THE HARISH-CHANDRA ISOMORPHISM FOR sl(2, C)
SEAN MCAFEE
1. summary
Given a reductive Lie algebra g and choice of Cartan subalgebra h, the HarishChandra map gives an isomorphism between the center z of the universal enveloping
algebra U(g) of g and the algebra of Weyl group invariant symmetric tensors of h,
denoted S(h)W . We define these objects and give a sketch of the proof, using
sl(2, C) as a motivating example. The proofs and examples mirror those found in
Cohomological Induction and Unitary Representations by Knapp and Vogan.
2. Basic Objects
Given any Lie algebra g, we define the tensor algebra T (g) of g as
∞
L
T (g) :=
xi1 ⊗ · · · ⊗ xik , (xj ∈ g).
k=0
We can then define the universal enveloping algebra U(g) by
U(g) := T (g)/I,
where I is the ideal in T (g) generated by elements of the form x⊗y−y⊗x−[x, y] for
x, y ∈ g. This has the universal property of being the smallest associative algebra
containing g. We denote the center of U(g) by z.
Given a Cartan subalgebra h of g, we may consider the universal enveloping
algebra U(h) of h. Since h is abelian, we have U(g) ∼
= S(h), the symmetric tensors
over h. Given the Weyl group of the root system Φ determined by g, h, we write
S(h)W for the Weyl group invariants of S(h).
For each root α ∈ Φ, let eα be a nonzero root vector. Pick a positive root system
Φ+ in Φ. We define
P
P
P :=
U(g)eα and N :=
e−α U(g).
α∈Φ+
α∈Φ+
Given a choice of positive roots β1 , ..., βk , and given a basis h1 , ..., h` of h, the
Poincaré-Birkoff-Witt Theorem gives us a basis of U(g) of monomials of the form
pk
1
k
` p1
eq−β
. . . eq−β
hm1 . . . hm
` eβ1 . . . eβk .
1
k 1
In the case of g = sl(2, C), we have g = CH ⊕ CX ⊕ CY , where
1 0
0 1
0 0
H=
,X =
, and Y =
.
0 −1
0 0
1 0
In this case, we have h = CH, and Φ = {±α}, where α ∈ h∗ is given by α(H) = 2.
Choosing X as our eα and Y as our e−α , we have monomials of the form
Y `H mX n
1
2
SEAN MCAFEE
as a basis for U(g). We also have
P = U(g)X and N = Y U(g).
Theorem 2.1. Let g = sl(2, C). Then, with P, N as above, we have
a) U(g) = U(h) ⊕ (P + N )
b) For z ∈ z, the P + N part of z lies in P.
Proof.
a) By Poincaré-Birkhoff-Witt, we have that monomials of the form Y ` H m X n
form a basis for U(g). This gives us
U(g) = U(h) + P + N .
Let p ∈ P. Then p is a linear combination of terms of the form (Y ` H m X n )X =
Y ` H m X n+1 , where `, m, n ∈ Z≥0 . Thus, each monomial term of p has a
tensor power of X greater than 0. Similarly, each monomial term of n ∈ N
has a tensor power of Y greater than 0. This shows that any element of
P + N is a linear combination of terms Y ` H m X n with either ` > 0 or
n > 0. Since each element of U(h) is a linear combination of tensor powers
of H, we have that ` = n = 0 for each such term. This proves a).
b) (Remark: we will abuse notation below and use `, m, n repeatedly to describe tensor powers in different monomials contained in U(g)). Again, we
have that the monomials Y ` H m X n are a basis of U(g); they are also weight
vectors of ad(h) with weight −2` + 0m + 2n = −2` + 2n. Let z ∈ z. Then
[H, z] = Hz − zH = 0, hence for each monomial term Y ` H m X n of z we
have −2` + 2n = 0. By the proof of part a), each linear term of the P + N
part of z has ` > 0 or n > 0. Thus, since −2` + 2n = 0 we must have both
`, n > 0. Hence the P + N part of z lies in P, as needed.
3. Construction of the Harish-Chandra map
Although the discussion below applies to any reductive Lie algebra g, we will only
consider g = sl(2, C) in the construction of the Harish-Chandra homomorphism γ
and in the proof of the isomorphism z ∼
= S(h)W induced by γ. We will use the basis
H, X, Y of g as outlined in section 2. Note that the Weyl group W for g = sl(2, C)
consists of two elements, one trivial and one sending α → −α. This gives an action
of W on H sending H → −H. Let z ∈ z. By the previous section, we have
z ∈ U(h) ⊕ P. Let γ 0 : z → U(h) be the projection onto the U(h) component of z.
Define ρ := 21 α, where α is as in section 2. Now define a linear map τ : h → U(h)
by
τ (h) = h − ρ(h).
By the natural embedding of h into U(h), we can extend τ to a map U(h) → U(h).
Finally, we define
γ = τ ◦ γ0.
This is the Harish-Chandra map.
3
Example 3.1. It can be shown that the Casimir element Ω = 12 H 2 + XY + Y X
is contained in z by checking that it commutes with H, X, and Y . As in section 2,
X corresponds to the positive root α. Using the fact that H = [X, Y ] = XY − Y X,
we may rewrite
Ω = ( 21 H 2 + H) + (2Y X) ∈ U(h) ⊕ P.
We apply the Harish-Chandra map, using the fact that ρ(H) = 21 α(H) = 1:
γ(Ω) = τ (γ 0 (Ω)) = τ ( 21 H 2 + H) = 12 (H − 1)2 + (H − 1) = 12 H 2 − 12 .
Notice that, since W maps H → −H, the projection 12 H 2 + H is not W -invariant,
but after composing with τ , we arrive at the W -invariant 12 H 2 − 21 .
4. Proof of the isomorphism z ∼
= S(h)W for g = sl(2, C).
We will show the following:
a) Im(γ) ⊆ S(h)W .
b) γ does not depend on the choice of Φ+ .
c) γ is multiplicative.
d) γ is 1 − 1.
e) γ is onto.
We will make frequent use of the following observation: given λ ∈ h∗ , we may
extend λ to U(h). Thus by definition of the Harish-Chandra map we have (for any
z ∈ z) the equality λ(γ(z)) = (λ − ρ)(γ 0 (z)).
Proof that Im(γ) ⊆ S(h)W : The idea of the proof is as follows: given some λ ∈ h∗ ,
we create a certain U(g) module V (λ) such that z ∈ z acts on V (λ) by the scalar
(λ − ρ)(γ 0 (z)) = λ(γ(z)). We then show that, by the same construction (and for λ
dominant integral), V (sα λ) is isomorphic to a submodule of V (λ). This will imply
that λ(γ(z)) = (sα λ)(γ(z)) for all dominant integral λ and all z ∈ z. Hence, since
W is generated by the simple reflections sα , and since h∗ is generated by dominant
integral weights, Im(γ) ∈ S(h)W .
Let λ ∈ h∗ . We define a one-dimensional representation Cλ−ρ of b = CH + CX
by (for w ∈ C)
H · w = ((λ − ρ)(H))(w) and X · w = 0.
We can then define the U(g)-module
V (λ) = U(g) ⊗U (b) Cλ−ρ .
Let z ∈ z. Then we have z = h + p, where h ∈ U(h) and p ∈ P. Let u ⊗ 1 be an
arbitrary element of V (λ). Then we have
z(u ⊗ 1) = zu(1 ⊗ 1) = uz(1 ⊗ 1) = u(h(1 ⊗ 1) + p(1 ⊗ 1))
= u((1 ⊗ h · 1) + (1 ⊗ p · 1)) = u((1 ⊗ (λ − ρ)(h)) + (1 ⊗ 0))
= (λ − ρ)(h)u(1 ⊗ 1) = (λ − ρ)(h)(u ⊗ 1)
Thus we have shown that z acts on V (λ) by the scalar (λ − ρ)(γ 0 (z)) = λ(γ(z)).
Let us require that λ ∈ h∗ be dominant integral (thus 2(λ,α)
(α,α) ∈ Z≥0 ).We now
create a submodule M of V (λ) and show that it is isomorphic to V (sα λ). Define
4
SEAN MCAFEE
M to be the submodule of V (λ) generated (over U(g)) by the element v = Y m (1⊗1).
Define a map φ : Csα λ−ρ → M by φ(1) = v. We claim this map is U(b) equivariant.
We do this by showing that an element b = h + n ∈ b acts on 1 ∈ Csα λ−ρ and
v ∈ M by the same scalar (sα λ − ρ)(h) and extending the action to U(b).
First, observe that
H·v = H·Y m (1⊗1) = H·Y m ⊗1+Y m ⊗H·1 = −mα(H)+(λ−ρ)(H) = (sα λ−ρ)(H),
thus H acts equivariantly with respect to φ. We now check that X acts by 0 on v,
completing the claim that φ is U(b) equivariant. We have
X · v = XY m (1 ⊗ 1) = [X, Y m ](1 ⊗ 1) + Y m X(1 ⊗ 1) = [X, Y m ](1 ⊗ 1).
By repeated application of the property [X, Y ] = XY − Y X and the observation
that [Y, [Y, [Y, X]]] = 0, we have
[X, Y m ] = XY m − Y m X = XY (Y m−1 ) − Y m X
= ([X, Y ] + Y X)Y m−1 − Y m X = · · · = Y m X + mY m−1 (H − (m − 1)I) − Y m X
= mY m−1 (H − (m − 1)I).
Thus we have
X·v = [X, Y m ](1⊗1) = mY m−1 (H−(m−1)I)(1⊗1) = mY m−1 H(1⊗1)−m(m−1)Y m−1 (1⊗1)
= mY m−1 (H(1⊗1)−(m−1)(1⊗1)) = mY m−1 ((λ(H)−ρ(H))(1⊗1)−(m−1)(1⊗1))
= mY m−1 ((m − 1)(1 ⊗ 1) − (m − 1)(1 ⊗ 1)) = 0,
hence, by the above discussion, φ is a U(b) equivariant map, as needed.
We can view V (sα λ) = U(g) ⊗U (b) Csα λ−ρ as the induced representation of U(g)
over the U(b) module Csα λ−ρ . Thus by Frobenius Reciprocity we have
Homb (Cs λ−ρ , M ) ∼
= Homg (V (sα λ), M ).
α
e of V (sα λ) into M which takes 1 ⊗ 1 to v.
This gives us a U(g) map (call it φ)
This means that
V (sα λ) = U(g)(1 ⊗ 1) maps to U(g)v = M,
hence φ̃ is onto. To see that φe is one to one, we first observe there is a U(n− )
isomorphism
V (sα λ) ∼
= (U(n− ⊗C U(b)) ⊗U (b) Csα λ−ρ ∼
= U(n− ) ⊗C (U(b) ⊗U (b) Csα λ−ρ )
∼
= U(n− ) ⊗C Cs λ−ρ .
α
So, letting u ⊗ 1 be an arbitrary element of V (sα λ), where u 6= 0 ∈ U(n− ), we
have that
uv = uY m (1 ⊗ 1) 6= 0
since uY m is a nonzero element of U(n− ). Thus the kernel of φe is trivial, hence the
map is one to one. Thus, we have shown that V (sα λ) is isomorphic to a submodule
of V (λ). We have already seen that z acts on V (λ) and V (sα λ) by the scalars
5
λ(γ(z)) and (sα λ)(γ(z)), respectively. Therefore, since V (sα λ) is a submodule of
V (λ), we have that
λ(γ(z)) = (sα λ)(γ(z))
for all z ∈ z and dominant integral λ. Thus, since h∗ is generated by the dominant
integral weights, and since any element of U(h) (in particular γ(z)) is determined
by the effect of all λ ∈ h∗ on it, we must have that sα (γ(z)) = γ(z) for all z ∈ z,
hence Im(γ) ⊆ S(h)W , as needed.
Proof that γ does not depend on the choice of Φ+ : The root system of sl(2, C) contains α : H 7→ 2 and −α : H 7→ −2. Thus, we have two choices of positive root
system:
∆+ = {α} and ∆+
# = {−α} .
Using weight space vectors X and Y as before, we define
P = U(g)X and P# = U(g)Y.
Fix z ∈ z. Then, by Theorem 2.1 b), we can write z as an element in U(h) ⊕ P or
0
U(h) ⊕ P# . We can thus define the Harish-Chandra maps γ 0 and γ#
according to
our choice of positive root. We may also define ρ and ρ# accordingly. The Weyl
group of sl(2, C) has two elements:
and w : α 7→ −α. We can realize w
the identity
0 1
explicitly as Ad(g)|h , where g =
∈ SL(2, C):
−1 0
w · α(H) = α(w−1 · (H)) = α(Ad(g −1 )H)
0 −1
1 0
0
= α(
1 0
0 −1
−1
1
) = α(−H)
0
= −α(H).
We wish to show that γ# (z) = γ(z) for each z ∈ z. First, we prove the following
lemma:
0
Lemma 4.1. With w as above, we have wγ 0 (z) = γ#
(z) for every z ∈ z.
Proof. Let z ∈ z. We may write z = h + p ∈ U(h) ⊕ P. Thus, z − γ 0 (z) = p ∈ P.
Writing p = uX for some u ∈ U(g), we have
Ad(g)(z − γ 0 (z)) = Ad(g)(uX) = Ad(g)(u) Ad(g)(X)
= Ad(g)(u)gXg −1 = Ad(g)(u)(−Y ) = −u0 Y ∈ P# .
Now, for any z ∈ z and any X ∈ g we have ad(X)(z) = 0, thus Ad(x)(z) = z for
any x ∈ G. In particular, Ad(g)(z) = z. Thus we have
Ad(g)(z − γ 0 (z)) = z − Ad(g)γ 0 (z) ∈ P# .
0
Since Ad(g)|h = w, this shows that wγ 0 (z) = γ#
(z), as required.
Now, observe that wρ = ρ# ; then for any λ ∈ h∗ we have
0
λ(γ# (z)) = (λ − ρ# )(γ#
(z)) = (λ − wρ)(wγ 0 (z))
= w−1 (λ − wρ)(γ 0 (z)) = (w−1 λ − w−1 wρ)(γ 0 (z))
= (w−1 λ − ρ)(γ 0 (z)) = w−1 λ(γ(z)) = λ(w−1 γ(z))
= λ(γ(z)),
6
SEAN MCAFEE
the last equality following from the fact that γ(z) ∈ U(h)W . Since our choice of λ
was arbitrary, we have shown that γ# (z) = γ(z) for every z ∈ z, as needed.
Proof that γ is multiplicative: We claim that, for z1 , z2 ∈ z,
γ 0 (z1 z2 ) = γ 0 (z1 )γ 0 (z2 ).
Indeed, consider
z1 z2 − γ 0 (z1 )γ 0 (z2 ) = z1 z2 − z1 γ 0 (z2 ) + z1 γ 0 (z2 ) − γ 0 (z1 )γ 0 (z2 )
= z1 (z2 − γ 0 (z2 )) + γ 0 (z2 )(z1 − γ 0 (z1 )) ∈ P + P = P.
Thus we have
z1 z2 = γ 0 (z1 )γ 0 (z2 ) + (z1 z2 − γ 0 (z1 )γ 0 (z2 )) ∈ U(h) ⊕ P,
so by uniqueness of the direct sum, we have γ 0 (z1 )γ 0 (z2 ) = γ 0 (z1 z2 ), as claimed.
Furthermore, we have that the translation τ is an algebra isomorphism, hence the
composition γ = τ ◦ γ 0 is multiplicative, which completes the proof.
Proof that γ is one to one: We consider G = SU (2), which has g as its Lie algebra.
For any f ∈ C ∞ (G) and X ∈ g, we have X act as a left-invariant differential
operator on f by
d
X · f (g) := (f (g exp(tX))|t=0 .
dt
We can extend this to a U(g) action on f ∈ C ∞ . The idea of the proof is to show
that any z ∈ z is identically zero as a differential operator. Then we show that U(g)
is injective into the space of differential operators on C ∞ (G), hence z = 0 as an
element of U(g). Suppose that γ(z) = 0. Then we have γ 0 (z) = 0, hence for any
highest weight λ of a finite dimensional representation of U(g) we have λ(γ 0 (z)) = 0;
z acts by this scalar in the representation (this can be seen by looking at the action
of λ on individual basis elements Y ` H m X n of U(g)). This implies that π(z) = 0
for every finite dimensional representation of G, since any such representation is
the direct sum of irreducibles.
Since G is compact and connected, any finite dimensional representation π ofG
can be regarded as unitary. Given such a π, let (π(g)v1 , v2 ) be a matrix coefficient,
and let X ∈ g. Then X acts as a differential operator by
d
d
(π(g exp(tX)v1 , v2 )|t=0 = (π(g)π(exp(tX))v1 , v2 )|t=0
dt
dt
d
∗
= (π(exp(tX))v1 , π(g) v2 )|t=0 = (π(X)v1 , π(g)∗ v2 )
dt
= (π(g)π(X)v1 , v2 ).
X(π(g)v1 , v2 ) =
As mentioned above, we can extend this action to an arbitrary element u ∈ U(g),
hence
u(π(g)v1 , v2 ) = (π(g)π(u)v1 , v2 ).
In particular, for z as above we have
z(π(g)v1 , v2 ) = (π(g)π(z)v1 , v2 ) = 0,
thus z acts as zero on any matrix coefficient of π. We then see that z acts as zero on
any matrix coefficient of any finite dimensional representation of G. This extends
to sums of tensor products of matrix coefficients, which by Peter-Weyl are dense in
7
C ∞ (G). Thus we have that z is identically zero as a differential operator. Finally,
we check that U(g) is injective into the space of left-invariant differential operators
on C ∞ (G), which will show that z = 0 as an element of U(g), completing the proof.
Suppose that, for X, Y ∈ g, we have
d
d
(f (g exp(tX))|t=0 = (f (g exp(tY ))|t=0
dt
dt
for every f ∈ C ∞ (G) and g ∈ G. In particular, this holds for g = 1 and for all
polynomials f . Hence f (X) = f (Y ) for every polynomial, therefore X = Y .
Proof that γ is onto: First, recall that the Weyl group W corresponding to (g =
s(2, C), h = CH) consists of two elements: the identity and w : H → −H. This
means that the W invariant elements of U(h) are precisely the polynomials in H 2 .
Observe that
1
1
1 1
1
1
γ(Ω + ) = γ(Ω) + γ( ) = H 2 − + = H 2 .
2
2
2
2 2
2
Thus, by the multiplicativity and additivity of γ, we have that Im(γ) = S(h)W , as
needed.
5. The center of U(g) for g = sl(2, C)
We can now completely describe the center z of U(g) in the case of g = sl(2, C).
By the proof that γ is onto, we see that the image under γ of the Casimir element
Ω generates S(h)W . By the injectivity of γ, then, we have that Ω generates z, i.e.
z is the polynomial algebra of Ω.