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Let $X,Y$ be varieties over $\mathbb{C}$, and $D(X),D(Y)$ be the derived categories of bounded complex of coherent sheaves. Let $U \subset X, V \subset Y$ be open subvarieties, and let $X-U, Y-V$ have codimensions $c_1, c_2$ respectively. Assume $U $ is isomorphism to $V$. I heard a result that says: when the codimension $c_1, c_2$ high enough, then $D(X)$ is derived equivalent to $D(Y)$.

Does anyone know any references or the precise statement of any results along this direction, I am especially interest in the upper bound of codimension.

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The highest codimension is reached for $U=\varnothing=V$, so I'd say the 'result' is not true. – Fernando Muro Apr 30 '13 at 1:02
Do you mean to say that the codimensions of the complements of $U$ and $V$ are high? – Benjamin Antieau Apr 30 '13 at 5:00
This reminds me of A. Usnich's triangulated categories, which are birational invariants . – S. Okada Apr 30 '13 at 12:42
@Benjamin Antieau Yes, that is what I mean. – Li Yutong Apr 30 '13 at 14:57
up vote 6 down vote accepted

In your assumption the varieties $X$ and $Y$ are birational. In particular, there is a smooth variety $Z$ with two proper birational maps $p:Z\to X$ and $q:Z\to Y$ which are isomorphisms over $U$ and $V$ (for $Z$ you can take an appropriate resolution of singularities of the graph of your birational isomorphism).

There is a conjecture saying that if $p^*K_X = q^*K_Y$ then $D(X) \cong D(Y)$ (for short K-equivalence implies D-equivalence). Its condition holds if both $codim(Z\setminus p^{-1}(U))$ and $codim(Z\setminus q^{-1}(V))$ are at least 2. But I am afraid this cannot be formulated just in terms of $codim(X\setminus U)$ and $codim(Y\setminus V)$.

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Could you let me know the reference for the result you mentioned above? Thank you! – Li Yutong Apr 30 '13 at 14:59, Conjecture 1.2. – Sasha Apr 30 '13 at 17:31
@Sasha thank you! Do you know any reference about the result: "$codim(Z-p^{-1}(U))$ at least 2..." implies D-equivalent? – Li Yutong Apr 30 '13 at 19:31
If two line bundles on a smooth variety agree on an open subset with complement of codimension at least two, then they are isomorphic. This is the Hartogs theorem. – Sasha Apr 30 '13 at 19:37
One could perhaps add that without the assumption on the canonical divisor there is no hope to prove derived equivalence. For instance, let $V$ and $W$ 2 vector spaces such that $2 < \dim V < \dim W$. Consider $X$ the total space of W(-1) over $\mathhbb{P}(V)$ and $Y$ be the total space $V(-1)$ over $\mathbb{P}(W)$. Then $X$ and $Y$ are isomorphic in codimension $1$, but $D(X) \neq D(Y)$ (though there is a fully faithful embedding of one in the other) – Johan May 9 '13 at 18:00

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