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



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


