Computing locally free resolutions of a filtration of projective space

The title is unfortunately not as informative as it could be. What I want to do in this short blog is compute some locally free resolutions of certain sheaves. The sheaves I have in mind are structure sheaves for certain subvarieties of projective space \mathbb{P}^n. More specifically, I want to compute the structure sheaves of the subschemes in the filtration

\text{Spec}(k)=\mathbb{P}^0\subset \mathbb{P}^1\subset\cdots \subset \mathbb{P}^{n-1}\subset \mathbb{P}^n.

This turns out to be equivalent to computing the Koszul complex, and proving it’s a locally free resolution in the particular case of a polynomial ring k[x_0,...,x_n] with regular sequence (x_0,\ldots, x_{n-1}). Here I’m thinking of k as a field but any base ring will do. For a long time this was really confusing to me but, it turns out it really didn’t need to be. I’ll give an application at the very end of the post which relates the K-theory of projective space to the G-theory of projective space. This complements a set of notes which I wrote today, (and will include here once I post them).

Some Preliminaries

There really isn’t much we’ll need for this. But, I do want a couple of statements and reminders before starting.

(1.1) Definition: Let R=\bigoplus_{i\in \mathbb{Z}} R_i be a graded ring, and M=\bigoplus_{i\in \mathbb{Z}} M_i a graded R-module. We define the nth shift or the nth twist of M to be the module M(n) whose ith graded piece is the i+nth graded piece of M. I.e. M(n)_i:=M_{n+i}.

The reason for introducing graded rings and modules in the first place is because of their relation to projective space. This is essentially the analogy: modules and rings are to affine space as graded modules and graded rings are to projective space. This can actually be made more precise by the fact any projective variety corresponds to an affine cone, and vice-versa. I’m making this up on the spot, so I won’t actually go into details so the next paragraph will have to suffice.

There’s an action of \mathbb{G}_m:=\mathbb{A}_{\mathbb{Z}}^1\setminus \{0\} on affine space \mathbb{A}^{n+1} given by scaling. Ring theoretically this is the map \varphi : k[x_0,...,x_n]\rightarrow k[x_0,...,x_n]\otimes_k \mathbb{Z}[t,t^{-1}] defined by x_i\mapsto x_i\otimes t. This gives k[x_0,...,x_n] a \mathbb{Z}-graded ring structure by defining the ith graded piece to be those elements p(x_0,...,x_n) such that \deg_t (\varphi(p(x_0,...,x_n)))=i. Any projective variety in \mathbb{P}^n comes from an affine subvariety of \mathbb{A}^{n+1} which is stable under this action. Said otherwise, these are exactly the quotients k[x_0,...,x_n]/I that have a grading compatible with the above one. It can be shown if I has homogeneous generators then this holds. I’ll stop here, before mentioning why one wants to remove the origin, constructing quotient varieties, and proving all of the theorems.

One nice result we will need, which follows the lines of this analogy, is related to how we construct sheaves on projective space.

(1.2) Proposition: Given a graded module M over a graded ring R, we can construct a sheaf \widetilde{M} of \mathcal{O}_{\text{Proj}(R)}-modules. This association defines a functor from graded R-modules to sheaves on \text{Proj}(R). Moreover, this functor is exact and commutes with direct sums and tensor products.

Proof. This is Lemma 26.8.4 of the Stacks Project, tag 01M3. The surrounding section is also beneficial for the construction and more. The last statement about direct sums and tensor products is at least partially at the Stacks project. For the direct sum part wikipedia cites EGA I, Ch. I, Corollaire 1.3.9.

(1.3) Proposition: Let \mathcal{M} be a graded module over a graded ring R. The sheaves \mathcal{O}_{\text{Proj}(R)}(n) are locally free of rank 1 and \widetilde{M(n)}\simeq \widetilde{M}(n)=\widetilde{M}\otimes_{\mathcal{O}_{\text{Proj}(R)}} \mathcal{O}_{\text{Proj}(R)}(n)

Proof. Lemma 26.10.3 of the Stacks Project, tag 01MM. The surrounding section is also beneficial for details.

This should be enough to allow us to compute some examples now.

Some first examples

(2.1) Example: Let’s compute locally free resolutions for the structure sheaves of a point, a line, and a plane all embedded in a plane. That is, we’ll compute locally free resolutions for the structure sheaves of the varieties appearing in the filtration

\mathbb{P}^0\subset \mathbb{P}^1\subset \mathbb{P}^2.

The structure sheaves of these varieties depend on how they are embedded in \mathbb{P}^2.  I’ll pick the easiest case, which could also be reached from any other through an automorphism of \mathbb{P}^2. This reduces our problem to the following statement: compute locally free resolutions of:

\mathcal{O}_{\mathbb{P}^2}\simeq \widetilde{k[x_0,x_1,x_2]}, \mathcal{O}_{\mathbb{P}^1}\simeq \widetilde{k[x_0,x_1,x_2]/(x_0)}, \mathcal{O}_{\mathbb{P}^0}\simeq \widetilde{k[x_0,x_1,x_2]/(x_0,x_1)}.

It’s actually really easy, if we just focus on computing free resolutions of these modules (forgetting about the gradings), and then we twist or shift the grading appropriately to get a morphism of graded modules so that we can apply the functoriality statement of (1.2).

(2.2) Lemma: Let R=k[x_0,x_1,x_2]. Some free resolutions of the modules R, R/(x_0), R/(x_0,x_1) are respectively


0\rightarrow R\xrightarrow{\cdot x_0} R \rightarrow R/(x_0)\rightarrow 0

and 0\rightarrow R\xrightarrow{(x_1,-x_0)} R\oplus R \xrightarrow{\cdot (x_0,x_1)^T} R\rightarrow R/(x_0,x_1)\rightarrow 0.

Proof. The first one is trivial, since R is free. For the second note that the natural surjection R\rightarrow R/(x_0) has kernel (x_0). Define the map R\rightarrow (x_0) by 1\mapsto x_0. This is an isomorphism since the ideal is principal, and concatenating these maps gives the second resolution. The last one comes from a similar method. The natural surjection R\rightarrow R/(x_0,x_1) has kernel (x_0,x_1). Define a map R\oplus R\rightarrow (x_0,x_1) by (1,0)\mapsto x_0 and (0,1)\mapsto x_1. The kernel of this map is generated by (x_1,-x_0), so define the last map taking R\rightarrow R\oplus R by 1\mapsto (x_1,-x_0) which is an isomorphism because of principality again.\square

(2.3) Corollary: Some locally free resolutions of the sheaves \mathcal{O}_{\mathbb{P}^2},\mathcal{O}_{\mathbb{P}^1}, \mathcal{O}_{\mathbb{P}^0} are given respectively by


0\rightarrow \mathcal{O}_{\mathbb{P}^2}(-1)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{O}_{\mathbb{P}^1}\rightarrow 0

0\rightarrow \mathcal{O}_{\mathbb{P}^2}(-2)\rightarrow \mathcal{O}_{\mathbb{P}^2}(-1)\oplus \mathcal{O}_{\mathbb{P}^2}(-1)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow \mathcal{O}_{\mathbb{P}^0}\rightarrow 0

Proof. The proof is literally: we computed free resolutions above, so apply the tilde operation, which is exact, and gives the above sequences up to shifting. The reason we have to shift is because we need to have the same degree going to the same degree in order to apply tilde. Looking at the maps for the resolutions and how the shift works, along with the natural grading on k[x_0,x_1,x_2] giving the linear terms degree 1 gives the result.\square

Now let’s generalize this, because I don’t actually care about planes and lines but instead about projective n-space and arbitrary dimensional subvarieties.

(2.4) Proposition: Consider the filtration of \mathbb{P}^n given by \mathbb{P}^0\subset \cdots \subset \mathbb{P}^n. Let m+1 be an integer with 0\leq m+1 \leq n. Then there is a locally free resolution of \mathcal{O}_{\mathbb{P}^{m+1}} by \mathcal{O}_{\mathbb{P}^n}-modules

0\rightarrow  \cdots \rightarrow \mathcal{O}_{\mathbb{P}^n}^{\oplus {m \choose i}}(-i)\rightarrow \cdots \rightarrow \mathcal{O}_{\mathbb{P}^n}^m(-1)\rightarrow\mathcal{O}_{\mathbb{P}^n} \rightarrow \mathcal{O}_{\mathbb{P}^{n-m-1}}\rightarrow 0.

Proof. As before we’ll assume the structure sheaf of \mathbb{P}^{n-m-1} is given by the tilde of the graded k[x_0,...,x_n]-module k[x_0,...,x_n]/(x_0,...,x_m). Note this module is isomorphic to the tensor product \otimes_{i=0}^{m} k[x_0,...,x_n]/(x_i) where we tensor with respect to the k[x_0,...,x_n]-module structure. Again call k[x_0,...,x_n]=R for brevity. For each i we have a free resolution of graded modules

0\rightarrow R(-1)\xrightarrow{\cdot x_i} R\rightarrow R/(x_i)\rightarrow 0.

Since the modules to the left of R/(x_i) are free, taking the tensor product of this complex as i ranges from 0 to m is exact. Taking the total complex of this tensor product gives us a free resolution of R/(x_0)\otimes_R \cdots \otimes_R R/(x_{m}), and all that’s left is to describe this total complex and then take the tilde operation.
Since I don’t actually need to know what the maps are, I won’t describe them. It’s really explicit but I’d have to keep track of a bunch of (-1)’s. All I’ll need for what I want to do below are the objects in the locally free resolution of this sheaf. But writing this out gives me

0\rightarrow R(-m)\rightarrow R^{\oplus m}(-m+1) \rightarrow \cdots \rightarrow R^{\oplus {m\choose i}}(-i) \rightarrow \cdots \rightarrow R^{\oplus m}(-1) \rightarrow R \rightarrow R/(x_0,\ldots, x_{m})\rightarrow 0.

Taking the tilde operation gives the result.\square

An Application

Earlier today I computed:

(3.1) Proposition: K^0(\mathbb{P}^n)=\mathbb{Z}\{[\mathcal{O}_{\mathbb{P}^n}(-n)],\ldots,[\mathcal{O}_{\mathbb{P}^{n}}]\} and G_0(\mathbb{P}^n)=\mathbb{Z}\{[\mathcal{O}_{\mathbb{P}^0}],\ldots, [\mathcal{O}_{\mathbb{P}^n}]\} where the notation is meant to read “the free abelian group on the set…”.

For the definitions of these groups, and the above computations, see these notes (will be linked eventually).

Our work in section 2 allows us to explicitly describe an isomorphism between the two groups K^0(\mathbb{P}^n) and G_0(\mathbb{P}^n). Sometimes this is called the Cartan isomorphism, or Poincare duality.

(3.2) Proposition: The isomorphism G_0(\mathbb{P}^n)\xrightarrow{\sim} K^0(\mathbb{P}^n) defined by taking the class of a coherent sheaf \mathcal{C} to the alternating sum \sum_{i=0}^n(-1)^i \mathcal{L}^i, where \mathcal{L}^{\bullet}\rightarrow \mathcal{C} is a locally free resolution of \mathcal{C}, has an explicit description in terms of integral matrices.

Proof. Identify \mathbb{Z}^{n+1} with G_0(\mathbb{P}^n) by e_i\mapsto [\mathcal{O}_{\mathbb{P}^{n+1-i}}] where e_i=(\delta_{ji})_{j=1}^{n+1} is the standard basis vector. Similarly identify \mathbb{Z}^{n+1} with K^0(\mathbb{P}^n) by the identification e_i\mapsto [\mathcal{O}_{\mathbb{P}^n}(-n+i-1)]. Then the matrix describing this isomorphism is given by

A_{ij}=\begin{cases} (-1)^i {{j-1} \choose {i-1}}  & \text{if}\,\,\, i\leq j \\  0 & \text{if}\,\,\,j<i \end{cases}. \square

(3.3) Example: Let n=2. The matrix in (3.2) is

\begin{pmatrix} 1 & 1 & 1\\ 0 & -1 & -2 \\ 0 & 0 & 1\end{pmatrix}.

(3.4) Example: Let n=3. The matrix in (3.2) is

\begin{pmatrix} 1& 1 & 1 & 1\\ 0 & -1 & -2 & -3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & -1\end{pmatrix}


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