AN ELEMENTARY PROOF OF THE INJECTIVITY OF THE
HARISH-CHANDRA HOMOMORPHISM FOR sl(2, C)
SEAN MCAFEE
1. Idea of proof
Suppose we are given a nonzero z ∈ z such that γ(z) = 0; this implies that
γ 0 (z) = 0 as well. We have shown previously that z ∈ U(h) ⊕ U(g)X, hence the
above implies that z lies entirely in U(g)X. We may then write z = uX for some
u ∈ U(g). Now, for any v ∈ U(g) we have
zv = vz = vuX ∈ U(g)X,
that is, multiplying z on the left or right by an element of U(g) will result in an
element completely contained in U(g)X. The proof will amount to multiplying z on
the right by Y and on the left by X sufficiently many times to arrive at an element
of U(h) ⊕ U(g)X with a nontrivial U(h) component. This contradiction will imply
that no such nonzero z exists, completing the proof.
2. Proof of Injectivity
We have that z ∈ U(g)x. By the Poincaré-Birkhoff-Witt Theorem, then, we have
that z is a sum of monomial terms that each look like cY ` H m X n , where Y, H, X
are the usual basis of sl(2, C) and c is some scalar in C. Since z ∈ U(g)X, we have
that n > 0 for each such term. Also, since z is central, we have that each of these
terms is a zero weight for the adjoint representation, hence we must have ` = n in
each term (for details, see the proof that central elements lie in U(h) ⊕ U(g)X). We
look at the collection of terms such that n is minimal. The choice will not matter;
as we will see by the following proposition, our process of multiplication will effect
all such terms equally for our purposes.
Proposition 2.1. Given an arbitrary monomial Y ` H m X n , n > 0, consider the
monomial X k (Y ` H m X n )Y k . Then
a) for k < n, we have that X k (Y ` H m X n )Y k ∈ U(g)X.
b) for k = ` = n, we have that
X k (Y ` H m X n )Y k = X n (Y n H m X n )Y n ∈ U(h) ⊕ U(g)X,
with a nontrivial U(h) component.
Assume this proposition holds, and consider X n zY n with n minimal as described
above. Distributing the left and right multiplication over all terms of z will give us a
collection of terms in U(h)⊕U(g)X, some of which have nontrivial U(h) component.
To show that there is no cancellation among these U(h) components, we will see
during the course of the proof that the degree of the U(h) component (which will
be a tensor power of H) in X k (Y k H m X k )Y k is equal to m + 2. We can choose
m to be minimal among monomials with X of minimal degree, thus since there
can be no cancellation among different tensor powers of H, the U(h) component of
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SEAN MCAFEE
X n zY n will be nontrivial. By the discussion in section 1, then, we will have proved
injectivity.
Proof. Given Y ` H m X n , we calculate X(Y ` H m X n )Y . The proposition will then
follow by a simple induction. We will use the following properties, which can be
shown by brute calculation and repeated use of the Lie bracket in sl(2, C).
1)
2)
3)
4)
5)
X n H = −2nX n + HX n
X n Y = (n − n2 )X n−1 + nHX n−1 + Y X n
H m Y = c0 Y +c1 Y H +c2 Y H 2 +· · ·+cm−1 Y H m−1 +Y H m , for some cj ∈ Z
XY ` = (` − `2 )Y `−1 + `Y `−1 H + Y ` X
XH m = c0 X + c1 HX + · · · + cm−1 H m−1 X + H m X, for some cj ∈ Z.
We now calculate (Y ` H m X n )Y :
(Y ` H m X n )Y = (Y ` H m X n−1 (H + Y X) = Y ` H m X n−1 H + Y ` H m X n−1 Y X
= Y ` H m (−2(n − 1)X n−1 + HX n−1 ) + Y ` H m X n−1 Y X
= −2(n − 1)Y ` H m X n−1 + Y ` H m+1 X n−1 +
h
i
Y ` H m ((n − 1) − (n − 1)2 )X n−2 + (n − 1)HX n−2 + Y X n−1 X
= −2(n − 1)Y ` H m X n−1 + Y ` H m+1 X n−1 +
(−n2 + 3n − 2)Y ` H m X n−1 + (n − 1)Y ` H m+1 X n−1 + Y ` H m Y X n
= (n−n2 )Y ` H m X n−1 +nY ` H m+1 X n−1 +Y ` (c0 Y +· · ·+cm−1 Y H m−1 +Y H m )X n
= (n − n2 )Y ` H m X n−1 + nY ` H m+1 X n−1 +
c0 Y `+1 X n + c1 Y `+1 HX n + · · · + cm−1 Y `+1 H m−1 X n + Y `+1 H m X n .
We also calculate X(Y ` H m X n ):
h
i
X(Y ` H m X n ) = (` − `2 )Y `−1 + `Y `−1 H + Y ` X H m X n
= (` − `2 )Y `−1 H m X n + `Y `−1 H m+1 X n + Y ` XH m X n
= (`−`2 )Y `−1 H m X n +`Y `−1 H m+1 X n +Y ` (c0 X+c1 HX+· · ·+cm−1 H m−1 X+H m X)X n
= (` − `2 )Y `−1 H m X n + `Y `−1 H m+1 X n +
c0 Y ` X n+1 + c1 Y ` HX n+1 + · · · + cm−1 Y ` H m−1 X n+1 + Y ` H m X n+1 .
To summarize (for our purposes), given a PBW monomial Y ` H m X n :
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1) Multiplication on the right by Y gives us a linear combination of PBW
monomials with X degree either n or n − 1 and corresponding Y degree of
either ` + 1 or `, respectively.
2) Multiplication on the left by X gives us a linear combination of PBW
monomials with Y degree either ` or ` − 1 and corresponding X degree of
either n or n + 1, respectively.
Therefore, by iterating this process, the product (Y ` H m X n )Y k is a linear combination of PBW monomials, all of which have X degree ≥ n−k, hence (Y ` H m X n )Y k ∈
U(g)X n−k . Since multiplication on the left by X k preserves U(g), this proves a).
Now, consider the monomial Y n H m X n . We have
h
X(Y n H m X n )Y = X (n − n2 )Y n H m X n−1 + nY n H m+1 X n−1
i
+ {PBW terms with X degree = n}
= (n − n2 )XY n H m X n−1 + nXY n H m+1 X n−1 + {PBW terms with X degree ≥ n}
h
i
= (n−n2 ) (n−n2 )Y n−1 H m X n−1 +nY n−1 H m+1 X n−1 +{PBW terms with Y degree = n}
h
i
+n (n−n2 )Y n−1 H m+1 X n−1 +nY n−1 H m+2 X n−1 +{PBW terms with Y degree = n}
+ {PBW terms with X degree ≥ n}
= (n − n2 )2 Y n−1 H m X n−1 + 2n(n − n2 )Y n−1 H m+1 X n−1 + n2 Y n−1 H m+2 X n−1
+ {PBW terms with either X or Y degree ≥ n} .
To prove b), it will suffice to prove the following:
Claim 2.1. The product X n (Y n H m X n )Y n has a nontrivial term cm+2 H m+2 ,
cm+2 ∈ Z. Furthermore, this is the lowest degree term that is a power of H.
Proof. We proceed by induction. By the above, we have that
X(Y n H m X n )Y = (n − n2 )2 Y n−1 H m X n−1
+ 2n(n − n2 )Y n−1 H m+1 X n−1 + n2 Y n−1 H m+2 X n−1
+ {PBW terms with either X or Y degree ≥ n} .
Thus, if n = 1, we have
X(Y H m X)Y = H m+2 + {PBW terms with either X or Y degree ≥ 1} .
Now, write A = X n−1 (Y n−1 H m X n−1 )Y n−1 , with n > 1. Suppose that A has a
nontrivial term cm+2 H m+2 which is the lowest degree term which is a power of H.
Now, consider
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X n (Y n H m X n )Y n = X n Y (Y n−1 H m X n−1 )XY n
h
i
h
i
= (n−n2 )X n−1 +nHX n−1 +Y X n (Y n−1 H m X n−1 ) (n−n2 )Y n−1 +nY n−1 H+Y n X
h
i
= (n − n2 )2 A + (2n2 − n3 ) + nY X AH
h
i
h
i
+ (n − n2 ) + nH + Y X AY X + n(n − n2 )H + (n − n2 )Y X A.
∗
We now consider each term separately. We have by hypothesis that A is a sum of
cm+2 H m+2 along with higher powers of H and some PBW terms in U(g)X. Thus
(since n > 1) we have (n − n2 )2 A is a nonzero multiple of that sum. We show that
m+2
this is the only
.
h term in (*) whichicontains H
Consider (2n2 − n3 ) + nY X AH. First, note that for a generic monomial
Y ` H m X n we have
Y ` H m X n H = Y ` H m (−2nX n + HX n ) = −2nY ` H m X n + Y ` H m+1 X n .
This shows that multiplication on the right by H will preserve the terms of A
which lie in U(g)X. Left multiplication by anything will also keep these terms in
U(g)X. Now, consider the terms of A which are scalar multiples of powers of H.
Multiplication on the right by H simply raises the powers by one, and multiplication
on the left by nY X gives us
nY XH k = nY (c0 X + · · · + ck−1 H k−1 X + H k X)
= n(c0 Y X + · · · + ck−1 Y H k−1 X + Y H k X),
h
i
where k is some integer ≥ m + 2. Therefore we see that (2n2 − n3 ) + nY X AH
does not contain H m+2 or anyhlower power of H. i
Next, we consider the term (n − n2 ) + nH + Y X AY X. By definition, this lies
in U(g)X, so does not contain anyhpowers of H.
i
Finally, we consider the term n(n − n2 )H + (n − n2 )Y X A. We have that
multiplying A on the left by n(n − n2 )H will raise the monomial powers of H in A
by one and preserve elements of U(g)X, thus the product does not contain H m+2
or lower powers of H. Now, multiplying terms of A which lie in U(g)X on the left
will keep these terms in U(g)X, so all that is left is to consider powers of H that
lie in A when multiplied on the left by (n − n2 )Y X. We have
(n − n2 )Y XH k = (n − n2 )Y (c0 X + · · · + ck−1 H k−1 X + H k X)
= (n − n2 )(c0 Y X + · · · + ck−1 Y H k−1 X + Y H k X).
These terms all lie in U(g)X.
Thus we have shown that X n (Y n H m X n )Y n contains a nontrivial term cm+2 H m+2
which is of minimal degree as a power of H, completing the induction and proving
the claim.
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Part b) of the proposition immediately follows from this claim, and the proof is
complete.
3. Summary
To summarize what we have done:
1) We observed that if given a nonzero z ∈ z, with γ(z) = 0, then z ∈ U(g)X.
2) We observed that, for such a z, multiplication on the right or left by any
element in U(g) results in an element still contained in U(g)X.
3) We looked at the set of terms in z which (when written in a PBW basis)
had minimal X degree n.
4) Among these terms, we chose the one with minimal H degree m. Note that
m may be zero.
5) We observed that, since these monomials are zero weights of the adjoint
representation, the chosen term must look like Y n H m X n .
6) We then considered the product X n (z)Y n .
7) By part a) of the proposition, each of the nonminimal (with respect to the
degree of X) terms in z will remain in U(g) under this multiplication.
8) Each of the minimal (with respect to the degree of X) terms in z is of
the form Y n H k X n , with k = m being the minimal degree of H among
these terms. By part b) of the proposition, X n (Y n H k X n )Y n is a linear
combination of PBW terms containing a nonzero multiple of H k+2 , which
is the lowest power of H in the product.
9) Since m was chosen to be minimal, we then have that X n (z)Y n can be
written in a PBW basis as some nonzero multiple of H m+2>0 plus other
terms in U(g)X, plus (possibly) multiples of higher powers of H.
10) This contradicts the fact that z must lie entirely in U(g)X, implying that
z must be zero.