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This question arised when I was trying to use this answer to understand Reid's "Young Person's guide to Canonical Singularities". In particular page 352 when computing the blow-up $Y\rightarrow A^2/\mu_3$, the affine plane quotient the cyclic group of order 3, arises to the conclusion that the exceptional divisor is $E\sim P^1$, (no problems there) and $\mathcal{O}_E(-E)\sim \mathcal{O}(3)$ (problems here).

Given a variety $Y$ and an effective Cartier divisor $D$ on it, there seems to be a pretty standard exact sequence:

$$0 \longrightarrow \mathcal{O}_Y \longrightarrow \mathcal{O}_Y(D) \longrightarrow\mathcal{O}_D(D)\longrightarrow 0 $$ As far as I understand, if $U$ is an open set in $S$ and $D\cap U = div(g)_U$ (for $D$ a hypersurface, if you want, and extend by linearity), then

$$ \mathcal{O}_Y(D)(U)= \{g \in \mathcal{O}_Y(U) \vert div(g)\geq D \}$$

or equivalently $g/f$ is regular. The first map must be something like $g\rightarrow gf$ maybe with some order. A good answer to my question would include:

  1. Is this correct?
  2. What is $\mathcal{O}_D(D)$?
  3. What is the second map?
  4. What does $\mathcal{O}_D(-D)$ mean
  5. Why $\mathcal{O}_E(-E) \sim \mathcal{O}(3)$? I understood the RHS is generated by polynomials of degree 3?

I am aware this is a simple question and probably everyone knows why, but I could not find a proper answer for it.

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1. Yes, the sequence is correct. 2. and 4. are dual, so it's really the same question. $\mathcal{O}_X(-D)$ is the ideal sheaf of D which is invertible. So $\mathcal{O}_D(-D)$ is its restriction to D as an invertible sheaf. It can also be understood as the cononormal sheaf. So then $\mathcal{O}_D(D)$ would be the normal sheaf and so on. I have go, but I hope that helps a little –  Donu Arapura Apr 26 '11 at 18:55
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2 Answers

up vote 15 down vote accepted

Expanding the comment of Donu Arapura, let $X$ be a variety and $Y\subset X$ a subvariety. Then, you have a short exact sequence of sheaves $$ 0\to\mathcal I_Y\to\mathcal O_X\to\mathcal O_X/\mathcal I_Y\to 0, $$ where $\mathcal I_Y$ is the ideal sheaf of $Y$. By definition, $\mathcal O_X/\mathcal I_Y=\mathcal O_Y$ is the structure sheaf of $Y$.

If $\mathcal F$ is any invertible sheaf, then tensoring by $\mathcal F$ leaves the sequence exact, so that you have a short exact sequence $$ 0\to\mathcal I_Y\otimes\mathcal F\to\mathcal F\to\mathcal O_Y\otimes\mathcal F\to 0 $$ and $\mathcal O_Y\otimes\mathcal F$ is just the restriction of $\mathcal F$ to $Y$.

Now, suppose that your $Y=D$ is a (Cartier) divisor, and $\mathcal F=\mathcal O_X(D)$ is its associated (invertible) sheaf of sections (meromorphic functions with poles allowed along $D$). In this case, $\mathcal I_D=\mathcal O_X(-D)$ and the above-mentioned short exact sequence becomes $$ 0\to\mathcal O_X\to\mathcal O_X(D)\to\mathcal O_D(D)\to 0, $$ and $\mathcal O_D(D)$ is nothing but the restriction $\mathcal O_X(D)\otimes\mathcal O_D$ of the invertible sheaf $\mathcal O_X(D)$ to the hypersurface $D$.

You can argue dually for $\mathcal O_D(-D)$, which is thus just the restriction to $D$ of the invertible sheaf $\mathcal O_X(-D)$.

So your "second map", is just the restriction map.

For your last question, an heuristic explanation is the following: blow-up a smooth point on a surface to obtain a new surface $\widetilde X$, and call the exceptional divisor $E$. Then $\mathcal O_{\widetilde X}(-E)$ restricted to $E$, which is precisely $\mathcal O_E(-E)$, can be easily shown to be isomorphic to the (anti)tautological line bundle $\mathcal O(1)$ over $\mathbb P^1\simeq E$ (you can find that on every introductory book in algebraic geometry). Now, you are blowing up a singular point which is an isolated quotient singularity of order three, thus in some sense you are "counting three times" your point, so that $\mathcal O_E(-E)$ now becomes isomorphic to $\mathcal O(3)$.

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Diverietti's answer is exhaustive with respect to the first four questions, so let me say something just about the last one.

Miles Reid's example is actually a particular case of the following more general situation, which is explained for instance in Barth-Peters-Van de Ven book on compact complex surfaces.

Let $n$ and $q$ be natural numbers with $0 < q < n$, $(n,q)=1$ and let $\xi_n$ be a primitive $n-$th root of unity. Let us consider the action of the cyclic group $\mu_n=\langle \xi_n \rangle$ on $\mathbb{C}^2$ defined by

$\xi_n \cdot (x,y)=(\xi_nx, \xi_n^qy)$.

Then the analytic space $X_{n,q}=\mathbb{C}^2 / \mu_n$ has a cyclic quotient singularity of type $\frac{1}{n}(1,q)$, and $X_{n,q} \cong X_{n', q'}$ if and only if $n=n'$ and either $q=q'$ or $qq' \equiv 1$ (mod $n$). The exceptional divisor on the minimal resolution $\tilde{X}_{n,q}$ of $X_{n,q}$ is a H-J string (abbreviation of Hirzebruch-Jung string), that is to say, a connected union $E=\bigcup_{i=1}^k Z_i$ of smooth rational curves $Z_1, \ldots, Z_k$ with self-intersection $\leq -2$, and ordered linearly so that $Z_i Z_{i+1}=1$ for all $i$, and $Z_iZ_j=0$ if $|i-j| \geq 2$. More precisely, given the continued fraction

$\frac{n}{q}=[b_1,\ldots,b_k]:=b_1- \cfrac{1}{b_2 -\cfrac{1}{\dotsb - \cfrac{1}{\,b_k}}}$,

we have

$(Z_i)^2=-b_i, \quad i=1, \ldots, k.$

In the case considered by Miles Reid, we have $n=3$, $q=1$, i.e. the action is

$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3y)$,

hence the resolution is a unique smooth curve $E:=Z_1$ with self-intersection $(-3)$; this explain why $\mathcal{O}_E(-E)=\mathcal{O}(3)$.

Notice that there is another possible action of $\mu_3$ on $\mathbb{C}^2$, namely

$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3^2y)$.

The corresponding continued fraction is

$\frac{3}{2}=[2,2]=2- \frac{1}{2}$,

so the resolution is this case is given by two smooth rational curves of self-intersection $(-2)$ intersecting in a single point (this is a Rational Double Point of type $A_2$).

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Thanks, this is quite interesting. However: a) you say before that the irreducible components of the minimal resolution will have self-intersection $\leq -2$, but in the case of $\mu_3$ they have -3. Why is this? b) According to Castelnuovo-Enriques contractibility criterion, when you blow-up or blow-down, the only divisor involved is a -1 line. I follow Beauville book. I suspect this because it only works on complex smooth surfaces c) I like the continues fraction bit. I fail to find it in BPV, though. Do you know a source which talks about this? –  Jesus Martinez Garcia Apr 27 '11 at 9:58
2  
a) Well, $-3 < -2$, right? :-) b) You can contract any smooth rational curve with negative self-intersection (by Artin contractibility criterion), but in general the resulting surface is singular, unless you are contracting a $(-1)$-curve. c) Look at BPHV (Barth-Peters-Hulek-Van de Ven - the new edition) p. 99-105 –  Francesco Polizzi Apr 27 '11 at 10:09
    
You are absolutely right, thank you very much for your answer –  Jesus Martinez Garcia May 11 '11 at 12:11
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