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.


2 Answers 2


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...

  • $\begingroup$ Thats exactly what I was looking for. Thank you. $\endgroup$
    – Dhruv
    Aug 23, 2011 at 23:46
  • 1
    $\begingroup$ @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) $\endgroup$ Aug 24, 2011 at 15:05
  • $\begingroup$ 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. $\endgroup$ Aug 24, 2011 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.)

  • $\begingroup$ 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 $D$. $\endgroup$ Aug 24, 2011 at 4:39
  • $\begingroup$ 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. $\endgroup$
    – Dhruv
    Aug 24, 2011 at 5:59
  • $\begingroup$ Oh whoops, you're right about the signs of course. Fixed. $\endgroup$ Aug 24, 2011 at 12:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.