This question arise from the comparision of the reconstruction theorems of Bondal-Orlov and Balmer and is inspired by Shizhuo Zhang's mathoverflow question: [How to unify various reconstruction theorems (Gabriel-Rosenberg, Tannaka,Balmers)](https://mathoverflow.net/questions/16257/how-to-unify-various-reconstruction-theorems-gabriel-rosenberg-tannaka-balmers)


On the one hand, A. Bondal and D. Orlov in <http://arxiv.org/abs/alg-geom/9712029> proved their celebrated reconstruction theorem: Let $X$ be a smooth projective variety such that the canonical bundle $\omega\_X$ is either ample or anti-ample, and let $Y$ be any projective variety. If $D^b\_{\text{coh}}(X)\cong D^b\_{\text{coh}}(Y)$ as triangulated categories, then $X\cong Y$.


In fact, they reconstruct $X$ from $D^b\_{\text{coh}}(X)$ and their construction heavily use the $\textit{Serre functor}$ S. In general, for a $k$-linear category $\mathcal{C}$, an equivalence $S: \mathcal{C}\rightarrow \mathcal{C}$ is called a Serre functor for $\mathcal{C}$ if their is a natural, bifunctorial isomorphisms
$$
\varphi\_{A,B}: \text{Hom}\_{\mathcal{C}}(A,B)\xrightarrow{\sim} \text{Hom}\_{\mathcal{C}}(B,SA)^{\vee}
$$
for every $A,B$. Here $(\bullet)^\vee$ denotes the $k$-dual. When $\mathcal{C}=D^b\_{\text{coh}}(X)$, the Serre functor $S\_X$ is given by 
$$
S\_X\mathcal{E}=\mathcal{E}\otimes \omega\_X[\dim X]
$$
i.e tensor product with the canonical bundle and shift by $\dim X$ in $D^b\_{\text{coh}}(X)$. We can refer to Section 4 of A. Caldararu's lecture notes <http://arxiv.org/abs/math/0501094> for an excellent introduction. We notice that Bondal-Orlov reconstruction does not involve the tensor structure of $D^b\_{\text{coh}}(X)$ besides the definition of the Serre functor.


(As pointed out by Piotr and Qiaochu in the comments, the Serre functor is unique, hence is part of the data of $D^b\_{\text{coh}}(X)$ and is not an extra data.)


On the other hand, P. Balmer in <http://arxiv.org/abs/math/0111049> proved another reconstruction theorem: Let $X$ be a noetherian scheme, then we can reconstruct $X$ from the $\textit{tensor triangulated category}$ $(D^{\text{perf}}(X),\otimes^L\_{\mathcal{O}\_X})$. This reconstruction uses the extra structure of $D^{\text{perf}}(X)$: the tensor product, and it applies for more general $X$ than Bondal-Orlov.


Now we can look at the case when the two theorems overlap: If $X$ is smooth projective with $\omega\_X$ ample or anti-ample, then $D^{\text{perf}}(X)\cong D^b\_{\text{coh}}(X)$. Now from $D^b\_{\text{coh}}(X)$ either using the Serre functor or using the tensor structure. It seems that there are some redundancy here. In fact we have
$$
D^b\_{\text{coh}}(X) \xrightarrow [\text{reconstruction}]{\text{Bondal-Orlov}}X\rightarrow (D^b\_{\text{coh}}(X),\otimes^L\_{\mathcal{O}\_X})\xrightarrow [\text{reconstruction}]{\text{Balmer}} X\rightarrow D^b\_{\text{coh}}(X) \ldots
$$
Hence one may expect a direct construction of the tensor structure $(D^b\_{\text{coh}}(X),\otimes^L\_{\mathcal{O}\_X})$ just from $D^b\_{\text{coh}}(X)$ itself, considered as a triangulated category.


$\textbf{My question}$ is: If $X$ is smooth projective variety with $\omega\_X$ ample or anti-ample, could we define the tensor product structure on $D^b\_{\text{coh}}(X)$ just from the triangulated cateogry structure on $D^b\_{\text{coh}}(X)$? 


In the other direction we have $\textbf{a related question}$: What is the role of the Serre functor $S\_X$ and the canonical bundle $\omega\_X$ in the the tensor triangulated category $(D^b\_{\text{coh}}(X),\otimes^L\_{\mathcal{O}\_X})$?



Can we define the tensor product in the derived category $D^b_{\text{coh}}(X)$ just from $D^b_{\text{coh}}(X)$ in certain cases?
It seems like the answer to your question is no, at least without further clarification. If you could define the tensor product structure on $D^b(X)$ just from the triangulated structure and the Serre functor, then any $k$-linear derived autoequivalence $F:D^b(X)\rightarrow D^b(X)$, since it commutes with the Serre functor, would be a $\otimes$-autoequivalence as well. Now, consider $\mathbb{P}^1\_k$ and the autoequivalences $F(n,k)(\mathscr{F})=\mathscr{F}(n)[k]$. These are not $\otimes$-autoequivalences unless $n=k=0$. More or less, they induce new $\otimes$-structures on $D^b(X)$ where the unit is $\mathscr{O}\_{\mathbb{P}^1\_k}(-n)[-k]$. So, I don't see how to make this work.


That being said, it's possible that you can reconstruct all of these $\otimes$-structures at once. Here is what I have in mind. Pick any invertible object $U$ in $D^b(X)$. Recall that the invertible objects are those complexes such that for every point-like object $P$ there exists $n\_p\in\mathbb{Z}$ such that $Hom(U,P[i])=k(P)$ when $i=n\_P$ and $0$ otherwise. Bondal and Orlov proved that the point-like objects are just the shifted sheaves $k(P)[n]$ for $P\in X$ (here on out I'll work over an algebraically closed field for simplicity). Then, they proved that the invertible objects are precisely the shifted line bundles $\mathscr{L}[n]$. So, our $U$ is equivalent to $\mathscr{L}[n]$ for some $\mathscr{L}$ and $n$. This can be ignored for time being; I'll come back to it later.


I claim that there is a tensor product $\otimes\_U$ on $D^b(X)$ for which $U$ is the unit. Given two complexes $\mathscr{F}$ and $\mathscr{G}$, resolve $\mathscr{F}$ by terms of the form $S\_X^n(U)$. The fact that we can do this is because the $S\_X^n(U)$ form a spanning class of $D^b(X)$, which follows easily from the fact that the powers of $\omega\_X$ form a spanning class, and the fact that $U=\mathscr{L}[n]$. Then, define $S\_X^n(U)\otimes\_U\mathscr{G}$ as $S\_X^n(G)$. Then use these to define $\mathscr{F}\otimes\_U\mathscr{G}$. Note that we need ampleness or anti-ampleness in order to perform this construction. It looks like the details work out: uniqueness follows from $S^n\_X(U)\otimes\_UU=S^n\_X(U)$.


Now, recall that Bondal and Orlov also proved that the group of derived $k$-linear autoequivalences of $D^b(X)$ fits into an exact sequence $$0\rightarrow\mathbb{Z}\times Pic(X)\rightarrow Aut(D^b(X))\rightarrow Aut(X)\rightarrow 1.$$ We see that by fixing the invertible object $U$, i.e. the shifted line bundle $\mathscr{L}[n]$ we are picking exactly an element in $\mathbb{Z}\times Pic(X)$.


