# Local cohomology and cohomology of line bundles on projective spaces

Posted on December 4, 2017 by Dima

tags: cohomology

all this is terribly unpolished, and arguments are ridden with holes, as is customary on this blog

### Flatness

Frist things first: Nakayama’s lemma.

Statement 1: Let $$R$$ be a ring, $$I$$ an ideal Let $$M$$ be a finitely generated module, and assume $$IM=M$$. Then there exists $$x \in 1 + I$$ such that $$xM=0$$.

Proof: let $$x_1, \ldots, x_n$$ be the generators of $$M$$. Then there exists a matrix $$A=(a_{ij}) \in I^{n \times n}$$ such that $$x_i = \sum a_{ij} x_j$$, or, $$(I-a)x=0$$. Multiply by the matrix formed of minors on the left and get $$\det (Id-A) \cdot I \cdot x = 0$$ where $$x=(x_i)$$. Note that $$\det (Id-A) \in 1 + I$$

Statement 2: Let $$R$$ be a ring, $$I$$ an ideal that annihilates all simple $$R$$-modules (equivalently: if it lies in the intersection of all maximal ideals, i.e. in a local ring, in the maximal ideal), then $$M=0$$.

Proof. Follows from Statement 1, since $$x$$ is invertible. Indeed, otherwise $$(x)$$ would be contained in a maxmial ideal ideal that intersects trivially with $$I$$ or otherwise $$1 \in I$$, but $$I \cap \mathfrak{m}\neq 0$$ for any maxmial $$\mathfrak{m}$$.

But now there is also a more direct proof. Let $$x_1, \ldots, x_n$$ be the minimal set of generators of $$M$$ and assume $$\mathfrak{m}M = M$$. Then $(1-a_n)x_n = \sum_{i=1}^{n-1} a_i x_i,$ and since $$1-a_n$$ is invertible, $$x_1, \ldots, x_{n-1}$$ generate $$M$$, which contradicts the minimality of the set of generators.

A module $$M$$ is flat iff $$-- \otimes M$$ is an exact functor.

Lemma. $$M$$ is flat if for all ideals $$\mathfrak{p}\subset R$$ $0 \to \mathfrak{p}\otimes M \to M \to M/\mathfrak{p}M \to 0$ is exact (and for that it suffices to check that the first arrow is inclusion).

Proposition. Assume $$R$$ is a Dedekind domain (i.e. every ideal is a product of prime ideals; equivalently, 1-dimensional integrally closed). Then an (integral) $$R$$-algebra $$S$$ is flat iff $$R \hookrightarrow S$$ is injective.

Proposition. Let $$f: X=\mathrm{Spec}S \to Y=\mathrm{Spec}R$$ be a morphism. Then $$S$$ is a flat $$R$$-algebraic iff $$\mathcal{O}_{X,x}$$ is flat over $$\mathcal{O}_{Y,y}$$ for all $$x \in X$$, all $$y = f(x)$$.

Lemma. Let $$R$$ be a local ring withe the maximal ideal $$\mathfrak{m}$$. Let $$M$$ be a finitely generated $$R$$-module. Then $$M$$ is free iff $$\mathrm{Tor}_1(M,k) = 0$$ (and moreover both conditions are equivalent to being projective and flat).

Proof. Assume $$\mathrm{Tor}_1(M,k)$$ vanishes. Since $$M$$ is finitely generated, there is a surjective morphism $$f: R^n \to M$$ where $$n = \dim k$$. Tensoring the short exact sequence $$\mathrm{Ker}f \to R^n \to M$$ with $$k$$ we get that $$\mathrm{Ker}f \otimes k = 0$$ which by Nakayama lemma implies $$\mathrm{Ker}f = 0$$.

Lemma. Setting as above. $$M$$ has a projective resolution of length $$n$$ if and only if $$Tor^n(M,k) = 0$$.

Proof. The non-trivial direction is right-to-left. Assume $\ldots \to P_{n+1} \to P_n \to P_{n-1} \to \ldots \to P_0 \to M \to 0$ is a resolution. Let $$Z_i = \mathrm{Ker}d_i \hookrightarrow P_i$$. Then we have short exact sequences $0 Z_i \to P_i \to Z_{i-1} \to 0$ Let $$i=0$$ and tensor with $$k$$, then $\ldots\mathrm{Tor}_n(M,k)=0 \to \mathrm{Tor}_{n-1}(Z_0,k) \to \mathrm{Tor}_{n-1}(P_0, k)=0 \to \ldots$ So $$\mathrm{Tor}_{n-1}(Z_0,k)=0$$. Arguing inductively, we get $$\mathrm{Tor}_{n-i-1}(Z_i,k) = 0$$. In particular, $$\mathrm{Tor}_{1}(Z_{n-2}, k)=0$$ and by the previous lemma it’s free and hence projective. Then $Z_{n-2} \to P_{n_2}\to P_{n-1} \to \ldots$ is a length $$n$$ projective resolution.

### Koszul complex

Let $$R$$ be a local ring with the maximal ideal $$\mathfrak{m}$$ and residue field $$k=R/\mathfrak{m}$$. Then there exists a particularly interesting projective (in fact, free) resolution of $$k$$. Let $$x_1, \ldots, x_n$$ be a set of generators of $$\mathfrak{m}$$. Let $$K\langle\xi_1, \ldots, xi_n \rangle$$ be the ring of grassamanian polynomials in $$\xi_1, \ldots, x_n$$ graded by degree, where $$n$$-th graded piece is isomorphic to $$\wedge^n R$$. Define the differential map to be $d(\xi_i) = x_i$ on the 1-homogeneaus piece, and extend it linearly and applying Leibniz’s rule $d(a \wedge b) = da \wedge b + (-1)^{|a||b|}a \wedge db$ The graded pieces of $$K\langle\xi_1, \ldots, \xi_n \rangle$$ with this differential are element of the Koszul complex.

Koszul complex can be obtained as a product of $$n$$ complexes of the form $0 \to R \to^{x_i} \to R \to 0$

(some posts of Akhil Mathew on the subject 1 2 3 4 5 )

### Cohomology of sheaves

A module $$I$$ is called injective if for any injective morphism $$M \to N$$ the diagram $\begin{array}{ccc} & & I \\ & \nearrow & \\ M & \to & N \\ \end{array}$ admits completion.

If $$\mathcal F$$ is a sheaf of $$k$$-vector spaces on $$X$$ and $$i_x: \{x\} \to X$$ is an inclusion of a point then $$\mathcal F$$ naturally embeds into $$\oplus_{x \in X} (i_x)_* (\oplus i_x^* \mathcal F)$$, and each $$i_x^* \mathcal F$$ is injective, hence the whole direct sum is injective.

Let $$R$$ be a local ring with maximal ideal $$\mathfrak{m}$$.

An injective hull of the residue field $$k=R/\mathfrak{m}$$ is an injective module $$I$$ such that

• $$\mathrm{Ann}\,I = \{ x \in R \mid x \cdot I = 0 \} = \mathfrak{m}$$
• $$\mathrm{Hom}(k,I) \cong k$$.

Assume that the ring $$R$$ contains the field $$k$$. Then the injective hull of $$k$$ is the direct limit $I := \lim\ \mathrm{Hom}_k(R/\mathfrak{m}^n, k)$ which, as one observes, is isomorphic to “dual” associated graded $\bigoplus_{n \geq 0} \mathrm{Hom}_k(\mathfrak{m}^n/\mathfrak{m}^{n+1}, k)$ In particular, the injective hull depends only on the completion of $$R$$, and if $$R$$ is regular, is (as a $$k$$-module) just the ring of polynomials $$k[x_1, \ldots, x_n]$$ where $$n$$ is the minimal number of generators of $$\mathfrak{m}$$.

Indeed, for any $$M$$ of finite length (supported at $$\mathfrak{m}$$) $\begin{array}{c} \mathrm{Hom}_R(M, I) = \lim_l \mathrm{Hom}_R(M, \mathrm{Hom}_R(R/m^l, k)) = \\ = \lim \mathrm{Hom}_R(R/\mathfrak{m}^l, \mathrm{Hom}_R(M, k)) = \mathrm{Hom}_R(R, \mathrm{Hom}_k(M, k)) = \mathrm{Hom}_k(M, k)\\ \end{array}$

Essential injection of modules $$M \hookrightarrow N$$ is one such that for any submodule $$N' \subset N$$, $$M \cap N' \neq 0$$.

Definition #2 of an injective hull. An injective hull of $$k$$ is an injective module $$I$$ that admits an essential injection $$k \hookrightarrow I$$.

Lemma. The two definitions are equivalent

Proof. (???)

Exercise 18.7 in Matsumura’s “Commutative ring theory”. Let $$k$$ be a field, $$S=k[x_1, \ldots, x_n]$$, $$P=(x_1, \ldots, x_n)$$, $$R=S_P$$, $$\hat{R}=k[[x_1, \ldots, x_n]]$$, and $$I = k[x_1^{-1}, \ldots, x_n^{-1}]$$.

Make $$I$$ into a module by multiplying by $$x_1, \ldots, x_n$$ in the usual way, but “cutting off” the powers at 0: $$x_1^2 \cdot x_1^{-1} = 0$$.

Then $$I$$ is the injective hull of $$k=R/\mathfrak{m}$$.

### Local cohomology

Let $$X = \mathrm{Spec}R$$ and let $$U = X \setminus \mathfrak{m}$$. Define the functor of sections with support in $$\mathfrak{m}$$ on the category $$R-mod$$ to be $$\Gamma_\mathfrak{m}(M) := \lim \mathrm{Hom}(R/\mathfrak{m}^n, M)$$.

Lemma. There is an exact triangle of derived triangles $0 \to R^\bullet \Gamma_\mathfrak{m}\to M \to R^\bullet\Gamma(U, M) \to$

Proof. The sum of images of $$M$$ in injective hulls of points naturally fits into ext sequenc of sum of images of $$M$$ in the injective hull of $$k$$, $$M$$, and sum of images of $$M$$ in the injective huls on the complement of $$\mathfrak{m}$$.

There should be discussion of homological characterization of regularity and depth here. I leave it for a future post.

Lemma. Let $$R$$ be regular of dimension $$n$$. Then $$H^{n}_{\mathfrak{m}}(R)$$ is the injective hull of $$k$$.

Proof. From regularity we know that $$R$$ has no local cohomology apart from degree $$n$$. Let $$I^\bullet$$ be an injective resolution of $$R$$. Then $$\Gamma_\mathfrak{m}(I^\bullet)$$ has cohomology only in $$n$$-th degree, and so is a resolution of $$H^n_\mathfrak{m}(R)[-n]$$. In derived category $\mathrm{Hom}(k, R[i]) = \mathrm{Hom}(k, I[i]) = \mathrm{Hom}(k,H^{n}_\mathfrak{m}[-n+i]$ In particular, $$\mathrm{Hom}(k, H^{n}_\mathfrak{m}) = \mathrm{Ext}^n(k,R)$$. The latter is isomorphic to $$n$$, if we compute it using Koszul resolution for $$k$$.

### Cohomology of projective space

Work over a field $$k$$. Let $$Y = \mathbb{A}^{n+1}, Y'=Y \setminus \{0\}$$, $$X = \mathbb{P}^n$$, and let $$\pi: Y' \to X$$ be the natural projection with fibres the orbits of the natural $$\mathbb{G}_m$$ action on $$Y'$$. Morphism $$\pi$$ is affine and hence acyclic.

Proposition. Cohomology $$H^q(X, \mathcal{O}(l))$$ can be described as follows:

• it vanishes for $$q\neq 0, n$$, $$-n-1 < l < 0$$;
• $$H^n(X, \mathcal{O}(-n-1-l))$$ is naturally dual, as a $$k$$-vector space to $$H^0(X, \mathcal{O}(l))$$.

Proof.

From the acyclicity of $$\pi$$ it follows that $\Gamma(X, R^\bullet\pi_* \mathcal{O}_{y'}) \cong R^\bullet\Gamma(X, \pi_* \mathcal{O}_{Y'})$

Now notice that $$\pi_* \mathcal{O}_{Y'} = \bigoplus_{k \geq 0} \mathcal{O}_{X}(l)$$. Applying the exact triangle of the local cohomology we conclude that for $$n > 0$$, $H^{q+1}_{0}(Y, \mathcal{O}_Y) = \oplus_{l \geq 0} H^q(X, \mathcal{O}(l))$ From which it follows that there is no cohomology for $$q \neq n$$. The rest of the statement will follow from duality, so that’s what we need to prove. And indeed it follows from the previous lemma, only we compute the local cohomology over the ring $$R=k[x_1, \ldots, x_{n+1}]$$ and moreover the graded ring. It would suffice to show that $Ext^{n+1}(k, R)$ is concentrated in degree $$n+1$$ piece. Writing down Koszul resolution we see that $$Ext^{n+1}(k, R)$$ is reperesented by elements dual to $$\xi_1 \wedge \ldots \wedge \xi_n$$, and so we can conclude.

(here’s a post of Akhil Mathew that uses a different technique to reduce the computation of cohomology of $$\mathcal{O}(l)$$ to a Koszul complex computation)

### Bott vanishing

A couple more steps, and we can compute the cohomology of $$\Omega^p$$ on the projective scheme.

Let $$V$$ be a vector space and consider the grassmanian $$\mathrm{Gr}(n,V)$$, and let $$W \in \mathrm{Gr}(n,V)$$. Then the tangent space $$T_W \mathrm{Gr}(n,V)$$ can be canonicaly associated with the vector space $$Hom(W, V/W)$$.

Think of tangent vector to $$W$$ as a “vector space over $$k[e]/e^2$$”. We have $$S_k^* V^\vee] \to S_k^* W^\vee$$, now we have to extend it to $$S_{k[e]/e^2}^* V^\vee \to S_{k[e]/e^2}^* K^\vee$$ such that reduction modulo $$e$$ gives the original. Such morphisms are determined by degree 1 elements of the kernel, so we are looking at a subspace $$K$$ of $$V \oplus e \cdot V$$ such that projection on $$V$$ is $$W$$. But such a subspace is a graph of a linear morphism from $$W$$ to $$V$$. Moreover, if $$K_1$$, $$K_2$$ are graphs of $$f_1, f_2$$ such that $$f_1 - f_2 \in W$$ then $$f_1(W)=f_2(W)$$ and the deformations are equivalent. Therefore, finally, the deformations are parametrized by the linear space $$Hom(V, W/V)$$.

Now in the context of $$\mathbb{P}^n$$, the deformations of a point $$l \in \mathbb{P}^n$$ are $$Hom(l, V/l)$$, and on the level of sheaves this corresponds to $$\mathrm{Hom}(\mathcal{O}(-1), \mathcal{O}^{n+1}/\mathcal{O}(-1))$$, where $$\mathcal{O}(-1)$$ is the tautological bundle which naturally embeds into $$\mathcal{O}^{n+1}$$. Now applying the functor $$\mathrm{Hom}(\mathcal{O}(-1), -)$$ to the sequence $0 \to \mathcal{O}(-1) \to \mathcal{O}^{n+1} \to \mathcal{O}^{n+1}/\mathcal{O}(-1) \to 0$ we get the Euler sequence $0 \to \mathcal{O}\to \mathcal{O}^{n+1}(1) \to \mathrm{Hom}(\mathcal{O}(-1), \mathcal{O}^{n+1}/\mathcal{O}(-1)) = T_{\mathbb{P}^n} \to 0$ Dualizing, we get $0 \to \Omega \to \mathcal{O}^{n+1}(-1) \to \mathcal{O}\to 0$

A trick from Hirzebruch’s “Topological mehods in algebraic geometry”, Theorem 4.1.3: take a sequence of locally free sheaves $0 \to L \to^f V \to W \to 0$ where $$L$$ is a line bundle. Then $$\Lambda^p V \to \Lambda^p W$$ and $$\Lambda^{p-1} V \otimes F\to \Lambda^p V$$ ($$w \otimes f \mapsto w \wedge f$$) are naturally induced. Now observe that $$Ker(L \otimes \Lambda^{p-1} V \to \Lambda^p V$$ consists of vectors $$l \otimes v_1 \wedge \ldots \wedge v_{p-1}$$ such that at least one of $$v_i$$ is in $$L$$, and that quotient by the subspace of such vectors is isomorphic to $$L \otimes \Lambda^{p-1} W$$ via $$l \otimes v_1 \wedge \ldots \wedge v_{p-1} \mapsto l \otimes [v_1] \wedge \ldots \wedge [v_{p-1}]$$. The image of this map consists of vectors $$v_1 \wedge \ldots \wedge v_{p}$$ such that at least one $$v_i$$ belongs to $$L$$, which is precisely the kernel of $$\Lambda^p V \to \Lambda^p W$$. We have, therefore, an exact sequence $0 \to L \otimes \Lambda^{p-1}W \to \Lambda^p V \to \Lambda^p W \to 0$ The dual of this statement is: from exact sequnce $0 \to W \to V \to L \to 0$ one gets the exact sequence $0 \to \Lambda^{p} W \to \Lambda^{p} V \to \Lambda^{p-1} W \otimes L \to 0$

Applying the last fact to the Euler sequence, then twisting by $$\mathcal{O}(p)$$, we get $0 \to \Omega^p(p) \to \Lambda^p \mathcal{O}^{n+1} \to \Omega^{p-1}(p) \to 0$

Serre duality: $$H^p(X, \mathcal{F})^\vee \cong H^{n-p}(X, \Omega^n \otimes \mathcal{F}^\vee)$$. On $$\mathbb{P}^n$$ the canonical sheaf is $$\Omega_X^n = \mathcal{O}(-n-1)$$. Therefore, $H^q(X, \Omega^p(k)) = H^{n-q}(X, (\Omega^p)^\vee(-k - n - 1) )^\vee = H^{n-q})X, (\Omega^n)^\vee \otimes \Omega^{n-p}(-k-n-1)) =H^{n-q}(X, \Omega^{n-p}(-k))$

Plus we know from before that $$H^p(X, \mathcal{O})$$ vanishes if $$p > 0$$. From the long exact sequence associated to the Euler sequence we get $$H^n(X, \Omega)=0$$ for $$n > 0$$, and to its $$p$$-the exterior power version, we get $$H^{q+1}(X, \Omega^{p+1})=H^q(X, \Omega^p)$$.