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Assume I am blowing up an algebraic variety $X$ in an ideal sheaf $\mathcal{I}$, write $Y:=\mathrm{Bl}_\mathcal{I}(X)$. Now, also assume I have a globally generated line bundle $\mathcal{O}_X^k\twoheadrightarrow\mathcal{L}$. Denote by $h_i\in\mathcal{L}(X)$ the images of the canonical base vectors, i.e. the global generators of $\mathcal{L}$. I can consider the strict transform of the vanishing set $Z_i:=Z(h_i)$ under the blowing-up $Y\to X$.

Edit: We assume that there is a connected component $Y_i$ of $Z(\mathcal{I})$ such that $Y_i\subseteq Z_i$ for all $i$.

I am now wondering if anything like the "strict transform" of $\mathcal{L}$ exists - I would want this to be a globally generated line bundle $\mathcal{O}_Y^k\twoheadrightarrow\mathcal{M}$ with generators $g_i\in\mathcal{M}(X)$ such that $Z(g_i)$ is the strict transform of $Z_i$.

The question is, does such a construction exist and is it well-known? If no, before I start and try to construct it, am I missing some obvious example where this does not even work?

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  • $\begingroup$ Hi Jesko, let me say something about why the most obvious thing doesn't work. (Maybe you already know this...) Let X be P^2, f:Y ->X its blowup in a point p, and L the line bundle O(1) on X. Then we can choose a basis for H^0(L) consisting of two sections s_1 and s_2 which don't vanish at p, and one section s_3 that does. Then the proper transforms of Z(s_1) and Z(s_2) will be (zero sets of) sections of the line bundle $H = f^\ast O(1)$, but Z(s_3) is a section of a different line bundle, namely $H \otimes E^\ast$ (where is E is the line bundle corresponding to the exceptional divisor). $\endgroup$
    – user5117
    Jun 18, 2011 at 16:44
  • $\begingroup$ Ah. Yes I have clearly not added enough assumptions for this to work. However, in my situation it will also always be the case that $Z(h_i)\cap Z(\mathcal{I}) \ne \emptyset$, i.e. the strict transforms never equal the total transform. $\endgroup$
    – Jesko Hüttenhain
    Jun 18, 2011 at 17:08
  • $\begingroup$ Do you have freedom to choose the generators? For a general choice, their intersection with $Z(\mathcal{I})$ would be "the same" and probably you can get what you want. But for special choices you can get the same problem Artie mentions with every blowup, even with $Z(h_i)\cap Z(\mathcal{I}) \ne \emptyset$. Think of X=P^3 blown up along a line, L=O(1). If one of the generators vanishes on the line, you loose. $\endgroup$
    – quim
    Jun 19, 2011 at 0:31
  • $\begingroup$ BTW, $Z(h_i)\cap Z(\mathcal{I}) \ne \emptyset$ does not seem to guarantee that the strict transforms never equal the total transform. Does it? $\endgroup$
    – quim
    Jun 19, 2011 at 0:32
  • $\begingroup$ True - as discussed below, I think I have found the correct formulation now. $\endgroup$
    – Jesko Hüttenhain
    Jun 20, 2011 at 11:30

1 Answer 1

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There are a number of issues with what you want mostly along the lines of the comments.

The main problem is that the strict transform is an operation on a divisor and not on a divisor class. A line bundle corresponds to a divisor class. So, basically you want to figure out a way to define the strict transform for a divisor class hoping that for a general choice in a basepoint-free system you would get the same.

I think that actually happens, but it will simply be the pull-back.

The condition that $Z(h)\cap Z(\mathscr I)\neq\emptyset$ does not mean that the strict transform of $Z(h)$ is different from its pull-back. As long as $Y\not\subseteq Z(h)$ for any irreducible component $Y$ of $Z(\mathscr I)$, the strict transform of $Z(h)$ is the same as its pull-back. Of course, if $Z(\mathscr I)$ is a point, then $Z(\mathscr I)\subseteq Z(h)$ if and only if $Z(h)\cap Z(\mathscr I)\neq\emptyset$. So, to see the difference, try blowing up a curve in a threefold and see what happens to a surface that intersects the curve, but does not contain it. Its pull-back cannot contain any exceptional divisor, so it will be the same as its strict transform.

Assuming that for all $h$ there exists an irreducible component $Y$ of $Z(\mathscr I)$ such that $Y\subseteq Z(h)$ is still not enough. I see at least two problems with this approach.

1) You would have to guarantee that the strict transforms are linearly equivalent. In other words, the pre-image of $Y$ in the pre-image of $Z(h)$ would have to be equivalent to the pre-image of $Y'$ in the pre-image of $Z(h')$ including multiplicities. However, their pre-images would be different exceptional divisors and hence not comparable in the Picard group.

2) A perhaps even bigger problem is that you don't have enough irreducible components to do this for the whole linear system. Since you want your line bundle to be generated by global sections, you can't have one irreducible component contained in all elements of the linear system, but then if you take a general member, it will contain none of them. You're back to the previous case: the "strict transform" of the general element of the linear system is the same as its pull-back.

I don't think you can escape this. If you have a basepoint free linear system, then the general member will be transversal to your $Z(\mathscr I)$ and hence its strict transform will be the same as its pull-back.

So, I think this still stands:

Conclusion: If your requirement is that you want to define the strict transform of a globally generated line bundle such that this strict transform is the line bundle corresponding to the strict transform of a general member of the linear system associated to the original line bundle, then the only way to do this is to take the pull-back. However, that already exists and is easier to define, so there is no point to bend over backwards to define the strict transform. Then again, it satisfies your goal requirement: the pull back will also be globally generated.

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  • $\begingroup$ I ment to consider $\mathcal{L}$ as an invertible subsheaf of the sheaf of total quotient rings $\mathcal{K}_X\cong\mathcal{K}_Y$. Together with this embedding, a line bundle should correspond to an actual Cartier Divisor, not just a class. $\endgroup$
    – Jesko Hüttenhain
    Jun 19, 2011 at 5:28
  • $\begingroup$ Assume I am blowing up $\mathbb{A}^2$ in two points $p$ and $q$. The strict transform of a line that passes through $p$ (but not $q$) would not equal its total transform, right? I do not see why $Z(h)\cap Z(\mathcal{I})$ is not enough to assume. $\endgroup$
    – Jesko Hüttenhain
    Jun 19, 2011 at 5:33
  • $\begingroup$ Concerning the basis, I have probably expressed myself not clearly enough - I am assuming that I have chosen a basis $h_1,\ldots,h_k\in\mathcal{L}(X)$. I just expressed this as a surjection $\mathcal{O}_X^k\twoheadrightarrow\mathcal{L}$ mapping $e_i$ to $h_i$, where $\mathcal{O}_X^k(X)=\bigoplus_{i=1}^k \mathcal{O}_X(X) \cdot e_i$. It is the $e_i$ which I referred to as canonical base vectors. I do not see why that is a faulty term. $\endgroup$
    – Jesko Hüttenhain
    Jun 19, 2011 at 5:34
  • $\begingroup$ Jesko, 1) If you are talking about a Cartier divisor, then I don't know what your question is. Strict transforms are defined for (Cartier) divisors. But in that case there are no other sections. Just the one defining your divisor. 2) Obviously, if $Z(\mathscr I)$ is a point, then containment is the same as non-trivial intersection. Try a case when $\dim Z(\mathscr I)>0$. 3) You are right. The $e_i$ are indeed a canonical basis. I misread what you wrote. Nonetheless, the word "canonical" is not the point. That makes no difference with respect to the construction. $\endgroup$
    – Sándor Kovács
    Jun 19, 2011 at 7:49
  • $\begingroup$ You are right concerning 1) of course, that was just silly confusion on my part. I think we really have to require that for some (connected!) component $Y_i$ of $Z(I)$, we have $Y_i\subseteq Z(h_i)$, which I believe is the correct condition for strict and total transform being different. Now this does not necessarily keep the $h_i$ from globally generating $\mathcal{L}$ (i.e. $\mathcal{I}$ is the ideal sheaf of a union of disconnected subvarieties of the $Z(h_i)$) - and I would gladly be willing to add this requirement. $\endgroup$
    – Jesko Hüttenhain
    Jun 20, 2011 at 11:19

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