# Elementary transformations of ruled surfaces as maps of vector bundles

This comes as a question in Beauville's Algebraic surfaces book (III.24 (2)). We work over $\mathbb{C}$.

All geometrically ruled surfaces (grs) $p:S\longrightarrow C$ over a curve $C$ can be seen as $S=\mathbb{P}(E)$ where $E$ is a vector bundle of rank $2$ over $C$, i.e. a locally free sheaf of rank $2$ over $C$. They are always minimal, or relatively minimal, depending on your book.

An elementary transformation $S\dashrightarrow S'$ with centre $s\in S$ corresponds to blowing up $s$ and contracting the strict transform of a fibre. This gives another grs $S'$ over $C$.

That point of view which is explicit and algebraic is the one I understand and usually work with. But you can also think in the following way. Take $s\in S$ and consider the pushforward of the skyscraper sheaf $\mathbb{C}(s)$, $F=p_*\mathbb{C}(S)$. Since $s$ can be seen as $(p(s),D_s)$ where $D_s\in E_{p(s)}$ is a line at the stalk of $E$ at $p(s)$, we get a map $$u_s:E\longrightarrow F$$ which I understand it can be defined on the stalks and sends $(c,v)$ to $(c,0)$ if $c\neq p(s)$ and to $(c,v+D_s)$ otherwise. Here we identify $F_s$ with $E_s/(D_s)$ or some similar quotient. Beauville did not write $F$ as a pushforward but as the skyscraper sheaf itself. I think that is an unimportant typo. Define $E'=ker(u_s)$.

The question is:

1) $E'$ is a vector bundle of rank $2$.

2) $S'=\mathbb{P}(E')$ corresponds to the elementary transformation with centre $s$ $S\dashrightarrow S'$.

3) That transformation corresponds to the inclusion $E\rightarrow E'$.

I find hard to believe 1) is true although it must be because it is written in the book and an article by Hartshorne says it is 'easy' to see. I fail to see how the rank of $E'$ is going to be $2$ over $p(s)$. It seems to me that in the short exact sequence $$E'\longrightarrow E \longrightarrow F,$$ if you restrict to stalks you necessarily have $E'_s\cong \mathbb{C}$, but I must be being somewhat naive.

Probably the problem is that I do not understand $u_s$. I probably can do 2) and 3) once I solve 1). Any ideas?

-

## 2 Answers

The kernel $E'$ of $u_s \colon E \to F$ is a torsion free sheaf over the curve $C$, hence it is necessarily a vector bundle (torsion free sheaves over curves are locally free).

Since the generic rank of $E'$ is $2$, it follows that $E'$ is a rank $2$ vector bundle.

The point you are missing is that the inclusion $E' \to E$ is not an inclusion of vector bundles, but only an inclusion of sheaves.

In fact, we can have a similar situation in rank one, for instance

$0 \to \mathcal{O}_C(-s) \to \mathcal{O}_C \to \mathbb{C}_s \to 0.$

Also in this case, the kernel of $v_s \colon \mathcal{O}_C \to \mathbb{C}_s$ is locally free.

-
Mmmh, good point. It seems I have to look more careful at the relationship between free sheaves and vector bundles. I'll do so and have another look at this. It seems more reasonable now that the map in 2) is not a morphism, and I bet I can get both, but I'll ask here again if not. thank you :) – Jesus Martinez Garcia Jun 30 '11 at 17:06
I do not think torsion-free sheaves over curves are always locally free. As far as I understand locally free means that for every point $p$ there is a $U_i$ neighbourhood of $p$ such that $E(U_i)\cong (O_C)^r$ and $r$ is the same for all $U_i$ (note this does not mean free). Also, torsion-free can be thought of as 'no stalk has torsion', or 'no $E(U_i)$ has a torsion element. But then, the skyscraper sheaf on $p(s)$ is torsion-free and it is not locally free, since any neighbour of $p$ has $r=1$, and any neighbour out from $p$ has $r=0$. – Jesus Martinez Garcia Jul 1 '11 at 10:34
The singular locus of torsion free sheaves on regular schemes have always codimension $\geq 2$, in particular torsion free sheaves on smooth curves are always free. The skyscraper sheaf is indeed a torsion sheaf, since its support is a proper subvariety of the variety (in other words, the general stalk of a torsion-free sheaf must be not zero). I suggest you the first two chapters of Friedman's book "Algebraic surfaces and holomorphic vector bundles" for a readable introduction to these important concepts and the proofs of statements I mentioned above. – Francesco Polizzi Jul 1 '11 at 11:42
Thank you. The book seems to have exactly what I did not know and in a more accessible way that I would have expected. I always wanted a reference which explained vector bundles from an algebraic point of view, since I saw the differential one. I also was using a 'wrong' definition of torsion sheaf, as one which has torsion in some of the stacks. In bit.ly/kNdaID you made me remember the right one. – Jesus Martinez Garcia Jul 1 '11 at 16:53
Using the observation that $E'\rightarrow E$ is an inclusion of sheaves and your sequence in rk 1, I solved the whole exercise by constructing the sheaf $E'$ and its associated vector bundle as well as working out what the map had to be. Thanks again – Jesus Martinez Garcia Jul 1 '11 at 16:54

1) To see that the rank of $E'_s$ is 2 you can tensor the short exact sequence $$0 \to E' \to E \to O_s \to 0$$ by $O_s$. You will get $$0 \to Tor_1(O_s,O_s) \to E'_s \to E_s \to O_s\otimes O_s \to 0.$$ Now it is easy to see that $Tor_1(O_s,O_s) = O_s\otimes O_s = O_s$ (for example you can use the locally free resolution $0 \to O_C(-s) \to O_C \to O_s \to 0$ to compute the Tor and the tensor product), so one obtains $$0 \to O_S \to E'_s \to E_s \to O_s \to 0,$$ whereof the rank of $E'_s$ is 2.

2,3) Both answers are "yes".

-