Lectures on Chevalley groups
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last update: 30 July 2013
§2. A basis for
Let be a Lie algebra over a field and an associative algebra over
We say that is a homomorphism if:
(1) |
is linear.
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(2) |
for all
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A universal enveloping algebra of a Lie algebra is a couple such that:
(1) |
is an associative algebra with 1.
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(2) |
is a homomorphism of into
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(3) |
If is any other such couple then there exists a unique homomorphism
such that
and
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For the existence and uniqueness of see, e.g., Jacobson, Lie Algebras.
Birkhoff-Witt Theorem: Let be a Lie algebra over a field and
its universal enveloping algebra. Then:
(a) |
is injective.
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(b) |
If is identified with its image in and if
is a linear basis for the monomials
form a basis for (where the are nonnegative integers).
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The proof here too can be found in Jacobson.
Theorem 2: Assume the basis elements
of are as in Theorem 1 and are arranged in some order. For each choice of numbers
form the product, in of all and
according to the given order. The resulting collection is a basis for the
generated by all
Remark: The collection is a for by the Birkhoff-Witt Theorem.
The proof of Theorem 2 will depend on a sequence of lemmas.
Lemma 4: Every polynomial over in variables
which takes on integral values at all
integral values of the variables is an integral combination of the polynomials
where and
degree of the polynomial in (and conversely, of course).
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Proof. |
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Let be such a polynomial. We may write
each being a polynomial in
We replace by and take the difference.
If we do this times we get Assuming the lemma true for polynomials
in variables (it clearly holds for polynomials in no variables), hence for
we may subtract the term
from and complete the proof by induction on
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Lemma 5: If is a root and we write
for
then
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Proof. |
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The case
together with induction on
yield
This equation and induction on yield the lemma,
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Corollary: Each is in
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Proof. |
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Set in Lemma 5, write the right side as
then use induction on and Lemma 4.
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Lemma 6: Let be the of the basis
of
Then under the adjoint representation, extended to every
preserves
and the same holds for
any number of factors.
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Proof. |
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Making act on the basis of
we get
if (see the definition of
in Theorem 1),
and 0 in all other cases, which proves is preserved.
The second part follows by induction on the number of factors and:
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Lemma 7: Let and be and and
additive subgroups thereof. If and are preserved by every
then so is
(in
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Proof. |
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Since acts on as
it follows from the binomial expansion that
acts as
whence the lemma.
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Lemma 8: Let be a set of roots such that (a)
and (b)
(e.g. the set of positive roots), arranged in some order. Then
is a basis for the generated by all
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Proof. |
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By the Birkhoff-Witt Theorem applied to the Lie algebra for which
is a basis we see that every is a complex combination of the given elements. We must show all corn efficients are
integers. Write
terms of at most the same total degree. We make act on
copies) and look for the component of
in Any term of
other than the first leads to a zero component since there are either not enough factors (at least one is needed for each
or barely enough but with the wrong distribution (since
is a nonzero element of
only if while the first leads to a non-zero component
only if the and the
are matched up, in all possible permutations. It follows that the component sought is
Now each is a primitive element of (to see this imbed
in a simple system of roots and then use Lemma 1). Since preserves
by Lemma 6 it follows that whence Lemma 8.
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Any formal product of elements of of the form
or
will be called a monomial and the total degree in the its degree.
Lemma 9: If and
then
is an integral combination of
and monomials of lower degree.
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Proof. |
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This holds if obviously and if
by Lemma 5. Assume By Lemma 8 applied to the set
of roots of the form
arranged in the order
we see that
is an integral combination of terms of the form
The map
leads to a grading of the algebra with values in the additive group generated by The
left side of the preceding equation has degree
Hence so does each term on the right, whence are restricted by the
condition
hence also by
Clearly the ordinary degree of the above term, can be as
large as only if and
by the last condition, and then
and by the first, which proves Lemma 9.
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Lemma 10: If and are roots and is any polynomial, then
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Proof. |
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By linearity this need only be proved when is a power of and then it easily follows by
induction on the two exponents starting with the equation
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Observe that each is an integer.
Now we can prove Theorem 2. By the corollary to Lemma 5 each
is in hence so is each of the proposed basis elements. We must show that each element of
is an integral combination of the latter elements, and for this it suffices to show that each monomial is.
Any monomial may, by induction on the degree, Lemma 9, and Lemma 10, be expressed as an integral combination of monomials such that for each
the terms all come together and in the order of the roots prescribed by Theorem 2, then also such that each
is represented at most once, because
The terms may now be brought to the front (see Lemma 10), the resulting polynomial expressed as an integral combination of
by Lemma 4, each term shifted to the position prescribed by Theorem 2, and Lemma 4 used for each
separately, to yield finally an integral combination of basis elements, as required.
Let be a semisimple Lie algebra having Cartan subalgebra Let
be a representation space for We call a vector a
weight vector if there is a linear function on such that
for all
If such a exists, we call the
corresponding a weight of the representation.
Lemma 11: If is a weight vector belonging to the weight then for
a root we have is a weight vector belonging to the weight
if
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Proof. |
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If then
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Theorem 3: If is a semisimple Lie algebra having Cartan subalgebra then
(a) |
Every finite dimensional irreducible contains a nonzero vector
such that is a weight vector belonging to some weight
and
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(b) |
It then follows that if is the subspace of consisting of weight vectors belonging to
then
Moreover, every weight has the form
where the are positive roots. Also,
a weight).
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(c) |
The weight and the line containing are uniquely determined.
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(d) |
for
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(e) |
Given any linear function satisfying (d), then there is a unique finite dimensional
in which is realized as in (a).
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Proof. |
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(a) There exists at least one weight on since acts as an Abelian set of endomorphisms. We introduce a partial
order on the weights by if
a positive root). Since the weights are finite in number, we have a maximal weight
Let be a nonzero weight vector belonging to
Since is not a weight for
we have by Lemma 11 that
(b) and (c) Now let
Let be the Lie subalgebra of
generated by with
Let
and
be the universal enveloping algebras of
and respectively. By the Birkhoff-Witt theorem,
has a basis
has a basis
has a basis
and the universal enveloping algebra of has a basis
where
Hence,
Now is invariant under Also,
since is irreducible,
and
Hence and (b) and (c) follow.
(d) is in the 3-dimensional subalgebra generated by
Hence, by the theory of representations of this subalgebra,
(See Jacobson, Lie Algebras, pp. 83-85.)
(e) See Séminaire "Sophus LIE," Exposé 17.
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Corollary: If is a weight and a root, then
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Proof. |
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This follows from (b) and (d) of Theorem 3 and
for
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Remark: are called the highest weight, a highest weight vector, respectively.
By Theorem 2, we know that the
generated by
has a
Now if
and
denote the generated by
and
respectively, then
Lemma 12: If and is a highest
weight vector, then the component of in
is for some
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Proof. |
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We know that and
Hence the component is nonzero only if Now
acts as an integer on
by Theorem 3 (d), so
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Lemma 13: Let be a point of and a finite subset
of not containing Then there is a polynomial
in variables such that:
(a) |
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(b) |
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(c) |
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Proof. |
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Let
with Set
We see that and
takes the value zero at all other points of
within a box with edges and center For
sufficiently large, this box contains
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If is a vector space over and is a finitely generated (free Abelian) subgroup of
which has a which is a
for we say is a lattice in
We can now state the following corollaries to Theorem 2.
Corollary 1:
(a) |
Every finite dimensional contains a lattice invariant
under all
i.e., is invariant under
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(b) |
Every such lattice is the direct sum of its weight components; in fact, every such additive group is.
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Proof. |
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(a) By the theorem of complete reducibility of representations of semisimple Lie algebras over a field of characteristic 0 (See Jacobson, Lie Algebras, p. 79),
we may assume that is irreducible. Using Theorem 3, we find and set
is finitely generated over since only finitely many monomials in
fail to annihilate
Since and since
spans over
we see that spans over
Before completing the proof of (a), we will first show that if
with
and then there exist
such that
To see this, let be such that the component of
in is nonzero. Then
implies
where is the component of
in We have
by Lemma 12 and by choice of Finally,
suppose a basis for is not a basis for Let be minimal such
that there exist
linearly independent over but linearly dependent over Suppose
Then there exist
such that
We see that
Since
are linearly
independent over we have a contradiction to the choice of
Hence, is a lattice in
(b) Let be any subgroup of the additive group of invariant under
If is a weight, set
For a fixed let
Let be as in Lemma 13 with If
then and acts on like the
projection of onto Thus, if
the projection of to
is in and is the direct sum of its weight components.
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Corollary 2: Let be faithfully represented on a finite dimensional vector space Let
be a lattice in invariant under Let
be the part of which preserves
Then is a lattice, and
where
for all weights of the given In particular,
independent of (But, of course,
is not independent of the representation.)
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Proof. |
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We recall that associated with the representation on there is a representation on the dual space
of called the contragredient representation given by
where
and where
denotes the value of the linear function at If
is the dual lattice in of
i.e.,
then clearly preserves if and only if
preserves We know that
is isomorphic with and that the tensor product of the two representations
corresponds to the representation
of in (See Jacobson, Lie Algebras, p. 22).
Now
is a lattice in since the tensor product of two lattices is a lattice. Also,
is a lattice in since
and
because all and are in
Since preserves
and preserves
by Lemma 7, and hence
preserves the lattice in under the adjoint representation. By Corollary 1 (b),
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Now implies
If spans
over for then
Hence
Thus,
which implies and
Hence,
Example: Let be the 3 dimensional Lie algebra generated by
and with
and
Let
and let be the lattice in spanned by
and Then since the only weights
of the adjoint representation are 0 with
Now is isomorphic with
on a 2 dimensional vector space Here corresponds to
and the weights are with
Hence
and
We are now in a position to transfer our attention to an arbitrary field We have already defined the lattices
and Considering these lattices as
and considering as a
we can form the tensor products,
and
We then have:
Corollary 3:
(a) |
(direct sum) and
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(b) |
(direct sum), each
and
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Proof. |
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This follows from Corollaries 1 and 2.
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Notes and References
This is a typed excerpt of Lectures on Chevalley groups by Robert Steinberg, Yale University, 1967. Notes prepared by John Faulkner and Robert Wilson.
This work was partially supported by Contract ARO-D-336-8230-31-43033.
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