# Restriction of the exceptional divisor to itself as a line bundle

If we take a complex projective variety $X$ and blow it up at a point, we get an exceptional divisor $E\cong \mathbb{P}^{n-1}$, where $n=dim(X)$.

My question basically regards $\mathcal{O}_{\tilde X}(E)$ or to be more precise $\mathcal{O}_{\tilde X}(E)|_E$. This will be a line bundle on $E$ and I think it should be just $\mathcal{O}_E(-(n-1))$.

The question is, how can I see this claim? We can consider e.g. the simple case of the blow up of $\mathbb{P}^2$, which looks like this $\{[a:b:c],[x:y]|ay-bx=0\}\subset \mathbb{P}^2\times\mathbb{P}^1$. Then $E$ will be of the form $E=\{a=b=0\}\subset X$.

My attempt: I tried to represent the divisor $E$ by a line bundle through choosing a covering and then calculating coycles for that covering (which was just the product of the standard affine coverings). If needed I can elaborate on this, but the cocycles that I got gave me the trivial bundle as restriction on $E$, which is not what I should get, but rather the tautological bundle!

How do I actually end up with the tautological bundle?

For your simple example, it might be easier to concentrate on the blow-up of $\mathbb A^2$ (instead of $\mathbb P^2$), since the question is local around the exceptional divisor. The blow-up of $\mathbb A^2$ looks like $$X= \left\{ (a, b), [x : y] \mid ay = bx \right\} \subset \mathbb A^2 \times \mathbb P^1,$$ and the exceptional divisor is of the form $$E =\{ a=b=0\} \subset X.$$

Let's use the following open covering: $$U_0 = \left\{(a , av), [1 : v] \mid (a, v) \in \mathbb A^2 \right\}$$ $$U_1 = \left\{(bu, b), [u : 1] \mid (b, u) \in \mathbb A^2 \right\}$$ so $$E \cap U_0=\left(\left\{ a = 0\right\} \subset U_0 \right)\cong\{[1:v] | v \in \mathbb A^1 \}$$ $$E \cap U_1=\left(\left\{ b = 0\right\} \subset U_1\right) \cong \{[u:1] | u \in \mathbb A^1 \}$$ [Excuse the poor notation - I hope it's clear what is meant by this.]

The key point is that $E$ is the vanishing locus of the regular function $a$ on $U_0$, and it is the vanishing locus of the regular function $b$ on $U_1$.So the invertible sheaf $\mathcal O_X(E)$ trivialises on the open cover $\{U_0, U_1\}$, and the transition function from $U_0$ to $U_1$ on the overlap $U_0 \cap U_1$ is $$b / a = 1 / u = v.$$

Restricting to $E$, we observe that $1/u = v$ is precisely the transition function for the sheaf $\mathcal O_E(-1)$ on the open cover $\{ E\cap U_0, E \cap U_1 \}$.

For the higher-dimensional version of this calculation, see Griffiths and Harris, Chapter 1.4, p184.

• Hi Kenny, do you know if there is a coordinate free approach? Jun 13, 2018 at 9:41
• @user347489 I don't know I'm afraid. Griffiths and Harris works with complex manifolds, so using local Euclidean-like coordinates is valid. Maybe it's in Hartshorne? (I suspect the general case in the algebraic category boils down to finding generators $a, b$ of the local ring at the blow-up point, which play the same role as $(a,b)$ in the answer above... which isn't exactly coordinate free...) Jun 13, 2018 at 10:23
• @user347489 You can prove first that $E^2=-1$, but you also have $deg(\mathcal(O(E)|_E)=E^2$. So $deg(\mathcal(O(E)|_E)=-1$. But a line bundle on $E\cong\mathbb{P}^1$ is uniquely determined by its degree and thus you can conclude. Jun 13, 2018 at 11:03
• @Notone Do you know a way to prove that $E^2 = -1$ without first proving that $O(E)|_E = O_E(-1)$? Jun 13, 2018 at 14:41
• @KennyWong www-math.sp2mi.univ-poitiers.fr/~sarti/corso_Perego.pdf Proposition 3.1.13. I believe Jun 13, 2018 at 15:04