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Book3264Examples > Intersection Theory Section 5.4.3

Intersection Theory Section 5.4.3 -- Tangent bundles to hypersurfaces

Subsection 5.4.3

To treat tangent bundles to hypersurfaces in Schubert2, we have to be a little more careful. If $X$ is a hypersurface in ${\mathbb P}^n$, we cannot hope to construct the Chow ring to $X$. Even for the case of an elliptic curve $E$ (a degree-3 hypersurface in $\mathbb{P}^2$), the construction of $A^1(E)$ amounts to completely understanding the group law on $E$ and all points of $E$ (so in particular, this ring is never finitely generated over $\mathbb{C}$), and the situation quickly gets worse for higher dimensions and degrees.

However, for classes on $X$ which are obtained by restricting classes on ${\mathbb P}^n$ to $X$, we can easily understand a great deal via the projection formula, which in this particular case tells us that if $i:X \rightarrow {\mathbb P}^n$ is the inclusion, then

$$i_*(\alpha|_X) = \alpha \cap [X]$$

So, if for example we are interested in calculating the degree of $\alpha|_X$, we can equivalently calculate the degree of $\alpha \cap [X]$. In this way we ``push the problem forward'' to ${\mathbb P}^n$.

As an example, if we want to calculate the degree of the top chern class of the tangent bundle to a hypersurface $X$ of degree $4$ in ${\mathbb P}^3$, we can compute:

i1 : P3 = flagBundle({1,3})

o1 = P3

o1 : a flag bundle with subquotient ranks {1, 3}
i2 : O1 = dual(P3.Bundles#0)

o2 = O1

o2 : an abstract sheaf of rank 1 on P3
i3 : T = tangentBundle(P3)

o3 = T

o3 : an abstract sheaf of rank 3 on P3
i4 : NX = O1^**4 -- the fourth tensor power of O(1), i.e. O(4)

o4 = NX

o4 : an abstract sheaf of rank 1 on P3
i5 : X = chern(1,NX) -- the fundamental class [X] of X

o5 = 4H
       2,1

                          QQ[][H   , H   ..H   ]
                                1,1   2,1   2,3
o5 : ----------------------------------------------------------------
     (- H    - H   , - H   H    - H   , - H   H    - H   , -H   H   )
         1,1    2,1     1,1 2,1    2,2     1,1 2,2    2,3    1,1 2,3
i6 : TX = chern(T - NX) * X

o6 = 4H    + 24H
       2,1      2,3

                          QQ[][H   , H   ..H   ]
                                1,1   2,1   2,3
o6 : ----------------------------------------------------------------
     (- H    - H   , - H   H    - H   , - H   H    - H   , -H   H   )
         1,1    2,1     1,1 2,1    2,2     1,1 2,2    2,3    1,1 2,3
i7 : integral TX -- The Euler characteristic of a quartic surface

o7 = 24

This works because we have $$c(T_X) = \frac{c(T_P)|_X}{c(N_X)} = \frac{c(T_P)}{c(O_P(X))}|_X.$$

More generally, we can compute the Euler characteristic of a degree-$d$ hypersurface in $\mathbb{P}^n$ as in the book. We can even compute a closed formula for all $d$ and fixed $n$ using base.

i8 : eulerChar = n -> (
          S := base d;
          Pn := flagBundle({1,n},S);
          TPn := tangentBundle(Pn);
          O1 := dual(Pn.Bundles#0);
          NX := O1^**d;
          TX := chern(TPn - NX)*chern(1,NX);
          integral TX)

o8 = eulerChar

o8 : FunctionClosure
i9 : eulerChar(4) -- The Euler characteristic of a degree-d hypersurface in P4

        4     3      2
o9 = - d  + 5d  - 10d  + 10d

o9 : QQ[d]
i10 : sub(eulerChar(4),{d=>4/1}) -- The Euler characteristic of quartic threefold

o10 = -56

o10 : QQ

And now we can similarly calculate a formula for the middle Betti number of such a hypersurface:

i11 : middleBetti = n -> (
           euC := eulerChar(n);
           ((-1)^(n-1)) * (euC - 2*ceiling((n-1)/2)))

o11 = middleBetti

o11 : FunctionClosure
i12 : middleBetti(4) -- The middle Betti number of a degree-d hypersurface in P4

       4     3      2
o12 = d  - 5d  + 10d  - 10d + 4

o12 : QQ[d]
i13 : sub(middleBetti(4), {d => 5/1}) -- The middle Betti number of a quintic threefold

o13 = 204

o13 : QQ

Using this, we easily reproduce the table given in the text:

i14 : for n from 3 to 5 do (
           for e from 2 to 5 do (
                euC := sub(eulerChar(n),{d=>e/1});
                midB := sub(middleBetti(n),{d=>e/1});
                << "n: " << n << " d: " << e << " chi: " << euC << " middle Betti: " << midB << endl))
n: 3 d: 2 chi: 4 middle Betti: 2
n: 3 d: 3 chi: 9 middle Betti: 7
n: 3 d: 4 chi: 24 middle Betti: 22
n: 3 d: 5 chi: 55 middle Betti: 53
n: 4 d: 2 chi: 4 middle Betti: 0
n: 4 d: 3 chi: -6 middle Betti: 10
n: 4 d: 4 chi: -56 middle Betti: 60
n: 4 d: 5 chi: -200 middle Betti: 204
n: 5 d: 2 chi: 6 middle Betti: 2
n: 5 d: 3 chi: 27 middle Betti: 23
n: 5 d: 4 chi: 188 middle Betti: 184
n: 5 d: 5 chi: 825 middle Betti: 821

Exercise 5.11: Betti numbers of smooth complete intersections

In the same way as for hypersurfaces, we compute that if $X$ is a complete intersection of hypersurfaces of degrees $d_1, \ldots, d_k$ in $P = {\mathbb P}^n$, then $$c(T_X) = c(T_P)/(\prod_{i=1}^k c(O_P(d_i)))|_X$$ We can use then Schubert2 to produce a closed-form formula for the degree of the top Chern class of the tangent bundle to a complete intersection of $k$ hypersurfaces in ${\mathbb P}^n$:

i15 : eulerChar = (n,k) -> (
           S := base(e_1 .. e_k);
           Pn := flagBundle({1,n},S);
           TPn := tangentBundle(Pn);
           O1 := dual(Pn.Bundles#0);
           N := sum(1..k, i-> O1^**(e_i)); --the denominator in the above formula
           X := product(1..k, i->chern(1,O1^**(e_i))); --fundamental class of X
           TX := chern(TPn - N) * X;
           integral TX)

o15 = eulerChar

o15 : FunctionClosure
i16 : eulerChar(4,2) -- Euler char of a complete intersection surface in P4

       3      2 2      3     2         2
o16 = e e  + e e  + e e  - 5e e  - 5e e  + 10e e
       1 2    1 2    1 2     1 2     1 2      1 2

o16 : QQ[e ..e ]
          1   2

And from here we can compute the middle Betti numbers:

i17 : middleBetti = (n,k) -> (
           euC := eulerChar(n,k);
           ((-1)^(n-k)) * (euC - 2*ceiling((n-k)/2)))

o17 = middleBetti

o17 : FunctionClosure

Now our particular problem is easy:

i18 : sub(middleBetti(4,2),{e_1=>2,e_2=>3/1}) -- complete intersection of a quadric and cubic in P4

o18 = 22

o18 : QQ
i19 : sub(middleBetti(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5

o19 = 22

o19 : QQ

For good measure, we'll also compute the Euler characteristics:

i20 : sub(eulerChar(4,2),{e_1=>2,e_2=>3/1}) -- complete intersection of a quadric and cubic in P4

o20 = 24

o20 : QQ
i21 : sub(eulerChar(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5

o21 = 24

o21 : QQ

Exercise 5.12: Betti numbers of the quadric line complex

The only interesting Betti number is the middle one, which we compute immediately from the above:

i22 : sub(middleBetti(5,2),{e_1=>2,e_2=>2/1})

o22 = 4

o22 : QQ

Exercise 5.13: Calculate the Euler characteristic of a smooth hypersurface of bidegree $(a,b)$ in ${\mathbb P}^m \times {\mathbb P}^n$

Working on ${\mathbb P}^m \times {\mathbb P}^n$ in Schubert2 is easy using relative flag bundles (or relative projective spaces): this space is the same as the projectivization of a trivial rank-$n+1$ bundle on ${\mathbb P}^m$. So, for example, to build ${\mathbb P}^2 \times {\mathbb P}^3$:

i23 : P2 = flagBundle({1,2})

o23 = P2

o23 : a flag bundle with subquotient ranks {1..2}
i24 : P2xP3 = flagBundle({1,3},P2,VariableNames => K)

o24 = P2xP3

o24 : a flag bundle with subquotient ranks {1, 3}
i25 : intersectionRing(P2xP3)

                   QQ[][H   , H   ..H   ]
                         1,1   2,1   2,2
       ---------------------------------------------[K   , K   ..K   ]
       (- H    - H   , - H   H    - H   , -H   H   )  1,1   2,1   2,3
           1,1    2,1     1,1 2,1    2,2    1,1 2,2
o25 = ----------------------------------------------------------------
      (- K    - K   , - K   K    - K   , - K   K    - K   , -K   K   )
          1,1    2,1     1,1 2,1    2,2     1,1 2,2    2,3    1,1 2,3

o25 : QuotientRing

Note that if we didn't use the VariableNames options this ring would be horrible to look at, since classes pulled back from ${\mathbb P}^2$ and ${\mathbb P}^3$ would both be named $H$.

We can calculate a closed-form formula for the Euler characteristic of a smooth hypersurface of bidegree $(a,b)$ once we have fixed $m$ and $n$, but we cannot use $m$ and $n$ as base parameters themselves.

i26 : eulerChar = (n,m) -> (
           S := base(a,b);
           Pn := flagBundle({1,n},S);
           PnxPm := flagBundle({1,m},Pn);
           T := tangentBundle(PnxPm);
           O1Pn := dual(Pn.Bundles#0);
           f := PnxPm / Pn; -- the first projection map from P2xP3 to P2
           O10 := f^* O1Pn; -- we pull back O_P2(1) to get O(1,0)
           O01 := dual(PnxPm.Bundles#0); -- O(0,1)
           NX := (O10^**a)**(O01^**b); -- O(a,b)
           X := chern(1,NX); -- Chow class of divisor of type (a,b)
           TX := chern(T - NX) * X; -- pushed-forward total chern class of tangent bundle to X
           integral TX) -- chi of a smooth hypersurface of bidegree (a,b) in PnxPm

o26 = eulerChar

o26 : FunctionClosure
i27 : eulerChar(4,4) -- chi of a smooth hypersurface of bidegree (a,b) in P4xP4

           4 4       4 3       3 4       4 2       3 3       2 4      4   
o27 = - 70a b  + 175a b  + 175a b  - 150a b  - 500a b  - 150a b  + 50a b +
      -----------------------------------------------------------------------
          3 2       2 3        4     4       3        2 2         3     4  
      500a b  + 500a b  + 50a*b  - 5a  - 200a b - 600a b  - 200a*b  - 5b  +
      -----------------------------------------------------------------------
         3       2          2      3      2               2
      25a  + 300a b + 300a*b  + 25b  - 50a  - 200a*b - 50b  + 50a + 50b

o27 : QQ[a..b]
i28 : sub(eulerChar(2,3),{a=>1,b=>0/1}) -- is P1xP3, should be 8 by Kunneth

o28 = 8

o28 : QQ
i29 : sub(eulerChar(1,1),{a=>1,b=>1/1}) -- a conic in P2, should be 2

o29 = 2

o29 : QQ
i30 : sub(eulerChar(1,1),{a=>2,b=>1/1}) -- a twisted cubic, should be 2

o30 = 2

o30 : QQ