Download §13. Abstract theory of weights

§13. Abstract theory of weights
69
§13. Abstract theory of weights
In this section we describe that part of the representation theory of semisimple Lie algebras
which depends only on the root system. (None of this is needed until Chapter VI.) Let Φ be a
root system in a euclidean space E, with Weyl group W.
§13.1. Weights
Definition 13.1 (Weight). λ ∈ E is called a weight if hλ, αi ∈ Z for all α ∈ Φ. Denote by Λ
the set of all weights.
2(λ, α)
depends linearly on λ, hence Λ is a subgroup of E including Φ. Thanks
(α, α)
2α
| α ∈ Φ} is a
to Exercise 10.1, λ ∈ Λ if and only if hλ, αi ∈ Z for all α ∈ ∆. Φ∨ = {α∨ = (α,α)
2α
∨
∨
∨
root system of E and ∆ = {α = (α,α) | α ∈ ∆} is a base of Φ . hλ, αi ∈ Z for all α ∈ ∆, i.e.,
(λ, α∨ ) ∈ Z for all α ∈ ∆ implies (λ, α∨ ) ∈ Z for all α ∈ Φ, i.e, hλ, αi ∈ Z for all α ∈ Φ. Denote
l
L
by Λr the root lattice (= subgroup of Γ generated by Φ)= Zαi .
Since hλ, αi =
i=1
Definition 13.2 (dominant weights). Fix a base ∆ ⊆ Φ, λ ∈ Λ is called dominant if all the
integers hλ, αi (α ∈ ∆) are nonnegative, strongly dominant if these integers are positive. Let
Λ+ be the set of all dominant weights.
T
T
In the language of §10.1, Λ+ = Λ C(∆) and Λ C(∆) is the set of strongly dominant
i
weights. If ∆ = {α1 , . . . , αl }, then the vectors (α2α
again form a basis of E. Let λ1 , . . . , λl be
i ,αi )
2(λ ,α )
the dual basis (relative to the inner product on E): (αji,αjj) = δij . Since all hλi , αi (α ∈ ∆) are
nonnegative integers, then λi are dominant
weights. We call them the fundamental
dominant
P
P
weights (relative to ∆).
If
λ
=
m
λ
∈
E
is
arbitrary,
then
hλ,
α
i
=
m
hλ
,
αj i = mj ,
i
i
j
i
i
P
which implies that λ = hλ, αi iλi . Therefore, Λ is a lattice with basis (λi | 1 ≤ i ≤ l), and
λ ∈ Λ+ if and only if all mi ≥ 0. (Cf. Figure 1, for type A2 .)
α2
λ2
λ1
α1
Figure 1.
It is an elementary fact about lattices that Λ/Λr must be a finite group (called the fundal
P
mental group of Φ). We can see this directly as follows. Write αi =
mij λj (mij ∈ Z).
j=1
70
Lie algebras
Then hαi , αk i =
l
P
mij hλj , αk i = mik . In other words, the Cartan matrix expresses the change
j=1
of basis. To write the λj in terms of the αi , we have only to invert the Cartan matrix; its
determinant (cf. Exercise 11.2) is the sole denominator involved, so this measures the index of
Λr in Λ. For example, in type A1 , α1 = 2λ1 . (This is the only case in which a simple root
is
dominant,for reasons which will later become apparent.) In type A2 , theCartanmatrix is
1 2 1
2 −1
, so α1 = 2λ1 − λ2 and α2 = −λ1 + 2λ2 . Inverting, we get
, so that
−1 2
3 1 2
λ1 = 13 (2α1 + α2 ) and λ2 = 31 (α1 + 2α2 ). By computing determinants of Cartan matrices one
verifies the following list of orders for the fundamental groups Λ/Λr in the irreducible cases:
Al : l + 1;
Bl , Cl , E7 : 2;
Dl : 4;
E6 : 3;
E8 , F4 , G2 : 1.
With somewhat more labor one can calculate explicitly the λi in terms of the αj . This information is listed in Table 1, for the convenience, although strictly speaking we shall not need
it in what follows. The exact structure of the fundamental group can be found by computing
elementary divisors, or can be deduced from Table 1 once the latter is known (Exercise 4).
Table 1. fundamental dominant weights.
Al :
Bl :
Cl :
Dl :
E6 :
E7 :
h
(l − i + 1)α1 + 2(l − i + 1)α2 + · · · + (i − 1)(l − i + 1)αi−1 + i(l − i + 1)αi
i
+i(l − i)αi+1 + · · · + iαl
λi = α1 + 2α2 + · · · + (i − 1)ai−1 + i αi + αi+1 + · · · + αl (i < l)
λl = 21 α1 + 2α2 + · · · + lαl
λi = α1 + 2α2 + · · · + (i − 1)ai−1 + i αi + αi+1 + · · · + αl−1 + 21 αl
λi = α1 + 2α2 + · · · + (i − 1)ai−1 + i αi + · · · + αl−2 + 21 i(αl−1 + αl ) (i < l − 1)
λl−1 = 12 α1 + 2α2 + · · · + (l − 2)αl−2 + 21 lαl−1 + 12 (l − 2)αl
λl = 12 α1 + 2α2 + · · · + (l − 2)αl−2 + 21 (l − 2)αl−1 + 21 lαl
P
( qi αi is abbreviated (q1 , . . . , ql ) in the following lists.)
λi =
1
l+1
λ1 = 13 (4, 3, 5, 6, 4, 2),
λ2 = (1, 2, 2, 3, 2, 1) ,
λ3 = 13 (5, 6, 10, 12, 8, 4)
λ4 = (2, 3, 4, 6, 4, 2),
λ5 = 31 (4, 6, 8, 12, 10, 5) ,
λ6 = 13 (2, 3, 4, 6, 5, 4)
λ1 = (2, 2, 3, 4, 3, 2, 1),
λ2 = 21 (4, 7, 8, 12, 9, 6, 3) ,
λ4 = (4, 6, 8, 12, 9, 6, 3), λ5 = 21 (6, 9, 12, 18, 15, 10, 5) ,
λ3 = (3, 4, 6, 8, 6, 4, 2)
λ6 = (2, 3, 4, 6, 5, 4, 2)
λ7 = 12 (2, 3, 4, 6, 5, 4, 3)
E8 :
λ1 = (4, 5, 7, 10, 8, 6, 4, 2), λ2 = (5, 8, 10, 15, 12, 9, 6, 3) , λ3 = (7, 10, 14, 20, 16, 12, 8, 4)
λ4 = (10, 15, 20, 30, 24, 18, 12, 6),
λ6 = (6, 9, 12, 18, 15, 12, 8, 4),
F4 :
λ1 = (2, 3, 4, 2),
G2 :
λ1 = (2, 1),
λ7 = (4, 6, 8, 12, 10, 8, 6, 3),
λ2 = (3, 6, 8, 4) ,
λ2 = (3, 2)
λ5 = (8, 12, 16, 24, 20, 15, 10, 5)
λ3 = (2, 4, 6, 3),
λ8 = (2, 3, 4, 6, 5, 4, 3, 2)
λ4 = (1, 2, 3, 2)
§13. Abstract theory of weights
71
§13.2. Dominant weights
The Weyl group W of Φ preserves the inner product on E, hence leaves Λ invariant. (In
fact, we already made the more precise observation that σαi λj = λj − δij αi . Orbits of weights
under W occur frequently in the study of representations. In view of Lemma 10.16 and Exercise
10.14, we can state:
Lemma 13.3. Each weight is conjugate under W to one and only one dominant weight. If λ
is dominant, then σ(λ) ≺ λ for all σ ∈ W, and if λ is strongly dominant, then σ(λ) = λ only
when σ = idE .
T
Proof. Since λ ∈ Λ, then there is σ ∈ W such that σ(λ) ∈ C(∆), i.e., σ(λ) ∈ Λ C(∆) = Λ+ .
For γ ∈ E, let σ ∈ W such that (σ(γ), δ) is maximal. Then for any α ∈ ∆, we have
(σ(γ), δ) ≥ ((σα σ)(γ), δ) = (σ(γ), δ − α) = (σ(γ), δ) − (σ(γ), α)
T
i.e., (σ(γ), α) ≥ 0. This implies that σ(γ) ∈ C(∆). If σ1 (λ), σ2 (λ) ∈ Λ C(∆) = Λ+ , then by
Lemma 10.16 we have σ1 (λ) = σ2 (λ). If λ is dominan. Let σ = σαi1 · · · σαit ∈ W with αik ∈ ∆
and t = `(σ). Then by Lemma 10.11 or Corollary 10.12, we have σαi1 · · · σαit−1 (αit ) ∈ Φ+ . Now
by induction on `(σ) we have
λ − σ(λ) = λ − σαi1 · · · σαit−1 λ − hλ, αit iαit = λ − σσαit (λ) + hλ, αit iσαi1 · · · σαit−1 (αit ) 0.
Assume σ(λ) = λ and σ 6= idE , then there is α ∈ ∆ such that β = σ(α) ∈ Φ− . Now we have
0 < hλ, αi = hσ(λ), σ(α)i = hλ, βi =
X 2(λ, bi αi )
i∈J
(α, α)
< 0.
As a subset of E, Λ is partially ordered by the relation: λ µ if and only if λ − µ is a sum
of positive roots see §10.1. Unfortunately, this ordering does not have too close a connection
with the property of being dominant; for example, it is easy to have µ dominant, µ ≺ λ, but λ
not dominant (Exercise 2).
Examples 13.4. For A2 , we take ∆ = {α1 = 2λ1 − λ2 , α2 = 2λ2 − λ1 } as in §13.1, then the
fundamental dominant weights are λ1 = 13 (2α1 + α2 ) and λ2 = 13 (α1 + 2α2 ). Take λ = 3λ1 − λ2 ∈
Λ \ Λ+ , but we have λ − α1 = 3λ1 − λ2 − (2λ1 − λ2 ) = λ1 ∈ Λ+ .
Our next lemma shows, however, that dominant weights are not too badly behaved relative
to ≺.
Lemma 13.5. Let λ ∈ Λ+ . Then the number of dominant weights µ ≺ λ is finite.
Proof. Since λ + µ ∈ Λ+ and λ − µ is a sum of positive roots, 0 ≤ (λ + µ, λ − µ) = (λ, λ) − (µ, µ).
Thus µ lies in the compact set {x ∈ E | (x, x) ≤ (λ, λ)}, whose intersection with the discrete set
Λ+ is finite.
§13.3. The weight δ
Recall (Corollary 10.10) that δ =
1
2
P
α, and that σi (δ) = δ − αi (1 ≤ i ≤ l). Of course, δ
α0
may or may not lie in the root lattice Λr (cf. type A1 ); but δ does lie in Λ. More precisely:
Lemma 13.6. δ =
l
P
j=1
λj , so δ is a (strongly) dominant weight.
72
Lie algebras
Proof. Since δ − hδ, αi iαi = σi (δ) = δ − αi , hδ, αi i = 1 for all 1 ≤ i ≤ l. But δ =
l
P
hδ, αi iλi (see
i=1
§13.1), so the lemma follows.
The next lemma is merely an auxiliary result, needed in §13.4.
Lemma 13.7. Let µ ∈ Λ+ and σ(ν) = µ for some σ ∈ W and ν ∈ Λ. Then (ν + δ, ν + δ) ≤
(µ + δ, µ + δ), with equality only if ν = µ.
Proof. We have
(ν + δ, ν + δ) = (σ(ν + δ), σ(ν + δ)) = (µ + σδ, µ + σδ) = (µ + δ, µ + δ) − 2(µ, δ − σδ).
Since µ ∈ Λ+ , and δ − σδ is a sum of positive roots (Lemmas 13.3 and 13.6), the right hand
side is ≤ (µ + δ, µ + δ), with equality only if (µ, δ − σδ) = 0, i.e., (µ, δ) = (µ, σδ) = (ν, δ), or
(µ − ν, δ) = 0. But µ − ν is a sum of positive roots (Lemma 13.3) and δ is strongly dominant
(Lemma 13.6), so µ = ν.
Lecture 21. May 25th, 2016
§13.4. Saturated sets of weights
Certain finite sets of weights, stable under W, play a prominent role in representation theory.
Definition 13.8. (1). We call a subset Π of Λ saturated if for all λ ∈ Π, α ∈ Φ, and i
between 0 and hλ, αi, the weight λ − iα also lies in Π.
F Notice first that any saturated set is automatically stable under W, since σα (λ) = λ −
hλ, αiα and W is generated by reflections.
(2). We say that a saturated set Π has highest weight λ if λ ∈ Π and µ ≺ λ for all µ ∈ Π.
+
Indeed, we can
there is σ ∈ W such that σ(λ) ∈ Λ+ and
prove λ ∈ Λ . By Lemma 13.3,
−1
+
λ=σ
σ(λ) ≺ σ(λ) ≺ λ, i.e., λ = σ(λ) ∈ Λ
Examples 13.9. (1). The set consisting of 0 alone is saturated, with highest weight 0.
(2). The set Φ of all roots of a semisimple Lie algebra, along with 0, is saturated. In case
Φ is irreducible, there is a unique highest root (relative to a fixed base ∆ of Φ)
(Lemma 10.19) so Π has this root as its highest weight (why?). For α, β ∈ Φ, since
σα (β) = β − hβ, αiα ∈ Φ and the α-string through β is unbroken, hence for i between 0
and hβ, αi, we get β − iα ∈ Φ. Moreover, suppose β is the highest root, then β + αi ∈
/Φ
l
P
for 1 ≤ i ≤ l. Hence by Lemma 9.8 hβ, αi i ≥ 0, and this implies β =
hβ, αi iλi ∈ Λ+ .
i=1
Lemma 13.10. A saturated set Π of weights having highest weight λ must be finite.
T
T
Proof. By Lemma 13.5, we have |Π Λ+ | < +∞. Now |Π| = |W · (Π Λ+ )| < +∞ follows from
Lemma 13.3.
Lemma 13.11. Let Π be saturated, with highest weight λ. If µ ∈ Λ+ and µ ≺ λ, then µ ∈ Π.
§13. Abstract theory of weights
Proof. Suppose µ0 = µ +
P
73
kα α ∈ Π (kα ∈ Z+ ). (Important: We do not assume that µ0 is
α∈∆
dominant.) We shall show how to reduce one of the kα by one while still remaining in Π, thus
eventually arriving at the conclusion that µ ∈ Π. Of course, our startingP
point is P
the fact that
0
0
λ itself is such a µ . Now suppose µ 6= µ, so some kα is positive. From (
kα α,
kα α) > 0,
α∈∆
α∈∆
P
P
we deduce that (
kα α, β) > 0 for some β ∈ ∆, with kβ > 0. In particular, h
kα α, βi is
α∈∆
α∈∆P
positive. Since µ is dominant, hµ, βi is nonnegative. Therefore, hµ0 , βi = hµ, βi + h
kα α, βi
α∈∆
is positive. By definition of saturated set, it is now possible to subtract β once from µ0 without
leaving Π, thus reducing kβ by one.
From Lemma 13.11 emerges a very clear picture of a saturated set Π having
highest weight λ: Π consists of all dominant weights lower than or equal to λ in the partial
ordering, along with their conjugates under W. In particular, for given λ ∈ Λ+ , at most one
such set Π can exist. Conversely, given λ ∈ Λ+ , we may simply define Π to be the set consisting
of all dominant weights below λ, along with their W-conjugates. Since Π is stable under W,
it can be seen to be saturated (Exercise 10), and thanks to Lemma 13.3, Π has λ as highest
weight.
To conclude this section, we prove an inequality which is essential to the application of
Freudenthal’s formula in §22.
Lemma 13.12. Let Π be saturated, with highest weight λ. If µ ∈ Π, then (µ + δ, µ + δ) ≤
(λ + δ, λ + δ), with equality only if µ = λ.
Proof. In view of Lemma 13.7, it is enough to prove this when µ is dominant. Write µ = λ − π,
where π is a sum of positive roots. Then
(λ + δ, λ + δ) − (µ + δ, µ + δ) =(λ + δ, λ + δ) − (λ + δ − π, λ + δ − π)
=(λ + δ, π) + (π, µ + δ) > (λ + δ, π)
>0,
the inequalities holding because µ + δ and λ + δ are dominant. Equality holds only if π = 0,
since λ + δ is strongly dominant.