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Is there a good way to find the fan and polytope of the blow-up of $\mathbb{P}^3$ along the union of two invariant intersecting lines?

Everything I find in the literature is for blow-ups along smooth invariant centers.

Thanks!

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If you don't get a good answer here, I would contact Howard Thompson, homepages.umflint.edu/~hmthomps –  Karl Schwede Jul 12 '11 at 16:34
    
I was going to supplement my answer by a description of how to do the case of a general blowup on a torus invariant subscheme, but it got too long. Instead, I'll just say "see arxiv.org/abs/math/0310336 " –  David Speyer Jul 13 '11 at 22:27
    
Thank you. I am taking a look at the paper. Do you think there is a good way in the general case to know if the resulting scheme will be normal? As you mention in a comment below, is the subscheme is not reduced then you might get something which is not normal. –  Enrique Jul 15 '11 at 15:51
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4 Answers

up vote 7 down vote accepted

The question is local near the intersection of the two lines, so the more basic question is to work this out for $\mathbb{A}^3$.

So we want to blow up $k[x,y,z]$ at $\langle z, xy \rangle$. There are two maximal charts:

$$\mathrm{Spec} \ k[x,y,z, (xy)/z] \ \mbox{and} \ \mathrm{Spec} \ k[x,y,z,z/(xy)].$$

Note that the first chart is isomorphic to $k[a,b,c,d]/(ad-bc)$, so we expect it to correspond to a cone with four rays in the fan, while the second chart is isomorphic to $k[u,v,w]$, so we expect it to give an ordinary simplicial cone.

Let's work out exactly how these fit together. Our first semigroup is the integer points in the cone $\mathrm{Span}_{\mathbb{R}_{+}} ((1,0,0), (0,1,0), (0,0,1), (1,1,-1))$. The dual cone is $C_1 := \{ (u,v,w) : u,v,w \geq 0, \ u+v \geq w \}$. So we expect that cone to appear in the fan. Similarly, the second chart corresponds to the cone $C_2 := \{ (u,v,w) : u,v \geq 0, \ u+v \leq w \}$. We have omitted the inequality $w \geq 0$ because it is redundant.

So $\mathbb{A}^3$ blown up at two crossing lines in the toric variety for the fan which consists of the two cones $C_1$ and $C_2$ above (and their faces).

In order to do $\mathbb{P}^3$, you'll want to insert the picture I just described in place of the cone which corresponds to the torus invariant $\mathbb{A}^3$ around the intersection. You'll also have to modify the two other torus charts which contain one of the lines (each). Since that's a blow up at a smooth center, I'll leave it to you.

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To find the polytope associated to a toric variety directly you have to realize the variety as the closure of a map from the torus. In this case at least, it is not too hard to get such a description. Let the homogeneous coordinates of $\mathbb{P}^3$ be $[w:x:y:z]$ and the two lines be $C_1 := \lbrace w = x = 0 \rbrace$ and $C_2 := \lbrace w = y = 0 \rbrace$. Then the blow-up $B$ of $\mathbb{P}^3$ along $C_1 \cup C_2$ is the closure of
$\lbrace ([w:x:y:z], [w^2:wx:wy: wz: xy]) : [w:x:y:z] \in \mathbb{P}^3 \setminus (C_1 \cup C_2) \rbrace$
in $\mathbb{P}^3 \times \mathbb{P}^4$. If we identify $\mathbb{C}^3$ with $\mathbb{P}^3 \setminus V(w)$, and write $X, Y, Z$ respectively for $x/w, y/w, z/w$, then $\mathbb{C}^3$ is embedded in $B$ via the map
$(X, Y, Z) \mapsto ([1:X:Y:Z), (1: X: Y: Z: XY])$
Composing with the Segre embedding (and getting rid of duplicate coordinates), we get
$(X, Y, Z) \mapsto [1: X : Y: Z: X^2: XY: XZ: Y^2: YZ: Z^2: X^2Y: XY^2: XYZ]$
Therefore the polytope is the convex hull of the exponents of these monomials. I believe its vertices are (0,0,0), (2,0,0), (0,2,0), (2, 1, 0), (1, 2, 0), (0, 0, 2) and (1,1,1).

PS: There is a detail to be filled: blow-ups along singular subvarieties are not in general normal, so a priori $B$ might not be a normal toric variety (i.e. the polytope is associated not to $B$ but the normalization of $B$). But as David shows in his answer (and probably proved for a general torus invariant subspaces in the article he mentions in the comment), $B$ is indeed normal.

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See the discussion at mathoverflow.net/questions/37749/… for discussion of when blow ups are and are not normal. Karl S's example in that thread shows that blowing up a torus invariant subscheme can give a nonnormal result. I don't know an example where the locus being blown up is reduced and you get something non-normal, but I would guess that such examples exist. –  David Speyer Jul 14 '11 at 17:39
    
This way to get the polytope is very nice! Is there a place where I can read about it? Thanks! –  Enrique Jul 14 '11 at 19:04
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Since you already know how to blow up along either ${\mathbb P}^1$ individually, we can concentrate on what's happening nearby the intersection. Which means we can work affinely.

Then the polyhedron for ${\mathbb A}^3$ is the octant $({\mathbb R}_{\geq 0})^3$, and your two lines correspond to two of the three edges. To blow them up, take a carpenter's plane to those two edges, and shave them off exactly the same amount. The result will have an edge connecting two vertices, one of degree 3, one of degree 4 (the isolated singularity on the blowup).

In coordinates, the polyhedron is {$(x,y,z) : x,y,z \geq 0, x+y \geq 1, x+z \geq 1$}.

If you want the blowup of ${\mathbb P}^3$ not ${\mathbb A}^3$, also impose $x+y+z \leq N$ for some large $N$ (three will do).

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Thanks. Do you know if this works in general? (I mean, blowing up toric varieties coming from polytopes along a reduced union of invariant subvariaties corresponding to shaving off parts of the polytope "the same amount")? If so, is there a place where I can read about it? In this particular case, if you do not shave the edges off the same amount, then I think you get one of the iterated blow-ups. Thank you! –  Enrique Jul 20 '11 at 19:53
    
I don't know either a precise general statement or a place to read about it, alas. Another example worth knowing about is blowing up $\langle x,y^2 \rangle$ in the plane, which snips off the vertex of the quadrant, but unevenly, leaving an interval with slope $2$. –  Allen Knutson Jul 21 '11 at 6:46
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If you are interested not in the toric picture, but in the structure of the variety itself, the following approach is very useful. Note that the ideal of the pair of lines has the following resolution on $P^3$: $$ 0 \to O(-3) \to O(-2) \oplus O(-1) \to I \to 0. $$ Hence the graded algebra $\oplus I^k$ is the quotient of the symmetric algebra $\oplus S^k(O(-2) \oplus O(-1))$. This means that the blowup is a subscheme in $Proj(\oplus S^k(O(-2) \oplus O(-1))) = P_{P^3}(O(2) \oplus O(1))$. Moreover, the equation of the blowup in this projectivization of the vector bundle is given by the embedding $O(-3) \to O(-2) \oplus O(-1)$ in the resolution above. Namely, the map $O(-3) \to O(-2) \oplus O(-1)$ can be thought of as a global section of the line bundle $L\otimes \pi^*O(3)$, where $L$ is the Grothendieck relative $O(1)$ bundle on the projectivization and $\pi:P_{P^3}(O(2) \oplus O(1)) \to P^3$ is the projection.

So, we conclude that the blowup is the zero locus of a section of a line bundle $L\otimes \pi^*O(3)$ on $P_{P^3}(O(2) \oplus O(1))$.

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