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?


1 Answer 1


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.

  • $\begingroup$ Hi Kenny, do you know if there is a coordinate free approach? $\endgroup$
    – user347489
    Jun 13, 2018 at 9:41
  • $\begingroup$ @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...) $\endgroup$
    – Kenny Wong
    Jun 13, 2018 at 10:23
  • $\begingroup$ @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. $\endgroup$
    – Notone
    Jun 13, 2018 at 11:03
  • $\begingroup$ @Notone Do you know a way to prove that $E^2 = -1$ without first proving that $O(E)|_E = O_E(-1)$? $\endgroup$
    – Kenny Wong
    Jun 13, 2018 at 14:41
  • $\begingroup$ @KennyWong www-math.sp2mi.univ-poitiers.fr/~sarti/corso_Perego.pdf Proposition 3.1.13. I believe $\endgroup$
    – Notone
    Jun 13, 2018 at 15:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.