Kernels and Cokernels of differentials on a bounded long exact sequence.

Question

Given a bounded long exact sequence on a smooth projective variety where objects are direct sums of $\mathcal{O}(n)$ for different integer $n$'s i.e. $\bigoplus_i\mathcal{O}(n_i)$. Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles $\mathcal{O}(n)$ or they can be very complicated vector bundles?

Answer 1

Question: "Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"

Answer: Hartshorne Corr. II.5.18 says: If $X \subseteq \mathbb{P}^n_A$ is projective with $A$ noetherian, it follows any coherent sheaf $F$ is a quotient

$$ f_1:\oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

If $A$ is noetherian it follows $X:=Proj(A[x_1,..,x_n]/I)$ with $B:=A[x_i]/I$ noetherian. Hence if $F$ is a coherent module it follow any submodule $E \subseteq F$ is coherent. It follows $\oplus \mathcal{O}(n(1)_i)$ is coherent hence $ker(f_1)$ is a coherent module. Hence there is a surjection

$$ f_2: \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow ker(f_1)$$

and you get a long exact sequence of $\mathcal{O}_X$-modules

$$S1.\text{ }\cdots \rightarrow \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow \oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$

If $E$ is a finite rank locally free sheaf, it follows $E^*$ is coherent, hence there is a surjection

$$f: \oplus_{i=1}^m \mathcal{O}(n_i) \rightarrow E^* \rightarrow 0$$

Dualize to get

$$0 \rightarrow E \cong E^{**} \rightarrow \oplus_i \mathcal{O}(-n_i) \rightarrow \cdots .$$

Hence any finite rank locally free module can be realized as a submodule of a direct sum $\oplus_i \mathcal{O}(-n_i)$.

The "syzygy theorem" says that such a sequence is finite when $X$ has an open cover $U_i:=Spec(A_i)$ where $A_i$ is a polynomial ring over $k$.

In such cases we get an exact sequence with a finite number of terms

$$0 \rightarrow E_{k+1} \rightarrow \oplus_j \mathcal{O}(n(k)_j) \rightarrow \cdots \rightarrow E_1 \rightarrow 0.$$

Example: Projective space is such a variety: $\mathbb{P}_k^N$ has an open $D(x_i) \cong \mathbb{A}^N_k$. Hence on projective space it follows any coherent sheaf has a finite resolution of direct sums of invertible sheaves. A similar result holds for the grassmannian variety.

What is true is that by HH.II.5.18 for any coherent sheaf $F$ you always get a long exact sequence where the terms in the sequence are finite direct sums of linebundles. Note that for the grothendieck group $K_0(X)$ if there is a long exact sequence

$$ 0 \rightarrow F_k \rightarrow \cdots \rightarrow F_2 \rightarrow F_1 \rightarrow 0$$ you get a relation in $K_0(X)$:

$$[F_1]=[F_2]-[F_3]+ \cdots +(-1)^k[F_k].$$

We get

$$[\mathcal{O}(n)]=[\mathcal{O}(1)^{\otimes n}]=[\mathcal{O}(1)]^{ n}$$

hence if the $F_2,F_3...$ are direct sums of linebundles it follows

$$[F_1]\in \{ \sum_i [\mathcal{O}(n_i)]:\text{$n_i\in \mathbb{Z}$} \} \subseteq K_0(X).$$

If $F$ has a finite resolution of invertible sheaves for any finite rank vector bundle $F$ on $X$ it follows $K_0(X)$ is generated by the invertible sheaves $\mathcal{O}(n)$ for $n \in \mathbb{Z}$. Hence there is a surjection

$$ \rho: \mathbb{Z}[t, t^{-}] \rightarrow K_0(X)$$

defined by $\rho(f(t^i)):=f([\mathcal{O}(i)]), i=1,-1$, and an isomorphism

$$I1.\text{ }K_0(X) \cong \mathbb{Z}[t, t^{-1}]/(ker(\rho)).$$

There is an isomorphism $K_0(X)_{\mathbb{Q}}\cong CH^*(X)_{\mathbb{Q}}$

where $CH^*(X)$ is the Chow group of $X$ and $CH^*(X)$ is "infinite dimensional". But this does not immediately contradict the isomorphism I1.

Question: "Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"

Answer: As shown above: Any coherent module is the cokernel of such a map. The kernel of such a map is always coherent. If $E$ is finite rank and locally free, it follows $E$ can be realized as a subsheaf of such a finite direct sum. In the case of projective space it follows any coherent sheaf $E$ has a finite resolution by finite direct sums of invertible sheaves and you get an equality

$$ K_0(\mathbb{P}^N_k)\cong \mathbb{Z}\{[\mathcal{O}(n)]:n \in \mathbb{Z} \}.$$

The projective bundle formula confirms this calculation: It gives

$$K_0(\mathbb{P}^N_k)\cong \mathbb{Z}[t]/(t^{N+1})$$

where $t:=1-[\mathcal{O}(-1)]$.

https://en.wikipedia.org/wiki/Hilbert%27s_syzygy_theorem

Comment: "This doesn't answer the question as it is not known whether this resolution will end or not. If you re-read the question it is about a bounded long exact sequence on such objects."

Answer: It is well known the grothendieck group $K_0(X)$ is not generated by invertible sheaves in general, hence the answer to your question is: Not in general. For projective space it is true as shown above.