MORE NOTES FOR MATH 823, FALL 2007

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1 MORE NOTES FOR MATH 83, FALL 007 Prop 1.1 Prop 1. Lemma The Siegel upper half space 1.1. The Siegel upper half space and its Bergman kernel. The Siegel upper half space is the domain { U n+1 z C n+1 Im[zn+1 > z j }. (1.1) 1.1 The unit ball in C n+1 is the domain n+1 B n+1 {z C n+1 } z j < 1. (1.) 1. Proposition 1.1. U n+1 is biholomorphic to B n+1 via the mapping ( z1 z n F (z 1,..., z n+1 ),...,, i z ) n+1 (w 1,..., w n, w n+1 ). i + z n+1 i + z n+1 i + z n+1 Proof. Write z n+1 x n+1 + iy n+1, so that Then i z n+1 x n+1 + y n y n+1 z n Im[z n+1, i + z n+1 x n+1 + y n y n+1 z n Im[z n+1. n+1 w j 4 n z j + z n Im[z n+1 z n Im[z n+1 Thus n+1 w j < 1 if and only if Im[z n+1 > n z j, so F : U n+1 B n+1. The two domains are actually biholomorphic since the inverse mapping is given by F 1 (w 1,..., w n+1 ) This completes the proof. ( iw1 1 + w n+1,..., iw n 1 + w n+1, i 1 w n w n+1 Proposition 1.. The determinant of the Jacobian matrix of the mapping F is given by [ Fj n+1 i JF (z 1,..., z n+1 ) det (z 1,..., z n+1 ) z k (i + z n+1 ) n+. Proof. This is an easy calculation. Lemma 1.3. The Bergman kernel for the domain U n+1 is given by [ n. (n + 1)! i B U n+1(z, w) 4π n+1 (w n+1 z n+1 ) z j w j ). Date: November 11, 007.

2 MORE NOTES FOR MATH 83, FALL 007 Proof. The Bergman kernel for the unit ball is B B n+1(z, w) (n + 1)! π n+1 1 (1 z, w ) n+. Therefore, by the transformation rules for the Bergman kernel, we get B U n+1(z, w) JF (z)b B n+1( F (z), F (w) ) JF (w) (n + 1)! π n+1 JF (z) JF (w) (1 F (z), F (w) ) n+ 1 4 n+1 n (n + 1)![(i + z n+1 )(i + w n+1 ) π n+1 [1 (i + z n+1 )(i + w n+1 ) 1 [ 4 n z jw j + (i z n+1 )(i w n+1 ) n+ 1 4 n+1 (n + 1)! π n+1 [ (i + z n+1 )(i + w n+1 ) [ 4 n z jw j + (i z n+1 )(i w n+1 ) n+ 4n+1 (n + 1)! 1 π n+1 [ i(w n+1 z n+1 ) 4 n+ n z jw j [ n. (n + 1)! i 4π n+1 (w n+1 z n+1 ) z j w j This completes the proof. 1.. The Heisenberg group. We define a family of biholomorphic mappings of C n+1 which carry U n+1 into itself. These are the analogues of translation z z + a for a R when z C. For z C n+1, we use the notation z (z, z n+1 ) where z C n. For w C n+1, put T w (z) ( z + w, z n+1 + w n+1 + i z, w ). (1.3) 1.3 T w is a holomorphic mapping, and T 0 is the identity map. If u, w C n+1, we have T u ( Tw (z) ) T u ( z + w, z n+1 + w n+1 + i z, w ) ( z + w + u, z n+1 + w n+1 + i z, w + u n+1 + i z + w, u ) ( z + w + u, z n+1 + w n+1 + u n+1 + i w, u + i z, w + u ) T ζ (z) where ζ (w + u, w n+1 + u n+1 + i w, u ). then It follows that if we define a product (z, w) z w on C n+1 by z w ( z + w, z n+1 + w n+1 + i w, z ), (1.4) 1.4 T z w T z T w. (1.5) 1.5 Since composition of mappings is always associative, we have (z w) u z (w u) for all z, w, u C n+1. Clearly 0 z z 0 z, so 0 acts as an identity. Finally, if we

3 MORE NOTES FOR MATH 83, FALL put z 1 ( z, z n+1 + i z ), it follows from (1.4) that w 1 w w w 1 0. We have shown the following. Prop 1.4 Proposition 1.4. The product z w (z + w, z n+1 + w n+1 + i w, z ) makes C n+1 is a group with identity e 0, and inverse defined by w 1 ( w, w n+1 + i w ). Note that if dz denotes Lebesgue measure on C n+1, and if f L 1 (C n+1 ), then f(z w) dz f(w z) dz f(z 1 ) dz f(z) dz. C n+1 C n+1 C n+1 Thus the (left and right) Haar measure on this group is just Lebesgue measure. Define ρ : C n+1 R by ρ(z) Im[z n+1 z j, (1.6) 1.6 so that U n+1 {z C n+1 ρ(z) > 0}. Thus ρ is a height function on C n+1. Prop 1.5 Proposition 1.5. If z, w C n+1, then ρ(z w) ρ(z) + ρ(w). Proof. We make the following computation: ρ(w z) Im[z n+1 + w n+1 + i z, w z j + w j Im[z n+1 + Im[w n+1 + Re z, w ρ(z) + ρ(w). ( z j + w j ) Re z, w It follows that if ρ(w) 0, the mapping T w is a biholomorphic mapping of U n+1 to itself. However, the set {w C n+1 ρ(w) 0} is just the boundary of U n+1. Identify U n+1 with C n R via the correspondence C n R (z 1,..., z n, t) (z 1,..., z n, t + i z j ) U n+1. (1.7) 1.7 Then the point (w, s) C n R is identified with the biholomorphic mapping T w,s : U n+1 U n+1 where T w,s(z) (z + w, z n+1 + s + i w + i z, w ). Moreover, according to (1.5) and Proposition 1.5, the composition of two such maps is a map of the same form. Thus the set of mapping {T w } with ρ(w) 0 is a subgroup of the group structure we have put on C n+1. Explicitly, if z z, t + i z ), w (w, s + i w ) U n+1, the analogue of (1.4) is (z, t + i z ) (w, s + i w ) (z + w, t + s + Re z, w + i z + w ).

4 MORE NOTES FOR MATH 83, FALL Def 1.6 Definition 1.6. The set C n R with the product given by (z, t) (w, s) (z + w, t + s + Im z, w ) is called the n-dimensional Heisenberg group. The identity is e (0, 0), and the inverse is given by (z, t) 1 ( z, t). We now want to connect the new product on C n+1 with the Bergman kernel. For any z (z, z n+1 ) C n+1, put Then z (z, z n+1 iρ(z)). (1.8) 1.8 ρ( z) Im[z n+1 ρ(z) z ρ(z) ρ(z) ρ(z). Thus the operation z z just changes the sign of the height of z. We have while z w z w i(0, ρ(z w)) z w i(0, ρ(z) + ρ(w)) (z + w, z n+1 + w n+1 + i[ w, z ρ(z) ρ(w)), z w (z, z n+1 iρ(z)) (w, w n+1 iρ(w)) (z + w, z n+1 iρ(z) + w n+1 iρ(w) + i w, z ). Thus z w z w, so z z is a period automorphism of C n+1 which leaves the subgroup invariant. Prop 1.7 Proposition 1.7. Let z (z, z n+1 ) (z, t + i z + iρ(z)) and w (w, w n+1 ) (w, s + i w + iρ(w)). Then w 1 z (z w, z n+1 w n+1 i z, w ). Proof. According to (1.8), w (w, w n+1 iρ(w)), and so by Proposition 1.4, Thus w 1 ( w, w n+1 + iρ(w) + i w ) ( w, w n+1 ). w 1 z ( w, w n+1 ) (z, z n+1 ) (z w, z n+1 w n+1 i z, w ) as asserted. Lemma 1.3 and Proposition 1.7 now show the following connection between the Bergman kernel and the group structure on C n+1. Prop 1.8 Proposition 1.8. Let F (z) F (z 1,..., z n, z n+1 ) zn+1 n. Then ( (n + 1)! B U n+1(z, w) 4π n+1 ) i n F ( w 1 z ).

5 MORE NOTES FOR MATH 83, FALL The Szegö kernel. Def 1.9 Definition 1.9. The function S(z, w) n! 4π n+1 [ i (w n+1 z n+1 ) is called the Szegö kernel for U n+1. n 1 z j w j We show next that there is a reproducing formula for holomorphic functions using the Szegö kernel, but instead of integrating over all of U n+1, we only integrate over the boundary U n+1. Recall that we can identify U n+1 with C n R. With this identification, U n+1 f(w) dσ(w) just means integrating with respect to Lebesgue measure on C n R. Theorem 1.10 Theorem Let f be holomorphic in a neighborhood of the closure of U n+1, and belong to the Bergman space B (U n+1 ). Then for z U n+1, f(z) n! 4π n+1 Proof. Let U n+1 [ i (w n+1 z n+1 ) n 1 z j w j f(w1,..., w n, w n+1 ) dσ(w) ω n+1 dw 1 d w 1 dw n d w n dw n+1, ω dw 1 d w 1 dw n d w n dw n+1 d w n+1. Note that d w n+1 ω n+1 ω. Also, ω ( i) n+1 dw where dw is Lebesgue measure on C n+1. Then using the generalized Stokes theorem, we have U n+1 [ i (w n+1 z n+1 ) U n+1 w n+1 (n + 1)i ( i) n+1 (n + 1)i n 1 z j w j f(w) ω(w) [ i (w n+1 z n+1 ) U n+1 4πn+1 ( i) n+1 (n + 1)! F (z) n 1 z j w j f(w) ω [ i (w n+1 z n+1 ) n z j w j f(w) dw where the last equality follows from the reproducing property of the Bergman kernel. On the other hand, if we parameterize U n+1 as usual by C n R, G(w) ω n+1 (w) ( i) n G(w)dw dt. U n+1 U n+1 This gives the desired equality.

6 MORE NOTES FOR MATH 83, FALL We compute ( n! ) 4π n+1 C n R U n+1 S ( (z, t + i z + iρ 1 ), (w, s + i w + iρ ) ) dz dt i (s i w iρ(w) t i z iρ 1 ) ( 4) n+1 C n R n z j w j dz dt (t s + Re z, w + i[ z w + ρ 1 + ρ n dz dt 1.4. Informal discussion of H -spaces. The Szegö kernel arises as the integral kernel for the orthogonal projection from L ( U n+1 ) to a closed subspace denoted by H (U n+1 ). This can be thought of as the space of L functions on the boundary which are the boundary values of a holomorphic function on U n+1. The following discussion is an informal introduction to this topic. We first consider the classical case of the unit disk D or the upper half plane U in C. Thus if O(D) and O(U) denote the spaces of all holomorphic functions on the disk or the upper half plane, we define { H (D) f O(D) f H sup 1 π } f(re iθ ) dθ < + ; π { H (U) f O(U) f H 0 r<1 0 + sup 0<y } f(x + iy) dx < +. We would like to see that a function f in either of these spaces has boundary values (on the unit circle T in the case of H (D) and on the real line R in the case of H (U)). The case of the unit disk is easier, so we start with it. 1. The case of the unit disk If f O(D), then f is given by a power series f(z) a n z n, which converges absolutely and uniformly on any compact subset of D. Thus 1 π π 0 f(re iθ ) dθ 1 π π 0 m,n0 n0 ( a n r n e inθ)( ) a m r m e imθ dθ n0 a n a m r m+n 1 π a n r n. n0 m0 π 0 e i(n m)θ dθ It follows that f H (D) if and only if n0 a n < +, and in fact f H (D) a n. n0

7 MORE NOTES FOR MATH 83, FALL Now if f H (D), what are the boundary values of f? Given the representation of f as a power series, it is natural to guess that the boundary values should be the function f given by f (e iθ ) a n e inθ. n0 Of course, this sum need not converge uniformly, or even at any given point. However since n0 a n is finite, the sum does converge in L (T), and thus does define an element of L (T). In what sense is this function the boundary values of the original f? One can prove the following facts: (1) If f is holomorphic on D, let f r (e iθ ) f(re iθ ). Then each f r is infinitely differentiable on T and hence in L (T). Moreover, it follows from the definition that if f H (D), then sup 0 r<1 f r L (T) < +. One can prove that if f H (D), then lim f r f L r 1 (T) 0. () One can prove that if f H (D), then for almost every θ T, lim r 1 f(reiθ ) f (e iθ ). Thus one has radial convergence to the boundary values for almost all θ. In fact, one can show that for almost every θ lim f(z) f (e iθ ) z e iθ provided that z approaches e iθ within a non-tangential approach region. The reproducing formula for H (D) is so the Szegö kernel in this case is f(z) 1 π π 0 f (e iθ ) dθ 1 ze iθ S(z, w) 1 π (1 z w) 1.. The case of the upper half plane Instead of representations by power series, we need the following: Theorem: A function f H (U) if and only if there exists a function a L (R) such that f(x + iy) 0 a(ξ) e πi(x+iy)ξ dξ. (1.9) 1.9 (This is one of the Paley-Wiener theorems, and a proof is given on pages of the book Real and Complex Analysis by Walter Rudin). Let f y (x) f(x + iy). Then (1.9) shows that f y is the Fourier transform of the function a(ξ)χ(ξ)e πyξ, where χ is the characteristic function of the positive real axis. The Plancherel theorem then implies that + f(x + iy) dx f y L (R) f y L (R) a(ξ) e πyξ dξ, 0

8 MORE NOTES FOR MATH 83, FALL and hence that f H (U) sup y>0 + f(x + iy) dx We would expect that the boundary values of f are given by f (x) 0 a(ξ) e πix dξ 0 a(ξ) dξ. which is the Fourier transform of the function χa. Of course, this last integral does not converge absolutely or even pointwise, but the Fourier transform of a function in L (R) is in L (R). Then one can prove analogues of the two assertions for H (D): (1) lim y 0 f y f L (R) 0. () For almost every x R, lim f(x + iy) f (x). Moreover, for almost every y 0 x R, lim f(z) f (x) provided that z approaches x within a non-tangential z x region. The reproducing formula for H (U) is f(z) 1 π so the Szegö kernel in this case is + f (x) dx z x S(z, w) 1 π (z w) 1. The following are some reference for the classical theory of H p -spaces: [1 Theory of H p Spaces by Peter L. Duren, Academic Press [ Introduction to H p Spaces by Paul Koosis, Cambridge University Press, Some analysis on the Heisenberg group The left and right translation operators act on functions on as follows. If f : C, then L w [f(z) f(w 1 z) R w [f(z) f(z w) (left translation), (right translation). Note that with these definitions, L w1 L w L w1 w and R w1 R w R w1 w. We say that a space X of functions on is (left or right) translation invariant if whenever f X, it follows that L w [f or R w [f belongs to X for all w. If X and Y are (left or right) translation invariant spaces of functions, and if T : X Y is a linear operator, we say T is (left or right) translation invariant if T L w L w T or T R w R w T for all w. Suppose that T : X Y is a left-invariant linear operator, and consider the following informal argument. Suppose that the action of T is given by T [f(z) K(z, w) f(w) dw

9 MORE NOTES FOR MATH 83, FALL where K is the integral kernel of the operator T. (As always, dw is Lebesgue measure on, and hence is invariant under left and right translation.) Then we have L u T [f(z) T [f(u 1 z) K(u 1 z, w) f(w) dw. But L u T T L u, so we also have L u T [f(z) K(z, w) T u [f(w) dw K(z, w) f(u 1 w) dw H n K(z, u w) f(w) dw where the last equality follows by replacing the variable w by u w, and using the fact that dw is translation invariant. Thus for every z and every f X we have K(u 1 z, w) f(w) dw K(z, u w) f(w) dw. Under reasonable assumptions about the density of the space X, it should follow that K(u 1 z, w) K(z, u w) for all u, z, w, or replacing z by u z, that K(z, w) K(u z, u w). If we let u w 1, we see that K(z, w) K(w 1 z, e). Thus if we put k(z) K(z, e), we see that our left-invariant operator T can be written T [f(z) f(w) k(w 1 z) dw f(z w 1 )k(w) dw. Conversely, any operator of this form is left-invariant since L u T [f(z) T [f(u 1 z) f(w)k(w 1 u 1 z) dw f(w)k((u w) 1 z) dw H n f(u 1 w)k(w 1 z) dw L u [f(w)k(w 1 z) dw T L u [f(z). Note that if T [f(z) f(z w 1 ) k(w) dw, then H n [ T [f(z) g(z) dz k(w) f(z w 1 ) g(z) dz dw H [ n k(w) f(w z 1 ) g(z) dz dw H n k(w) ( f g)(w) dw, where f(z) f(z 1 ), and f g(z) f(z w 1 ) g(w) dw f(w) g(w 1 z) dw. A similar argument suggest that a right-invariant operator S can be written S[f(z) k(z w 1 ) f(w) dw k(w) f(w 1 z) dw.

10 MORE NOTES FOR MATH 83, FALL To state a precise result, we introduce the following notation. Let D( ) be the space of C functions with compact support on, and given the topology such that ϕ n 0 if and only if the supports of the {ϕ n } all lie in a fixed compact set, and the functions {ϕ n } and all their derivatives converge uniformly to zero. Let D ( ) be the space of continuous linear functionals on D( ). (Thus D ( ) is the space of distributions on ). The action of T D ( ) on ϕ D( ) is denoted by T, ϕ. We give D ( ) the topology such that T n 0 if and only if T n, ϕ 0 for every ϕ D( ). In a similar way, we define the space of test functions D( ) and the space of distributions D ( ) on. If ϕ, ψ D( ), define ϕ ψ D( ) by setting ϕ ψ(z, w) ϕ(z) ψ(w). Also define ϕ ψ D( ) by ϕ ψ(z) ϕ(z w 1 )ψ(w) dw ϕ(w) ψ(w 1 z) dw. Lemma.1. Suppose that T : D( ) D ( ) is a continuous linear operator. Then there is a distribution K D ( ) such that if ϕ, ψ D( ), then T [ϕ, ψ K, ϕ ψ. (.1) Moreover, if T is left-invariant, there is a distribution k D ( ) so that T [ϕ, ψ k, ϕ ψ. (.) if T is right-invariant, there is a distribution k D ( ) so that.1. Left-invariant vector fields. T [ϕ, ψ k, ψ ϕ. (.3) Recall that C n R R n+1 with multiplication given by (w, s) (z, t) (w + z, s + t + Im w, z ) (w, s) 1 (z, t) (z w, t s Im w, z ). If we use real coordinates, so that z j x j + iy j, and w j u j + iv j, then w j z j (u j + iv j )(x j iy j ) (x j u j + y j v j ) + i(x j v j y j u j ). Thus the multiplication on R n+1 is given by (u, v, s) (x, y, t) (x + u, y + v, t + s + (u, v, s) 1 (x, y, t) (x u, y v, t s (x j v j y j u j ), (x j v j y j u j ). Consider the vector fields {X 1, Y 1,..., X n, Y n, T } on given by X j + y j x j t, 1 j n, Y j x j y j t, 1 j n, T t.

11 MORE NOTES FOR MATH 83, FALL Proposition.. The vector fields { X 1, Y 1,..., X n, Y n, T } are left-invariant operators on, and they generate a finite dimensional Lie algebra with the following commutator identities: Proof. Let (u, v, s). Then L (u,v,s) X j [f(x, y, t) On the other hand, [X j, Y k 4δ j,k T 1 j, k n, [X j, X k 0 1 j, k n, [Y j, Y k 0 1 j, k n, [X j, T 0, 1 j n, [Y k, T 0 1 k n. [ f f L (u,v,s) (x, y, t) + y j x j t (x, y, t) f ( (u, v, s) 1 (x, y, t) ) + (y j v j ) f x j t X j L (u,v,s) [f(x, y, t) [ + y j f ( x u, y v, t s x j t ( (u, v, s) 1 (x, y, t) ). ) (x j v j y j u j f ( (u, v, s) 1 (x, y, t) ) f ( v j (u, v, s) 1 (x, y, t) ) x j t f ( + y j (u, v, s) 1 (x, y, t) ) t f ( (u, v, s) 1 (x, y, t) ) + (y j v j ) f ( (u, v, s) 1 (x, y, t) ), x j t so the two are the same. The argument for Y j is similar, and the argument for T is easy. The calculation of commutators is also easy... Tangential CR operators on U n+1. The defining function for U n+1 is ρ(z, z n+1 ) z n+1 z n+1 i z j z j. We look for operators L j, 1 j n of the form L j z j + A j z n+1 which are tangential along U n+1. Recall that this means we need L j [ρ 0 when ρ 0. But L j [ρ(z, z n+1 ) z j 1 i A j, and this is zero if A j iz j. Thus if we put L j i, (.4) z j z n+1

12 MORE NOTES FOR MATH 83, FALL then the collection { L1,..., L n } is a basis for the space of operators of type (0, 1) which are tangential along U n+1. In particular, these operators will annihilate the boundary values of functions which are holomorphic on U n+1. What are the operators { L j } in terms of the coordinates (z, t)? Given a function f on C n R, it defines a function f on U n+1 by setting f(z, t+i z ) f(z, t). To see what L j does to this function f, we extend it to a function on C n+1 by setting F (z, z n+1 ) f(z, Re[z n+1 + i z ). But then it is clear that L j [F (z, t + i z ) f f (z, t) iz j (z, t). z j t Thus on C n R, a basis for the CR differential operators is given by { Z 1,..., Z n } where Z j i z j z j t + i x j i(x j + iy j ) y j t [ [ + y j + i x j x j t y j X j + iy j. t Corollary.3. If we set Z j X j iy j, then the collection of differential operators {Z 1,..., Z n, z 1,..., Z n, T } are a basis for all first order differential operators on, and the satisfy the commutation relationships [Z j, Z k [ Z j, Z k [Z j, T [ Z k, T 0, and [Z j, Z k 8iδ j,k T..3. The b complex. In analogy with the problem on a domain in C n+1, we can now formulate a corresponding problem on the boundary of a domain. In the case of, we proceed as follows. Definition.4. If f C 1 ( ), define b [f If ω is a (0, q)-form on, that is, if ω Z j [f d z j. j 1< <j q ω j1,...,j q d z j1 d z jq, then b [ω Z j [ω j1,...,j q d z j d z j1 d z jq. j 1< <j q

13 MORE NOTES FOR MATH 83, FALL Proposition.5. If ω C ( ) (0,q), then b [ω 0. Proof. Let ω ω j1,...,j q d z j1 d z jq. Then j 1< <j q b [ω b Z j ω j1,...,j q d z j d z j1 d z jq j 1< <j q j 1< <j q k1 Z k Zj ω j1,...,j q d z k d z j d z j1 d z jq. But since [ Z k, Z j 0, we have Z k Zj [ω j1,...,j q d z k d z j Z j Zk [ω j1,...,j q d z j d z k Thus in the inner sum, each term appears twice, but with opposite signs. Thus the entire sum is zero. Since X j and Y j are real differential operators, integration by parts shows that if f, g C 1 ( ) and at least one of the two functions has compact support, then (X j [f, g) L ( ) (f, X j [g) L ( ) and (Y j [f, g) L ( ) (f, Y j [g) L ( ). It follows that ( Zj [f, g ) L(Hn) ( f, Z j [g ) L(Hn), We introduce the standard L -norm on the space of (0, q)-forms on. Thus if ω j 1< <j q ω j1,...,j q d z j1 d z jq and ψ j 1< j q ψ j1,...,j q d z j1 d z jq, we set ( ω, ψ )L ω j1,,j q (z, t)ψ j1,,j q (z, t) dz dt. j 1< <j q We want to compute the formal adjoint of b. Let ω as above be a (0,q)-form, and let θ k 1< <k q+1 θ k1,,k q+1 d z 1 d z q+1 be a (0,q+1)-form. Was assume that the coefficients are of class C 1 and either ω or θ has compact support. Then b [ω Z k [ω j1,...,j q d z k d z j1 d z q where j 1< <j q k1 m 1< m q+1 j 1< <j q k1 ɛ m1,m, mq+1 k,j 1, j q It now follows that ( b [ω, θ ) L ( ) (0,q+1) m 1< <m q+1 j 1< <j q k1 Z k [ω j1,...,j q ɛ m1,m, mq+1 k,j 1, j q d z m1 d z mq+1 { 0 if {m 1, m, m q+1 } {k, j 1, j q }, sgn(σ) if σ(m 1, m, m q+1 ) (k, j 1, j q ). j 1< <j q m 1< <m q+1 k1 ɛ m1,m, mq+1 k,j 1, j q ( Zk [ω j1,...,j q, θ m1,,m q+1 )L ( ) ) ɛ m1,m, mq+1 k,j 1, j q (ω j1,...,j q, Z k [θ m1,.m q+1. L ( )

14 MORE NOTES FOR MATH 83, FALL It follows that b [θ j 1< <j q m 1< <m q+1 k1 ɛ m1,m, mq+1 k,j 1, j q Z k [θ m1,.m q+1 d z j1 d z q. To eliminate the need for so many subscripts, we introduce the following notation. I q denotes the set of all strictly increasing q-tuples J (j 1 < j q ) of integers with 1 j 1 < j q n. If J 1 (j1, 1..., jq 1 ) and J (j1,..., jq ) are any ordered q-tuples of integers with 1 jk 1, j m n, we set 0 if either J 1 or J has a repeated element, ɛ J1 J 0 if the set J 1 is not equal to the set J, sgn(σ) if σ : J 1 J with σ S q. Then with this notation, if ω is a (0,q)-form and if θ is a (0,q+1)-form, b [ω [ ɛ K Z j,j j [ω J d z K, K I q+1 b [θ L Iq J I q [ K I q+1 l1 We now compute the operator b b b + b b. ɛ K l,l Z l [θ K d z L. Proposition.6. If ω J I q ω J d z J, then b [ω [ Z j Z j [ω J + Z j Zj [ω J d z J. j / J Proof. We have b b [ω b J I q [ K I q+1 L I q [ L I q [ j J [ J I q K I q+1 l1 J I q l1 [ [ ( n L I q J I q ɛ K Z j,j j [ω J d z K J I q ( l1 ɛ K l,l ɛ K j,j Z l Zj [ω J d z L K I q+1 ɛ K l,lɛ K j,j ɛ j,j l,l Z l Z j [ω J ) Z l Zj [ω J d z L ) d z L. Now in the inner sum, we can have J L and j l / J L, so that ɛ j,j l,l 1, which gives rise to the sum I [ Z l Zl [ω L d z L. L I q l / L The other possibility is that j l, j L, and l J, in which case there exists M I q 1 such that L M {j}, J M {l}. In this case ɛ j,j l,l ɛj,l,m l,j,m ɛj l,m ɛ j,m L ɛ J l,m ɛ j,m L

15 MORE NOTES FOR MATH 83, FALL and ω J ɛ J l,m ω l,m. Thus we can write ( n ɛ j,j l,l Z Z ) l j [ω J J I q l1 This gives rise to the sum II L I q [ M I q 1 M I q 1 j,l / M j l M I q 1 j,l / M j l j,l / M j l ɛ J l,m ɛ j,m L ɛj l,m Z l Zj [ω l,m ɛ j,m L Z l Z j [ω l,m ɛ j,m L Z Z l j [ω l,m d z L. On the other hand b b [ω b [ M I q 1 L I q [ [ J I q m1 M I q 1 l1 [ ( n L I q J I q ɛ J m,m Z m [ω J d z M [ l1 m1 J I q m1 ɛ L l,m ɛ J Z m,m l Z m [ω J d z L ɛ L l,m ɛ J Z ) m,m l Z m [ω J d z L M I q 1 Now for a given L I q, ɛ L l,m 0 unless l L and L M {l}. Moreover, ɛ J m,m 0 unless m / M and J M {m}. It can happen that l m, in which case J L, l m J L, and ɛ L l,m ɛj m,m 1. This gives rise to the sum III [ Z l Z l [ω L d z L. L I q l L The other possibility is that l m, in which case l, m / M, and L {l} M, J {m} M, and ω J ɛ J m,m ω m,m. Then l1 m1 M I q 1 ɛ L l,m ɛ J m,m Z l Z m [ω J This gives rise to the term IV L I q [ M I q 1 M I q 1 l, m / M l m M I q 1 l, m / M l m l, m / M l m ɛ L l,m ɛ J m,m ɛ J m,m Z l Z m [ω j,m ɛ L l,m Z l Z m [ω j,m. ɛ L Z l,m l Z m [ω j,m d z L.

16 MORE NOTES FOR MATH 83, FALL For l m, [ Z l, Z m 0, so the terms II and IV cancel. Thus b [ω [ [ Z l Zl [ω L d z L L Iq Z l Z l [ω L d z L. L I q This completes the proof. l/ L Remark.7. The operator b : C(0,q) C (0,q) is a system of second order partial differential operators, but in this case, the system is actually diagonal. Now We now write j 1 [ Zj Zj + Z j Z j 1 [ (X j iy j )(X j + iy j ) + (X j + iy j )(X j iy j ) 1 [ Xj + i[x j, Y j + Yj + Xj i[x j, Y j + Yj X j + Y j. l L [Z j, Z j [ (X j iy j ), (X j + iy j ) i[x j, Y j 4iT, Thus Z j Zj 1 [ Zj Zj + Z 1[ j Z j + Zj Zj Z j Z j j + 1 [Z j, Z j j it Z j Z j 1 [ Zj Zj + Z 1[ j Z j Zj Zj Z j Z j j 1 [Z j, Z j j + it. Finally, we let L j (Xj + Yj ). Then it follows that if L I q, j L Definition.8. For α R, put Z j Z j j / L Z j Zj L qit + (n q)it L α L + i (n q)t ). (Xj + Yj ) + iαt. Theorem.9 (Folland and Stein, 1974). Let φ α (z, t) ( z it ) n+α ( z + it ) n α. If δ 0 denotes the delta function supported at the origin of, then where γ α L α [φ α γ α δ 0 Γ ( n+α n π n+1 ) Γ ( n α ).