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This is a follow up to my previous question, and I have lowered my demands to a situation as follows:

Let $X$ be an algebraic variety, $\mathcal{I}$ a coherent sheaf of ideals and $\mathcal{L}$ a line bundle on $X$. Let $h_1,\ldots,h_k\in\mathcal{L}(X)$ be certain global sections with the property that under any local trivialization $\psi:\mathcal{L}|_U\xrightarrow{\ \sim\ }\mathcal{O}_U$, we have $\psi_U(h_i|_U)\in\mathcal{I}(U)$. In other words, $ Z(\mathcal{I})\subseteq Z(h_i)$ for all $i$.

Let $\beta:Y:=\mathrm{Bl}_\mathcal{I}(X)\to X$ be the blow-up of $X$ along $\mathcal{I}$. Now, I am looking for a line bundle $\mathcal{M}$ on $Y$ and global sections $g_0,\ldots,g_k\in\mathcal{M}(Y)$ such that $Z(g_i)$ is the strict transform of $Z(h_i)$.

From Artie's comment in my previous question, I assumed that the bundle $\beta^\ast\mathcal{L}\otimes \mathcal{E}^\vee$ should satisfy my demands, where $\mathcal{E}=\bigoplus_{d\ge 0} \mathcal{I}^{d+1}$ is the exceptional ideal sheaf of the blow-up and $\mathcal{E}^\vee$ its dual.

My question: Is this the $\mathcal{M}$ I am looking for, and if yes, why? I do no see what the corresponding global sections would be.

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    $\begingroup$ The condition in the second paragraph can be rewritten as $h_i \in \Gamma(X,{\mathcal I}\otimes{\mathcal L})$, the latter being considered as a subspace of $\Gamma(X,{\mathcal L})$. $\endgroup$
    – Sasha
    Jun 21, 2011 at 18:38
  • $\begingroup$ That's neat indeed! $\endgroup$
    – Jesko Hüttenhain
    Jun 21, 2011 at 21:07

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Jesko, you still need more. The condition $Z(\mathscr I)\subseteq Z(h_i)$ ensures that the strict transform of these sections will be different from their pull-back, but you also need a condition that guarantees that they will be linearly equivalent.

Here is the point more concretely: Let $D_i=Z(h_i)$ and $\sum E_j=\beta^{-1}(Z(\mathscr I))$. Now $$\beta^*D_i \sim \widetilde D_i+\sum a_{ij}E_j$$ where $\widetilde D_i$ is the strict transform of the divisor $D_i$ and the $a_{ij}$ are the appropriate numbers. In order for $\widetilde D_i\sim \widetilde D_{i'}$ to be true you need that $a_{ij}=a_{i'j}$ for all $j$. In other words, you need some condition that guarantees that the $D_i$ (in some complicated sense) have the same multiplicity along $Z(\mathscr I)$.

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  • $\begingroup$ I should first tell you that I am very grateful for the great deal of patience and help that you're providing. So, thanks a lot! I also understand your point now - but is there any way to control the $a_{ij}$? For instance, if $X$ is projective and $\mathcal{L}=\mathcal{O}_X(d)$, do I have a chance? At least for $d=1$, this should be possible. $\endgroup$
    – Jesko Hüttenhain
    Jun 21, 2011 at 21:07
  • $\begingroup$ Jesko, I would expect that it's more about $\mathscr I$. If that is relatively simple, say it defines a smooth irreducible subvariety and $X$ is smooth along $Z(\mathscr I)$, then there is only one $E_j$ and the $a_{ij}$ is the multiplicity of $D_i$ along $Z(\mathscr I)$. This is well-defined, so you could assume that they are equal. If $Z(\mathscr I)\subseteq X$ is more complicated it is difficult to tell the $a_{ij}$ from initial data. Then again, this construction is a way to define the $a_{ij}$ and they do mean some kind of multiplicity, but I suppose you would like something else $\endgroup$
    – Sándor Kovács
    Jun 21, 2011 at 21:42
  • $\begingroup$ I should have said that upfront: Everything I am dealing with may safely be assumed to be smooth (both $Z(\mathcal{I})$ and $Z(h_i)$ and also $X$ itself), so I'll probably get what I want, yet. Thanks again! $\endgroup$
    – Jesko Hüttenhain
    Jun 21, 2011 at 22:55

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