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Some background that is not necessary for answering the question:

Let $X = \mathbb{P}(\mathcal{O}_{\mathbb{P}^2}\oplus\mathcal{O}_{\mathbb{P}^2}(2))$ be a threefold. This is a $\mathbb{P}^1$-bundle over $\mathbb{P}^2$. Let $f$ be the cohomology class of the fiber. This bundle has a section whose image has normal bundle $\mathcal{O}(-2)$ as a hypersurface in $X$. Let $\beta \in H_2(X, \mathbb{Z})$ be the class of a line on this hypersurface. A localization computation gives me the Gromov-Witten invariant $GW^X_\beta\langle f \rangle = -1$. I would like to directly compute this invariant using obstruction theory.

My actual question:

The moduli space of lines in the hypersurface in question (which is isomorphic to $\mathbb{P}^2$) is the dual $(\mathbb{P}^2)^*$. The obstruction bundle on this dual space is the bundle whose fiber over each point (which is a line $i:L \hookrightarrow \mathbb{P}^2$) is $H^1(L, i^*\mathcal{O}(-2))$. How do I see directly that this bundle is in fact $\mathcal{O}(-1)?$

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The universal line (call it $Z$) is a divisor of type $(1,1)$ in the product $\mathbb{P}^2 \times (\mathbb{P}^2)^*$. Consequently, there is an exact sequence (the Koszul complex) $$ 0 \to \mathcal{O}_{\mathbb{P}^2 \times (\mathbb{P}^2)^*}(-1,-1) \to \mathcal{O}_{\mathbb{P}^2 \times (\mathbb{P}^2)^*} \to \mathcal{O}_{Z} \to 0. $$ You are asking about the computation of $R^1q_*(p^*\mathcal{O}(-2))$, where $p$ and $q$ are the projections of $Z$. Using the above resolution and the projection formula gives $$ R^1q_*(p^*\mathcal{O}(-2)) = R^2q_*(\mathcal{O}_{\mathbb{P}^2 \times (\mathbb{P}^2)^*}(-3,-1)) = H^2(\mathbb{P}^2,\mathcal{O}(-3)) \otimes \mathcal{O}_{(\mathbb{P}^2)^*}(-1) = \mathcal{O}_{(\mathbb{P}^2)^*}(-1) $$ (abusively, the projections of $\mathbb{P}^2 \times (\mathbb{P}^2)^*$ to the factors are also denoted by $p$ and $q$).

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  • $\begingroup$ Thank you, this is really a nice solution! $\endgroup$
    – AG learner
    Commented Jun 24, 2020 at 20:03
  • $\begingroup$ Thanks for the solution! Also, just to see if I have it right: the first equality comes from the long exact sequence associated to the short exact sequence above, and the second from the projection formula. One could also just skip the second equality and directly obtain the final answer using relative Serre duality, right? $\endgroup$
    – Nachiketa
    Commented Jun 24, 2020 at 20:49
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    $\begingroup$ @Nachiketa: Yes, there are many ways to compute, but I think this is the easiest. $\endgroup$
    – Sasha
    Commented Jun 25, 2020 at 9:13

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