# dimension of D^b(coh( P^1))

Hello,

I am trying to understand Orlov: Remarks on generators of triangulated categories.

Let E be a full subcategory of D^b(coh(P^1)). Let [E] be the smallest full subcategory of D^b(coh(P^1)) such that [E] is closed under direct summands, finite direct sums and shifts.

Now, let [E]x[E] be the full subcategory consisting of all objects C such that there is a dt

A->B->C with A,B in [E]. Identify an object F of D^b(coh(P^1)) with the subcategory consisting of that object alone. The claim is that if F contains two different line bundles as direct summands, [[F]x[F]] is equivalent to coh(P^1), in which case F is said to be a strong generator and the number of iterations gives the dimension of D^b(coh(P^1)), as dimension is defined to be the minimal numbers of iterations for E= object of D^b(coh(P^1)).

this is supposedly trivial but I fail to see it. Thank you very much.

Carsten

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Do you mean Orlov instead of Bondal? – ulrich Oct 23 '11 at 14:49
yes, thanks. I corrected this. – Carsten Oct 23 '11 at 16:07

Let me give a name to notion. If $T$ is a triangulated category and $G$ is an object, the minimal number of cones required to generate any object of $T$ starting with $G$ (and allowing for arbitrary finite sums, shifts, and splitting of summands) is called the generation time of $G$. So we want to check that the generation time of $\mathcal O(i) \oplus \mathcal O(l)$ is $1$ if $i \not = l$.

For specificity, assume $i < l$. As $- \otimes \mathcal O(-l)$ is an autoequivalence and generation time is invariant under autoequivalences, we can reduce to the case of $\mathcal O(-a)$ and $\mathcal O$ with $a > 0$. As ulrich mentioned, any object of $D^b(\operatorname{coh } \mathbb{P}^1)$ is isomorphic to a sum of shifts of twists, $\mathcal O(l)$, and torsion sheaves. Let us deal with getting $\mathcal O(j)$ first. We break it into two cases: $1-a \leq j \leq -1$ and $j < -a$ or $j > 0$. If $1-a \leq j \leq -1$, then we have an exact sequence $$0 \to \mathcal O(-a) \overset{( -y^{a-j} \ x^j)}{\to} \mathcal O(j) \oplus \mathcal O(-j+a) \overset{(x^j \ y^{a-j})}{\to} \mathcal O \to 0$$ showing that $\mathcal O(j)$ is a summand of the cone of a map $\mathcal O \to \mathcal O(-a)[1]$.

If $j < -a$ or $j > 0$, we use a resolution of the diagonal. On $\mathbb{P}^1 \times \mathbb{P}^1$ we have an exact sequence $$0 \to \mathcal O(-a-1,-1)^{\oplus a-1} \to \mathcal O(-a,-1)^{\oplus a} \to \mathcal O \to \mathcal O_{\Delta} \to 0$$ Coming from the exact sequence $$0 \to \mathcal O(-1,-1) \to \mathcal O \to \mathcal O_{\Delta} \to 0$$ and the exact sequence $$0 \to \mathcal O(-a-1,-1)^{\oplus a-1} \to \mathcal O(-a,-1)^{\oplus a} \to \mathcal O(-1,-1) \to 0$$ The last exact sequence is obtained from the exact sequence $$0 \to \mathcal O(-a-1)^{\oplus a-1} \to \mathcal O(-a)^{\oplus a} \overset{(x^{a-1} \ x^{a-2}y \ \cdots \ y^{a-1})}{\to} \mathcal O(-1) \to 0$$ by pulling back along the first factor of $\mathbb{P}^1 \times \mathbb{P}^1$ and tensoring with $\mathcal O(0,-1)$. Let us denote the projections, $\pi_1: \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$ and $\pi_2: \mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$. We can apply the functor $\mathbf{R}\pi_{1*}(\pi_2^*\mathcal O(j) \otimes -)$ to exact sequence $$0 \to \mathcal O(-a-1,-1)^{\oplus a-1} \to \mathcal O(-a,-1)^{\oplus a} \to \mathcal O \to \mathcal O_{\Delta} \to 0$$ giving a triangle $$\mathcal E \to \mathcal O(j) \to H^*(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[2]$$ where $\mathcal E$ is a cone over a map of sums of shifts of $\mathcal O$ and $\mathcal O(-a)$. $$H^*(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[2] =$$ $$H^0(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[2] \oplus H^1(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[1]$$ As there are no extensions of degree two or more, we only have to worry about a possible map to $H^1(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[1]$. There is no nonzero maps $$\mathcal O(j) \to H^1(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[1]$$ if $j < -a$ or $j > 0$. Thus, in the derived category, $$\mathcal E \cong \mathcal O(j) \oplus H^*(\mathbb{P}^1,\mathcal O(j-1)) \otimes_k \mathcal O(-a-1)[1]$$ So, in this case too, $\mathcal O(j)$ is a summand of a cone over a map between sums of shifts of $\mathcal O(-a)$ and $\mathcal O$. This gives any $\mathcal F$.

Next, we deal with the case of a torsion sheaf, $\mathcal T$. We can use the same method as in the case of $\mathcal O(j)$ with $j < -a$ or $j > 0$ as the map $$\mathcal T \to H^*(\mathbb{P}^1,\mathcal T(-1)) \otimes_k \mathcal O(-a-1)[2] = H^0(\mathbb{P}^1,\mathcal T(-1)) \otimes_k \mathcal O(-a-1)[2]$$ is zero.

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ok, thank you. I ll work through that. – Carsten Oct 24 '11 at 6:55

I will give the proof of a special case; presumably one can modify the proof to work in general but I have not thought about it much. I also work over an algebraically closed field.

Firstly, since $\mathbb{P}^1$ is a smooth curve, $Ext^i(F,G) = 0$ for all $i>1$ and all coherent sheaves $F$ and $G$. It follows that any $\mathcal{F}$ in the derived category is isomorphic to $\oplus H^i(\mathcal{F})[-i]$ (where $H^i$ denotes the cohomology sheaves). Thus, it suffices to consider $\mathcal{F} = F$ where $F$ is a coherent sheaf (in degree 0).

Any sheaf $F$ on a smooth curve is isomorphic to $F_1 + F_2$, where $F_1$ is locally free and $F_2$ is torsion. Any locally free sheaf on $\mathbb{P}^1$ is a direct sum of line bundles so it suffices to consider the case $F = \mathcal{O}(n)$, for $n \in \mathbb{Z}$, and $F = \mathcal{O}/\mathcal{I}^n$, where $\mathcal{I}$ is the ideal sheaf of a point and $n>0$.

I will now consider the special case $E = \mathcal{O} \oplus \mathcal{O}(1)$.

For any integer $n>0$ we have an exact sequence $$0 \to K \to \mathcal{O}^{n+1} \to \mathcal{O}(n) \to 0$$ where the second map is given by the global sections. A cohomology computation shows that $K \cong \mathcal{O}(-1)^n$. Tensoring this sequence with $\mathcal{O}(1)$ shows that $\mathcal{O}(n)$ is in the category for $n\geq 0$. Dualising the sequence gives us all $n<0$ as well.

Now consider $F = \mathcal{O}/\mathcal{I}^n$. We have an obvious surjection from $\mathcal{O}$ onto $F$. Consider the exact sequence $$0 \to K \to \mathcal{O}^n \to F \to 0$$ where the second map is the sum of the obvious surjection. A cohomology computation shows that $K \cong \mathcal{O}(-1)^n$. Tensoring this sequence by $\mathcal{O}(1)$ shows that $F$ is in the category as well, completing the proof.

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thanks! I'll think about your argument. – Carsten Oct 23 '11 at 19:47