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Let $X$ be a smooth projective surface over $\mathbb{C}$ and $E$ be a vector bundle on it. Let $F\subset E$ be a normal subvector bundle of $E$, that is, the quotient $E/F$ is torsion free. Any section $s\in H^0(F)$ naturally defines a section of $E$, that I call $s_E$. Does the zero locus of $s_E$ always coincide with that of $s$?

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I don't understand your question. The quotient of a vector bundle by a vector subbundle is always a vector bundle. Maybe you mean that you have a sheaf morphism $\mathcal O_X(F)\hookrightarrow\mathcal O_X(E)$? –  diverietti Jan 18 '13 at 12:47
    
Yes, it does, even as a subscheme. Ideal sheaf of the zero locus of a section $s\in H^0(E)$ is image of the natural morphisma $E^*\to \mathcal O_X$. If $s$ is a section of $F\subset E$, $F$ is also locally free and the quotient $E/F$ is locally free, the homomorphism from $E^*$ to $\mathcal O_X$ factors through $F^*$, and we are done. Maybe iI did not understand your question? –  Serge Lvovski Jan 18 '13 at 13:04
    
Perhaps you meant "sheaf inclusion of vector bundles" instead of "subvector bundle"? –  Qfwfq Jan 18 '13 at 15:36
    
Yes, I meant this. –  ginevra86 Jan 18 '13 at 17:56
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2 Answers

up vote 5 down vote accepted

If you mean that you have a morphism of sheaves $\mathcal O_X(F)\to\mathcal O_X(E)$ such that the quotient sheaf is torsion free, then in general the answer is no.

Think at the example of a holomorphic foliation $\mathcal F$ by curves on a smooth projective surface $X$. It is given by definition by the datum of a holomorphic line bundle $T_\mathcal F\to X$ together with a sheaf morphism $\varphi\colon\mathcal O_X(T_\mathcal F)\to\mathcal O_X(T_X)$, such that the quotient is torsion-free. Thus, you have a short exact sequence of sheaves $$ 0\to\mathcal O_X(T_\mathcal F)\to\mathcal O_X(T_X)\to\mathcal I_Z\cdot\mathcal O_X(N_\mathcal F)\to 0, $$ where $N_\mathcal F$ is a holomorphic line bundle and $\mathcal I_Z$ is an ideal sheaf supported on a zero dimensional cycle. This ideal sheaf detects exactly the set of point where the morphism you started with is not injective (and at these point it is in fact the zero morphism).

Thus a global holomorphic section $s\in H^0(X,T_\mathcal F)$ will vanish at those points, once sent in $H^0(X,T_X)$ by the morphism $\varphi$.

On the other hand, if you really meant subbundle, then the morphism is just the inclusion, the quotient is actually a vector bundle (hence torsion-free) and of course the zero loci always coincide.

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Thank you! This perfectly answers my doubt. –  ginevra86 Jan 18 '13 at 13:15
    
You are welcome Ginevra! –  diverietti Jan 18 '13 at 13:17
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Although there is already an accepted answer this example may still be of some interest:

Example Let $S=k[x,y,z]$ with the usual grading and consider the homomorphism of graded modules $$ S(-2)\to S(-1)\oplus S(-1) $$ given by taking the generator of $S(-2)$ to $(y,x)$. The cokernel of this morphism is the usual way to generate the ideal $(x,y)\subset S$. Considering the associated morphisms of locally free sheaves on $\mathbb P^2$ we get a short exact sequence as in the question: $$ 0\to \mathscr O_{\mathbb P^2}(-2) \to \mathscr O_{\mathbb P^2}(-1)\oplus \mathscr O_{\mathbb P^2}(-1) \to \mathfrak m\to 0, $$ where $\mathfrak m$ is the ideal of the point $[0:0:1]\in\mathbb P^2$. Next twist this sequence by $\mathscr O_{\mathbb P^2}(2)$ to get $$ 0\to \mathscr O_{\mathbb P^2} \to \mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1) \to \mathfrak m\otimes \mathscr O_{\mathbb P^2}(2)\to 0. $$ Now take the section $1\in H^0(\mathscr O_{\mathbb P^2})$ and observe that its image in $\mathscr O_{\mathbb P^2}(1)\oplus \mathscr O_{\mathbb P^2}(1)$ is $(y,x)$ and hence its zero set is the point $[0:0:1]\in\mathbb P^2$. $\square$

Observe that this is not a very special example and similar phenomena appear on any (smooth projective) variety of dimension at least $2$. For simplicity assume $X$ is defined over an algebraically closed field $k$. Let $x\in X$ be a smooth closed point and consider a locally free resolution of the sheaf $k$ supported at $x$. Since $X$ is smooth at $x$, we may choose a finite resolution and choose one of minimal length. Let $\phi:\mathscr F\to \mathscr E$ be the last morphism in the resolution and observe that $\phi$ is injective and the cokernel of $\phi$, say $\mathscr Q$, is a torsion-free sheaf that is not locally free. So we have a short exact sequence: $$ 0\to \mathscr F \to \mathscr E \to \mathscr Q \to 0 $$ Suppose $s\in H^0(X,\mathscr F)$ and that $y\in X$ is in the zero locus of $s$. This is equivalent to saying that $s\in H^0(X,\mathscr F\otimes \mathfrak n)$ where $\mathfrak n$ is the ideal sheaf of $y\in X$. (Note that since $\mathscr F$ is locally free, $\mathscr F\otimes \mathfrak n\subseteq \mathscr F$. By functoriality it follows that when considered as a section of $\mathscr E$ (using the notation in the question), $s_E\in H^0(X,\mathscr E\otimes \mathfrak n)$ and hence $y$ is still in the zero section of $s$. In other words:

Good news: The zero locus of $s_E$ contains the zero locus of $s$.

However, the above example shows that

Bad news: In general the zero locus of $s_E$ is strictly larger than the zero locus of $s$.

I think it is reasonable to expect that this will be the typical behaviour. I leave it to the reader to figure out some conditions that ensure this.

Anyway, at the end this is not surprising, since this happens all the time if we do not require the quotient to be torsion sheaf: If for instance a locally free sheaf of rank $1$ is contained in another locally free sheaf of rank $1$, that means that the corresponding divisors contain each other and if it is a proper containment, then one of the divisors is strictly bigger and hence if there is any non-zero section of the smaller sheaf, then its zero locus will be strictly smaller than that of the larger sheaf. The only "trick" here is to make the quotient torsion free, but with allowing higher rank and higher dimension it can happen.

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