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Background.

In section 27.5 of the Mirror Symmetry book, the following claim is made:

The normal bundle of any embedding $\mathbb{P}^1 \subseteq X$ (where $X$ is a Calabi–Yau threefold) is a rank $2$, degree $-2$ bundle.

The given reasoning is roughly:

  • $X$ is Calabi–Yau, so $K_X$ is trivial.
  • $\int_{\mathbb{P}^1} c_1(T\mathbb{P}^1) = 2$.

From this, the claim follows.


My thought process.

The first point follows from the definition of a Calabi-Yau threefold, and I understand how to obtain the second point from the Euler sequence:

$$ 0 \to \mathcal O \to \mathcal O(1)^{\oplus 2} \to T\mathbb{P}^1 \to 0, $$

which gives $c_1(T\mathbb{P}^1) = 2H$ and then $\int_{\mathbb{P}^1} c_1(T\mathbb{P}^1) = 2$.

  • There is the tangent–normal exact sequence

$$ 0 \to TC \to i^\ast TX \to N_{C\mid X} \to 0, $$

which comes from the map $i:\mathbb{P}^1\hookrightarrow X$, and $C$ is the image of $\mathbb P^1$.

  • $i^\ast TX$ is a rank $3$ bundle, since it is the restriction of $TX$, and $\dim X = 3$. $TC$ is a line bundle, so $N_{C\mid X}$ has rank $2$.
  • I know the degree of the bundle might be related to an integral so I expect the second point above to be relevant, but I’m blanking on the relation.

At this point I get stuck: how do we deduce $\deg N_{C\mid X} = -2$ and where am I using that $K_X \cong \mathcal O_X$? Does this have to do with Serre duality?


Also: do you have any recommendations for accessible references (something “hand-holding” like) that could help me work through the Mirror Symmetry book more effectively? My background: a couple of courses in AG and one in Riemann surfaces, no Hartshorne though :(

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1 Answer 1

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By degree they mean $c_1$ of the bundle, viewed in $H^2(\Bbb P^1,\Bbb Z)\cong\Bbb Z$. Then it’s just additivity of $c_1$ in the short exact sequence, since $c_1(TX)=-c_1(T^*X)=-c_1(K_X)=0$ implies $c_1(i^*TX)=i^*c_1(TX)=0$.

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    $\begingroup$ Ahhh, I believe I can follow it then. Let me see. From the tangent-normal sequence we have $$c_1(i^\ast TX)=c_1(TC)+c_1(N_{C\mid X}).$$ Using Calabi-Yau-ness we get that the left hand side is zero as you derive and then integrating both sides gives us $$\deg(N_{C\mid X})=\int_{X}c_1(N_{C\mid X})=-\int_X c_1(TC)$$ and via the second item we get the desired $-2$, right? $\endgroup$ Commented Sep 2 at 0:02
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    $\begingroup$ Yes, correct! … $\endgroup$ Commented Sep 2 at 0:37

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