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Let $X$ be a smooth algebraic variety, $A \subset X$ a smooth subvariety, $f:Y \to X$ the blow-up of $X$ along $A$ and $M$ a quasi-coherent $O_X$-module (in the case I'm interested in, $M$ is actually a $D_X$-module but it doesn't change a thing).

Question: Is there a natural resolution of $O_Y$ as a $f^{-1}O_X$-module to compute the inverse image $Lf^*M = O_Y \otimes^L_{f^{-1}O_X} f^{-1}M$?

Any reference where I could find such a computation? The elementary case of $A = 0$ inside $X = \mathbb{A}^n$ would already be a great help.

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I'd be surprised if there is an easy-to-compute one, but quite interested! –  Karl Schwede Mar 3 '11 at 16:04
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Even in the case of the blow-up of the origin in the affine plane?? –  YBL Mar 3 '11 at 17:40
    
There is a typo in the formula for $Lf^*M$ --- you mixed $X$ and $Y$. –  Sasha Mar 3 '11 at 18:39
    
@Sasha. Thanks I corrected it. –  YBL Mar 3 '11 at 19:21
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up vote 6 down vote accepted

The easiest way to compute is the following. Assume that $A$ is the zero locus of a regular section $s$ of a vector bundle $E$ on $X$ (this is always the case locally). Then there is an embedding $i:Y \to P_X(E)$ over $X$. Then $i_*O_Y$ has a nice resolution.

Indeed, consider the relative Grothendieck line bundle $O(-1)$ on $P_X(E)$. We have a natural embedding $O(-1) \to p^*E$, where $p:P_X(E) \to X$ is the projection. Consider the cokernel of this morphism, $Q$ (in fact $Q$ is isomorphic to the relative tangent bundle up to a twist). On the other hand, $p^*$ is a section of $p^*E$, so it gives a section $s'$ of $Q$. Then $i(Y)$ is the zero locus of $s'$ on $P_X(E)$. Moreover, this section is regular, so we have the following Koszul resolution $$ 0 \to \Lambda^{r-1}Q^* \to \dots \to \Lambda^2Q^* \to Q^* \to O_{P_X(E)} \to i_*O_Y \to 0. $$ This resolution can be used to compute $i_*Lf^*M$ for any quasicoherent sheaf $M$ on $X$. Indeed, we have $$ i_*Lf^*M = i_*Li^*Lp^*M = Lp^*M\otimes^L i_*O_Y. $$ Since $i$ is a closed embedding this gives a nice control over $Lf^*M$.

EDIT. In case $X = A^n$, $A = \{0\}$ we can take $E = O^n$, $s = (x_1,\dots,x_n)$, where $x_i$ are the coordinates. Then $P_X(E) = A^n\times P^{n-1}$ and $Q = T_{P^{n-1}}(-1)$. So, $i_*Lf^*M$ is quasiisomorphic to the following complex $$ M\boxtimes\Omega^{n-1}(n-1) \to \dots \to M\boxtimes\Omega^2(2) \to M\boxtimes\Omega^1(1) \to M\boxtimes O. $$ For example, if $M$ is a skyscraper in $0$ then the above complex is supported on the exceptional divisor $P^{n-1} \subset Y$ and all differential vanish, so one concludes that $L_pf^*M \cong j_*\Omega^p(p)$, where $j:P^{n-1} \to Y$ is the embedding.

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This looks great. Thanks. –  YBL Mar 18 '11 at 18:07
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