O(d))= \begin{cases} S_d,&q=0,\ d\ge0,\\ L_d,&q=n,\ d\le-n-1,\\ 0,&\text{otherwise}. \end{cases} \] All the nonzero modules are finite free over \(A\). For \(n=0\), instead, \[ H^0(\mathbf P_A^0,\mathcal O(d))=A\quad\text{for every integer }d, \qquad H^q=0\quad(q>0). \]

Proof. For positive dimension, Lemma 2.1 and the weight direct sum identify degree-zero classes with monomials having nonnegative exponents, and top-degree classes with monomials having strictly negative exponents. All other weights are contractible. The sum condition gives the stated ranges. For fixed \(d\), there are only finitely many vectors of nonnegative exponents summing to \(d\), or of strictly negative exponents summing to \(d\), proving finite freeness.

In dimension zero there is one chart, and its degree-\(d\) section module is \(A T_0^d\) for every integer \(d\). This is \(A\); geometrically \(\mathbf P_A^0=\operatorname{Spec}A\), and the trivializing generator of the twist is \(T_0^d\), including negative powers. There are no higher Čech terms. \(\square\)

The separate zero-dimensional clause prevents a false negative-twist vanishing statement when degree zero is also top degree. For \(n\ge1\), all groups vanish in the entire interval \(-n\le d\le-1\). The calculation is [Stacks, Tag 01XT], with the two cohomological degrees kept distinct when necessary.

3. A pairing with no factorials

The top group of \(\mathcal O(-r)\) is free of rank one. Fix its trace by \[ \operatorname{tr}\left[\frac1{T_0\cdots T_n}\right]=1. \] For \(m\ge0\), multiplication by global sections gives a bilinear pairing \[ S_m\times H^n(\mathbf P_A^n,\mathcal O(-m-r)) \longrightarrow H^n(\mathbf P_A^n,\mathcal O(-r)) \xrightarrow{\operatorname{tr}}A. \]

Theorem 3.1 (perfect monomial pairing). This pairing is perfect: each module is the \(A\)-linear dual of the other.

Proof. Index the monomial basis of \(S_m\) by vectors \(\alpha_i\ge0\) with \(\sum\alpha_i=m\). A corresponding basis of the other group is \[ u_\alpha=T_0^{-\alpha_0-1}\cdots T_n^{-\alpha_n-1}. \] The product \(T^\beta u_\alpha\) has exponent \(\beta_i-\alpha_i-1\). If \(\beta=\alpha\), it is the trace generator. If the vectors differ but have the same total, at least one \(\beta_i>\alpha_i\), so that product has a nonnegative exponent and represents zero in top cohomology by Lemma 2.1. The pairing matrix is therefore the identity. This proves perfection over every ring, including rings with positive characteristic or nilpotents. \(\square\)

The proof also applies to \(n=0\) using the generators of its one-chart section modules. There is no appeal to integration and no normalization by a factorial. This will be the local calculation behind Serre duality later in the course.

Proposition 3.2 (base change and polynomial multiplication). The preceding identifications commute with every ring map \(A\to B\), including nonflat ones. If \(f\in S_s\), multiplication by \(f\) on top cohomology is the dual of \[ S_{-d-s-r}\xrightarrow{\ f\ }S_{-d-r}, \] where negative homogeneous pieces are zero and \(n\ge1\).

Proof. The section complex over \(B\) is obtained by replacing the coefficient ring in every monomial factor by \(B\). The weight contractions use only signed identities, so they remain contractions after any coefficient change. The surviving bases also change by extension of scalars. The resulting isomorphism \[ H^q(\mathbf P_A^n,\mathcal O(d))\otimes_A B \simeq H^q(\mathbf P_B^n,\mathcal O(d)) \] is the natural one induced by the chart section complexes. No general nonflat base-change theorem is being assumed here.

For the multiplication statement, let \(u\) be a top class and \(g\) a homogeneous polynomial of degree \(-d-s-r\). Associativity of multiplication gives \[ \operatorname{tr}(g(fu))=\operatorname{tr}((fg)u). \] The perfect pairing identifies this equality with the claimed dual map. Products that leave the negative-exponent range are zero classes, as already proved. \(\square\)

This functorial form agrees with [Stacks, Tag 01XV]. In particular, the finite free modules in the calculation retain their ranks under every specialization of the base ring.

4. Higher direct images over any base

Let \(\pi:\mathbf P_S^n\to S\), for an arbitrary scheme \(S\). In positive dimension the formula becomes \[ R^q\pi_*\mathcal O(d)= \begin{cases} \operatorname{Sym}^d(\mathcal O_S^{r}),&q=0,\ d\ge0,\\ \mathcal H om(\operatorname{Sym}^{-d-r}(\mathcal O_S^{r}),\mathcal O_S), &q=n,\ d\le-r,\\ 0,&\text{otherwise}. \end{cases} \] The fixed ordered standard coordinates trivialize the determinant line, which is why it does not appear in this particular formula. For \(n=0\), the morphism is the identity and \(\pi_*\mathcal O(d)=\mathcal O_S\) for every \(d\).

Proof of the relative formula. On an affine open \(\operatorname{Spec}A\subset S\), the preceding lesson identifies higher direct images with the sheaves of the modules \(H^q(\mathbf P_A^n,\mathcal O(d))\). Theorem 2.2 computes these modules and Theorem 3.1 gives their dual descriptions. Proposition 3.2 makes the identifications compatible on distinguished opens; hence they glue across arbitrary affine charts of \(S\). The same proposition shows that the natural higher-direct-image base-change maps are isomorphisms for every \(S'\to S\), by checking locally on source and target affines. This supplies the relative proof of [Stacks, Tag 01XW]. \(\square\)

5. The determinant line of a projective bundle

Let \(\mathcal E\) be finite locally free of constant positive rank \(r\) on \(S\). We use the quotient convention \[ \mathbf P(\mathcal E)=\operatorname{Proj}_S\operatorname{Sym}(\mathcal E), \qquad\pi^*\mathcal E\twoheadrightarrow\mathcal O(1). \] Thus \(\pi_*\mathcal O(1)=\mathcal E\) when \(r\ge2\); the corresponding assertion for rank one follows below. This convention is important: defining projective space from \(\mathcal E^\vee\) would change the displayed bundles. The construction and tautological quotient are [Stacks, Tag 01OA]. Write \(D=\det\mathcal E=\bigwedge^r\mathcal E\).

Lemma 5.1 (canonical top trace). For \(r\ge2\), there is a canonical isomorphism \[ D\otimes R^{r-1}\pi_*\mathcal O(-r)\simeq\mathcal O_S. \] In a local ordered frame it takes the frame's exterior product tensored with the reciprocal-coordinate class to one.

Proof. Twist the tautological quotient down to obtain a surjection \[ F=\pi^*\mathcal E\otimes\mathcal O(-1)\longrightarrow\mathcal O. \] Its Koszul differential on \(\bigwedge^j F\) is contraction with this functional. The resulting sequence \[ 0\to\pi^*D\otimes\mathcal O(-r)\to\cdots\to \pi^*\mathcal E\otimes\mathcal O(-1)\to\mathcal O\to0 \] is exact. Indeed locally a vector \(v\in F\) maps to one; exterior multiplication by \(v\) is a contracting homotopy because \[ \iota(v\wedge w)+v\wedge\iota(w)=w. \] This also shows that its images are locally direct summands, so the sequence remains exact under any base change.

For \(1\le j\le r-1\), every higher direct image, including degree zero, of \(\pi^*(\bigwedge^j\mathcal E)\otimes\mathcal O(-j)\) is zero. This can be checked on a trivializing open: it is a finite direct sum of copies of the acyclic twist \(\mathcal O(-j)\), by Section 4. Breaking the Koszul sequence into short exact sequences of its images and composing their connecting maps therefore gives a canonical isomorphism \[ \mathcal O_S=\pi_*\mathcal O \xrightarrow{\delta}R^{r-1}\pi_*(\pi^*D\otimes\mathcal O(-r)). \] For a finite locally free base module \(G\), the natural projection map identifies \(R^q\pi_*(\pi^*G\otimes\mathcal F)\) with \(G\otimes R^q\pi_*\mathcal F\): on a trivializing open it is just commutation of cohomology with a finite direct sum. Applying this to \(D\) and inverting \(\delta\) gives the trace in the statement.

To check its normalization, choose a frame \(e_0,\ldots,e_{r-1}\), with corresponding coordinates \(T_i\). On the \(i\)-th chart put \(v_i=e_i/T_i\) in \(F\); its Koszul contraction is one. The Čech cochains \[ v_{i_0}\wedge\cdots\wedge v_{i_p} \] are successive lifts in the connecting-map computation: their Koszul differential is the alternating deletion sum of the preceding cochain. Thus \(\delta(1)\) is represented on the full intersection by \[ \frac{e_0\wedge\cdots\wedge e_{r-1}}{T_0\cdots T_{r-1}}. \] This proves the stated normalization, and the construction from the tautological quotient proves independence of the frame. \(\square\)

This is the canonical Koszul argument of [Stacks, Tag 01XX]. It determines the determinant factor without assuming that every invertible matrix over the base ring has a chosen elementary factorization.

Theorem 5.2 (cohomology of a projective bundle). If \(r\ge2\), then \[ R^q\pi_*\mathcal O(d)= \begin{cases} \operatorname{Sym}^d\mathcal E,&q=0,\ d\ge0,\\ (\operatorname{Sym}^{-d-r}\mathcal E\otimes D)^\vee, &q=r-1,\ d\le-r,\\ 0,&\text{otherwise}. \end{cases} \] These are canonical identifications compatible with isomorphisms of \(\mathcal E\) and arbitrary base change. For rank one, \(\mathbf P(\mathcal E)=S\), \(\mathcal O(d)=\mathcal E^{\otimes d}\), and all higher direct images vanish; negative powers mean powers of the dual line bundle.

Proof. Trivializing \(\mathcal E\) reduces vanishing to Section 4. The natural map \(\operatorname{Sym}^d\mathcal E\to\pi_*\mathcal O(d)\) is an isomorphism for \(d\ge0\) by the same local calculation. For \(d=-m-r\), multiplication and Lemma 5.1 give the canonical pairing \[ \operatorname{Sym}^m\mathcal E\otimes R^{r-1}\pi_*\mathcal O(-m-r)\longrightarrow D^\vee. \] In any ordered local frame, Lemma 5.1 identifies it with the perfect pairing of Theorem 3.1. It consequently induces an isomorphism to \((\operatorname{Sym}^m\mathcal E\otimes D)^\vee\). Perfection is local, so this identifies the global bundles. All maps used are natural in the tautological quotient; local monomial base change and the normalized trace show compatibility with arbitrary base change. This proves the positive-rank formula.

In rank one the tautological quotient of the pulled-back line is an isomorphism of line bundles. The quotient-line functor is represented by \(S\) itself: a surjection from a line to a line is an isomorphism, leaving one isomorphism class of quotient at every base change. Hence \(\mathbf P(\mathcal E)=S\) and \(\mathcal O(1)=\mathcal E\), giving the separate clause. \(\square\)

The dual on a symmetric power should be left as the dual of that symmetric power. Over general rings one must not silently replace it by the symmetric power of the dual using a characteristic-zero symmetrization argument. The coordinate pairing above works without such a replacement.

For a finite locally free sheaf of varying rank, apply the theorem on its open rank loci. A rank-zero locus has empty projective bundle, since its symmetric algebra has no positive-degree elements; all direct images there are zero. Thus the preceding statements cover arbitrary finite locally free \(\mathcal E\), including rank one and rank zero.

6. Ranks and Euler characteristics

On \(\mathbf P_A^1\), the top group of \(\mathcal O(-2)\) is \(A\), generated by \(1/(T_0T_1)\). With \(t=T_1/T_0\) and the frame \(T_0^{-2}\) on the first chart, its coefficient is \(t^{-1}\), agreeing with the two-chart calculation in Čech cohomology.

More generally, on \(\mathbf P_k^n\) over a field, define \[ \chi(\mathcal O(d))=\sum_q(-1)^q\dim_k H^q(\mathbf P_k^n,\mathcal O(d)). \] For \(d\ge0\), stars and bars counts \(\binom{d+n}{n}\) nonnegative exponent vectors. For \(d\le-n-1\), write \(e_i=-\alpha_i-1\); then \(\sum\alpha_i=-d-n-1\), giving \(\binom{-d-1}{n}\) top basis vectors. Their Euler contribution has sign \((-1)^n\). In the intervening range every group is zero. All three cases are expressed by the same polynomial: \[ \chi(\mathcal O(d))=\binom{d+n}{n} =\frac{(d+1)\cdots(d+n)}{n!}\qquad(d\in\mathbf Z). \] Here the binomial is the generalized binomial polynomial, not a convention that sets negative upper arguments to zero. In the negative range reversing the signs of its \(n\) factors gives exactly \((-1)^n\binom{-d-1}{n}\); in the intervening range a factor is zero. For \(n=0\), the empty product is one, agreeing with \(\chi=1\) for every twist.

The same numbers count the ranks of the finite free cohomology modules over an arbitrary ring. This is a statement about integral ranks, not division by \(n!\) inside the base ring.

7. Exercises with solutions

Exercise 7.1 (easy: every twist on a line). Compute \(H^q(\mathbf P_A^1,\mathcal O(d))\) for every integer \(d\), giving bases whenever it is nonzero.

Solution. For \(d\ge0\), the degree-zero basis is \(T_0^{d-j}T_1^j\), with \(0\le j\le d\), and the first group is zero. For \(d=-1\), both groups vanish. For \(d\le-2\), the degree-zero group vanishes and the first group has basis \[ T_0^{-i}T_1^{-j},\qquad i,j\ge1,\quad i+j=-d. \] There are \(-d-1\) such vectors. Degrees greater than one vanish. These conclusions follow from the two-chart complex: its first cohomology kills precisely Laurent monomials with at least one nonnegative homogeneous exponent. The groups are free \(A\)-modules on the listed bases, regardless of torsion in \(A\).

Exercise 7.2 (easy: the dual bases). For \(m=3\) on \(\mathbf P_A^1\), write the two dual bases in Theorem 3.1 and check every pairing entry.

Solution. The polynomial basis is \(T_0^3,T_0^2T_1,T_0T_1^2,T_1^3\). The reciprocal basis of \(H^1(\mathcal O(-5))\) is \[ T_0^{-4}T_1^{-1},\quad T_0^{-3}T_1^{-2},\quad T_0^{-2}T_1^{-3},\quad T_0^{-1}T_1^{-4}. \] Matching entries multiply to \(T_0^{-1}T_1^{-1}\), with trace one. For unequal entries the two exponent sums still total \(-2\); one exponent is nonnegative, so the product is a zero class. Thus every off-diagonal entry is zero and every diagonal entry is one. This identity matrix is invertible over any \(A\), not merely over fields.

Exercise 7.3 (medium: the projective plane). Compute all groups on \(\mathbf P_k^2\) and determine the rank of \(H^2(\mathcal O(-5))\). Explain why weights with only one or two negative exponents contribute nothing.

Solution. For \(d\ge0\), only \(H^0\) is nonzero, with dimension \(\binom{d+2}{2}\). For \(d=-1,-2\), every group vanishes. For \(d\le-3\), only \(H^2\) is nonzero, with basis \(T_0^{-i}T_1^{-j}T_2^{-\ell}\), where \(i,j,\ell\ge1\) and \(i+j+\ell=-d\). Its dimension is \(\binom{-d-1}{2}\). There is no first cohomology for any twist, and no groups in degrees greater than two.

For \(d=-5\), the positive triples are the three permutations of \((3,1,1)\) and the three permutations of \((2,2,1)\); there are six, agreeing with \(\binom42=6\). With one negative index the weight complex is \(A\to A^2\to A\), with maps \((-1,-1)\) and \((1,-1)\), and is exact. With two negative indices it is a signed identity \(A\to A\). These are the mixed-sign cancellations that eliminate intermediate classes.

Exercise 7.4 (medium: the binomial polynomial). Verify the Euler-characteristic formula for \(n=2\), including \(d=-1,-2,-3,-4\). What changes for projective dimension zero?

Solution. The polynomial is \((d+1)(d+2)/2\). At \(-1,-2\), it is zero, agreeing with total vanishing. At \(-3,-4\), it is one and three, the dimensions of the top groups; their Euler sign is positive because the top degree is two. For \(d\ge0\), it counts polynomial monomials. For arbitrary negative \(d\le-3\), it equals \(\binom{-d-1}{2}\), so these cases cover every integer. In dimension zero every twist is a trivial line on \(\operatorname{Spec}k\), and \(\chi=1\) for all \(d\). The generalized binomial \(\binom d0\) and its empty-product interpretation are also one. Treating a negative upper binomial argument as automatic zero would give the wrong answer.

Exercise 7.5 (challenging: a product of projective lines). Over any ring \(A\), compute the cohomology of \(\mathcal O(a,b)\) on \(\mathbf P_A^1\times_A\mathbf P_A^1\). Give a proof that justifies a tensor-product computation without an unproved Künneth assumption.

Solution. Put \(V_0(d)=H^0(\mathbf P_A^1,\mathcal O(d))\) and \(V_1(d)=H^1(\mathbf P_A^1,\mathcal O(d))\), with the free bases of Exercise 7.1. The answer is \[ \begin{aligned} H^0&=V_0(a)\otimes_A V_0(b),\\ H^1&=(V_0(a)\otimes_A V_1(b))\oplus(V_1(a)\otimes_A V_0(b)),\\ H^2&=V_1(a)\otimes_A V_1(b), \end{aligned} \] with all later groups zero. Tensoring the two monomial bases gives explicit bases. The ranks are obtained using \(\operatorname{rank}V_0(d)=\max(d+1,0)\) and \(\operatorname{rank}V_1(d)=\max(-d-1,0)\).

Here is the justification. Use the standard two-chart cover in each factor. The two-direction Čech complex on the product is the tensor product over \(A\) of their section complexes; each bi-intersection is affine and its section monomials are products of the two factor monomials. Its total complex computes sheaf cohomology by the same acyclic-cover double-complex proof used earlier: on stalks the augmented complexes in both cover directions have an insertion contraction, and on the bi-intersections positive quasi-coherent cohomology vanishes. The total differential is \(d\otimes1+(-1)^p1\otimes d\) in bidegree \((p,q)\).

Each two-term factor complex splits into its cohomology and contractible pieces already over \(A\). To see this explicitly, write it in the first chart frame as \[ A[t]\oplus t^d A[t^{-1}]\longrightarrow A[t,t^{-1}], \qquad(u,v)\longmapsto v-u. \] At an exponent present in both source modules the map is \(A^2\to A\), \((u,v)\mapsto v-u\); the diagonal is the degree-zero class and \((0,1)\) splits the map. At an exponent in exactly one source module the map is a signed identity and is contractible. At an exponent in neither source module there is just a copy of \(A\) in degree one. These decompositions exhaust every exponent. Tensoring a contraction with the other complex, using the total-complex signs, still gives a contraction. Hence only the tensor products of the two cohomology summands survive. This proves the displayed formulas without any flatness or field assumption.

Exercise 7.6 (medium: detect the determinant convention). Let \(\mathcal E=\mathcal L\oplus\mathcal M\), for line bundles on \(S\). Compute \(R^1\pi_*\mathcal O(-3)\). Then compare with the rank-one bundle \(\mathbf P(\mathcal L)\).

Solution. Here \(r=2\), \(-d-r=1\), and \(\det\mathcal E=\mathcal L\otimes\mathcal M\). Theorem 5.2 gives \[ R^1\pi_*\mathcal O(-3) =(\mathcal L^{\otimes2}\otimes\mathcal M)^\vee \oplus(\mathcal L\otimes\mathcal M^{\otimes2})^\vee. \] The zeroth direct image is zero and the groups in degrees at least two vanish. If both lines are trivial, the displayed bundle is \(\mathcal O_S^{\oplus2}\), matching the two reciprocal monomials of degree \(-3\) on a projective line. In rank one, instead, the projective bundle is \(S\) itself, \(\pi_*\mathcal O(-3)=\mathcal L^{-3}\), and the first direct image is zero. Applying the positive-dimensional negative-twist rule without the rank-one clause would lose this nonzero zeroth direct image.

References and construction prerequisites