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ANOTHER ELEMENTARY PROOF OF THE INJECTIVITY OF THE HARISH-CHANDRA HOMOMORPHISM FOR sl(2, C) SEAN MCAFEE 1. Overview This proof is taken directly from some lecture notes of Dragan Milicic. The idea of the proof is as follows: let Un (g) denote the set of elements of U(g) whose underlying monomials have degree at most n. Let z ∈ z ∩ Un (g), z ∈ / Un−1 (g). We let 1 1 0 0 1 0 0 H= ,Y = , Ȳ = . 0 0 −1 0 2 0 −1 We will prove some combinatorial lemmas and use them to show that (for some c 6= 0), z = cH n + linear combinations of monomials of the form Y i H k Ȳ i , (i ≥ 1) + { terms in Un−1 (g)} . Thus, for the untwisted Harish-Chandra homomorphism γ 0 , we will have γ 0 (z) = cH n + { lower powers of H } . Since the image of ρ lies in C, we will thus have shown that γ(z) is nonero for any nonzero z ∈ z, proving the injectivity of γ. 2. Preliminary lemmas Lemma 2.1. For arbitrary A, B ∈ g, we have n n X X n k n k n−k n (−1) B n−k ((ad B)k A). ((ad B) A)B = [A, B ] = − k k k=1 k=1 Proof. We define operators RY , LY : U(g) → U(g) by LY A = Y A and RY A = AY, (A ∈ U(g)), where Y is as defined in section 1. We clearly have RY LY = LY RY . Furthermore, (LY − RY )A = Y A − AY = [Y, A] = (ad Y )A, so LY = RY + ad Y and RY = LY − ad Y. We then have n LY = n X n k=0 and n RY = n X k RY n−k (ad Y )k n (−1) LY n−k (ad Y )k . k k k=0 1 2 SEAN MCAFEE This gives us (for A ∈ U(g)) [A, Y n ] = AY n − Y n A = RY n (A) − Y n A = X n n n−k k (−1) Y ((ad Y ) A) − Y n A k k k=0 = n X (−1)k k=1 n Y n−k ((ad Y )k A), k and n n [A, Y ] = AY − n X n k=0 =− k RY n−k (ad Y )k (A) n X k=1 n ((ad Y )k A)Y n−k , k as needed. Observe that (ad Ȳ )Y = [Ȳ , Y ] = 2H, (ad Ȳ )2 Y = [Ȳ , 2H] = 2Ȳ , and (ad Ȳ )k Y = 0 for k ≥ 3. Combining this result with the above lemma gives [Y, Ȳ n ] = −(2nH Ȳ n−1 + n(n − 1)Ȳ n−1 ). For future use, we also note that [H, Y ] = Y and calculate n n X X n n [Y, H n ] = (−1)k H n−k ((ad H)k Y ) = (−1)k H n−k Y. k k k=1 k=1 Lemma 2.2. Let X1 , ..., Xn ∈ g, and let σ ∈ Sn . Then X1 X2 · · · Xn − Xσ(1) Xσ(2) · · · Xσ(n) ∈ Un−1 (g). Proof. First, observe that X1 · · · Xi Xi+1 · · · Xn − X1 · · · Xi+1 Xi · · · Xn = X1 · · · [Xi , Xi+1 ] · · · Xn ∈ Un−1 (g). Thus, the lemma holds for any transposition in Sn . Now, suppose that σ and τ are any two permutations in Sn for which the lemma holds. We have X1 · · · Xn − X(στ )(1) · · · X(στ )(n) = (X1 · · · Xn − Xτ (1) · · · Xτ (n) ) + (Xτ (1) · · · Xτ (n) − X(στ )(1) · · · X(στ )(n) ) ∈ Un−1 (g). Therefore, the set of permutations for which the lemma holds is a subgroup of Sn . Since the lemma holds for transpositions (which generate Sn ), the proof is complete. 3 3. proof of injectivity We are ready to prove the injectivity of the Harish-Chandra homomorphism. As mentioned in the first section, we let z ∈ z ∩ Un (g) such that z ∈ / Un−1 (g) and z 6= 0. We show that γ(z) 6= 0, thus γ is injective. Using Y, H, and Ȳ as a basis for SL(2, R), we have by Poincaré-Birkhoff-Witt that X z= ciji Y i H j Ȳ i , (i, j ∈ Z≥0 , i + j + i ≤ n) Note that the exponents of Y and Ȳ are equal in any given monomial term of z. This is because z is a zero weight of the adjoint representation of U(g). We let j0 = max {j ∈ Z≥0 |ciji 6= 0 and 2i + j = n} . Such a j0 exists since z ∈ / Un−1 (g). We now apply the lemmata and observations from section 2 to arrive at the following series of congruences: X 0 = [Y, z] ≡ ciji [Y, Y i H j Ȳ i ] mod Un−1 (g) 2i+j=n = X ciji (Y i [Y, H j ]Ȳ i + Y i H j [Y, Ȳ i ]) 2i+j=n ≡ X ciji (−jY i+1 H j−1 Ȳ i − 2iY i H j+1 Ȳ i−1 ) mod Un−1 (g) 2i+j=n ≡ −2i0 · ci0 j0 i0 (Y i0 H j0 +1 Ȳ i0 −1 ) modulo terms involving lower powers of H. By hypothesis, ci0 j0 i0 6= 0, so the above implies that i0 = 0, thus j0 = n. Therefore, z = c0n0 H n + { other terms } , so γ(z) = c0n0 H n + {lower powers of H} − ρ( multiples of some powers of H ) 6= 0, hence γ is injective, as claimed.