Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
Last update: 10 December 2013
Lecture 1: February 7 1997
and acts on the de Rham cohomology space
Moreover, since maps to the classifying space we have a
of algebras. The map
is a smooth projective variety.
The de Rham cohomology can be used to answer the
following question: suppose that three subvarieties and
of are in general position, and that
What is the number of points in the intersection
The quantum cohomology answers a
more general question: what is the number of holomorphic maps
with a fixed degree such that
Some features of
There is no natural homomorphism
If is a filtration, sth. similar to
does not act on
So take equivariant cohomology
where acts on from the left by left translations.
Can define quantum cohomology
The affine Weyl group acts on
(the parameter inverted);
Have creation and annihilation operators;
Have Schubert basis for
Have "stable" Bruhat order on
Formula for multiplication by
Special for symmetric spaces of the form
Geometrical models - the variety
For each parabolic, have and
where is the group of
loops in Moreover,
for some and
Will express the Schubert basis elements as matrix entries of some representations. The variety lies in
where is the Langland dual of
Lecture 2: February 11, 1997
Kac-Moody root datum
Definition: A generalized Cartan matrix is a matrix
with integer entries for some finite set such that
A Kac-Moody root datum consists of
a generalized Cartan matrix
two finitely generated free and
end????? with a perfect pairing
Can form direct sums of root data
Can form the "dual" root data:
Remark: In the classical case, root datum comes from connected reductive algebraic groups over
Definition: We say that
is symmetrizable if
Assumption: Will assume that is symmetrizable.
The numbers Define, for
The Weyl group is the group with generators
The are called the simple reflections.
[red] means that this is a reduced expression, i.e., is the minimum number such that is a product of
simple reflections. Also write
If is finite, use to denote the longest element.
From now on, write
?????tions of on
acts on by
acts on by
The actions on and on
preserve the pairing ?????
The actions on and on are faithful
the symmetric algebra of (????? the action on
acts by algebra automorphisms. The action
of on will be denoted by
The Nil-Hecke ring
Definition: the Nil-Hecke ring associated to the root datum
is the associated ring with with generators
The grading on is defined to be
For and for any
Then it is clear that
is independent of the reduced expression
Clearly as a subring.
is an for
(Does this need a proof?)
Proposition: The map
defines an injective ring homomorphism.
Only need to check
for Injectivity is clear (?)
Proposition: The following defines an structure on
The induced action on is
as the usual one.
Remark: Suppose we need to check certain specified operators for
and on some space is an action. We first
Then this is how acts. If this gives a we are done.
Note: All of this proof was crossed out in the scanned notes.
Just need to check that
The anti-automorphism on
To check that this is an anti-automorphism, need to check only
This is easy. Now since
Definitions of on and
Assume that and are and are thus ?????odules.
want to define structures on
????? check that this is an action, we first need to show that the above operators are well-defined. The
operator is clearly OK. ?????
we have, by definition
Hence is well-defined.
This gives the 2nd expression for
Now for and we
need to show
From this, we see that
This clearly induces an action of on
Thus we have proved that we indeed have an action of on
Need to check that this is indeed an action. Clearly is o????? First, since
and since is we have
This shows that the two expressions for are equal. Now we show that
????? see that
????? shows that
????? need to check that
Finally, for we see that
This is certainly an action of on
Hence we have an action of on
All these proofs seem to be longer than necessary.
But anyway, we have showed that
?????ition Given modules and
(they are therefore also the following canonical
maps are also maps:
Definition: For an set
Proposition: For and
Example: Regard as an module by left multiplications. Then our previous constructions define an
For any two modules and since we have
Proposition: In the finite case,
It is easy to show by induction on that for any
for some So
with Now for any
Lecture 3: February 12, 1997
Recall that a Kac-Moody root datum consists of
a generalized Cartan matrix
two finitely generated free
and with a perfect pairing
The weight lattice is
The coweight lattice is
The root lattice is
The co-root lattice is
Say the datum is of the adjoint type if is an isomorphism.
Say the datum is of the simply connected type if is an isomorphism.
for the standard generators such that
Given a Kac-Moody root datum
and regard it as a commutative Lie algebra. Set
Theorem (see Kac?): For any Kac-Moody root datum, there exists a Lie algebra over (of Kac-Moody type)
and Lie algebra homomorphisms
For each as an
module via (using the adj. rep) is a direct sum of finite-dimensional
are another such system, then there exists a unique
such that and
over is integrable if it is a direct sum of finite-dim. modules.
An over is integrable if
A over is integrable if it is
(via for each
So the adjoint representation of on is integrable.
is injective, so call
the Cartan subalgebra
is injective for each Set
the center of
is contained in
Every ideal of contained totally in
Fact: is the Lie algebra over with generators
For the root lattice, set
The principle of
Let be the unique(?) element such that
For the integer
is called the height of For set
Thus is a for
The set of real roots:
Need to define the Weyl group first. To define the Weyl group, need to define the Kac-Moody group.
Compact involution of
This is the conjugation-linear automorphism of such that
?????l finite-dimensional representation of
is said to be ?????ational if its matrix entries are regular functions on
????? representation of
on a vector space over is said to be differentiable if it is a direct sum of
finitely many finite dimensional rational representations.
?????t: Integrable representations of
(This is because is an algebraic group).
The complex torus
Thus, for each such the map
is a homomorphism. Moreover
A representation of on is said to be differentiable if it is a direct sum
of rational representations of
Fact: Differentiable representations of integrable representations of
Next: The Kac-Moody group corresponding to the Kac-Moody root datum we started with at the beginning.
The Kac-Moody group
Given the Kac-Moody root datum, there is a group with homomorphisms
There exists a representation Ad of on such that under and
corresponding representations of and
on differentiate to the representations of
defined by ad.
is another system with above properties, then
there exists a unique such that
is generated by the images of and
A is said to be differentiable if it is differentiable as and
There exists a faithful differentiable (probably not Ad).
is injective. So we call
the Cartan subgroup. Have
The Weyl group
For each set
????? in general
????? the subgroup of generated by
Warning: can have
The real roots:
so permutes the root system.
and call elements in the real roots.
for some then
Lecture 4: February 19, 1997
I am moving the part on Bruhat decomposition of to the end of lecture 3. The main part of this lecture is
Equivariant Cohomology (due to Borel):
Let be a topological group. or in
our applications). A principal is a topological space equipped with
continuous free right action of
and a projection such that locally
where with acts on by
with the quotient topology.
The universal principal
is a principal such that is contractible. Set
called the Classifying space of
An is a topological space endowed with a continuous left
Given an form the space
is a free left The cohomology of
is by definition the singular homology of
It is a graded ring, where the grading is nothing but the grading on
(And so is the ring structure ?????
gives a graded ring homomorphism
Thus has a natural
Given an of
form the map
Have commutative diagram
have a graded ring homomorphism
which is also a
????? special case of with
????? just the one considered before.
This is the space of
Than any map
The restriction homomorphism (or the evaluation at 0):
This is the map
induced by the map
This is a graded homomorphism.
For any map
have commutative diagram
acts freely on Then
Have the following fibre bundle with contractible fibre
acts trivially on Then
If is a principal form the complex line bundle
gives a homomorphism
In particular, take Then get
One can then show that this is an isomorphism of graded rings if is
The second structure on
Set The map
is another ring homomorphism, which we will denote by for
reasons that will be clear next time;
together with the map
incuded by will be the source and target
maps for the Hopf algebroid structure on
that will be discussed next lecture.
It is a subset of
on freely, we have the identification
Under this identification, the action on
becomes the action
(easy to check this:
Thus we have
now is just the projection from
to the 2nd factor This will be used in the next lecture.
Proposition: For any
where the structure on
is via the second ring homomorphism
(The structure on
is the usual one).
Consider the following commutative square:
is the usual homo. (induced from ?????
is the second homomorphism. Now since the square is commutative, ie.
we get a ring homomorphism
To show that this is an isomorphism, we first notice that the fibration has fibre
which is a if only even dimensions. Thus Leray-Hitsch theorem tells us that
is a free module
over with basis coming from
The special case of
is a free
with basis coming from
Using a basis of we see that the map
is an isomorphism.
Definition: The morphism
is called the co-unit map.
Definition: For any the map
induced by the
is called the co-module map.
Proposition: For any we have
Definition:A groupoid scheme consists of two schemes
and and five morphisms:
(fibre produ????? refers to
and refers to
They must satisfy:
Lecture 5: February 25, 1997
Recall the concept of a groupoid:
A groupoid is a small category with every morphism invertible
Example: Let be a group acting on a space Then we can form a groupoid
Multiplication is given by
of a groupoid scheme on a scheme ?????
with structure morphism is one such that
The groupoid scheme
Let be the principal (and thus also
As a subset of the set
is invariant under the so
is a principal
Then it is easy to check that is a groupoid over
with the following structure maps: (This is a subquotient of the coarse groupoid over
Source and target maps:
We now pull back all the above structure maps on cohomology:
First note that
Under this identification, the action on
Thus we get an induced identification
Similarly, we have
where the action on is
so we have the identifications
where the last identification is due to the general fact we proved last time that for any
We also have
Therefore, the pull-backs on cohomology of all the structure maps for the groupoid over
give the groupoid structure on
Theorem: The above maps
into a groupoid scheme. Moreover, if is any the map
is the composition of an action of on
(assuming that is even)
Definition: A characteristic operator for is a rule that assigns to each
such that if is a then
Remark: When any
endomorphism of must be a multiplication operator by characters.
This is why the name characteristic operators.
Fact: The set of all characteristic operators
Definition: We say that a characteristic operator is of compact support if there exists a compact subset
which is such that
given any a
subset of and an element
the element must vanish in
Remark: In the finite case, can take and every characteristic operator is compact.
Proposition: For any characteristic operator and any
Corollary 1: For a characteristic operator we have
then for any
Corollary 2: has no
are such that
then for any
But since we know by Corollary 1 that
Since is a polynomial algebra, it has no Thus
This shows that
Corollary 3 (added by me) (of Corollary 1): The action of
on is expressed using
and the map by
Remark: Should think of as the dual of
Integration over the fibre
Assume that is a fibration over a pathwise connected base
Assume that this fibration is orientable. This means that the holonomy around acts trivially on
Since is pathwise
connected, the weak homotopy type of is independent of the choice of
Then we have, assuming for
obtained as follows by using the Serre spectral sequence:
This is just the identity map when
It is functorial over pullbacks.
It preserves certain Mayer-Vietoris sequences
Can do this for relative cohomology as well.
The on for any principal
If is a principal then we have a ring homomorphism
We call it the characteristic homomorphism. Using the characteristic homomorphism, we get an structure
Now assume that is also a principal so thus also a
Then we can use the to define the following
Because of the following
basic property of the characteristic map,
we have, for any and
as operators on Therefore
we have an action of the smashed product algebra on
Now for each consider the fibre bundle
which has fibre
so it has a preferred orientation
namely the fundamental cycle. Integration over the fibre gives
Proposition: For any
We will check this over (Why?). The fibration
structure on Since the
fibre is this is in fact a free
a basis of which is given by and
we use the same letter to denote the pull back
We will check for and
Thus holds for
(a calculation over
Assume Lemma. Then
On the other hand,
Hence holds for
It is strange to carry the around. Why necessary?
Therefore we have
Theorem: For any principal the following
define an on
Moreover, the characteristic morphism
is an where
acts on by
as before (see Lecture 2).
Example: with right action of by right multiplications. Then the
are the BGG-operators.
Example: If is a
is clearly an
Example: Let be a and let
be the universal principal bundle of Let
with the given by
so get an action of on
is a so
is an Finally, the
is clearly Thus we can think of elements of
as characteristic operators.
Property: For any the morphism
This is the same as the characteristic morphism. ?????
Proposition: For any the multiplication map
For a the
homology of is defined to be
Suppose that is a Then
is also an we know that
is then also a (see Lecture 2):
If is a
then we have shown that
is an Define
Then is an as well. Let's check
So let need to show
Since is an
we indeed have
Example: Suppose is a such that
for Then integration over the fibre for
gives a map
For each we have a map
is the composition
using integration over the fibre first for the bundle
Consequently we have a map
Now suppose that is a with
Assume that is a map.
we have so
On the other hand, we have
This fact will be used in the next lecture for a
Schubert variety in ?????
Notes and references
This is a typed version of Lecture Notes for the course Quantum Cohomology of by Dale Peterson. The course was taught at MIT in the Spring of 1997.