# Ample divisors on $\mathbb{P}^3$ blow-up along single point

Let $$\pi:X\to\mathbb{P}^3$$ be the blowing up at single point with $$E$$ be the exceptional divisor. Let $$H=\pi^\ast\mathcal{O}_{\mathbb{P}^3}(1)$$.

In Ample divisors on the blow up of $$\mathbb{P}^3$$ at points , we know that the divisor on $$X$$ of the form $$dH-E$$ with $$d\geq5$$ is ample. But in this case I am very much willing to know a easy and straightforward proof.

Let $[x:y:z:w]$ be the homogeneous coordinates on $\mathbb{P}^3$ and suppose that the point blown up is given by the vanishing of the first three coordinates. Introduce a projective plane with coordinates $[X:Y:Z]$ then the blowup is a closed subset in $\mathbb{P}^3 \times \mathbb{P}^2$ given by the equations $$xY = yX; xZ = zX; yZ = zY.$$ The bundle $\mathcal{O}(1) \boxtimes \mathcal{O}(1)$ on $\mathbb{P}^3 \times \mathbb{P}^2$gives the Segre Embedding, so it is very ample on the product of projective spaces. Its restriction to the blowup is also very ample and it must be of the form $dH - k E$. We can computed $d$ and $k$ by intersecting with (a) a line in $\mathbb{P}^3$ that avoids the point blown up, which gives $d = 2$; and (b) with a line in the exceptional fiber $[0:0:0:1] \times \mathbb{P}^2$, which gives $k=1$. So $2H - E$ is very ample.

• It is even very ample, since it separates points and tangent vectors. – IMeasy Feb 16 '14 at 10:10

I'll tell you the proof that's "easy and straightforward" to me, because I like toric varieties, $X$ is toric, and $dH-kE$ is an equivariant line bundle.

On a projective toric variety, an ample equivariant line bundle gives you a lattice polytope whose fan is that of the toric variety. For $\mathbb P^3$ the polytope is a tetrahedron, and $\mathcal O(d)$ gets you a tetrahedron with edge-length $d$. For $X$ you snip off one corner, obtaining something with two parallel triangular faces. If $dH-kE$ is ample, the smaller triangle has edge-length $k$, and the distance between the two triangles is $d-k$.

You want $k=1$ apparently, i.e. you just barely snip off the corner. If you took $d=1$ the snipped tetrahedron would degenerate to a triangle, and the line bundle would be only nef, not ample; the map to projective space would be the one collapsing $X$ to $\mathbb P^2$.

So you need $d\geq 2$ for this to be ample. I don't know where that $5$ comes from.

(It's fun to consider $k$ negative, where instead of the corner moving into the tetrahedron and becoming a triangle, it comes out as a sort of negative triangle, and the polytope is the size $d$ tetrahedron plus an extra size $-k$ "negative" tetrahedron. The extra one is telling you something about $H^1$ of this non-ample line bundle.)

The divisor $H-E$ (pull-back of the linear system of planes through the blown-up point $p$) does not separate points. Indeed if $q\in\mathbb{P}^{3}$ and $t\in\left\langle p,q\right\rangle$ then any plane through $q$ passing through $p$ contains $t$ as well. The divisor $H-E$ induces the resolution of the rational projection of $\mathbb{P}^{3}$ from $p$. Then $H-E$ is nef but not ample.

Now, we prove that $dH-E$ is ample (indeed very ample) for any $d\geq 2$. For any couple of point $q,t\in\mathbb{P}^{3}$ we can find a surface of degree $d\geq 2$ trough $p,q$ but not passing through $t$. Furthermore given a point $q\in\mathbb{P}^{3}$ and a tangent direction $v_q$ at $q$ we can find a surface $S$ of degree $d\geq 2$ through $p,q$ such that $v_q$ is not a tangent direction through $S$. Furthermore the base locus of $|\mathcal{I}_{p}(d)|$ is just $p$. Then $dH-E$ is base point free, separates points and tangent directions. Therefore it induces an embedding of $Bl_p\mathbb{P}^3$ to $\mathbb{P}(H^{0}(\mathbb{P}^3,\mathcal{I}_p(d)))$ and $dH-E$ is very ample for $d\geq 2$.

For instance $2H-E$ (the space of quadrics through a point has dimension $9$) induces an embedding $f:Bl_p\mathbb{P}^3\rightarrow\mathbb{P}(H^{0}(\mathbb{P}^3,\mathcal{I}_p(2)))\cong\mathbb{P}^8$ and $X = Im(f)\subset\mathbb{P}^8$ is a smooth $3$-fold of degree $8-1 = 7$.

The divisor $D=2H-E$ is ample because $D^3>0$ and $D$ intersects any irreducible curve $C$ positively, this last assertion is an easy exercise:

$a$) if $C$ is contained in the exceptional divisor, then $C$ is a multiple of a line $l$ of the plane obtained, and the line $l$ satisfies $l\cdot D=l\cdot E=1$.

$b)$ if $C$ is not contained in the exceptional divisor, then it is the strict transform of a curve of degree $d$ having multiplicity $r$ at the point. Then, $C\cdot D=2d-r$. Since $d\ge r$ we have $2d-r>0$.

By the way, $2H-E$ is the half of the anti-canonical divisor and the fact that the blow-up of one point in $\mathbb{P}^n$ is Fano (easy exercise, same as above).

Remark: the cone of curves of the the blow-up $X$ of one point in $\mathbb{P}^n$ is generated by $l$ and $f$, where $l$ is a line in the exceptional divisor and $f$ is a fibre of the map $X\to \mathbb{P}^{n-1}$ corresponding to the projection by the point. This cone is closed and both rays intersect $-K_X$ positively.

• I think your first sentence requires a bit more explanation, because it is not true in general. Perhaps you can add that the cone of curves of the blow up has rank two and both rays are generated by actual curves, so what you claim follows from Kleiman's criterion. – Sándor Kovács Feb 19 '14 at 0:48
• Actually, by the way, I don't understand your last sentence either. What does it mean that "the blow-up of one point in $\mathbb P^n$ is ample"? – Sándor Kovács Feb 19 '14 at 0:50
• Uh, did you mean that the blow up is "Fano" instead of "ample"? – Sándor Kovács Feb 19 '14 at 0:56