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Let $\mathcal{D}$ be a $k$-linear triangulated category and $\langle E_n,E_{n-1},\ldots, E_0\rangle$ be a sequence of objects of $\mathcal{D}$. We call $\langle E_n,E_{n-1},\ldots, E_0\rangle$ an $\textit{exceptional sequence}$ if it satisfies $$ Hom_{\mathcal{D}}(E_i,E_j[l])=0, ~\forall 0\leq i<j\leq n \text{ and } \forall l $$ and $$ Hom_{\mathcal{D}}(E_i,E_i[l])= \left\{ \begin{array}{l l} 0 & \quad \text{if }l\neq 0\\ k & \quad \text{if } l=0. \end{array} \right. $$

Moreover, we call an exceptional sequence $\langle E_n,E_{n-1},\ldots, E_0\rangle$ $\textit{full}$ if the $E_i$'s generate $\mathcal{D}$, i.e. any full triangulated subcategory containing all $E_i$'s is equivalent to $\mathcal{D}$. For example, A. Beilinson shows that $\langle\mathcal{O}(-n),\mathcal{O}(-n+1),\ldots,\mathcal{O}\rangle$ is a full exceptional sequence of $D^b_{\text{coh}}(\mathbb{P}^n)$. See DERIVED CATEGORIES OF SHEAVES: A SKIMMING, Lecture 3.

Of course the full exceptional sequence of $\mathcal{D}$ (if exists) is not unique. For example if $\langle E_n,E_{n-1},\ldots, E_0\rangle$ is a full exceptional sequence, then the shift $\langle E_n[1],E_{n-1},\ldots, E_0\rangle$ is also a full exceptional sequence.

$\textbf{My question}$ is: if $\mathcal{D}$ has more than one full exceptional sequence, do they all have the same length? By length I just mean the number $n$ for $\langle E_n,E_{n-1},\ldots, E_0\rangle$.

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    $\begingroup$ Yes: if there is a full exceptional sequence $\langle E_n,\ldots,E_1\rangle$ of length $n$, then $K_0(D)=\mathbb{Z}^n$. $\endgroup$ Commented Oct 7, 2014 at 22:53

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