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 ℙn, we cannot hope to construct the Chow ring to X. Even for the case of an elliptic curve E (a degree-3 hypersurface in ℙ2), the construction of A1(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 C), and the situation quickly gets worse for higher dimensions and degrees.
However, for classes on X which are obtained by restricting classes on ℙn to X, we can easily understand a great deal via the projection formula, which in this particular case tells us that if i:X →ℙn is the inclusion, then
So, if for example we are interested in calculating the degree of α|X, we can equivalently calculate the degree of α∩[X]. In this way we “push the problem forward” to ℙn.
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 ℙ3, we can compute:
i1 : P3 = flagBundle({1,3})
o1 = P3
o1 : a flag bundle with subquotient ranks {1, 3}
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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 , H ]
1,1 2,1 2,2 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
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i6 : TX = chern(T - NX) * X
o6 = 4H + 24H
2,1 2,3
QQ[][H , H , H , H ]
1,1 2,1 2,2 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
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i7 : integral TX -- The Euler characteristic of a quartic surface o7 = 24 |
This works because we have
More generally, we can compute the Euler characteristic of a degree-d hypersurface in ℙn 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
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i9 : eulerChar(4) -- The Euler characteristic of a degree-d hypersurface in P4
4 3 2
o9 = - d + 5d - 10d + 10d
o9 : QQ[d]
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i10 : sub(eulerChar(4),{d=>4/1}) -- The Euler characteristic of quartic threefold
o10 = -56
o10 : QQ
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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
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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]
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i13 : sub(middleBetti(4), {d => 5/1}) -- The middle Betti number of a quintic threefold
o13 = 204
o13 : QQ
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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
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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 d1, ..., dk in P = ℙn, then
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 ℙn:
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
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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
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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
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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
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i19 : sub(middleBetti(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5
o19 = 22
o19 : QQ
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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
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i21 : sub(eulerChar(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5
o21 = 24
o21 : QQ
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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
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Exercise 5.13: Calculate the Euler characteristic of a smooth hypersurface of bidegree (a,b) in ℙm ×ℙn
Working on ℙm ×ℙn 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 ℙm. So, for example, to build ℙ2 ×ℙ3:
i23 : P2 = flagBundle({1,2})
o23 = P2
o23 : a flag bundle with subquotient ranks {1..2}
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i24 : P2xP3 = flagBundle({1,3},P2,VariableNames => K)
o24 = P2xP3
o24 : a flag bundle with subquotient ranks {1, 3}
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i25 : intersectionRing(P2xP3)
QQ[][H , H , H ]
1,1 2,1 2,2
---------------------------------------------[K , K , K , K ]
(- H - H , - H H - H , -H H ) 1,1 2,1 2,2 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
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Note that if we didn’t use the VariableNames options this ring would be horrible to look at, since classes pulled back from ℙ2 and ℙ3 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
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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]
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i28 : sub(eulerChar(2,3),{a=>1,b=>0/1}) -- is P1xP3, should be 8 by Kunneth
o28 = 8
o28 : QQ
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i29 : sub(eulerChar(1,1),{a=>1,b=>1/1}) -- a conic in P2, should be 2
o29 = 2
o29 : QQ
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i30 : sub(eulerChar(1,1),{a=>2,b=>1/1}) -- a twisted cubic, should be 2
o30 = 2
o30 : QQ
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