These are my liveTeXed notes for the course Math 266: Intersection theory in algebraic geometry taught by Joe Harris at Harvard, Spring 2015.
Any mistakes are the fault of the notetaker. Let me know if you notice any mistakes or have any comments!
01/26/2015
Introduction
This is a course not only about intersection theory but intended to introduce modern language of algebraic geometry and build up tools for solving concrete problems in algebraic geometry. The textbook is EisenbudHarris, 3264 & All That, Intersection Theory in Algebraic Geometry. It is at the last stage of revision and will be published later this year.
We will fix a base field , an algebraically closed field of characteristic 0. The starting point of an enormous amount of mathematics (including all of cohomology) is the classical theorem of Bezout:
Let
be two curves of degree
intersecting transversely (i.e., with linearly independent differentials at intersection points), then
.
For
any variety, a
cycle is a formal linear combination
of subvarieties
. We say two cycles
are
rationally equivalent if there exists a family of subvarieties parametrized by
interpolated between them. Namely, there exists
a subvariety, not contained in one fiber (better way to say this:
is flat over
) such that
and
.
We define the
Chow group of
to be the group of cycles modulo rational equivalence (the equivalence relation generated by
). Notice the rational equivalence always preserves the dimension, so we can write
01/28/2015
In contrast to homology/cohomology, the Chow groups are not computed for 99 percent of algebraic varieties: even for simple cases like surfaces in of degree (for , the Chow ring is : every two points are connected by rational curves; we don't know if this is true or not for ). Nevertheless this is not the end of the world: a lot of calculation are performed on spaces whose Chow rings are known (e.g., projective spaces; product of projective spaces; Grassmannians); we also know many subrings of Chow rings (of certain special degrees).
Chow rings of projective spaces
We will compute that . Namely, if is of codimension , degree , then . The key to proving this is that every subvariety is rationally equivalent to a multiple of a linear subspace. We observe that has an affine stratification
A
stratification of
is the expression
as a disjoint union of locally closed subvarieties
, such that for any
, the closure of
is a disjoint union of certain strata. We say it is an
affine stratification if
is isomorphic to an affine space for each
.
The above assertion about follows from the following more general
If
has an affine stratification, then
is generated by the classes of
.
01/30/2015
Examples
(Product of projective spaces)
A product of projective spaces also has an affine stratification. It follows that
One can think of this formula as the Kunneth formula for Chow rings. But be careful that there is no such a thing in general: e.g., if
is a product of two curves of genus
, then
is very different from
(see Example
5).
(Veronese varieties)
Let
be the Veronese embedding, where
and
runs over all monomials of degree
in
variable
. Let
be its image. What is
? The degree of
is the size of its intersection with a general linear subspace
in
of dimension
. Since the preimage of a general hyperplane under
is a general hypersurface of degree
, it is equal to the intersection number of
hypersurfaces of degree
in
, which is
by Bezout.
(Segre varieties)
Let
be the Segre map, where
. Let
be its image. What is
? Write
Then
by the pushpull formula. Since
, we know that
(only the
term is nonzero).
02/02/2015
Today's motivating questions:
If
are 3 general homogeneous quadratic polynomials in 3 variables, how many (nonzero) solutions are there to the system of equations
?
If
are 3 general homogeneous cubic polynomials in 3 variables, for how many triples
does
factor?
(Kunneth does not hold for Chow rings)
Let
be a smooth curve of genus
. Let
be the diagonal. We claim its class
is not the linear combination of fibers (hence
). To see this, we use the usual Kunneth formula in topology: the cohomology class of
has nontrivial component in
, but
is not seen in Chow rings: Chow rings only see even degree cohomology.
Plane cubics
What are the degrees of
? (one can also ask the degree of the locus of singular cubics, but we will wait until we talk about Chern classes.)
Notice
is simply the answer to the second question in the beginning (the intersection of
with a generic codimension 2 subspace). To compute it, look at the regular map defined by multiplying a linear form and a quadratic form in 3 variables
If we write
, then
,
and thus
02/04/2015
To compute
, look at the map defined by multiplying three linear forms
If
, then we have
Be careful about the extra factor
arising from the fact that
is not generically onetoone: there are six ways of writing a product three linear forms as a triple of linear forms. By the pushpull formula, this is equal to
To compute
, we need to compute the class
of the subspace
consisting of concurrent triples of lines in
. If the three lines are given by
, for
. Then
is given by the vanishing locus of the determinant of the matrix
which is again a homogeneous trilinear form. Hence
. Thus
This agrees with
and one naturally wonders if
is a hyperplane section of
. The answer is no, however. Notice the image of
is a hyperplane section under Segre embedding
but the projection
is a regular map of degree 6 on the image of
and does not send this hyperplane section in
to a hyperplane section in
.
Here is another way of computing
. For any point
, we have a hyperplane
of cubics passing through
. Let
be general points, then
Be careful that one needs to verify that this intersection is transverse since
are no longer general hyperplanes and we cannot invoke Bertini's theorem directly. On the other hand, this intersection can be compute directly (without pushpull formula), it is the number of ways of grouping six points into three pairs of points, which is
.
Curves on surfaces
Let be a quasiprojective variety variety. Any line bundle on has a rational section . The vanishing locus and of two rational sections of are rationally equivalent since they can be connected by the family . Hence taking the vanishing locus of gives a welldefined Chern class map If is smooth, then is indeed an isomorphism since every codimension one cycle can be represented by a Cartier divisor. If is not smooth, can be neither injective nor surjective.
Now assume is smooth. We have the canonical bundle . By abuse of notation, we denote the canonical class again by . If is a smooth divisor, then we have an exact sequence The (Iwouldcall) adjunction formula says that the normal bundle . Therefore which is the usual adjunction formula
Consider
. Then
is a rational differential form, everywhere regular and nonzero on
and has a pole of order
on the boundary. Hence
, where
is the hyperplane section. For
a smooth hypersurface of degree
, the adjunction formula then implies that
.
02/06/2015
Now let be a smooth surface. Let and be curves on . If and intersect transversely, then is the number of intersection points. We will write this as for short.
Suppose is a smooth curve of genus . Then by the adjunction formula, Hence we deduce the genus formula
Let
and
is a smooth curve of degree
. Then
and
. Hence
Let
be a quadric surface in
. Then
and
, where
are the classes of the two rulings. We say
has bidegree
if
, or equivalently,
is the locus of bihomogeneous polynomial of degree
. By the adjunction formula, we have
Suppose
has bidegree
, then
,
and
Let
be a smooth curve. In general,
may not always be defined by exactly two equations. Suppose
, where
are smooth surfaces of degree
. We write
and say that
and
are
linked. We define
liaison to be the equivalence relation generated by linkage. The classification of liaison has been done by Hartshorne and Rao.
If has degree , genus , what can we say about and ? By Bezout, one easily sees that . To compute , we proceed as follows by using the genus formula on twice.
 Since , hence . By the genus formula, we find that
 Since , we have
 We compute that
 By the adjunction formula, we get Namely, So the genus of is indeed determined and is very easy to remember: the difference of the genera is proportional to the difference of degrees.
02/11/2015
Grassmannians
Today's motivating question is the first nontrivial question in enumerative geometry:
Given four general lines
in
, how many lines meet all these four?
To answer an enumerative problem like this, we are going to proceed as follows.
 Introduce a parameter space for the objects we are studying (in this case: the Grassmannian of lines in , which is 4 dimensional);
 Describe the Chow ring of this space;
 Find the classes of the loci of objects satisfying desired conditions (in this case: the locus of lines meeting a given line, which is codimension 1 in );
 Take the product of these classes and evaluate its degree (in this case: intersecting four such codimension 1 classed);
 Verify the transversality of the intersection!
We will begin with some generality on Grassmannians and come back to later. There are three stages of for understanding Grassmannians:
 Definition via Plucker embedding; local coordinates;
 Description of the tangent spaces. This is useful for verifying transversality);
 Functorial description. This also helps us get familiar with Hilbert schemes.
Let
be an
dimensional vector space. We denote by
the set of
planes in
. This is also denoted by
when viewing everything projectively.
Given , then we have an inclusion of the line . This gives an embedding The claim is that the image is closed, which gives the set the structure of an algebraic variety. Namely, we will exhibit polynomials of degree that cut out the image (however, this is the wrong way of doing things since it does not define the image schemetheoretically: the homogeneous ideal is in fact generated by quadratic polynomials — the Plucker relations).
Let
. We say that
if
for some
. Then
if and only if
in
.
Notice the map givens a linear map It follows that is totally decomposable (i.e., ) if and only if , and the last condition is cut out by polynomials of degree on (vanishing of minors).
In terms of matrices, can be viewed by the spaces of matrices of rank (up to action), since any plane can be viewed as the row space of such matrix. The Plucker embedding is simply taking the matrix to the set of its minors. This gives us a way of writing down the local coordinates concretely.
Now let us come back to . We have a stratification of by closed stratas. Choose a flag . We have the following subset of : These are again algebraic (with dimension 3,2,1,0 respectively), known as Schubert cycles. To be rigorous: depends on the choice of and better denoted by ; similar remarks apply to other Schubert cycles.
Notice that each open stratum (the complement in a closed stratum of all its substrata) is an affine space by the argument in Remark 13. We will denote the classes of these cycles by the with lower case symbols . By Lemma 1, these classes generate . We will compute the intersection product on case by case.
 First take two Schubert cycles of complementary dimension and compute their intersections: Notice by Kleiman for two generic points , . This intersection number is 1 since there is a unique line passing through . Similarly since two generic planes intersect at a unique line, and since two generic lines don't intersect. It follows the and are linearly independent (with their intersect matrix the identity). Finally, since there unique line passing through and the intersection point .
 By Kleinman, for a generic point and a generic line . This is the Schubert cycle Similarly,
 The only remaining intersection product is . It is given by the number of lines intersecting both general lines , . This is not any Schubert cycle, but we can use the method of undetermined coefficients: write Then Similarly Hence we conclude that
02/13/2015
Now we can answer the earlier motivating question: So there are exactly two lines intersecting all four given general lines.
More generally,
Tangent spaces of Grassmannians and specialization
Given four (not necessarily general) skew lines
and
be a line meeting all four. When is
a point of transverse intersection for
?
This question has a simple answer.
Let
and
. Then the intersection is not transverse if and only if the cross ratio of the four points
is equal the cross ratio of
.
In general, to test the transversality, we need to describe the tangent space of the cycles, which lie inside the tangent space of the Grassmannian.
Let
. For
, how to describe the tangent space
?
If we want to describe how
can move in
, we can use all oneparameters families
passing through
. Let
be a oneparameter family of
planes such that
. Let
and choose a oneparameter family
such that
and
for any
. One can check that the tangent direction
is welldefined modulo
. So
gives us a map
. This association provides a natural isomorphism
02/18/2015
Let
be a curve. Let
(Example
18). Let
be a line that meets
in just one point
and
. Then
is smooth at
if and only if
,
and
. Using the previous description of
in terms of oneparameter families, one can see that
In particular, when
is a line, this gives the tangent space of the Schubert cycle
.
Let us calculate
again using the fundamental idea of
specialization. We would like to compute
. But this time we take
to be special:
. Let
. Then we find that
This gives the right answer
but we need to check the transversality without Kleinman. To check this we simply compute the tangent space of
and
and see in fact they are
distinct in the tangent space of
. For example, suppose
and
,
. Then
and
are distinct.
02/20/2015
Let us calculate the number of common chords of two twisted cubics again using specialization. Every twisted cubic lies on a smooth quadric and has type
. We take special twisted cubics
on the same smooth quadric
of type
and
. Then there is no line on
that is a common chord. All common chords are then given by the lines through any two intersection points in
. There are 5 such intersection points (since
in
). Hence we can conclude the number of common chords is
(Example
19), after checking the transversality. Notice if we chose the types to be both
, then the intersection consisting of
points and a one dimensional component coming from one ruling. In fact, Fulton has a formula for computing the components even when the intersection is not transverse and we will come back to this later.
Digression on multiplicities
There are three levels of intersection theory. Suppose is smooth of dimension and are subvarieties of codimension .
 Suppose intersect transversely. Then , where is the schemetheoretical intersection.
 Suppose has the correct codimension but does not necessarily intersect transversely. Then we can look at the the components of the intersection and assign a multiplicity to each component such that This applies to much broader range than the first level.
 Fulton was able to drop all assumptions except a small one: assume is locally CohenMacaulay, then where is is the pushforward of a class under . The point is that there is a formula for the class . This puts the intersection theory on solid basis: one takes the formula as the definition of the intersection pairing. After one checks this intersection pairing is welldefined, the independence of choice of the cycles in first level follows automatically.
Suppose we are in the second level: if are locally CohenMacaulay. Then is simply the multiplicity of the component of supported at (and this was the reason for introducing the notion of CohenMacaulay in the first place). However, when the CohenMacaulay assumption is not satisfied, the intersection formula in terms of multiplicity no longer holds in general. Serre eventually found a general formula (without any assumptions) in terms of alternating sums of Tor groups (we are not going into that but it is good to know it is there).
If
, then
is simply the degree of
supported at the point
. This implies the stronger version of Bezout: if
are plane curves without common component, then
.
General Schubert cycles
Let . We introduce the flag , Suppose is a general dimensional subspace. We look at the intersection of with the flag , Notice the intersection will be empty for the first terms and jumps by one for each of the last terms. For an arbitrary dimensional subspace , we should specify how the jumps occur.
For any sequence
, we define the
Schubert cycle In other words, this consists of
dimensional subspaces whose
th jump occurs at least
steps earlier. In particular,
consists of all general
dimensional subspaces.
02/23/2015
We define
. Since the dimension
is nondecreasing, we will adopt the convention that
unless
is nondecreasing. We also drop the ending zeros from the notation (so
).
The role played by the Schubert cycles for Grassmannians is like the role played by linear subspaces for projective spaces. But there are differences. For example, the Schubert cycles may be singular: for , is singular at the line : can be viewed as a hyperplane section of under the Plucker embedding by the hyperplane tangent to ). It is known (but not easy) which Schubert cycles are singular. But in general it is not known which Schubert cycles are rationally equivalent to smooth cycles.
Also, there are algorithms for computing the intersection product between 's but we don't know in general which coefficient in the intersection product is zero or not. The situation in complimentary dimension is much better as in the following theorem.
If
, then
, unless
, in which case
.
Notice
. Suppose
is in the intersection. Then
So
This gives exactly
. Adding this inequality for all
we obtain that
. Hence the equality holds and
for each
. The previous intersection is exactly 1dimensional for each
and uniquely determines
, hence
.
¡õ
We end by discussing the method of dynamic specialization.
Our goal is calculate
. Here
We would like to compute
. If we take two general lines we would not be able to describe the intersection easily. If take two special lines
lying in a plane
with
. Then
This looks nice but the problem is that not only the intersection is not transverse, the component
has the wrong dimension! We are stuck now since there is no more options between "general" and "lying on a plane".
02/25/2015
The idea of dynamic specialization is to choose a oneparameter family of lines such that for but . Let Take to be the closure of in and let be the fiber of at (key point: this is the limit of the intersection, rather than the intersection of the limits). We have , but now has the correct dimension 2. In other words, must satisfies some additional assumption to be in the limit .
Let for and take be the limit of as . Then the additional assumption is . So The final claim is that this is indeed an equality and thus .
We can show geometrically that coincides with the support of . So it remains to show that is generically reduced. We are going to write down local equations to prove this claim. We can take , . Then , , . Let us choose the nice (= the limit does not go outside ) affine open of Then the equation of is given by , and is given by and . So the ideal of is given by . When , has dimension 2. But when , the ideal contains the hyperplane (this is what goes wrong when taking the intersection of the limit). Notice that we know that for any , so we should throw in as well. This is the limit of the intersection and gives the correct . From the equations we see that is reduced.
Let
be a curve. Let
. Then
is a closed subvariety of dimension
. This is easily seen by incidence correspondence
Then
, where
are the projections of
to
and
respectively. What is
? Let
be a general subspace of complimentary dimension, then
, which is equal to
. This is exactly the degree of
under the Plucker embedding
.
02/27/2015
Chern classes
Today's motivating questions:
If
is a general cubic surface, how many lines does
contain?
If
is a general pencil of quartic surfaces, how many contain lines?
Before we see how Chern classes help answer these two questions, we should first ask the following simpler question. If is a general hypersurface of degree , we introduce its Fano scheme, When is ? And what is ? This can again be answered using the incidence correspondence The projection is easier to understand: its fiber above consists of all hypersurfaces of degree containing , which is the the projectivization of the kernel of the surjection (An intrinsic description is that if , then ). It follows that is irreducible and has dimension since . So if , a general is expected to contain no line and when , we expect that .
In the first question, we have
and
, so there should be finitely many lines. In the second question,
and
, so a general quartic surface should contain no line and there should be a hypersurface of special quartic surfaces which contain lines. The second question is simply asking the degree of this hypersurface. Ultimately, we would like a polynomial in 35 variables (coefficients of homogeneous quartic polynomials in four variables) which exactly determines whether a quartic surface contains lines or not. It turns out the degree is 320 (and one shouldn't try to write it down!)
The key new idea is linearization: we replace such a complicated polynomial by a family of of system of linear equations.
For the second question, similarly we introduce vector bundle
over
of rank 5 such that
If
are general quartic polynomials on
, we again obtain two sections
of
. Then we are asking for how many points
,
are
linearly dependent. Now a section locally is a 5tuple of functions on a 4dimensional space, so two sections pin down finitely many such
's.
Let be a smooth variety. Let be a vector bundle of rank . Notice that is a trivial bundle is equivalent to saying that there are linearly independent sections. More generally, let be sections of . What is the locus where they becomes linearly dependent? If , the zero locus of the section should be of codimension . In general, these sections locally gives a matrix of functions and the locus where the matrix has rank is of codimension .
We define the
degeneracy locus to be the subscheme
of
cut out by the maximal minors of this
matrix. If
has the expected codimension
, we then define the
Chern class The Chern class does not depend on the choice of the sections (as long as the codimension is as expected). One issue with this definition is that it takes effort to extend the definition to general vector bundles without global sections satisfying the assumption (okay when
is generated by global sections). Define
What have we achieved? We gave (nice) names for the answers of the questions: the answer should be and . Apart from that we hardly accomplished anything. The point of the abstract construction of Chern classes is that it allows us to carry the information about vector bundles around: if a vector bundle is built up (via multilinear operations) from simpler vector bundles whose Chern classes we can calculate, then we can also calculate the Chern classes of itself.
03/02/2015
Let
be a rank 2 vector bundle on
such that
consists of linear functions on
. This is simply the dual of the universal subbundle over
. Then Chern class of
should reflect the number of lines on planes (instead of cubic surfaces), which is an easy linear problem. Given a linear form
on
, one obtains a section
of
by
. The zero locus of
is simply the Schubert cycle
. Hence
. Similarly, given
two linear forms on
, then
are linearly dependent if and only if
. So
. Our remaining task is to relate
and
using
.
The general tools for calculating Chern classes:
 Whitney formula. If , then . The same formula applies more generally we have an exact sequence This more general formula easily follows from the split case: since the extension group is a vector space, the class of does not depend on its extension class.
 Splitting principle. We say that is totally filtered if there exists a sequence of subbundles such that is a line bundle. In this case,
(Splitting principle)
Given a vector bundle
. There exists a morphism
such that
 is injective.
 is totally filtered.
It follows from the splitting principle that any formula for Chern classes that holds for direct sum of line bundles, holds in general.
Let
be a vector bundle of rank 2. Let us calculate
. Suppose first that
is a direct sum of two line bundles. Write
and
. Then
. In particular,
,
. Notice that
it follows that
which is expressible in terms of
and
since it is symmetric in
:
This formula makes sense and in fact is true for any (not necessarily split) rank 2 vector bundle. Similarly,
gives
Applying this to in Example 31, we obtain that In this way we obtain the wellknown number of lines on general cubic surfaces!
Let
be general polynomials of degree
in
. How many of
are singular? Let
be the space of hypersurfaces of degree
in
and let
be the singular locus. We are simply asking the degree of
. To linearize, instead of asking if a given hypersurface is singular, we ask if a given hypersurface is singular at
a given point. Let
be the vector bundle over
such that
where
consists of polynomials vanishing to degree 2 at
. Given
a degree d polynomial, we obtain a section
of
. The question now becomes how many points in
where
and
are linearly dependent. By definition of Chern classes, this is
since
(determined by the value and
partial derivatives).
03/04/2015
Let
be a vector bundle on
. What is the Chern class of its dual bundle
? If
with
. Then
Similarly,
, we have
since
. It follows that
.
Let
. We have the universal vector bundles
over
with
and
. Let us first compute the Chern class of
(it has many sections since it is a quotient of a trivial bundle). Let
, its image in
gives rise to a section
of
. Let
be linearly independent vectors and
. Where are
linearly dependent? It happens exactly when
. In other words,
Similarly, we can obtain directly that
Hence
.
On the other hand, we have the exact sequence Whitney's formula then implies that Comparing the quadratic terms recovers the intersection product .
Let us compute the Chern class of the tangent bundle of
. The easy way is to look at the Euler sequence
where the second map is simply
. (Notice that
is not a welldefined element of
but multiplying it by a linear form gives a welldefined element.) By Whitney's formula, we obtain that
Alternatively, we can use the general fact for
,
. Applying to
we know that
We can compute
The formula in Example
35 then gives the same answer for
.
Now let come back to the fundamental question:
Where does the Whitney formula come from?
Let
be vector bundles of rank
. The Whitney formula says that the top Chern class
This is complete straightforward: let
be sections of
, let
be the corresponding section of
, then the zero locus of
is simply the intersection of the zero loci of
and
.
The next graded piece of the Whitney formula says that
Suppose
and
be two general sections of
. Let
We have a map
so that
and
. The degeneracy locus
is the preimage of the diagonal
. The Whitney formula in this case boils down to the class of
in
. Making this argument work in general requires lots of technical work.
The preimage of the second factor gives . Similarly, the pullbuck And
03/06/2015
I owe you the proof of the following theorem.
The last item is how Chern classes were originally defined and is what many people use to do computation. But unlike in the differentiable or continuous category, there may not be enough global sections in the algebraic category. I will defer the proof as it is slightly more subtle than one may think. As we have seen in examples, the last item characterizes Chern classes and is useful for solving enumerative problems: its existence will be shown in due course.
Let us return to Example 33 .
Let be the space of all hypersurfaces of degree in . We would like to compute the degree the singular locus . To see is in fact of codimension one, we look at the incidence correspondence The fiber at along the second projection is a linear subspace (defined by the vanishing of the value and partial derivatives), hence is irreducible of dimension . It remains to show that the first projection is generically finite (in fact generically onetoone). For this it suffices to show that there is an isolated point in the fiber of the first projection, which is clear by Bertini, or by taking a singular cone (whose nearby points are all smooth).
Now we introduce a vector bundle on such that Let be a general pencil of hypersurface of degree . Then
 there are no elements singular at the same point (since by Bertini, the common zero locus of two general hypersurfaces is smooth).
 any element of is singular at at most one point (since one can show that is generally onetoone).
Therefore the number of singular is equal to
 the number of points such that is singular at , which is,
 the number of points such that viewing as general sections of , which is,
 by definition, the Chern class .
To calculate , we look at the exact sequence where In other words, we filter the vector bundle by the order of vanishing. Notice that is simply the the line bundle . The fiber of at is and hence Now we win because the Chern classes of and can be easily computed.
Suppose
. Write
. We know that
Suppose
, then
Hence
Reality check: when
, we get the answer 0 (no line is singular); when
, we get the answer 2 (a general pencil of conics has a base consisting of four points, there are exactly 3 pairs of lines passing through 4 points); when
, we get the answer 12 (the modular discriminant is of degree 12).
For a general , one can compute that .
03/25/2015
Space of complete conics and 3264
Today's motivating question:
Given five general conics
in
, how many conics are tangent to all 5?
We would like to ask this question only for smooth conics since double lines are trivial solutions. The key problem is that the parameter space for smooth conics is noncompact. We need to introduce a compactification for the space of smooth conics for doing intersection theory and moreover the intersection on the compactification does not contain extraneous points coming from the boundary.
To see how the boundary may cause trouble, let us consider a simpler problem.
Let us come back to the motivating question.
We define
where
are smooth conics dual to each other:
,
. We simply take the closure
of this locally closed subspace, and call it the space of
complete conics.
What happens for the dual conic when the conic becomes singular? The dual of two lines is a double line dual to the intersection point. And vice versa. Therefore , where consists of pairs , where is union of two distinct lines () and is the double line , consists of and , and consists of the common specialization: , , where . Notice that and are of codimension 1 and is of codimension 2.
One can check that:
 is smooth.
 If is smooth and is the closure in of the locus of smooth conics tangent to . Then and the intersection is transverse.
After checking these, we get a desired nice parameter space. The next step is to calculate the Chow ring. For our purpose we only need to calculate and the degree 5 map .
We claim that . In fact,
 . Since is the complement of a cubic hypersurface, any divisor on extends to a divisor on , which is of the form . If , the there exists a section of which vanishes to order along and nowhere zero in . Hence in . In general, the Picard group of the complement of a hypersurface in a projective space is always torsion.
 Let be any line bundle on . Then is trivial on . Find a section of nowhere vanishing on . Extend it to a section of on . Then . Hence has rank at most 2.
 To see the rank is equal to 2, we introduce We claim that and are linearly independent in . Let be a general pencil of conics in . Let be a general pencil of conics in . Let be their classes in . Then we can compute the intersection product since a general pencil has one member passing through a given point and has two members tangent to a given line. The intersection matrix is nonsingular, so are linearly independent.
Now we calculate . We see that by counting the number of conics passing through 5, 4, 3 points and tangent to 0,1,2 lines. The remaining three products are then given by symmetry.
The last step is to calculate the class . We compute that , hence that . Using the above results we see that
03/27/2015
Projective bundles
Today's motivating question:
Given 8 general lines
, how many (plane) conics in
meet all 8?
This question is significant for us because the parameter space in question will be a projective bundle and calculating the Chow ring of projective bundles is a useful tool and in fact can also be used to give a rigorous definition of the Chern classes. Moreover, the parameter space in question will be a simple example of Hilbert schemes.
As usual, we need to first construct a parameter space for conics in . We should have a map , sending a conic to the unique plane containing it. The fiber of this map consists of conics contained in a given plane and has a natural compacitifaction . So we define where is any homogeneous quadratic polynomial on .
The map has fiber and hence is irreducible of dimension 8. Let be the universal family We have two projection maps , For a given line . Define Then has codimension 1 in . Hence we expect that the above question has a finite number as its answer.
Next we would like to describe and and compute . is an example of an important class of varieties appearing in enumerative geometry: projective bundles. Let us first discuss projective bundles in general.
Let
be a smooth variety. Let
be a vector bundle of rank
. Define
One can verify directly that it is welldefined by looking at trivializations and transition functions. More intrinsically,
since homogeneous polynomial of degree
on
is
. Notice on
we have an exact sequence of universal sub and quotient bundles
The hyperplane class is then the first Chern class of
. Similarly, we have an exact sequence in families
where
. We then define
and let
. Its restriction to each fiber of
is the hyperplane class. We remark that there may not exist a divisor on
such that its restriction to each fiber is an actual hyperplane (rather than in the hyperplane class).
The following theorem provides a way of computing .
 is injective.
 Additively, This can be viewed as an additive Kunneth formula for Chow ring.
 In the case of a single vector space, , but there is no reason for this to be true in . In fact, as a ring,
To prove c), we use the exact sequence of bundles on
,
Then
. The Whitney formula gives
(we suppress
from the notation using a). So
In particular,
.
¡õ
03/30/2015
Recall that , a projective bundle over . To compute , we need to compute . Write Then
Another way: the restriction map from the trivial bundle of rank 10 gives an exact sequence where the kernel is of rank 4 (its fiber at consists of quadratic forms vanishing on the plane corresponding to , which can be identified with the space of linear forms). We know that .
It follows that Now it remains to find and compute . We do this by undetermined coefficients. Let . We need two curves in : fix a plane and let be a general pencil of conics in ; fix a quadric and take to be the intersection of with a general pencil of planes. Then one can compute the following intersection matrix
It follows that . After checking that the intersection is transverse (one needs to know more about the tangent space of Hilbert schemes, which we will do next time), the answer to the motivating question would be Since the pullback of is the class of the fiber and restricts to the hyperplane class in the fiber, we know that . Using the relation in , we know that Similarly, we obtain that , and . Finally we can conclude that
04/13/2015
Segre classes and trisecants
If
is a general rational curve of degree
, how many trisecant lines does it have?
Instead of looking at lines in and imposing condition that they intersect at three points, we look at triples of points on the curve and impose the condition that they are colinear, which becomes a linear problem. The compact parameter space we need is then the space of effect divisors on of degree 3, which is the same as (projectivization of homogeneous polynomials of degree 3 on ). To impose the condition that the three points are colinear (i.e. the failure of the three points to impose independent conditions), we introduce the vector bundle on given by In other words, the homogeneous degree polynomials modulo those vanishing on the divisor . More precisely, we introduce the universal divisor of degree 3 on , Notice is the vanishing locus of , hence is a closed subvariety of . Let be the projection maps to and respectively. Then
We have an evaluation map from the trivial bundle of rank 5 The colinear locus is the locus where fails to be surjective, whose class is by definition the Segre class . Let . We have an exact sequence To compute the Segre class of it suffices to compute that of .
Let which is given by a pair of homogeneous polynomials of degree 3 and such that . It follows that Under the map , , we have three hyperplane classes on ,
 the pullback of ,
 the pullback of (linear forms on a the 1dimensional space of degree polynomials vanishing on ).
 the pullback of .
In particular, . Also we know that , so , which gives a degree monic polynomial in satisfied by . Using Theorem 6, we know that Hence . Therefore the class of the locus of colinear degree 3 divisor is
Contact problems
If
is a general quintic surface, how many lines
meet
in only one point?
To linearize the problem, we look at the the space of lines together with a point Then the condition that the quintic polynomial restricted to vanishes to order 5 at the given point becomes a linear condition. We introduce the vector bundle of rank 5 the space of quintic polynomials on modulo the 1dimensional space of quintic polynomials on vanishing to order 5 at (5th power of a linear form). Now a quintic polynomials on gives a section of . The locus we want is simply the zero locus of . So the answer is the Chern class .
04/15/2015
We define an increasing filtration Then and for , , where is the relative tangent bundle, which is the same as the kernel of the differential . To find , we use the exact sequence on , where is the universal subbundle on and . Then Since is a bundle over , by Theorem 6, we obtain where (see Example 36). Then and hence . Thus Using the filtration on , it follows that in particular,
Porteous' formula
Let
be a smooth surface and
be the projection from a general
plane
. At how many points does
fail to be injective?
Let be a vector bundle of rank . Recall that the Chern class is the class of the locus where fails to be injective and the Segre class is the class of the locus where fails to be surjective. Given a general map between two vector bundles , we can ask for the class of the locus This class again does not depend on (any two maps can be interpolated) and only depends on the vector bundle and . The key fact is that these classes are all expressible in terms of and , which is the content of Porteous' formula.
04/17/2015
The three steps to obtain Porteous' formula:
 Linearize;
 Calculate;
 Get incredibly lucky.
Linearize.
The vanishing of minors is not a linear condition. To linearize, we specify a subspace of the source and require that this specific subspace lies in the kernel. So we introduce a parameter space It is a vector bundle of rank . We denote the universal subbundle of by and , and we have an exact sequence of vector bundles over , Composing with we obtain . Then the locus is exactly , where is viewed as a section of the vector bundle over . In particular, has codimension This is equal to the rank of . Therefore is the top Chern class and hence
Calculate.
Our next goal is to calculate this Chern class. Suppose are vector bundles of rank . Then . The splitting principle ensures (in theory) that the Chern classes of can be expressible in terms of the Chern classes of and : if , , then One case this can be calculated explicitly is the top Chern class:
Suppose is the space of degree polynomials on . Similarly for . Look at The fiber of over a point of is a disjoint union of hyperplanes. Hence is a hypersurface of bidegree .
What is this bidegree polynomial? Notice that have a common factor if and only if there exists a relation with , , if and only if the linear map between two dimension spaces has a kernel. So the bidegree polynomial is simply famous determinant of the Sylvester matrix of .
Let . Multiplying the Sylvester matrix by
we obtain the matrix where and . Hence have a common root if and only if .
Now let and . It follows that Hence Therefore we get the formula
04/20/2015
Get lucky.
Notice . So we can write , where . So the determinant is an matrix with entries a linear combination of with coefficients Chern classes of a pullback. We observe that if is any monomial in , then if for dimension reason. Hence the only nonzero contribution comes from By the pushpull formula, we finally obtain Porteous' formula
Excess intersection
What happens when the intersection is not of expected codimension?
Let
be surfaces of degree
such that
, where
is a smooth curve of degree
, genus
and
is a collection of isolated points. What is
?
Let
be smooth surfaces such that
, where
is a smooth curve of degree
, genus
and
is a collection of isolated points. What is
?
04/22/2015
Let us assume is smooth in the first question (though it is necessarily to do so: adding a generic linear combination of the ideals of can make smooth while keeping the intersection the same). Suppose , and . Let be the hyperplane class. Then Hence By adjunction, Hence We recognize that the term as the expected number of intersection points, and the remaining term comes from the excess intersection. We also observe the formula not only depends on the class of (i.e., its degree ) but also depends on its genus . We can use this method to compute the excess intersection whenever the objects in question are hypersurfaces (the intersection is reducible one step before the intersection has the wrong dimension), but not in general.
To answer the second question, we use deformations. Suppose there exist deformations that intersect transversely: i.e., families over a disk such that , and intersect transversely when . Therefore , where is finite over of degree . The question becomes: how many of the points of in the generic fiber specialize to ? To calculate this , we characterize the points of as the singular points of . If , then we would have In general, we only have the map The locus is precisely the locus where this map fails to be an isomorphism. Hence The first two bundles indeed does not depend on the choice of and gives The third term is equal to since Using and We obtain the general formula
(Excess intersection formula)
Let
be a smooth variety. Let
be smooth varieties. Suppose
, where
's are smooth. Let
be the expected dimension of intersection. Let
. Then
where
is a dimension
class in
, given by the dimension
component of
04/24/2015
From the beginning of our course, we assumed the ambient variety is smooth. This is absolutely essential for defining the intersection product.
Suppose
is a cone over a quadric surface
. Let
be a 2plane. Let
be a line in
. We can rotate
in the same ruling as
to get a line
. Then
, and
. However, if we rotate
in the other ruling to get another line
, then
but
intersect transversely at one point.
This examples destroys the hope to define a general intersection product on singular varieties. But the excess intersection formula helps: though the Chern class does not make sense (coherent sheaf may not have a resolution by vector bundles on a singular variety) and Segre class does make sense (Fulton's brilliant observation). We saw an alternative definition of the Segre class: for any coherent sheaf on (not necessarily smooth), let be it projectivization with the projection map . Let . Then Notice one cannot invert the Segre class to define the Chern class : as we already have seen in Example 43, the Chow group does not have a ring structure on the singular varieties.
Now under an extra assumption we can take the excess intersection formula as the definition of intersection product when the intersection is not transverse.
Let
such that
is locally complete intersection (which implies that
is a vector bundle, whose Chern class still makes sense). Suppose
, where
is
not necessarily smooth. The we define
where
is the dimension
class of
.
Let us come back to the situation where
is smooth. Suppose
is smooth of codimension
. Let
be subvarieties (of codimension
in
) such that they intersect generically transversely in
(so
has codimension
in
). Let
. We would like to relate
and
. Let
and
. Then it follows from the excess intersection formula that
In particular,
is simply the multiplication map by
. For example, if
has dimension
and
has dimension
, then
. Slightly more generally, suppose
is smooth of dimension
,
is smooth of dimension of
. Then a generic map
is an embedding except at a finite number of double points. The number of double points is then given by
, half of the different between the selfintersection of
and
. One recover the genus formula for plane curves from this easily.
Chow rings of blowups
One main application of the excess intersection formula is the computation of Chow rings of blowups. Suppose are both smooth. Consider the blowup diagram, The key fact is that the is generated by and . The remaining problem is to calculate the intersection product between them.
04/27/2015
Notice that By the pushpull formula, we also have Let . Since (the universal subbundle), we have $c_1(N_{E/W})
\zeta$. By the excess intersection formula, we get
As an application, we answer the following earlier question using blowups.
04/29/2015
GrothendieckRiemannRoch
The HirzebruchRiemannRoch formula relates the Euler characteristic of some sheaf and the degree of certain dimension zero class in the Chow group. There are three flavors of this formula: the basic package, the standard package and the deluxe package.
Let
be a smooth projective curve. In this case the basic package says
Let
be a line bundle
. The standard package the says
which can be deduced from the basic package using the exact sequence
The deluxe package says
which can be deduced from the standard package for any coherent sheaf
by induction on its rank.
Now let
a smooth projective surface. The basic package says that
Similarly, for any line bundle
, one deduce the standard package by combining the basic package and the formula for curves:
For any coherent sheaf
, we obtain
The formula looks more complicated and you may start to worry what to write down when . But it becomes simpler once we introduce some new language.
Let
be a vector bundle over
. Write
. We define the
Chern character For example,
,
, and
Let
be a vector bundle over
. Write
. We define the
Todd class by
For example,
,
and
.
Now let be a smooth projective variety of dimension . Then the basic package the simply says This formula in fact originally motivates the weird looking definition of the Todd class: Todd discovered his definition simply by reverse engineering using the case when is a dimension product of projective spaces. The deluxe formula is also simple: for any coherent sheaf ,
We summarize the three flavors of HirzebruchRiemannRoch formulas in the following table.

Basic 
Standard 
Deluxe 
dim 1 



dim 2 



dim 



If you know Grothendieck, anything that can be done for a single variety should also be done for a family of varieties. Given a family of coherent sheaves on a family of projective varieties , we should be able to fit together the cohomology on the fibers into a sheaf on . Moreover we should also be able to see the "twisting". Associated to this situation we have a sheaf on which is a best approximation of the th cohomology of the fibers, i.e., for a general ,
(GrothendieckRiemannRoch)