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I am interested in computing the (anti)-canonical class of the (total space of the) projective completion of the tautological bundle over $P^1\times P^1$. That is, the canonical class of $\mathbb P_{P^1\times P^1}(J \oplus \mathscr O)$, where $J$ is the tautological line bundle on $P^1\times P^1$.

I believe this can be done by computing the fan of the toric variety and summing the classes of the orbit closures of the one-skeleton? I was hoping for insight into perhaps a slicker/less cumbersome way of approaching this computation.

Thanks in advance.

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2 Answers 2

up vote 14 down vote accepted

Why not use the Leray-Hirsch theorem? That says that the integral cohomology ring of a projectivized rank $n$ vector bundle $\pi: PE \to B$ is generated, as an algebra over the cohomology of the base $B$, by the first Chern class $h$ of the relative $O(1)$, with relation $h^n + c_1 h^{n-1} + \dots + c_n$, where $c_i$ is the $i$th Chern class of $E$.

On the other hand, you have a short exact sequence $$0 \longrightarrow T_\pi \longrightarrow T_{PE} \longrightarrow \pi^* T_B \longrightarrow 0,$$ where $T_\pi$ is the relative tangent bundle. For a projectivized vector bundle, $T_\pi = \mbox{Hom }(O(-1),E/O(-1))$ (this is a special case of the formula for the tangent bundle of a Grassmannian). Note that $\mbox{Hom }(O(-1),E/O(-1)) \cong E(1)/O$, so $\Lambda^{n-1} T_\pi \cong (\Lambda^n E)(n)$ canonically.

Hence if $B$ is smooth of dimension $b$, $$K_{PE} = \Lambda^{b+n-1} T^*_{PE} \cong \Lambda^b \pi^* T^*_B \otimes \Lambda^{n-1} T^*_\pi \cong \pi^* K_B \otimes \pi^* (\Lambda^n E^*)(-n).$$

Taking Chern classes, if $k_{PE} := c_1(K_{PE})$ and $k_b := c_1(K_B)$, we get $$k_{PE} = \pi^* k_B - \pi^* c_1(E) - nh.$$

As a check, note that if we replace $E$ by $E \otimes L$ for some line bundle $L$, then $PE$ is unchanged, but $c_1(E)$ and $h$ are increased and decreased by $nc_1(L)$ and $c_1(L)$ respectively, so the formula above is unchanged.

In your case, the tautological line bundle $J \to P^1 \times P^1$ has $c_1(J) = -h_1-h_2$ where $h_i$ generate the cohomology of the two $P^1$ factors, so the canonical class, in terms of this basis, is $-2h_1-2h_2+h_1+h_2-2h = -h_1-h_2-2h$. Warning: as we saw, $h$ is NOT unchanged if we replace $E$ by $E \otimes L$! So if you replace $J \oplus O$ by, say, $O \oplus J^{-1}$, you will get an apparently different answer...

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Thats exactly what I was looking for. Thank you. –  Dhruv Aug 23 '11 at 23:46
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@Michael Thaddeus. Where is the Leray-Hirsch theorem actually used here ? Your computation from the second paragraph onwards gives the canonical class, but where do you need the structure of the cohomology of a projectivized bundle ? (other than for fixing ideas) –  Damian Rössler Aug 24 '11 at 15:05
    
It's not, except to describe the ring in which the expression for $k_{PE}$ lies. Since this is in $H^2$, I could have simply said that $H^2(PE,Z)$ is the abelian group freely generated by $H^2(B,Z)$ and $h$. I take your point. Anyway, the expression for $K_{PE}$ is what really matters, and for that, of course, none of this is necessary. –  Michael Thaddeus Aug 24 '11 at 16:45

I like the way you asked to avoid. Forgive me if I describe it in polytope rather than fan language.

Step 1: ${\mathbb P}^1 \times {\mathbb P}^1$'s polytope is a square (or any rectangle). The four edges, taken clockwise, correspond to the ${\mathbb P}^1$s giving the classes $h_1,h_2,h_1,h_2$ Michael mentions. (EDIT: I had signs there before, by overthinking the Danilov relations.)

I can only guess that by "tautological line bundle on ${\mathbb P}^1 \times {\mathbb P}^1$ you mean ${\mathcal O}(-1) \boxtimes {\mathcal O}(-1)$.

If we blow down that ${\mathbb P}^1 \times {\mathbb P}^1$, we get the affine cone over the Segre embedding of ${\mathbb P}^1 \times {\mathbb P}^1$. The polyhedron of that is also a cone, on a square.

Step 2: Blow the singular point back up, which corresponds to cutting the corner off that cone, leaving a square. So far we have an unbounded polytope that retracts to the square, just as the line bundle retracts to ${\mathbb P}^1 \times {\mathbb P}^1$.

Step 3: Projectively complete. This corresponds to bounding the cone. Combinatorially, we now have a square-based pyramid with the top corner cut off, so there's a big square on the bottom (whose class is Michael's $h$) and a little square on the top.

Step 4: Take the anticanonical class. On any toric variety, the boundary of the polytope defines an anticanonical divisor.

So far our anticanonical class is the bottom square $h$ plus the top square plus the other four faces. To calculate the linear relations between them, one needs to be precise about the locations of the vertices. I have the bottom square at $(0,0), (2,0), (0,2), (2,2)$ with $z=0$ and the top one at $(0,0), (1,0), (0,1), (1,1)$ with $z=1$. The Danilov relations from the $z$-axis vector says $$ (-1) \text{bottom} + (+1) \text{top} + 0 \text{west} + 0 \text{south} + (+1) \text{north} + (+1)\text{east} = 0 $$ so the total of the faces is $2\text{bottom} + \text{south} + \text{west}$, matching Michael's $2h+h_1+h_2$.

(As it ought, since I learned at least some of this from him.)

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I find the adjunction formula easiest to remember in terms of the rule in Step 4. Let $\partial X$ denote the anticanonical class of $X$. Then adjunction in general says that if $\partial X = [D \cup E]$, then $\partial D = [D \cap E]$. Now picture $X$ as a polytope, $\partial X$ as its boundary, $D$ as one facet, and $E$ as the rest of the facets (a hemisphere decomposition, topologically). This says that $D\cap E$ is the boundary of $D$. –  Allen Knutson Aug 24 '11 at 4:39
    
Thats a great response, thank you sir. I wanted to understand it free from the toric machinery via something like Leray-Hirsch. But thanks again, largely for the sake of insight, but thanks nonetheless. One comment I will make, I think on P1xP1 if you take the four edges of the polytope clockwise I believe you should get h1, h2, h1, h2 (no minus signs). As a quick check, the anticanonical class of P1xP1 is 2(h1+h2). Doesnt change the answer though. –  Dhruv Aug 24 '11 at 5:59
    
Oh whoops, you're right about the signs of course. Fixed. –  Allen Knutson Aug 24 '11 at 12:38

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