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Let $f:X\dashrightarrow Y$ be a birational map of smooth projective varieties, i.e., there exist open subsets $U_1, \subset X$ and $U_2 \subset Y$ such that $f|_{U_1} : U_1 \rightarrow U_2$ is an isomorphism.

There is an homomorphism induced by $f$ given by

$$ f_* : Div(X) \rightarrow Div(Y) , D\mapsto \overline{ f(D|_{U_1}) }$$ Is known that if $f$ is a small modification then $h^0(f_*(D))=h^0(D)$ for any divisor $D$ of X.

Exists a relation between $h^0(D)$ and $h^0(f_*(D))$ in general for a divisor $D$ of X?

If $D$ is a divisor of $X$ and $\mathcal{L}_D$ is the associated line bundle we consider the following notation $h^0(D) = h^0(X, \mathcal{L}_D)$.

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  • $\begingroup$ This is a weird question. How do you define $f_*$ without knowing the answer to this? $\endgroup$ Commented Feb 18, 2013 at 7:42
  • $\begingroup$ Let me ask for clarification: is $f$ a morphism or just a map? $\endgroup$
    – user5117
    Commented Feb 18, 2013 at 14:28
  • $\begingroup$ $f$ is a birational map, not necessarily a morphism. $f_*$ is the push-forwards homomorphism. $\endgroup$ Commented Feb 18, 2013 at 17:14
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    $\begingroup$ OK, so this is much better. Of course, now the answer you accepted does not answer your question. Which makes me wonder what you had in mind when you accepted it. Sasha did not know what you really asked so he answered what he assumed you did, but you should have realized that. I'm sorry to be so critical, but I feel that if we invest time and energy to answer questions, it is a reasonable expectation that the person asking the question would do the same. $\endgroup$ Commented Feb 18, 2013 at 23:54
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    $\begingroup$ As you define it, $f_*$ depends on the choice of $U_1$. $\endgroup$ Commented Feb 19, 2013 at 7:57

3 Answers 3

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So, the latest formulation is much better, but there are still some problems with this setup. You can push-forward cycles via morphisms, but rational maps are trickier.

First of all if the inverse of $f$ contracts a divisor, then $f_*$ does not (necessarily) respect linear equivalence. Here is an example. Let $X$ be your favorite smooth projective variety and $\pi:Y\to X$ the blow-up of a (closed) point $x\in X$. Let $f:X\dashrightarrow Y$ be the inverse of $\pi$ (as a rational map). Now let $\mathfrak d$ be a very ample linear system on $X$. If $D\in\mathfrak d$ is such that $x\not\in X$, then $f_*D\sim\pi^*D$ and if $D'\in\mathfrak d$ is such that $x\in D'$ and, for simplicity, the multiplicity of $D'$ at $x$ is $1$ (this happens for instance if $X=\mathbb P^n$ and $\mathfrak d$ is the hyperplane class), then $f_*D'=\pi^*D'-E \sim \pi^*D-E\sim f_*D-E$.

Therefore there is a short exact sequence of sheaves on $Y$: $$ 0\to \mathscr O_Y(f_*D')\to \mathscr O_Y(f_*D) \to \mathscr O_E \to 0. $$ Taking global sections one obtains another short exact sequence: $$ 0\to H^0(Y,\mathscr O_Y(f_*D'))\to H^0(Y,\mathscr O_Y(f_*D)) \to H^0(E,\mathscr O_E) \to 0. $$ Here the non-obvious exactness follows from the fact that $f_*D\cap E=\emptyset$ and hence the restriction of the corresponding section to $E$ is a non-zero global section of $\mathscr O_E$.

This shows that (using the OP's notation): $$ h^0(f_*D')= h^0(f_*D)-1. $$ On the other hand, from the projection formula it follows that $$ h^0(D)=h^0(f_*D) $$ and hence $$ h^0(D')=h^0(f_*D')+1 $$


OK, so we see that we better assume that the inverse of $f$ does not contract divisors. I assume you meant this to be part of the assumption that $f$ is a small modification, but you also seemed to be asking without that. Also, the above should warn you that using notation like $h^0(f_*D)$ is dangerous. The usage of $h^0$ suggests using the linear equivalence class of $D$, but $f_*D$ is not well-defined for that. Of course, this can still be done right, you just have to make sure to emphasize that $D$ is an explicit divisor, not a divisor class. (So for instance this is another reason why your original $D\in\mathrm{Pic}\\, X$ was bad).

Assuming that $f^{-1}$ does not contract divisors, there is of course a problem if $f$ contracts a divisor. Just take $D''=f_*D'$ from the above example. Clearly, $\pi_*D''=D'$ and we already saw that $$ h^0(D'')=h^0(\pi_*D'')-1. $$


So, we're left with the case when neither $f$ nor $f^{-1}$ contract any divisors. In this case it is indeed true what you want. Here is why:

The assumption means that in this case we have open sets $U\subseteq X$ and $V\subseteq Y$ such that $\mathrm{codim}_X(X\setminus U)\geq 2$, $\mathrm{codim}_Y(Y\setminus V)\geq 2$, and $f:U\to V$ is an isomorphism. In this case clearly we have that $$ H^0(X,\mathscr O_X(D)) = H^0(U,\mathscr O_U(D|_U)) = H^0(V,\mathscr O_V(f(D|_U))) = H^0(Y,\mathscr O_Y(f_*D)). $$ The middle equality is obvious the other two follows from the fact that $X$ and $Y$ are $S_2$. This is sometimes called the Hartog property. See this MO answer for more.


Note that this does not need $X$ and $Y$ to be smooth, only $S_2$. In order to deal with divisors you're probably better off assuming that they are normal. For some musings about that see this MO answer.

Also, to be fair, you asked for a relation in general, not equality so I assume you are aware of some of the above. I think that in general the relationship between $h^0(D)$ and $h^0(f_*D)$ will be very complicated and has to do with how $D$ relates to the exceptional divisor(s).

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For any birational map $f:X \to Y$ the projection formula gives $h^0(X,f^*L) = h^0(Y,L)$ for any line bundle $L$ on $Y$. If the morphism is small then the map $f^*:Pic(Y) \to Pic(X)$ is an isomorphism, so for any $L' \in Pic(X)$ there is $L \in Pic(Y)$ such that $L' = f^*(L)$. Consequently $h^0(X,L') = h^0(Y,L)$. It remains to note that $f_*(L') = f_*(f^*(L)) = L$.

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    $\begingroup$ This seems way too complicated. You don't need anything else but the definition of $f_*$.... $\endgroup$ Commented Feb 18, 2013 at 8:36
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    $\begingroup$ The projection formula does not give what you say in the first line. It only gives $h^0(X,f^*L)=h^0(Y,L\otimes f_*\mathcal{O}_X$. So, for example if $X$ is the normalizatiion of a non-normal $Y$, we do not get what you say. Of course this is irrelevant in the `small' case above. $\endgroup$
    – Mohan
    Commented Feb 18, 2013 at 15:16
  • $\begingroup$ @Mohan: why is it irrelevant in the small case? If $Y$ is regular in codimension 2, but not normal, then $H^0(X,f^*\mathscr O_Y)\neq H^0(Y,\mathscr O_Y)$. $\endgroup$ Commented Feb 18, 2013 at 16:10
  • $\begingroup$ @Mohan: Yes, I assumed implicitly that $Y$ is normal. On the other hand, note that the higher direct images of the structure sheaf (which may be nonzero if the singularity of $Y$ is not rational) do not change the global sections. $\endgroup$
    – Sasha
    Commented Feb 18, 2013 at 16:23
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[EDIT] In view of the latest edit to the question this is no longer an answer. However, I will not delete it as I think it contributed to the revision of the question, so it should be part of the record.


If $D\in\mathrm{Pic}\\, X$, in fact if $D$ is any sheaf on $X$ and $f:X\to Y$ is a continuous map, then for any $V\subseteq Y$ open $H^0(V,f_*D):=H^0(f^{-1}V,D)$ by definition, so in particular $H^0(Y,f_*D):=H^0(X,D)$ and you don't need any of the assumptions.

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    $\begingroup$ The problem is that a priori the pushforward of a line bundle is not a line bundle. In other words, the pushforward of sheaves is not the same as the pushforward of divisors. $\endgroup$
    – Sasha
    Commented Feb 18, 2013 at 9:31
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    $\begingroup$ That's what I thought first as the reasonable interpretation, but even though the OP used "$D$", he never said it was a divisor. The only stated property is that $D\in\mathrm{Pic} X$, which says it is a sheaf. Just because the OP uses a letter that's usually means divisors, it does not make it one. At best this is a very sloppy question. $\endgroup$ Commented Feb 18, 2013 at 15:20
  • $\begingroup$ p.s.: By the way, if $D$ were a divisor, then WTH is $h^0(D)$? $\endgroup$ Commented Feb 18, 2013 at 15:21
  • $\begingroup$ p.p.s.: In fact, how do you define the notion of a divisor under the assumptions? $\endgroup$ Commented Feb 18, 2013 at 16:11
  • $\begingroup$ Undoubtedly, the question is very sloppy. But it seems that I was correct in guessing what the question was. $\endgroup$
    – Sasha
    Commented Feb 18, 2013 at 16:20

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