Lets say we're on a smooth surface, and we blow up at a point. Is there a simple explicit computation that shows to me the fact that the exceptional divisor E has self intersection -1 ? I don't consider the canonical divisor explicit (but am open to it). I do consider power series hacking to be explicit.

I'm quite unnerved by this -1. Is E effective (seems to be, by definition?). Is E ample (seems to not be, by Nakai-Mozeishon type things)? More generally, I used to think of effective and ampleness as both being measures of "positivity"; but perhaps this is wrong - what do effectiveness and ampleness have to do with each other.

What happens locally at a point of -1 intersection? I thought two irreducible curves on a surface should intersect either in 0 points, or in a positive number of points. To find E.E I would have tried to move E to some other divisor, and then I would get E.E = 0 or nonnegative.

Sorry for the multiple questions, but I'm really distressed :(


5 Answers 5


Dear Fellow,

You can't move $E$ (!), hence there is no contradiction with it having self-intersection -1. Indeed, if you take a normal vector field along $E$, it will necessarily have degree -1 (i.e. the total number of poles is one more than the total number of zeroes), or (equivalently), the normal bundle to $E$ in the blown-up surface is $\mathcal O(-1)$.

[Added:] Here is a version of the argument given in David Speyer's answer, which is rigorous modulo basic facts about intersection theory:

Choose two smooth very ample curves $C_1$ and $C_2$ passing through the point $P$ being blown-up in different tangent directions. (We can construct these using hyperplane sections in some projective embedding, using Bertini; smoothness is just because I want $P$ to be a simple point on each of them.) If the $C_i$ meet in $n$ points away from $P$, then $C_1\cdot C_2 = n+1$.

Now pull-back the $C_i$ to curves $D_i$ in the blow-up. We have $D_1 \cdot D_2 = n + 1$. Now because $C_i$ passes through $P$, each $D_i$ has the form $D_i = D_i' + E,$ where $D_i'$ is the proper transform of $C_i$, and passes through $E$ in a single point (corresponding to the tangent direction along which $C_i$ passed through $P$). Thus $D_1'\cdot D_2' = n$ (away from $P$, nothing has changed, but at $P$, we have separated the curves $C_1$ and $C_2$ via our blow-up).

Now compute $n+1 = D_1\cdot D_2 = D_1'\cdot D_2' + D_1'\cdot E + E\cdot D_2' + E\cdot E = n + 1 + 1 + E\cdot E$, showing that $E\cdot E = -1$. (As is often done, we compute the intersection of curves that we can't move into a proper intersection by adding enough extra stuff that we can compute the resulting intersection by moving the curves into proper position.)

  • $\begingroup$ @David Speyer and Emerton: See the last two paragraphs of Andrea Ferretti's version of the computation: it's simpler to move one curve instead of two. Maybe somebody should consolidate all these answers into one community wiki answer? $\endgroup$ Mar 2, 2010 at 2:50
  • $\begingroup$ Dear Bjorn, I saw that, but thought it might be helpful to have a computation that didn't use anything (counting the push-pull formula as something). $\endgroup$
    – Emerton
    Mar 2, 2010 at 3:33
  • $\begingroup$ All Andrea is really using is that E.f^* C = 0, which is easy because C can be replaced by a linearly equivalent divisor disjoint from P. Aren't you implicitly using the same kind of argument to know that D_1.D_2 = C_1.C_2 ? $\endgroup$ Mar 2, 2010 at 5:56
  • $\begingroup$ Dear Bjorn, I think you're right. $\endgroup$
    – Emerton
    Mar 2, 2010 at 15:01

First, there is no contradiction. If you take another representative of the same linear system, this will need to have some negative coefficients. You can compute intersections counting points only when the intersection is transverse (or at least proper if you count multiplicities) and certainly $E$ is not transverse to itself.

From another point of view, the tangent space to the space deformations of $E$ in $S$ in the point $[E]$ is $T_{[E]} Def = H^0 ( N_{E/S})$, and the latter is zero. Indeed by adjunction $N_{E/S} = \mathcal{O}_S(E) |_E = \mathcal{O}_E(-1)$. So not only your linear system only contains the point $[E]$; there is no way to deform $E$ at all (even in a nonlinear way).

As for the computation, say $S$ is the blowup of $T$ in the point $p$, let $f \colon S \to T$ be the blowup. Take any curve $C$ passing through $p$ with multiplicity $1$; then $f^{*}(C) = \widetilde{C} + E$, where $\widetilde{C}$ is the strict transform.

By the push-pull formula $E \cdot f^{*}(C) = f_{*}(E) \cdot C = 0$, hence $E \cdot \widetilde{C} = - E^2$. But $E \cdot \widetilde{C} = 1$ because they intersect transversely in one point and you're done!


That $E$ has self-intersection $-1$ is Hartshorne Proposition V.3.1. The relation between effectivity and ampleness is more clear in the case of divisors on a curve where a (non linearly trivial) divisor has degree greater than $0$ if and only if it is ample (Hartshorne IV.3.3). So certainly for curves effectivity implies ampleness. On the other hand, even on curves, there exist ample divisors which aren't effective, for example consider a smooth non-hyperelliptic curve of genus at least $3$ and take the divisor $2p-q$ for $p,q$ points on the curve. This is ample but not effective.

As for how to think about ampleness in general, a divisor is ample if and only if some tensor power of it is very ample (Hartshorne II.7.6), and very ampleness is convenient to think about since it basically says you have an embedding into projective space and that the locally free sheaf of rank $1$ associated to the divisor is the pullback of $\mathcal{O}(1)$ from the projective space. Finally, a curve might have negative self-intersection only with itself, so you can still rely on Bezout's theorem working as your intuition does! Also, as you can see, all the above statements are in Hartshorne!


The intuitive argument is the following: Let $D$ be a local chart on your smooth surface, with coordinates $(z,w)$, and with $(0,0)$ the point being blown up. Let $\pi: D' \to D$ be the blow up, with $E$ the curve $\pi^{-1}{\large (}(0,0){\large )}$. Consider the intersection of $X_t := \pi^{-1} \left( \{ x = t \} \right)$ and $Y_u := \pi^{-1} {\large (}\{ y = u\} {\large )}$.

When $t$ and $u$ are not zero, these are smooth curves, meeting transversely at $(t,u)$, so they intersect with multiplicity $1$. When $t$ becomes $0$, $X_0$ splits up into two pieces $X' \cup E$. For $u$ nonzero, $Y_u$ misses $E$ and meets $X'$ transversely, so the intersection is still $1$. Similarly, $Y_0 = E \cup Y'$.

Now, let $t=u=0$. We want to compute $$\langle X_0, Y_0 \rangle = \langle X'+E,\ Y'+E \rangle = \langle X', Y' \rangle + \langle X', E \rangle + \langle E, Y' \rangle + \langle E, E \rangle.$$

By continuity, the left hand side should be $1$. $X_0$ and $Y_0$ miss each other, and $E$ meets $X_0$ and $Y_0$ transversely. We get $$1 = 0 + 1 + 1 + \langle E, E \rangle$$ so $$ \langle E,E \rangle = -1.$$


Write $\mathbb{CP}^1$ as two copies of $\mathbb{C}$, with coordinates $z_1$ and $z_2$ respectively, glued along the map $z_1 \mapsto z_2=\frac{1}{z_1}$ on $z_1\neq 0$.

Write the line bundle $\mathcal{O}(-1) \rightarrow \mathbb{CP}^1$ as two copies of $\mathbb{C} \times \mathbb{C}$, with coordinates $(z_1, v_1)$ and $(z_2,v_2)$ respectively, glued along the map $(z_1,v_1) \mapsto (z_2, v_2) = (\frac{1}{z_1}, z_1 v_1)$ on $z_1\neq 0$. Denote the zero-section by $Z$.

By definition of a blow-up, there is a holomorphic isomorphism from a small neighborhood of E to a neighborhood of $Z$, and this isomorphism sends $E$ to $Z$.

Therefore, it is enough to compute the self-intersection of $Z$. This is a topological notion, so a natural thing to do is to find a cycle $\gamma$ homologous to $Z$ and intersecting $Z$ transversely. (You can't ask $\gamma$ to be a divisor: $Z$ is the only compact divisor in $\mathcal{O}(-1)$.)

Construct such a $\gamma$ as continuous section of $\mathcal{O}(-1)$: on $\vert z_1 \vert \leq 1$, take $z_1 \mapsto (z_1,v_1=1)$; on $\vert z_2 \vert \leq 1$, take $z_2 \mapsto (z_2,v_2= \overline{z_2})$. On the overlap $\vert z_1 \vert=1= \vert z_2 \vert$, we have $v_2= \overline{z_2} = \frac{1}{z_2} = z_1 = z_1 v_1$, as needed.

A homotopy from $Z$ to $\gamma$ is given by $z_1 \mapsto (z_1,t)$ and $z_2 \mapsto (z_2, t\ \overline{z_2})$, for $t\in [0,1]$. In particular (the image of) $\gamma$ is homologous to $Z$.

The only intersection point of $\gamma$ and $Z$ is at $(z_2,v_2)=(0,0)$. There, the orientation of $Z$ given by its complex structure is represented by the vectors $(1,0)$ and $(i,0)$. Pushing this orientation with the above homotopy gives the orientation for $\gamma$ represented by $(1,1)$ and $(i,-i)$.

The $\mathbb{R}$-basis of $\mathbb{C} \times \mathbb{C}$ given by $(1,0)$, $(i,0)$, $(1,1)$ and $(i,-i)$ has same orientation as $(1,0)$, $(i,0)$, $(0,1)$ and $(0,-i)$, which is negative. Conclusion: $Z.\gamma = -1$, so $Z.Z=-1$, so $E.E=-1$.


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