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Let $f:X \to Y$ be a proper, birational morphism with connected fibers, $X$ is non-singular and $Y$ is normal. Does there exist a moduli space parametrizing all invertible sheaves $\mathcal{L}$ on $X$ such that $f_*\mathcal{L}$ is isomorphic to the trivial sheaf $\mathcal{O}_Y$? If so, is there any literature on this topic, for example how can I compute the dimension of such a space, is it non-singular, etc.?

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    $\begingroup$ I'm not sure a moduli space would be the right thing to expect. However, it should be easy to exactly characterize what sheaves these are. Up to isomorphism, these are exactly the sheaves of the form $O_X(D)$ where $D$ is effective and exceptional. For instance if $Y$ is $\mathbb{A}^2$ and $X$ is the blowup of $Y$ at the origin, then the set of such sheaves up to isomorphism is in bijection with the natural numbers. $\endgroup$ Commented Nov 10, 2015 at 18:49
  • $\begingroup$ Also, you should clarify your problem. Do you want $f_* \mathcal{L}$ to be equal to the trivial sheaf, or isomorphic to it? $\endgroup$ Commented Nov 10, 2015 at 18:51
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    $\begingroup$ @KarlSchwede I have edited the question to isomorphic to the trivial bundle. Could you please suggest some reference for your first comment. $\endgroup$
    – Ron
    Commented Nov 10, 2015 at 18:58

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I don't know a reference but I think what I said in the comment is more or less obvious. Let me prove it though in detail. Again, I can't imagine one can make this into any reasonable moduli space.

Setting: Suppose that $\pi : X \to Y$ is a proper birational map with connected fibers with $X$ nonsingular and $Y$ normal.

Lemma: Suppose that $L$ is a line bundle on $X$ and $\pi_* L \cong O_Y$. Then $L \cong O_X(D)$ for a uniquely determined $D$, an effective $\pi$-exceptional divisor.

Proof: Embed $L \subseteq K(X)$ so that $L = O_X(F)$ for some divisor $F$. Then $\pi_* L \subseteq K(Y)$. Let $M = \pi_* L \subseteq K(Y)$. We know $M \cong O_Y$ is a line bundle and hence $M = O_Y(G)$ for some divisor $G$ (in fact $G = \text{div}_Y(h)$ where $h$ is in $K(Y)$ but that doesn't matter for what follows).

Next consider $D := F - \pi^* G \sim F$ and observe that $O_X(F - \pi^*G) \cong O_X(F) = L$. Also not the $\pi_* O_X(D) = O_Y \subseteq K(Y)$. Since $F$ and $G$ agree where $\pi$ is an isomorphism by construction, we see that $D$ is exceptional. Next observe that if $D$ is non-effective, say $D = \sum a_i D_i$ and $a_1 < 0$, then $1 \notin \Gamma(X, O_X(D))$ (since $1$ does not vanish at $D_1$). But then $\Gamma(X, O_X(D)) = \Gamma(Y, \pi_* O_X(D)) = \Gamma(Y, O_Y)$ and so $1 \notin \Gamma(Y, O_Y)$ which is absurd. This proves existence of an effective $D$ satisfying the property.

For uniqueness, suppose that $O_X(D) \cong O_X(D')$ so that $D - D' = \text{div}_X u$ for some $u \in K(X)$. Also suppose $\pi_* O_X(D) = O_Y = \pi_* O_X(D')$. Then $D - D'$ is $\pi$-exceptional but also the divisor of a rational function in the fraction field. Since $\text{div}_Y(u) = 0$ by construction, we see that $u$ is a unit in $\Gamma(Y, O_Y)$. But then $u$ is also a unit in $\Gamma(X, O_X) = \Gamma(Y, O_Y)$ and hence $\text{div}_X(u) = 0$ as well. Thus $D - D' = 0$ and we have proven uniqueness.

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