-- -*- coding: utf-8 -*-
--To do:
-- -Add functionality to multiply elements of the "Schubert Ring", perhaps by
-- overloading the * operator
newPackage(
"Book3264Examples",
Version => "0.1",
Date => "July 20, 2010",
Authors => {{Name => "Charley Crissman",
Email => "charleyc@math.berkeley.edu",
HomePage => "http://math.berkeley.edu/~charleyc/"}},
PackageExports => {"Schubert2", "SchurRings"},
Keywords => {"Intersection Theory"},
Headline => "examples to accompany the eponymous book by Eisenbud and Harris"
)
export {"grassmannian", "placeholderSchubertCycle", "diagrams", "placeholderToSchubertBasis"}
protect schuberttoh
protect htoschubert
protect intersectionmap
protect schubertring
grassmannian = method(TypicalValue => FlagBundle)
grassmannian(ZZ,ZZ) := (k,n) -> flagBundle({k,n-k}) --The grassmannian of k-dimensional subspaces of
--an n-dimensional space
placeholderSchubertCycle = method()
giambelli = (r,E,b) -> (
p := matrix for i from 0 to r-1 list for j from 0 to r-1 list chern(b#i-i+j,E);
if debugLevel > 15 then stderr << "giambelli : " << p << endl;
det p
)
placeholderSchubertCycle(List,FlagBundle) := (b,X) -> (
if #X.BundleRanks != 2 then error "expected a Grassmannian";
E := last X.Bundles;
r := rank E;
n := X.Rank;
r' := n-r;
if r' != #b then error("expected a list of length ", toString r');
for i from 0 to r'-1 do (
bi := b#i;
if not instance(bi,ZZ) or bi < 0 then error "expected a list of non-negative integers";
if i>0 and not (b#(i-1) >= bi) then error "expected a decreasing list of integers";
if not (bi <= r) then error("expected a list of integers bounded by ",toString(r));
);
giambelli(r',E,b))
diagrams = method()
diagrams(ZZ,ZZ) := (k,n) -> ( --diagrams {k>=a_1>=...>=a_n>=0}
if n==1 then apply(k+1, i->{i})
else flatten apply(k+1, i -> apply(diagrams(i,n-1), l -> flatten {i,l})))
diagrams(ZZ,ZZ,ZZ) := (k,n,d) -> (--partitions of d of above form
select(diagrams(k,n), i -> (sum(i) == d)))
placeholderToSchubertBasis = method()
placeholderToSchubertBasis(RingElement,FlagBundle) := (c,G) -> (
if #G.BundleRanks != 2 then error "expected a Grassmannian";
if not ring c === intersectionRing G then error "expected first input to be a ring element
in the intersection ring of the second input";
R := intersectionRing G;
B := intersectionRing (G.Base);
(k,q) := toSequence(G.BundleRanks);
P := diagrams(q,k);
M := apply(P, i-> placeholderSchubertCycle(i,G));
E := flatten entries basis(R);
local T';
if R.cache.?htoschubert then T' = R.cache.htoschubert else (
T := transpose matrix apply (M, i -> apply(E, j-> coefficient(j,i))); --matrix converting from schu-basis
--to h-basis
T' = T^-1; --matrix converting from h-basis to s-basis
R.cache.schuberttoh = T;
R.cache.htoschubert = T');
c2 := T'*(transpose matrix {apply (E, i-> coefficient(i,c))}); --c in the s-basis
local S;
if R.cache.?schubertring then S = R.cache.schubertring else (
--should maybe add option to choose variable name
s := getSymbol "s";
S = B[apply(P, i-> s_i)]; --poly ring with generators <=> schubert basis elts
R.cache.schubertring = S;
S.cache = new MutableHashTable;
S.cache.intersectionmap = map(R,S,M));
rez := (vars S)*(lift(c2,B));
rez_(0,0)
)
beginDocumentation()
doc ///
Key
placeholderToSchubertBasis
(placeholderToSchubertBasis,RingElement,FlagBundle)
Headline
Express cycles on G(k,n) in terms of the Schubert basis
Usage
placeholderToSchubertBasis(c,G)
Inputs
G:FlagBundle
Any grassmannian (i.e. one-step flag variety) of $k$-dimensional subspaces of a rank-$n$
bundle
c:RingElement
An element of the intersection ring of G
Outputs
:
An element $c'$ of a polynomial ring $B[s_\lambda]$ where $B$ is the base ring of G and
$\lambda$ runs over all diagrams in a $k\times n$ rectangle. The element $c'$ is the
representation of $c$ in terms of the Schubert basis of the intersection ring of G over B.
Description
Example
A = flagBundle({3,3},VariableNames => H)
S = A.Bundles#0
G = flagBundle({1,2},S,VariableNames => K)
c = H_(2,3)*((K_(2,1))^2) + H_(1,1)*K_(2,2)
placeholderToSchubertBasis(c,G)
///
doc ///
Key
diagrams
(diagrams,ZZ,ZZ)
Headline
Ferrers diagrams contained in a rectangle
Usage
diagrams(k,n)
Inputs
k:ZZ
maximum size of each entry in diagram
n:ZZ
number of entries in diagram
Outputs
:
a list of lists of integers $\{a_1, \dots, a_n\}$ such that
$k \geq a_1 \geq ... \geq a_n \geq 0$
///
doc ///
Key
(diagrams,ZZ,ZZ,ZZ)
Headline
Partitions contained in a rectangle
Usage
diagrams(k,n,d)
Inputs
k:ZZ
maximum size of each entry in partitions
n:ZZ
number of entries in partition
d:ZZ
number being partitioned
Outputs
:
a list of lists of integers $\{a_1, \dots, a_n\}$ such that
$k \geq a_1 \geq ... \geq a_n \geq 0$ and
$\sum_{i=1}^n a_i = d$
///
doc ///
Key
Book3264Examples
Headline
Examples using M2 and Schubert2 to do intersection theory
Description
Text
This package consists almost entirely of example code for the main text and exercises of the book
'3264 \& All That: Intersection Theory in Algebraic Geometry' by Eisenbud and Harris. Most of the
example code relies on the package @TO Schubert2@. The information in this package is best
accessed via the @TO help@ or @TO viewHelp@ commands.
Conventions in effect throughout:
-Blackboard bold is represented in code by doubling a letter, so for example {\tt G(2,4)} and
{\tt GG(1,3)} are both the Grassmannian of lines in ${\mathbb P}^3$.
Subnodes
:Chapter 4
"Intersection Theory Section 4.1"
"Intersection Theory Section 4.2"
"Intersection Theory Section 4.3"
:Chapter 5
"Intersection Theory Section 5.2"
"Intersection Theory Section 5.3"
"Intersection Theory Section 5.4.1-2"
"Intersection Theory Section 5.4.3"
"Intersection Theory Section 5.4.4"
///
doc ///
Key
"Intersection Theory Section 4.1"
Headline
The coordinate ring of the Grassmannian
Description
Text
Subsection 4.1.1
We can use Macaulay2 to build the coordinate ring of $G(k,n)$
using the Plücker embedding. Exercise 4.4 is the simplest
interesting case, $G(2,4) = {\mathbb G}(1,3)$. We'll start there before
writing more general code for this.
Exercise 4.4:
Example
kk = ZZ/32003 --Our base field
R = kk[x_1 .. x_8]
M = genericMatrix(R,x_1,2,4) -- A generic 2x4 matrix in the x_i
I = minors(2,M) -- The ideal of 2x2 minors of M
P5 = kk[p_0 .. p_5] -- The coordinate ring of PP^5
f = map(R,P5, gens I) -- The Plücker map for GG(1,3)
J = saturate ker f -- The ideal of GG(1,3) in PP^3
Text
We see that the ideal $J$ of ${\mathbb G}(1,3)$ in ${\mathbb P}^5$ is indeed generated by the
single relation given in the text.
More generally, we can build $G(k,n)$ in its Plücker embedding for any $n$ and $k$:
Example
kk = ZZ/32003
pluckerIdeal = (k,n) -> (
assert (k <= n);
N := k*n; --number of variables in our generic matrix
R := kk[x_1 .. x_N];
M := genericMatrix(R,x_1,k,n); --the generic k-by-n matrix
s := binomial(n,k) - 1; --the dimension of PP(Wedge^k(kk^n))
Ps = kk[p_0 .. p_s];
f := map(R,Ps, gens minors(k,M)); --the Plücker map
J = saturate ker f) --the kernel of the Plücker map is the ideal we want
Text
Now we can do Exercise 4.4 in one line:
Example
pluckerIdeal(2,4)
Text
The reader is invited to try running {\tt pluckerIdeal(4,7)}. On our machine, this computation
had not terminated after 15 minutes of runtime.
We can do a little better by using the built-in function @TO Grassmannian@, which computes
the Plücker ideal in a more efficient way:
Example
Grassmannian(1,4)
Text
The reader should try running {\tt Grassmannian(4,7)} (which runs very quickly) to see just
how large this ideal is. Running {\tt Grasmannian(4,10)}, on the other hand, is likely to
never terminate.
Given how large these rings are and how difficult it is to compute in them, we need to
simplify our computational system if we want to get answers to harder questions.
Subsection 4.1.3
It is possible to use Macaulay2 to build the universal sub- and quotient- bundles of
the Grassmannian using explicit equations. However, as above, computations very
quickly become intractable. We need some simplifications if we hope to compute anything.
Schubert2 makes two major simplifications that allow us to do intersection theory
with computers:
1) Varieties are replaced by their Chow rings
2) Bundles are replaced by their (total) Chern classes (see Ch. 5)
Here is Schubert code that will build the Grassmannian and its universal sub- and quotient
bundles.
Example
grass = (k,n) -> flagBundle({k,n-k}) --In Schubert, we build Grassmannians as special cases
G = grass(2,4) -- Our favorite GG(1,3)
(S,Q) = G.Bundles -- S is the universal subbundle, Q is the universal quotient bundle
S -- Schubert tells us that S is an abstract sheaf of rank 2
Q -- And so is Q.
///
doc ///
Key
"Intersection Theory Section 4.2"
Headline
The Chow ring of GG(1,3)
Description
Text
Subsection 4.2.2
Schubert2 identifies ${\mathbb G}(1,3)$ with its Chow ring. We can see this directly using the
command {\tt intersectionRing}.
Example
G = flagBundle({2,2})
intersectionRing G
Text
The generators $H_{i,j}$ of the above ring are defined by the formula $H_{i,j} = c_j(B_i)$ where
$B_i$ is the {\tt i}-th bundle in the list G.Bundles (numbered starting with 1) and $c_j$ is the
{\tt j}-th Chern class, defined in Ch. 5. The relationship with the Schubert classes on
${\mathbb G}(1,3)$ is as follows:
$H_{2,1} = \sigma_1$ @BR{}@ $H_{2,2} = \sigma_2$ @BR{}@ $H_{1,1} = -\sigma_1$ @BR{}@
$H_{1,2} = \sigma_{1,1}$
The Schubert classes can also be accessed directly using the {\tt placeholderSchubertCycle}
command -- see Section 4.3 for details.
As an example, we can compute $(\sigma_1)^2$:
Example
sigma_1 = H_(2,1)
c = (sigma_1)^2
Text
Oops! This just gave us $H_{2,1}^2$ back! Schubert2 actually uses $\sigma_1^2$ and
$\sigma_{1,1}$ as its "preferred basis" for the codimension-2 part of the Chow ring of
${\mathbb G}(1,3)$. To convert to the Schubert basis, we use the function
@TO placeholderToSchubertBasis@:
Example
placeholderToSchubertBasis(c,G)
Text
We recover the formula of Theorem 4.13: $\sigma_1^2 = \sigma_2 + \sigma_{1,1}$.
Subsection 4.2.4
How many lines in ${\mathbb P}^3$ meet four general lines?
After phrasing the problem in terms of Schubert calculus, this is easy to calculate both by hand
and in Schubert2:
Example
sigma_1 = H_(2,1)
integral (sigma_1)^4
Text
The command {\tt integral} here returns the degree of the zero-cycle $(\sigma_1)^4$, which is
the number we want (namely, 2).
Lines meeting a curve
We can easily build a function which, given the degree $d$ of a space curve $C$, returns the
cycle of lines in ${\mathbb P}^3$ meeting $C$:
Example
sigma_1 = H_(2,1)
linesMeetingCurve = d -> d*sigma_1
Text
And now we can calculate, for example, how many lines meet four general conics:
Example
integral (linesMeetingCurve(2))^4
Text
But we really want to answer the question once and for all: how many lines meet four general
curves of degree $d$? To do this, we use the @TO base@ command, which allows us auxiliary
parameters:
Example
S = base d --Our base variety, with one "auxiliary parameter" d
G' = flagBundle({2,2},S,VariableNames => K) --GG(1,3) with our extra parameter
intersectionRing G' --note the additional parameter d
sigma_1 = K_(2,1)
linesmeetingcurve = d*sigma_1
integral linesmeetingcurve^4
Text
And we get back the answer $2d^4$, solving the problem once and for all.
Chords to a Space Curve
For each $d$ and $g$ we build the cycle in ${\mathbb G}(1,3)$ of lines secant
to a general curve of degree d and genus g:
Example
S = base(g,d') --We use d' to avoid the d from the last example
G'' = flagBundle({2,2},S,VariableNames => L)
sigma_2 = L_(2,2)
sigma_(1,1) = L_(1,2)
cycleofchords = ((d'-1)*(d'-2)/2 - g)*sigma_2 + (d'*(d'-1)/2)*sigma_(1,1)
Text
The keynote question was: how many lines are secant to two general twisted cubics? But
we can do better, and answer the question: how many lines are secant to two general curves
of degree $d$ and genus $d$?
Example
chordstotwocurves = integral cycleofchords^2
Text
Now if we want to answer our specific question, we just substitute in the desired values
for $d$ and $g$:
Example
sub(chordstotwocurves, {d' => 3, g => 0/1})
Text
WARNING: because of some ugly M2 design decisions, if you don't make at least one of $d'$ or
$g$ a rational number, this substitute will return the wrong answer! Hopefully this design
will be changed in the future.
Exercise 4.25 (a):
If $C$ is a smooth, nondegenerate space curve and $L$ and $M$ are general lines in
${\mathbb P}^3$, how many chords to $C$ meet both $L$ and $M$? Using our work above,
we immediately compute:
Example
sigma_1 = L_(2,1)
integral (cycleofchords*(sigma_1)^2)
Text
Tangent Lines to a Surface
Exercise 4.28:
Using our Grassmannian {\tt G'} with an extra base parameter $d$, we build the cycle
of tangent lines to a general surface of degree $d$:
Example
sigma_1 = K_(2,1)
tangentcycle = d*(d-1)*sigma_1
Text
Now we can compute the number of lines tangent to four general surfaces of degree $d$:
Example
tangentlines = integral tangentcycle^4
Text
In particular, we calculate the number of lines tangent to four general quadric surfaces:
Example
sub(tangentlines, d => 2/1)
///
doc ///
Key
"Intersection Theory Section 4.3"
Headline
Schubert calculus in general
Description
Text
Subsection 4.3.1
We build arbitrary Schubert cycles using the command {\tt placeholderSchubertCycle}.
For example, on ${\mathbb G}(2,4)$, we can build the cycle $\sigma_{2,1,1}$ as follows:
Example
G24 = flagBundle({3,2})
sigma_(2,1,1) = placeholderSchubertCycle({2,1,1},G24)
Text
Subsection 4.3.2
Exercise 4.34
How many lines meet 6 general 2-planes in ${\mathbb P}^4$?
The cycle of lines meeting a 2-plane in the Grassmannian ${\mathbb G}(1,4)$ is the Schubert
cycle $\sigma_1$, so the number of lines meeting 6 general 2-planes is the degree of
$(\sigma_1)^6$:
Example
G14 = flagBundle({2,3})
sigma_1 = placeholderSchubertCycle({1,0},G14)
integral (sigma_1)^6
Text
Note that this is the degree of ${\mathbb G}(1,4)$ in the Plücker embedding, since $\sigma_1$
is the hyperplane class.
Exercise 4.36 (a)
How many lines meet four general $k$-planes in ${\mathbb P}^{2k+1}$?
The cycles of lines meeting a $k$-plane in ${\mathbb G}(1,2k+1)$ is the Schubert cycle
$\sigma_k$. We can build a function that calculates this value for any $k$, but we
cannot use $k$ as a base parameter, since we need to build a different Grassmannian and
Schubert cycle for each $k$.
Example
numOfLines = k -> (
G := flagBundle({2,2*k});
sigma := placeholderSchubertCycle({k,0}, G);
integral sigma^4)
Text
Now we can calculate to our hearts' content:
Example
for k from 1 to 5 do (
<< numOfLines(k) << " lines meet four " << k << "-planes in P" << 2*k+1 << "\n")
Text
Calculations slow down pretty quickly as $k$ gets large (the bottleneck is building the
Chow ring), but we suspect the reader will have guessed the correct formula from the
above data.
Linear Spaces on Quadrics
Exercise 4.43
A 2-plane in ${\mathbb P}^6$ is the same as a 3-plane in a 7-dimensional space. According
to Proposition 4.42, the cycle of 3-planes contained in the zero-locus of a nondegenerate
quadratic form on a 7-dimensional space is $2^3\sigma_{3,2,1}$ in $G(3,7)$. Hence we
calculate:
Example
G37 = flagBundle({3,4})
A37 = intersectionRing G37
sigma = 8*placeholderSchubertCycle({3,2,1},G37)
integral sigma^2
Text
More generally, we can ask: given 2 general quadrics in ${\mathbb P}^{2k+2}$, how many
$k$-planes are contained in their intersection? We calculate:
Example
numOfPlanes = k -> (
G:= flagBundle({k+1,k+2});
schubertlist := apply(k+1,i-> k+1-i); --the list {k+1,k,...,1}
sigma := (2^(k+1))*placeholderSchubertCycle(schubertlist, G);
integral sigma^2)
numOfPlanes(2) --This was Exercise 4.43
for k from 2 to 4 do (
<< numOfPlanes(k) << " " << k << "-planes in two quadrics in P" << 2*k+2 <<"\n")
Text
Exercise 4.44:
Compute $\sigma_{2,1}^2$ in the Chow ring of $G(3,6)$.
This is easy with the function @TO placeholderToSchubertBasis@, which we already saw in
@TO "Intersection Theory Section 4.2"@:
Example
G36 = flagBundle({3,3})
c = placeholderSchubertCycle({2,1,0},G36)
placeholderToSchubertBasis(c^2,G36)
Text
We see that $\sigma_{3,2,1}$ occurs with coefficient $2$ in $\sigma_{2,1}^2$.
///
doc ///
Key
"Intersection Theory Section 5.2"
Headline
Basics of vector bundles and Chern classes
Description
Text
In Schubert2, a vector bundle (or more generally, an @TO AbstractSheaf@) is given by two pieces
of data: its Chern classes and its rank. Schubert2 has many built-in bundles for common
varieties. For example, a Grassmannian {\tt G} comes with its universal subbundle and
quotient bundle stored in {\tt G.Bundles}:
Example
G = flagBundle({2,3})
(S,Q) = G.Bundles
S
Q
Text
The Chern classes of a vector bundle are accessed using the @TO chern@ command:
Example
chern(1,Q) -- The first Chern class of Q
chern Q -- The total Chern class of Q, defined as the sum of the Chern classes of Q.
Text
If we want to specify a bundle directly by its Chern classes, we can use the
@TO abstractSheaf@ command:
Example
Q = abstractSheaf(G,ChernClass=>1+H_(2,1)+H_(2,2)+H_(2,3),Rank=>3)
chern Q
///
doc ///
Key
"Intersection Theory Section 5.3"
Headline
Operations on vector bundles
Description
Text
Schubert2 has all of the basic operations on vector bundles and their Chern classes built in.
A full list of all of the available operations can be found in the documentation for
@TO AbstractSheaf@. A few examples:
Direct Sums:
Example
GG24 = flagBundle({3,2})
(S,Q) = GG24.Bundles
B1 = S + Q --direct sum of S+Q
chern B1
Text
Note that the Chern class of $S+Q$ is the same as that of the trivial bundle, since $S$ and $Q$
fit into an exact sequence whose middle term is trivial (see Prop 5.5).
Tensor Products:
Example
B2 = S ** Q --tensor product of S and Q
chern B2
Text
Duals:
Example
B3 = dual(S) ** Q
chern B3
Text
Note that {\tt B3} is the tangent bundle to $\mathbb{G}(2,4)$.
Pullbacks:
Currently Schubert2 has few morphisms implemented, but given a morphism of abstract varieties,
it is easy to pull back vector bundles:
Example
GG13 = flagBundle({2,2})
f = GG13 / point -- The structure map of G13
B = abstractSheaf(point,Rank=>2) -- The trivial vector bundle of rank 2 over point
f^* B --Pulls back to a trivial bundle of rank 2 on G13
///
doc ///
Key
"Intersection Theory Section 5.4.1-2"
Headline
Chern class computations on projective space
Description
Text
Subsection 5.4.1 - Universal bundles on projective space
We have two different methods in Schubert2 for producing projective spaces. We have already
seen one method: build $\mathbb{P}^n$ as a Grassmannian:
Example
P3 = flagBundle({1,3})
(S,Q) = P3.Bundles
Text
In this setting, the the bundle $O(1)$ is the dual of the universal subbundle
$S$.
Example
O1 = dual(S)
chern O1
Text
Now, Schubert2 also comes with a built-in function @TO abstractProjectiveSpace@ for making projective
spaces. Using {/tt abstractProjectiveSpace} to build $\mathbb{P}^n$ is nice, because the resulting
Chow ring is presented as a truncated polynomial ring in one variable, rather than as a ring
with $n+1$ generators. {\bf But, be careful}: this built-in actually produces the projective
space of 1-{\em quotients}. For example:
Example
P3' = abstractProjectiveSpace 3
(S',Q') = P3'.Bundles
chern S'
chern Q' -- Q' is O(1) on P3'
Text
For the rest of this section, we will use the {\tt flagBundle} method to produce $\mathbb{P}^n$,
in order to be consistent with the choices in the book.
Subsection 5.4.2
The tangent bundle to projective space comes built-in in Schubert2. It can be accessed
via the @TO tangentBundle@ method:
Example
T = tangentBundle(P3)
chern T
Text
We can also produce the tangent bundle to $\mathbb{P}^n$ ourselves by using the Euler exact
sequence:
Example
TP3 = (4 * O1) - 1
chern T == chern TP3
rank T == rank TP3
Text
Note how Schubert2 treats integers in a bundle computation as copies of a trivial bundle. See
@TO "AbstractSheaf * AbstractSheaf"@ and @TO "AbstractSheaf - AbstractSheaf"@, for example,
for more information.
///
doc ///
Key
"Intersection Theory Section 5.4.3"
Headline
Tangent bundles to hypersurfaces
Description
Text
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:
Example
P3 = flagBundle({1,3})
O1 = dual(P3.Bundles#0)
T = tangentBundle(P3)
NX = O1^**4 -- the fourth tensor power of O(1), i.e. O(4)
X = chern(1,NX) -- the fundamental class [X] of X
TX = chern(T - NX) * X
integral TX -- The Euler characteristic of a quartic surface
Text
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 @TO base@.
Example
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)
eulerChar(4) -- The Euler characteristic of a degree-d hypersurface in P4
sub(eulerChar(4),{d=>4/1}) -- The Euler characteristic of quartic threefold
Text
And now we can similarly calculate a formula for the middle Betti number of such a
hypersurface:
Example
middleBetti = n -> (
euC := eulerChar(n);
((-1)^(n-1)) * (euC - 2*ceiling((n-1)/2)))
middleBetti(4) -- The middle Betti number of a degree-d hypersurface in P4
sub(middleBetti(4), {d => 5/1}) -- The middle Betti number of a quintic threefold
Text
Using this, we easily reproduce the table given in the text:
Example
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))
Text
Exercise 5.11: {\bf 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 ${\mathbb P}^n$:
Example
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)
eulerChar(4,2) -- Euler char of a complete intersection surface in P4
Text
And from here we can compute the middle Betti numbers:
Example
middleBetti = (n,k) -> (
euC := eulerChar(n,k);
((-1)^(n-k)) * (euC - 2*ceiling((n-k)/2)))
Text
Now our particular problem is easy:
Example
sub(middleBetti(4,2),{e_1=>2,e_2=>3/1}) -- complete intersection of a quadric and cubic in P4
sub(middleBetti(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5
Text
For good measure, we'll also compute the Euler characteristics:
Example
sub(eulerChar(4,2),{e_1=>2,e_2=>3/1}) -- complete intersection of a quadric and cubic in P4
sub(eulerChar(5,3),{e_1=>2,e_2=>2,e_3=>2/1}) -- three quadrics in P5
Text
Exercise 5.12: {\bf Betti numbers of the quadric line complex}
The only interesting Betti number is the middle one, which we compute immediately from the
above:
Example
sub(middleBetti(5,2),{e_1=>2,e_2=>2/1})
Text
Exercise 5.13: {\bf Calculate the Euler characteristic of a smooth hypersurface of bidegree}
$(a,b)$ {\bf 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
${\mathbb P}^2 \times {\mathbb P}^3$:
Example
P2 = flagBundle({1,2})
P2xP3 = flagBundle({1,3},P2,VariableNames => K)
intersectionRing(P2xP3)
Text
Note that if we didn't use the {\tt 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.
Example
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
eulerChar(4,4) -- chi of a smooth hypersurface of bidegree (a,b) in P4xP4
sub(eulerChar(2,3),{a=>1,b=>0/1}) -- is P1xP3, should be 8 by Kunneth
sub(eulerChar(1,1),{a=>1,b=>1/1}) -- a conic in P2, should be 2
sub(eulerChar(1,1),{a=>2,b=>1/1}) -- a twisted cubic, should be 2
///
doc ///
Key
"Intersection Theory Section 5.4.4"
Headline
Bundles on Grassmannians
Description
Text
We already know everything necessary to calculate chern classes of bundles on Grassmannians.
As an example, we can do:
Exercise 5.17: {\bf Calculate the chern classes of the tangent bundle to} ${\mathbb G}(1,3)$
{\bf in two different ways.}
We calculate directly:
Example
G13 = flagBundle({2,2})
T = tangentBundle(G13)
chern T
Text
The above amounts to using the splitting principle.
We also can calculate the total Chern class of the tangent bundle of $G = {\mathbb G}(1,3)$
by realizing $G$ as a smooth quadric in ${\mathbb P}^5$. The plan is the following: first, we'll
calculate the total Chern class of the tangent bundle in terms of powers of the hyperplane
section of $G$ in ${\mathbb P}^5$. Then, we'll substitute $\sigma_1$ into this polynomial,
since we know $\sigma_1$ is the hyperplane section.
Example
P5 = flagBundle({1,5})
TP5 = tangentBundle(P5)
O1 = dual(P5.Bundles#0)
O2 = O1^**2
TG = chern(TP5 - O2) -- total Chern class of TG in terms of the hyperplane section
-- the coefficient of H_(2,i) is the coefficient of sigma_1^i
sigma_1 = chern(1,G13.Bundles#1)
1 + sum(1..4, i -> coefficient(chern(i,P5.Bundles#1),TG) * ((sigma_1)^i))
///