Fundamental Solutions on the Heisenberg Group
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(2) 謝詞. 首先謝謝我的指導教授陳瑞堂老師收我為學生，並幫助我各方面 的大小事。感謝陳老師一直以來細心的指導，指出問題所在，而且相 當有包容力和耐心，一步一步帶領我完成論文，也提供我各類機會見 識不同人物，接觸學術環境，讓我增廣見聞。也謝謝清華大學的邱鴻 麟教授，在討論班中與陳老師共同提出寶貴的建議，都使我認清自己 有所不足，並推進我更加進步。陳老師和邱老師對我的提攜，對我有 莫大的幫助，我必將加倍用功，才得以回報師恩。最後謝謝陳瑞堂老 師、邱鴻麟老師、林惠娥老師，在百忙之中撥空擔任我的口試委員； 以及一路上幫助過我的所有師長，沒有他們就不會有今日的我。. 羅阡豪 謹誌於 臺灣師範大學數學所 中華民國一○八年五月.
(3) Fundamental Solutions on the Heisenberg Group. May 30, 2019. Contents 1 An introduction to CR manifolds. 2. 2 An introduction to Heisenberg groups. 3. 3 Computation of @ b and. 9. b. 4 Construction of the fundamental solutions. 12. 5 Solutions of L f = g. 27. 6 References. 32 Abstract. This paper is an expository note mainly concentrated on section 2,4,5,6 and 7 of Folland and Stein’s work [4]. We …rst review some de…nitions and properties of CR manifolds, @ b complex and Heisenberg groups. We then derive the explicit form of the Laplacian. b. and deduce the problem of. b. to the. operator L . The major part of the paper is to …nd the fundamental solutions for L on the Heisenberg group and to compute the constant c . Last we analyze the solutions of L f = g to …nish the paper.. Keywords: CR manifold, Heisenberg group, fundamental solution. 1.
(4) 1. An introduction to CR manifolds. In order to introduce the @ b complex, we start with the formal de…nition of CR manifolds. De…nition 1 A CR manifold is a real oriented C 1 manifold M of dimension 2n+1; n = 1; 2; 3;. ; together. with a subbundle T1;0 of the complex tangent bundle CT M satisfying: (a) dimC T1;0 = n; (b) T1;0 \ T 1;0 = f0g; (c) T1;0 is integrable in the sense of Frobenius, i.e., if Z1 and Z2 are sections of T1;0 , then their Lie bracket [Z1 ; Z2 ] is also a section of T1;0 , T 1;0. (d) the line bundle T1;0. ?. CT M has a global section.. Remark 2 We don’t consider the trivial case n = 0. We set T0;1 = T 1;0 and de…ne q. T0;1 : Note that. p;q. is a quotient bundle of. De…nition 3 The operator @ b : C 1 (. p;q. ; Wq+1 2 C 1 (T0;1 ), and ' 2 C 1 (. W1 ;. @ b '; (Z1 ^. ^ Zp ). (W1 ^. ) p;q. p+q. =. ! C1. p+q. =. p. T1;0. CT M .. p;q+1. ; Zp 2 C 1 (T1;0 ),. is de…ned by: if Z1 ;. ), then. ^ Wq+1 ). q+1 D E 1 X j+1 cj ( 1) Wj '; (Z1 ^ ^ Zp ) W1 ^ W ^ Wq+1 q + 1 j=1 D X 1 i+j ci ( 1) '; (Z1 ^ ^ Zp ) [Wi ; Wj ] ^ W1 ^ W + q+1. =. p;q. 0 i<j q+1. cj means that Wj is omitted from the product. where W. cj W. ^ Wq+1. E. ;. Remark 4 The @ b complex is also called the tangential Cauchy-Riemann complex. As usual di¤erential forms, @ b has the following properties: 2. (i). @ b = 0,. (ii). if ' 2 C 1 (. (iii). if f 2 C 1 = C 1. p;q. ), then @ b (' ^ ) = @ b ' ^ 0;0. p+q. + ( 1). ' ^ @b. ,. and W 2 C 1 (T0;1 ), then @ b f; W = W f .. Note that although we de…ne @ b in global terms, we can also restrict our attention to an open set U. M. without changing anything. Now we impose a Hermitian metric on M such that T1;0 ? T0;1 , then we can form the formal adjoint #b : C 1 (. p;q. ) ! C1. p;q 1. of @ b , and we have the “Laplacian” b. = @ b #b + #b @ b : C 1 ( 2. p;q. ) ! C1 (. p;q. ):.
(5) Let. be a nonvanishing real one-form which annihilates T1;0 T0;1 . The Levi form h ; iL is the Hermitian. form on T1;0 de…ned by hZ1 ; Z2 iL =. i d ; Z1 ^ Z 2 ;. here h ; iL is de…ned up to a nonvanishing real factor since it only depends on the choice of . We say M is nondegenerate if h ; iL is nondegenerate at every point, i.e., if there is no nonzero Z 2 T1;0 such that hZ; Z 0 iL = 0 for all Z 0 . Replacing. by. if necessary, we may assume that the (constant). dimension k of a maximal positive de…nite subspace for h ; iL is at least. n 2,. and say that M is k-strongly. pseudoconvex. If k = n, we simply say that M is strongly pseudoconvex. From the de…nition of h ; iL , we have hZ1 ; Z2 iL. = because. 2. annihilates T1;0. i Z1 2. = i 2. ; Z2. Z 2 h ; Z1 i. ; Z1; Z 2. ; Z1; Z 2. T0;1 .. An introduction to Heisenberg groups. From now on, we study the CR structures on the Heisenberg group. De…nition 5 The Heisenberg group (of degree n) is the Lie group Hn whose underlying manifold is Cn with coordinates (z1 ;. ; zn ; t) = (z; t) and whose group law is given by (z; t) (z 0 ; t0 ) = (z + z 0 ; t + t0 + 2 Im z z 0 ) ;. z z0 =. n P. j=1. Remark 6 If we set zj = xj + iyj ; then (x1 ;. ; xn ; y1 ;. zj z 0j :. ; yn ; t) form a real coordinate system for Hn :. De…nition 7 In the real coordinate system, we de…ne the vector …elds: Xj = Proposition 8 fX1 ;. @ @ + 2yj ; @xj @t. ; Xn ; Y1 ;. Yj =. @ @yj. 2xj. @ ; @t. T =. @ : @t. ; Yn ; T g forms a basis for the left-invariant vector …elds on Hn :. Proof. We only prove the case for n = 1: For any g = (u; v; w) 2 H1 ; the left translation is give by Lg (x; y; t) = (u; v; w) (x; y; t) = (u + x; v + y; w + t + 2(vx. 3. uy)):. R.
(6) Let A be a left-invariant vector …eld on H1 , then we have Ag = Lg Ae for all g 2 H1 , where e is the identity of H1 . In local coordinates, we write Ae = a. @ @ @ @ @ @ +b +c and Ag = A1g + A2g + A3g : @x @y @t @x @y @t. To evaluate Aig , apply both sides of Ag to the coordinate functions: A1g = Ag (x) = Lg Ae (x) = Ae (x Lg ) (by de…nition of the di¤erential), similarly, A2g = Ae (y Lg ) and A3g = Ae (t Lg ). For h = (u0 ; v 0 ; w0 ) 2 H1 , we have (x Lg )(h) = x(Lg h) = x(gh) = u + u0 = x(g) + x(h); (y Lg )(h) = y(Lg h) = y(gh) = v + v 0 = y(g) + y(h); (t Lg )(h) = t(Lg h) = t(gh) = w + w0 + 2(vu0. uv 0 ) = t(g) + t(h) + 2y(g)x(h). 2x(g)y(h):. Viewed as functions of h, we get x Lg = x(g) + x; y Lg = y(g) + y; t Lg = t(g) + t + 2y(g)x. 2x(g)y:. Therefore, A1g = Ae (x(g) + x) = a; A2g = Ae (y(g) + y) = b; A3g = Ae (t(g) + t + 2y(g)x. x(g)y) = c + 2y(g)a. 2x(g)b:. Hence, Ag. @ @ @ +b + (c + 2y(g)a 2x(g)b) @x @y @t @ @ @ @ @ = a + 2y(g) + b( 2x(g) ) + c : @x @t @y @t @t = a. Again, viewed as a functions of g, we get A = a. @ @x. @ @ + 2y @t + b( @y. @ @ 2x @t ) + c @t. = aX + bY + cT: That is, A is generated by X; Y and T: Since X; Y and T are obviously linearly independent, fX; Y; T g is a basis for the left-invariant vector …elds on H1 : Proposition 9 We have the commutation relations: [Yj; Xk ] = 4. jk T;. [Xj; Xk ] = [Yj; Yk ] = [Xj; T ] = [Yj; T ] = 0:. 4.
(7) Proof. By de…nition, we have [Yj; Xk ]. = =. h. @ @yj. @ @ 2xj @t ; @x@ k + 2yk @t. @ @yj. @ 2xj @t. @ @ @yj @xk. =. 2. @ jk @t. =. 4. jk T:. @ + 2yk @t. @ @xk. @ + ( @y@ j 2yk ) @t + 2yk @y@ j @ ( @x@ k 2xj ) @t. @ @ @xk @yj. =. @ @xk. i. @ kj @t ). ( 2. @ @t. @ + 2yk @t. @ @yj. @ 2xj @t 2. @ @ 2xj @t @xk. @ 4xj yk @t 2. @ @ @ 2xj @x@ k @t + 2yk @t @yj. 2. @ 4yk xj @t 2. since Hn is C 1 ,. Similarly, we obtain [Xj; Xk ]. = =. h. @ @xj. @ @ + 2yj @t ; @x@ k + 2yk @t. @ @xj. @ + 2yj @t. @ @ @xj @xk. =. @ + 2yk @x j. @ @ @xk @xj. =. @ @xk @ @t. i. @ + 2yk @t. @ @xk. @ + 2yk @t. @ @xj. @ + 2yj @t. 2. @ @ @ + 2yj @t @xk + 4yj yk @t2 2. @ @ @ @ + 2yj @x@ k @t + 2yk @t @xj + 4yk yj @t2. 0;. and [Xj; T ]. = =. h. @ @xj. @ @ + 2yj @t ; @t. @ @xj. @ + 2yj @t. =. @ @ @xj @t. =. 0:. i. @ @t 2. @ @t. @ @xj. @ @ @t @xj. @ + 2yj @t 2. @ + 2yj @t @ 2 @t yj. 2. @ @t. @ 2yj @t 2. The others are similar, and thus omitted. De…nition 10 We further de…ne the complex vector …elds Zj , Z j and the one-form Zj =. 1 (Xj 2. Zj =. 1 1 (Xj + iYj ) = 2 2. iYj ) =. 1 2. @ @ + 2yj @xj @t. i. @ @ + 2ixj @yj @t. @ @ @ + 2yj + i @xj @t @yj n X = dt + 2 (xj dyj. 2ixj. @ @t. to be:. =. @ @ + iz j ; @zj @t. =. @ @z j. izj. @ ; @t. yj dxj ) :. j=1. Proposition 11 The commutation relations of Zj and Zj are Zj ; Z k =. 2i. jk T;. [Zj ; Zk ] = Z j ; Z k = [Zj ; T ] = Z j ; T = 0:. 5.
(8) Proof. We only verify the …rst case since the other cases are similar and can be easily checked. h i @ @ @ @ Zj ; Z k = + iz ; iz j k @zj @t @z k @t @ + iz j @t. =. @ @zj. =. @ @ @zj @z k. @ i( @z@ j zk ) @t. @ @z k. i. =. 2i. @ @zj. @ + iz j @t. @ @ @ + iz j @t @z k + z j zk @t2. @ jk @t. i. @ izk @t. 2. @ @t. izk @z@ j. @ @ + i( @z@k z j ) @t + iz j @z@k @t. 2. @ @ @ izk @t @zj + zk z j @t2. @ kj @t. jk T:. ; Zj = 0 for all j.. Proof. In real coordinates, we …nd that * n P h ; Zj i = dt + 2 (xj dyj. yj dxj ) ; 12. j=1. ixj dyj @y@ j. = yj + ixj =. Similarly,. @ izk @t. @ @ @z k @zj. =. Proposition 12 h ; Zj i =. @ @z k. @ + 2yj @t. @ @xj. @ i @y@ j + 2ixj @t. +. @ yj dxj @x j. 0:. ; Z j = 0.. One can see that the subbundle T1;0 of CT Hn spanned by Z1 ; the levi form is hZj ; Zk iL =. i 2. ; Zj; Z k. =. i 2. h ; 2i. jk T i. =. ; Zn de…nes a CR structure on Hn ; while. jk .. Hence, Hn is strongly pseudoconvex.. We now discuss some geometrical interpretation. For = ( 0 ; 1 ; ( n P 2 Dn+1 = 2 Cn+1 : < Im j j=1. and. Mn = @Dn+1 =. (. 2C. n+1. :. n P. 2 j. j=1. 0. ; n ) 2 Cn+1 , we set ). = Im. 0. ). :. One sees that Dn+1 is holomorphically equivalent to the unit ball in Cn+1 . Hn acts on Cn+1 by holomorphic a¢ ne transformations which preserve Dn+1 and Mn as follows: for (z; t) 2 Hn and action (z; t). =. 0. by 0 0 0 j. Since for (z1 ;. 2 Cn+1 , we de…ne the. ; zn ; t) 2 Hn and (z1 ;. ; zn ; t). =. =. j. 2. 0. + t + i jzj + 2i. n P. j=1. + zj ;. j = 1;. j zj ;. ; n:. 2 Dn+1 , one has =(. 2. 0. + t + i jzj + 2i. n P. j=1. 6. j zj ; 1. + z1 ;. ;. n. + zn );.
(9) and from Im. 0. n P. >. 2 j. j=1. , we know that 2. Im(. 0. + t + i jzj + 2i. n P. j zj ). j=1. =. 2. Im n P. >. 2. = =. 2. j=1 n P. =. j. + zj. j. + zj. (z; t) ((z 0 ; t0 ). 2. 0. + t0 + i jz 0 j + 2i 2. 0. j=1 j. 2. j zj. j zj. +. +. j zj j zj. + zj. ;. = . Secondly, for (z; t) and (z 0 ; t0 ) 2 Hn , one has. ). (z; t) ( (. zj z j +. j=1. j=1 n P. j zj. Dn+1 ! Dn+1 given by the above formulae is indeed a group action. For. the identity 0 2 Hn , we obviously have 0. =. 1 2. j=1 n P. is in Dn+1 . Therefore, this transformation preserves the structure of Dn+1 and Mn . Next,. we check that the operation Hn. =. n P. +. j=1. and thus (z; t). 2. + jzj + 2. j j. j=1 n P. =. 2. j. j=1 n P. j zj j=1 n P. + jzj + 2 Re. j. j=1 n P. n P. + jzj + 2 Im i. 0. + t0 + i jz 0 j + 2i. j=1. n P. 0 j zj. j=1. on the other hand,. n P. 0 j zj ; 1. + z10 ; 2. ;. + t + i jzj + 2i. n. + zn0 ). j. + zj0 z j ;. n P. j=1. 1. + z10 + z1 ;. ;. n. + zn0 + zn );. ;. n. ((z; t) (z 0 ; t0 )) = =. (z + z 0 ; t + t0 + 2 Im (z + z 0 ; t + t0 + i. n P. j=1. j=1. =. (. 0. 0. +t+t +i. n P. j=1. n P. zj z 0j ) zj z 0j ). z j zj0. 2. zj z 0j + i jz + z 0 j + 2i. z j zj0. n P. j=1. Comparing the two, we see that (z; t) ((z 0 ; t0 ). j. zj + zj0 ;. ) = ((z; t) (z 0 ; t0 )). 1. + z1 + z10 ;. .. Proposition 13 Hn acts simply transitively on Mn , which means that for any =. 0. 2 Mn , then Im. 0. unique u 2 Hn such that u Proof. Let ; 2i. n P. j=1. j zj ; 1. 0. + z1 ;. It follows that. 0. ;. n. +t+i. j=1. t=. 0 j. 0 0. and. 0. 2 Mn , there is a. . =. n P. j=1. 2. j. 0. 0 j n X i. 0 0. and Im. j. 0 0 0; 1;. + zn ) =. n P. + zn + zn0 ):. ;. 0 n. n P. j=1 0 j. n P. j=1. 0 2 j .. If (z; t). 0. =. , then we may take zj to be zj =. + 2i. j. =. j. 0 j. j. 0 j. j. j. =. 0 0.. n X 2i j=1. j=1. 7. j. , that is, ( 0 j. j. Hence, if we take 0 j. j. ;. 2 0 + t + i jzj +. for j = 1;. ; n..
(10) then (z; t) is uniquely determined. It remains to check that t is real. To this end, we take the imaginary part of t and show that Im t = 0. Im t. =. Im. 0 0. Im. n X. 0. 0 j. 0 j. j. n X 2 Im i. j. j=1. j=1. =. n X. 0 2 j. j=1. n X. 2 Re. 0 j j. =. 0 j j. +. 0:. n X. +. 0 j j. +. j=1. n X. 0 j j. j=1. n X +2. j j. j=1. n X. 2 j. j=1. 2 j. j=1. n X. n X. 0 j j. j=1. j=1. =. 0 2 j. j=1. n X +2. j=1. n X. n X. 2 j. j=1. n X. 0 j j. 0 j j. 0 j j. +. j=1. Therefore, (z; t) is in Hn . So Hn may be identi…ed with Mn via the correspondence 2. (z; t) $ (z; t) 0 = t + i jzj ; z1 ;. As for k-strongly pseudoconvex case,. for k + 1. j. j. k. n, we de…ne a left-invariant CR structure on Hn by taking. ; Zk ; Z k+1 ;. T1;0 to be the bundle spanned by Z1 ; on Hn , where wj = zj for 1. n 2. ; zn :. ; Z n . Take (w; t) = (w1 ;. k, and wj = z j for k + 1. ; wn ; t) as new coordinates. j. n. Let "j = 1 for 1 j k, and "j = 1 Pn n, and de…ne the Hermitian form Qk on Cn by Qk (w; w0 ) = j=1 "j wj w0j . Then in the. coordinates (w; t), the group law is. (w; t) (w0 ; t0 ) = (w + w0 ; t + t0 + 2 Im Qk (w; w0 )) : The left-invariant vector …eld agreeing with @=@wj at the origin is Wj = and T1;0 is the bundle spanned by W1 ;. ; Wn . In fact, Wj = Zj for j. The Levi form is given by hWj ; Wl iL = "j If j; l If j. k, hWj ; Wl iL =. i 2. ; Zj; Z l. k and l > k, hWj ; Wl iL =. If j > k and l. k, hWj ; Wl iL =. i 2 i 2. @ @ + "j iwj ; @wj @t. =. jl .. jl. k, and Wj = Z j for j > k.. A veri…cation is as follows: = "j. jl .. h ; [Zj; Zl ]i = 0 = ; Z j; Z l. =0= 8. jl. = "j. jl. jl .. = "j. jl ..
(11) If j; l > k, hWj ; Wl iL =. i 2. ; Z j; Zl. =. = "j. jl. jl .. Analogously, Hn may be embedded in Cn+1 as the hypersurface 2 Cn+1 : Qk. 0. ;. 0. = Im. ;. 0. where. 0. = ( 1;. ;. n) ;. through the map (w; t) 7! (t + iQk (w; w) ; w1 ;. Computation of @ b and. 3. ; wn ) :. b. We continue to use the notation in the previous section. Now we impose the left-invariant metric on Hn which makes W1 ,. ,Wn ,W 1 ,. ,W n ,T orthonormal,. which is independent of the choice of CR structure. The dual frame for the cotangent bundle is dw1 , ,dwn , Pn dw1 , ,dwn and , where = dt + 2 j=1 (xj dyj yj dxj ). Hence hdxj ; dxj i = hdyj ; dyj i = 12 and the volume element is. dV = 2. n. dx1. dxn dy1. dyn dt:. From now on, we identify T0;1 with the subbundle of CT Hn spanned by dw1 ; P ; jq ) with 1 C 1 0;q can be written as ' = J 'J dwJ , where 'J 2 C 1 , J = (j1 ;. ; dwn . Then ' 2 j1 <. < jq. n,. dwj1 ^. ^. and dwJ = dwj1 ^. ^ dwjq . We denote the set fj1 ; ; jq g by fJg. P 2 n If f 2 C 1 , then @ b f = j=1 W j f dwj . Since @ b (dwj ) = @ b (wj ) = 0, we have P 'J dwJ. @b. =. W j 'J dwj ^ dwJ :. J j=1. J. From (#b '; ) = '; @ b. n PP. and integration by parts, we …nd #b. P 'J dwJ. n PP. =. J m=1. J. (Wm 'J ) dwm ydwJ ;. where the interior product y is de…ned by dwm ydwJ = 0 if m 2 = fJg and dwm ydwJ = ( 1) dwji. 1. ^ dwji+1 ^. i 1. ^ dwjq if m = ji . Therefore, #b @ b. P 'J dwJ. = #b. J. =. P. n PP. J m=1 n n P P. J m=1l=1. @ b #b. P 'J dwJ. = @b. J. =. n PP. J l=1 n P n PP J l=1m=1. 9. W m 'J dwm ^ dwJ Wl W m 'J dwl y dwm ^ dwJ ; (Wl 'J ) dwl ydwJ W m Wl 'J dwm ^ dwl ydwJ :.
(12) Lemma 14 For m 6= l, dwl y dwm ^ dwJ = dwm y dwm ^ dwJ. dwm ^ dwl ydwJ . And for m = l, we have 8 < 0 if m 2 fJg; = : dwJ if m2 = fJg; 8 < dwJ = : 0. dwm ^ dwm ydwJ Proof. For m 6= l, one has 8 < 0 dwl y dwm ^ dwJ = : ( 1)i dw ^ dw ^ m j1 and. dwm ^ dwl ydwJ. 8 < dwm ^ 0 = 0 = : dw ^ ( 1)i. 1. m. Thus dwl y dwm ^ dwJ =. ^ dwji. dwj1 ^. 1. if. m 2 fJg;. if. m2 = fJg:. ^ dwji+1 ^. ^ dwji. 1. ^ dwjq. ^ dwji+1 ^. if. l2 = fJg (l 6= m) ;. if. l = ji 2 fJg;. ^ dwjq. if. l2 = fJg;. if. l = ji 2 fJg:. dwm ^ dwl ydwJ for m 6= l.. For m = l, one has dwm y dwm ^ dwJ. 8 < 0 = : ( 1)1. 1. J J [ dw m ^ dw = dw. and dwm ^ dwm ydwJ = 0 if m 2 = fJg, and dwm ^ dwm ydwJ =( 1). if. m 2 fJg;. if. m2 = fJg;. i 1. i 1. ( 1). dwj1 ^. ^ dwji ^. ^. dwjq =dwJ if m = ji 2 fJg. Hence the result. Since Wl W m = W m Wl , we …nd ! ! X X J J 'J dw = @ b #b 'J dw + #b @ b b J. J. n XX. =. =. J. Note that. b. 20 4@. 'J dw. J. J. !. Wm W m 'J dwm y dwm ^ dwJ + W m Wm 'J dwm ^ dwm ydwJ. J m=1. X. X. X. W m Wm +. X. m2fJg =. m2fJg. 1. 3. Wm W m A 'J 5 dwJ :. is in diagonal form: it does not mix up the components.. Since Wm ; W m = Zm ; Z m = when m > k, we have Wm ; W m = CR structure. Let K = f1; J;k. 2i. mm T. =. k, and Wm ; W m = Z m ; Zm = 2iT. 2i"m T . Now suppose we are working with the k-strongly pseudoconvex. ; kg, K 0 = fk + 1;. = # (K. 2iT when m. ; ng, fJg = fj1 ;. fJg) + # (K 0 \ fJg) 10. # (K \ fJg). ; jq g and # (K 0. fJg) ;.
(13) where # denotes cardinality. Lemma 15. P. P. Wm ; W m. Wm ; W m = 2i. J;k T:. m2fJg =. m2fJg. Proof. Suppose j1 ;. ; ja 2 K, and ja+1 ;. J;k. ; jq 2 = K. Then. fJg) + # (K 0 \ fJg). =. # (K. =. (k. a) + (q. =. 2k. 4a + 2q. a). a. (n. # (K 0. # (K \ fJg). k. (q. fJg). a)). n:. On the other hand, since Wm ; W m =. 2i"m T , where "m = 1 if m. k and "m =. 1 if m > k, it follows. that P. P. Wm ; W m. Wm ; W m. m2fJg =. m2fJg. = a ( 2iT ) + (q. a) (2iT ). =. (( 2a + q). =. ( 4a + 2q + 2k. =. J;k. [(k. a) ( 2iT ) + (n. [ 2k + 2a + n. q. (k. a)) (2iT )]. q]) (2iT ). n) (2iT ). (2iT ) :. Then we have P 'J dwJ b J. P P P W m Wm + Wm W m )'J ]dwJ [(. =. J. m2fJg =. m2fJg. P P = [ ( W m Wm + Wm W m m. J. =. P [( J. +. 1 2. P. P. W m Wm +. m2fJg =. =. P [( J. + 12. =. P [(. P [( J. P. P 1 2. 1 2. m. P m. 1 2. P. Wm W m )'J ]dwJ. m2fJg. W m Wm + Wm W m. m. Wm W m )'J ]dwJ. W m Wm + Wm W m. W m Wm +. P. 1 2. m2fJg. m. m2fJg =. J. =. 1 2. P. P. W m Wm. m2fJg =. W m Wm + Wm W m. m. P. P. 1 2. P. Wm W m. m2fJg =. Wm W m )'J ]dwJ. m2fJg. W m Wm + Wm W m +. 1 2. W m Wm + Wm W m + i. P. Wm ; W m. m2fJg J;k T )'J ]dw. 11. J. :. 1 2. P. W m Wm. m2fJg. 1 2. P. m2fJg =. Wm ; W m )'J ]dwJ.
(14) Note that. J;k. (a = q), we have For any. is one of the numbers n; n J;n. =n. q+0. q. 2;. 0 (= 2n. ; n + 2; n. In the strongly pseudoconvex case k = n 4q + 2q. n) = n. 2q for all J.. 2 C, we de…ne the operator n. 1X Zj Z j + Z j Zj + i T: 2 j=1. L =. Finally, since Wj is either Zj or Z j , we can write b. X. 'J dwJ. J. 4. b. !. as. =. X J. L. J;k. 'J dwJ :. Construction of the fundamental solutions. De…nition 16 For 0 < r < 1 and (z; t) 2 Hn , we de…ne the dilation of (z; t) by r to be r(z; t) = (rz; r2 t). 4. And we de…ne the (Heisenberg) norm of (z; t) to be j(z; t)j = jzj + t2. 1=4. . 2. Thus de…ned, we have jr(z; t)j = r j(z; t)j. Note that the Euclidean norm of (z; t) is jzj + t2 we denote it by k(z; t)k. Lemma 17 For u = (z; t) 2 Hn , we have the following inequalities: 1. juj < kuk 2 when juj < 1; and kuk < juj when juj <. 1 : 2. Proof. If juj < 1, we have jzj < 1. 4. 2. ). jzj < jzj. ). jzj + t2. ). juj < kuk. ). juj < kuk 2 :. 1=2. 4. 2. 1. 12. 2. < jzj + t2. 1=2. 1=2. and.
(15) 2. If juj < 12 , we have jzj <. 1 2 2. 1 4 2 ,. and t2 <. then 1 4. 2. 2. jzj + t2. kuk =. 4. 1 4. 2. =. jzj + 2 jzj t2 + t4. =. jzj + t2 2 jzj + t2. <. jzj + t2 2. <. jzj + t2. 1 4. 2. 4. 4. 1 4. 4. 1 4. 1 1 + 4 16. = juj :. We want to search for a fundamental solution ' for L . FoIland[3] showed that L0 j(z; t)j for some c0 6= 0, where form ' (z; t) = j(z; t)j j(z; t)j Hn. 2n. 2n. = c0. is the Dirac -function (concentrated at 0). Inspired by this, we consider ' of the 2n. f (!), where ! = j(z; t)j. 2. t, for such a ' is a dilation-invariant function times. and is invariant under unitary transformations in the z-variable. In order to solve L ' = 0 on. f0g, we need to compute some lengthy partial derivatives. First note that 0 1 @! @t. =. =. @ B @ @t. t. 1 4. jzj +. jzj + t2 0. t2. 1. =. 2. 1. 1 2. 4. = j(z; t)j. C A. B 4 2 @ jzj + t. jzj + t2. and for j = 1;. 1 2. 4. 1 2. t. 2. 4. t2 4. jzj + t2. jzj + t2 !. 1 2. 1 C A. !2 ;. 1. ; n, @! @z j. =. = =. 0. @ B @ @z j. t jzj + t2. 1. 4. t. t2. jzj + 1 4. jzj + t2 = j(z; t)j. 1 2. 4. 6. 1 C A. 1 4 jzj + t2 2. 1 2. t 3 1 4 jzj jzj 2 2. 3 2. 2. t jzj zj : 13. @ 4 jzj @z j 1. zj.
(16) For later reference, we compute @ j(z; t)j = @t = =. @ j(z; t)j = @zj. @ 4 jzj + t2 @t 1 4 jzj + t2 4 ! ; 2 j(z; t)j. 3 4. 1 4 jzj + t2 4 2 jzj z j 3; 2 j(z; t)j. =. 3 4. 2t. 1 4. @ 4 jzj + t2 @zj. =. 1 4. 3. 4 jzj. 1 jzj 2. 1. zj. together with @ f (!) @t. = f 0 (!). @ ! @t 2. = f 0 (!) j(z; t)j @ f (!) @zj. = f 0 (!). !2 ;. 1. @ ! @zj. = f 0 (!) j(z; t)j. 6. = f 0 (!) j(z; t)j. 4. 2. t jzj z j 2. ( !) jzj z j :. Now @ ' @t. =. @ j(z; t)j @t. =. 2n j(z; t)j. =. n j(z; t)j. =. j(z; t)j. 2n. f (!). 2n 1 2n 2. 2n 2. @ j(z; t)j f (!) + j(z; t)j @t. !f (!) + j(z; t)j. 2n. n!f (!) + ! 2. 1 f 0 (!) ;. 2n. f 0 (!) j(z; t)j. @ f (!) @t 2. 1. !2. and @ ' @zj. =. @ j(z; t)j @zj. =. 2n j(z; t)j. =. n j(z; t)j. =. j(z; t)j. 2n. f (!) @ j(z; t)j f (!) + j(z; t)j @zj. 2n 1 2n 4. 2n 4. 2. jzj z j f (!) + j(z; t)j 2. 2n. jzj z j [nf (!) + !f 0 (!)] : 14. 2n. + j(z; t)j. @ f (!) @zj 2n. f 0 (!) j(z; t)j. 4. 2. ( !) jzj z j.
(17) It follows that @2 ' @t2. = =. @ @t. 2n 2. j(z; t)j. @ j(z; t)j @t. 2n 2. n!f (!) + ! 2. 1 f 0 (!). n!f (!) + ! 2. 1 f 0 (!). @ @ (!f (!)) + ! 2 1 f 0 (!) @t @t 2n 3 @ j(z; t)j n!f (!) + ! 2 1 f 0 (!) = (2n + 2) j(z; t)j @t @ @ @ 2n 2 j(z; t)j n( ! f (!) + ! f (!)) + ! 2 1 f 0 (!) + ! 2 @t @t @t ! 2n 3 n!f (!) + ! 2 1 f 0 (!) = (2n + 2) j(z; t)j 2 j(z; t)j j(z; t)j. j(z; t)j. = j(z; t)j. 2n 2. 2n 2. 2n 4. n. [n j(z; t)j. 2. 1. ! 2 f (!) + !f 0 (!) j(z; t)j. +2! j(z; t)j. 2. 1. ! 2 f 0 (!) + ! 2. = j(z; t)j. 2n 4. ! 2 f (!) + !f 0 (!) 1. 1. fn (n + 1) ! 2. = j(z; t)j. 2n 4. !2. fn (n + 2) ! 2. + !2. 1. 2. !2. 1 f 0 (!). + (n + 1) ! ! 2 + 1. 1. 2. 1. @ 0 f (!) @t. !2. 1 f 00 (!) j(z; t)j. 2. !2 ]. 1. 1 f 0 (!). f(n + 1) ! n!f (!) + ! 2. [n. 2. 1. !2. + 2! 1. ! 2 f 0 (!). 1. f (!) n!f 0 (!) 1. !2. 2! 1. f 00 (!)g. 1 f (!) + (2n + 3) ! ! 2. f 00 (!)g;. and. 15. 1 f 0 (!). ! 2 f 0 (!). !2. 2. f 00 (!)]g.
(18) @2 ' @z j @zj. = =. @ 2n 4 2 j(z; t)j jzj z j [nf (!) + !f 0 (!)] @z j @ 2n 4 2 j(z; t)j jzj z j [nf (!) + !f 0 (!)] @z j @ 2 2n 4 jzj j(z; t)j z j [nf (!) + !f 0 (!)] @z j @ 2n 4 2 z j j(z; t)j jzj [nf (!) + !f 0 (!)] @z j @ 2n 4 2 [nf (!) + !f 0 (!)] j(z; t)j jzj z j @z j 2. =. 2 jzj 1 +. 1 jzj 2. j(z; t)j n. 1. jzj zj. 2n 5. (2n + 4) j(z; t)j. 2. jzj z j [nf (!) + !f 0 (!)]. 3. 2 j(z; t)j zj. j(z; t)j. 2n 4. 2n 4. z j [nf (!) + !f 0 (!)]. 2. jzj [nf (!) + !f 0 (!)] @ ! f 0 (!) @z j. @ f (!) @z j. @ 0 f (!) @z j. !. j(z; t)j. 2n 4. 2. jzj z j. 2. =. 2 jzj 1. 1 jzj 2. j(z; t)j. 1. zj. j(z; t)j. 2n 4. j(z; t)j. 2n 4 2n 4. =. 2n 4. z j [nf (!) + !f 0 (!)]. 2. jzj [nf (!) + !f 0 (!)]. 4. 2. ! jzj zj + j(z; t)j. +!f 00 (!) j(z; t)j (n + 2) j(z; t)j. 2. jzj z j [nf (!) + !f 0 (!)]. 3. 2 j(z; t)j. +[nf 0 (!) j(z; t)j. =. jzj zj. 2n 5. (2n + 4) j(z; t)j. 2n 8. 6. 2. 2. 6. ! j(z; t)j jzj zj ]. 4. 2. 2. ! j(z; t)j jzj zj f 0 (!) j(z; t)j. 2n 4. 2. jzj jzj j [nf (!) + !f 0 (!)]. 2. jzj j [nf (!) + !f 0 (!)] 2. j(z; t)j jzj [nf (!) + !f 0 (!)] h 4 2 + (n + 1) f 0 (!) j(z; t)j ! jzj zj + ! 2 f 00 (!) j(z; t)j (n + 2) j(z; t)j. 2n 8. 4. jzj j [nf (!) + !f 0 (!)]. j(z; t)j. 2n 4. jzj [nf (!) + !f 0 (!)]. 2. jzj zj. 2. 2n 4. 4. 4. jzj jzj j [nf (!) + !f 0 (!)]. j(z; t)j. + j(z; t)j. 2. jzj z j. i. j(z; t)j. 2. 2. 2. jzj zj (n + 1) f 0 (!)! + ! 2 f 00 (!). 16. j(z; t)j. 2n 4. 2. jzj z j. 2n 4. 2. jzj z j.
(19) =. 2n 8. (n + 2) j(z; t)j. 4. 2. jzj jzj j [nf (!) + !f 0 (!)]. j(z; t)j. 2n 4. jzj j [nf (!) + !f 0 (!)]. j(z; t)j. 2n 4. jzj [nf (!) + !f 0 (!)]. + j(z; t)j. 2n 8. jzj jzj j. = j(z; t)j. 2n 4. + j(z; t)j = j(z; t)j. + j(z; t)j. 2. 4. 2. (n + 1) !f 0 (!) + ! 2 f 00 (!). [nf (!) + !f 0 (!)] (n + 2) j(z; t)j. 2n 8. 2n 4. 2. h. 4. 2. 4. 4. 2. jzj jzj j. 2. jzj j. 2. jzj. (n + 1) !f 0 (!) + ! 2 f 00 (!). jzj jzj j. (n + 2) j(z; t)j. 2n 8. 4. 2. 4. 4. 2. jzj jzj j. 2. 2. jzj j. jzj. i. [nf (!) + !f 0 (!)]. (n + 1) !f 0 (!) + ! 2 f 00 (!) :. jzj jzj j. Write. n. L. 1X 2 j=1. =. n X. =. j=1. Since zj @z@ j. @ @ + iz j @zj @t. @ @z j. 2 @2 2 @ + jzj j @z j @zj @t2. i. @ @t. izj. @ @t. zj. @ @zj. +. zj. izj. @ @z j. +i. 2. 2. jzj 6. h. (n + 2) j(z; t)j. 4. +i. 6. jzj. n!f (!) + ! 2. @ @t. @ : @t j(z; t)j. 2n 2. 1 f 0 (!) + ! 2. 1. 1 f 0 (!). 2. o f 00 (!). 0: !2 = 1 1. j(z; t)j !2. 1 2. + 1. !2. + 1. !2. +i =. +i. i (n + 1) [nf (!) + !f 0 (!)]. 4. jzj. (n + 1) !f 0 (!) + ! 2 f 00 (!) n 2 2 + j(z; t)j jzj n (n + 2) ! 2 1 f (!) + (2n + 3) ! ! 2. + j(z; t)j. Because 1. @ @ + iz j @zj @t. z j @z@ j ' = 0, The equation L ' = 0 implies that (after dividing it by j(z; t)j. =. @ @t. @ @z j. 4 2. t = j(z; t)j. (n + 2) 1 3 2. 1 2. 4. !2. 4. jzj , we have (n + 1) [nf (!) + !f 0 (!)]. (n + 1) !f 0 (!) + ! 2 f 00 (!) n n (n + 2) ! 2 1 f (!) + (2n + 3) ! ! 2. n!f (!) + ! 2. 1 f 0 (!). 0:. 17. 1 f 0 (!) + ! 2. 1. 2. o f 00 (!). ).
(20) It follows that f (!). n. 1 2. !2. 1. +f 0 (!)f 1. !2. !2. (n + 2) 1 1 2. !2. (n + 2) 1. + 1 ! 2 (2n + 3) ! ! 2 1 + i ! 2 o n 1 3 2 +f 00 (!) 1 ! 2 2 ! 2 + 1 ! 2 2 1 ! 2. 1 g. (2n + 3) 1. !2 ]. = f (!) fi n!g + f 0 (!)f 1. 1 2. !2. ![. ! 2 (n + 2) + 1 + 1. o !2 + 1 !2 1 h = (i n!) f (!) + 1 ! 2 2 ( n 1) ! i 1 h = i n!f (!) 1 ! 2 2 (n + 1) ! + i 1 +f 00 (!). =. !2. (n + 1) ! + 1. 1 2. n. !2. 1. !2 1i. 1 !2. f 00 (!) =. 1 sin2. g 00 ( ). g0 ( ) sin. 2. cos. i n cos g( ) + sin [(n + 1) cos + i sin ]. =. d + n cos d. sin. i. 1 2. i. d d. +i. 1. f 0 (!) + 1. f 0 (!) + 1. !2 g. d +i d. g0 ( ) 1 + sin3 sin sin2. g( ). g( ) = 0 : =). g0 ( ) = 0. =). g( ) = g( ). =). ln g( ) =. =). g( ) = e. 18. i g( ) i i i. !2. !2 3 2. 3 2. f 00 (!). f 00 (!). +c +c. f 0 (!) cos ; it. and the ODE for f may be written as. 0:. We …rst solve. (n + 1) !. o. ! 2 (n + 1). = i n cos g( ) + [n cos + i sin ] g 0 ( ) + sin g 00 ( ) =. 3 2. 1 + i n!. f 0 (!) sin , g 00 ( ) = f 00 (!) sin2. , then g 0 ( ) =. Set ! = cos , f (!) = g( ); 0 g0 ( ) sin ,. n (n + 2) ! 2. 3 2. 0:. follows that f 0 (!) =. 1 2. !2. (n + 1) n + 1. = c0 e. i. :. g 00 ( ). g0 ( ) cos sin.
(21) And then solve sin. Now let. d d. d d. +i. + n cos. h( ) = 0 : cos h( ) sin cos n sin. =). h0 ( ) =. =). h0 ( ) = h( ). =). ln h( ) =. =). ln h( ) = ln (sin ). =). h( ) = (sin ). g( ) = c1 (sin ). n. n. n ln sin + c0 n. n c0. + c0. = c1 (sin ). e. i. . Assume g( ) = c2 e. n. :. ( ). We have c2 ( i ) e. n. i. ( )+ n. ( ) + i c2 e i ( ) = c2 e i dd ( ) = c1 (sin ) . From dd ( ) = c1 c2 1 ei (sin ) , we R R n n obtain ( ) = c1 c2 1 ei (sin ) d + c3 . Therefore, g( ) = c2 e i c1 c2 1 ei (sin ) d + c3 = R i n n e (sin ) d + c2 c3 e i . Since (sin ) ! 1 as ! 0 or , the only bounded solutions of c1 e i. c2 e. i. d d. this ODE for 0. i. are g( ) = ce. . Hence,. f (!) = g( ) = c (cos. i sin ) = c !. i 1. !. 2. 1 2. =c. t. 2. i jzj. 2. j(z; t)j. !. :. Choose c = i , we have ' (z; t). = j(z; t)j =. 2. jzj. 2n 2. 2. it + jzj. n 2. it. 2. n 2+2. 2. jzj + it. 2. (We use the principal branch of the power functions since jzj 2. Theorem 18 For ' (z; t) = jzj. n 2. it. 2. c = Lemma 19 For " > 0, let. 2. ". (z; t) = jzj + "2. 2. jzj + it 22. n 2. 4. = jzj + t2. n 2+2. 2. 2. jzj + it. :. it lies in the right half-plane.). , we have L ' = c. , where. 2n n+1. n+ 2. it and '. :. n 2. =. n+ 2. n. ;". ! ' as " ! 0 in. ! ' (u) as " ! 0 for all u 2 Hn . Let. 2 C01 (Hn ) and. ;". ". ". 2. , then '. distribution sense. Proof. Clearly, ' M. ;". is C 1 and '. ;" (u). = sup fj (u)j : u 2 supp( )g > 0. For any ; s 2 C, one has j s j = es log. 19. = es(lnj. j+i arg ). =.
(22) eRe s lnj. j Im s arg. Re s. =j j '. e. ;" (u). Im s arg. , where arg 2 [. (u). M. '. ;" (u). n. M j " (u)j 2. = M. jzj + "2. = M. 2. jzj + ". n 2. 2. 2. +t. e. " (u). ejIm. 2. j(z; t)j. 2n. dV. Z. =. supp( ). Z. supp( )\fj(z;t)j. Z. =. supp( )\fj(z;t)j. Z. n 2. 1 2. 1 2. Z. 1 2. e. jIm j. jIm j. e. :. 2n. Z. dV +. g. j(z; t)j. 2n. dV. supp( )\fj(z;t)j> 12 g. j(z; t)j. 2n. dV + A. j(z; t)j. 2n. 2. for some A > 0. g n. dxdydt + A. g k(z; t)k. supp( )\fj(z;t)j. <. 1 2. n 2. 2. jzj + "2 + it. 2. jIm j. j(z; t)j. supp( )\fj(z;t)j. 2. j. On the other hand, for any u = (z; t) = (x; y; t) 2 Hn , Z. n. jIm j. it. 2n. M j(z; t)j. n. j " (u)j. e. n+ 2. " (u). =M n+ 2. M j " (u)j. ; ). For u 2supp( ), we have. 2n. 2. n. dxdydt + A. g k(z; t)k. 2n. 2. n. dxdydt + A. supp( )\fj(z;t)j 1g. =. 2. Z. n. 0. =. 2. 1. Z. 2n. d (z; t) dr + A. k(z;t)k=r. 2. n. r n+ 12. n+. 1 2. +A 1. since the area of the unit sphere in Cn R is 2 n+ 2 = n + 12 . R 2n 2n Therefore supp( ) M j(z; t)j e jIm j dV < 1, i.e., M j(z; t)j e jIm j is integrable. Hence by R R Lebesgue dominated convergence theorem, lim"!0 ' ;" dV = ' dV and the lemma is proved. Lemma 20 We now set. ;". =L '. Proof. Fix " for now and write Zj (. a. )=. =. ;" ,. ". @ @ + iz j @zj @t. then. ;". !c. as " ! 0.. for simplicity. For any a 2 C, 2. jzj + "2. a. it. 20. = az j. a 1. + az j. a 1. = 2az j. a 1. :.
(23) Similarly, Zj (. a. ). =. 2azj. T(. a. ). =. ia. a 1 a 1. a. ;. Zj (. ;. T(. = 2b. a b 1. a. a. ) = Zj (. ) = ia. a 1. ) = 0;. :. Thus for a; b 2 C,. n X. Zj Z j. a b. = Zj 2bzj. a b 1. Zj Z j. a b. =. 2nb. a b 1. + 4ab jzj. T. a b. =. ia. a 1 b. + ib. j=1. From Zj ; Z j = Zj Z j. Z j Zj =. 2. a b 1. a 1 b 1. =i. 2. + 4ab jzj j =. a 1 b 1. a 1 b 1. a 1 b 1. ; 2. 2nb + 4ab jzj. ( a + b ):. 2iT , we know that n. L. =. 1X Zj Z j + Z j Zj + i T 2 j=1 n. =. =. =. 1X Zj Z j + Z j Zj + Zj Z j 2 j=1. n X. j=1 n X. Zj Z j. inT + i T. Zj Z j + i (. n) T:. j=1. 21. Z j Zj. 2iT. +i T. ;.
(24) Set a =. n+ 2. n. and b =. ;". , we have. 2. = L '. ;" n. n+ 2. = L n+ 2. = +i ( = =. 2. n. 1. 1. 2. n+ 2. n) i. n+ +2 2. n. n+ +2 2. n. +2. +2. + n+ +2 2. =. n. 2. ) jzj + "2. 2. [jzj. = ". n. 2. 2. ) jzj. 2. 2. 2. jzj. it ] 2. ). +2 2. 2. jzj + " + it. it. 2. ). 2. 2. n. n+ +2 2. 1 (n 2. 2. 1 2 2 (n ) jzj + "2 2 1 2 1 2 + n (n 2 2 1 2 (n ) 2 1 2 + (n ) ] 2. 2. 1 2 n 2 1 2 )+ n 2. n (n. jzj + ". n2. ) )+. 2. n2. it. jzj + "2 + it. n (n. 2. 2. 1 (n + ) (n 4 n 2 1 2 2 2 jzj + n 2. 2. 2. +"2 n (n. 2. n2. ). [n (n. +it. n+ 2. 1. 2. 1 2 n 2. +2 2. n. 1. ) +4. 2. n (n. 2. 2. n. 2n(. :. Since ;1. ". 1. (z; t). = =. ;1. 1. n2. 1. = ". ". 2. n+ +2 2. 2(. 2. z; ". ;". 2n 2. (z; t) = " Z. ;1. ;". ". n. +2 2. 2. 1. 1. ". = "2n+4 n2 we have. t z. 2. + 12. ) n2. 2. i". 2. jzj + "2 n+ +2 2. it. n. 1. ". 2. 2. jzj + "2. n+ +2 2. t. (z; t) dV (z; t). =. n+ +2 2. it. Z. n. ". 1. (z; t) ". 2n 2. Z. ". 1. ;1. (z; t) dV ". +2 2. dV (z; t). =. Z. ;1. (z; t) dV (z; t) :. Hn. . We now need to show that for any ;" ;. !c. 2 C01 (Hn ),. (0) = (c 22. ; ) as " ! 0:. 1. +2 2. t n. 2. 2. ;1. 2. jzj + "2 + it. jzj + "2 + it. Hn. ;1 dV. + 12 + i". Hn. =. R. 2. (z; t) , and hence. Hn. Let c =. z. (z; t). ;. +2 2.
(25) From the fact that ". 1. (V ) ! Hn as " ! 0 for any neighborhood V of 0 and the fact that c < 1, we have ;" ;. Z. =. ;". (z; t) (z; t) dV (z; t). Hn. Z. =. ". 1. ;1. (z; t). (z; t) ". ". 1. ;1. (z; t). " ". 2n 2. dV (z; t). Hn. Z. =. 1. (z; t). dV ". 1. (z; t). Hn. Z. =. ;1. ZHn. !. (z; t) (" (z; t)) dV (z; t) (0) dV = c. ;1. (0) = (c. ; ) as " ! 0:. Hn. Thus the lemma is proved. R1 Lemma 21 The Mellin transform of e xs : 0 e xs x 1 dx = ( ) s for Re s > 0 and 1. R 1 xs R R R N a N Proof. Write 0 e x 1 dx = lim 0 e xs x 1 dx = lim e xs x 1 dx + a e xs x 1 dx with 0 N !1 N !1 Ra Ra N > a > 0. Since 1, we know that 0 e xs x 1 dx = lim " e xs x 1 dx exists and de…nes an analytic "!0. function everywhere. Note that Z. 1 xs se. d dx. N. e. xs. 1. x. 1 e s. dx =. a. 1 e s. = This shows that. R1 a. e. xs. 1. x. xs. =e. , so from integration by parts: Z. N xs. 1. x 1. N. 1 e s. a. a Ns. N. 1 + e s. as. a. 1. xs. 1. +. s. ( Z. 1. e. e. 1. x. dx =. 0. Z. 1. 1 e (xs )s xs. 1. 1. xs. 2. x. d xs. R1 0. e. xs. 1. 0. =. Z. 1. e. x. x. 1. s. +1. e. x. x. 1. dx s. s. 1. dx. 0. =. Z. 1. 0. = Hence by analytic continuation,. R1 0. e. xs. x. 1. ( )s. dx =. :. ( )s. 23. dx:. a. dx de…nes an analytic function on Re s > 0. Thus. xs. dx. N. on Re s > 0. For s positive, Z. 2. 1) x. on the half-plane Re s > 0.. x. 1. dx is analytic.
(26) Proof. (of the Theorem) We now prove L ' = c (L ' ; ). = = = = =. (' ; L. ). lim '. ;" ; L. "!0. lim L '. "!0. lim. (c. = Z. 1 2 1. (n +. + 2) ;. (A. it). =. 1 2. (n. (A + it). ;" ;. ;" ;. "!0. Hence the result. Therefore it remains to compute c = Let us set. 2 C01 (Hn ),. . For any. ; ): R. ;1 dV. .. + 2). For any A > 0, Z. dt = A. 1. = A. n 1. Z. 1 1 1. 1 (1. i. t A. i ). 1+i (1 + i ). t A. dt. d :. 1. 2. Put A = jzj + 1, then c. =. Z. ;1 dV =. Hn. =. 2. n. n2. 2. Z. Z. 2. n. n2. 2. =. 2. n. n2. 2. 2. (A. it). (A + it). dV. Z. 1. (A it) (A + it) dtdxdy Z Z 1 A n 1 (1 it) (1 + it) dtdxdy Cn 1 Z Z 1 n 1 2 jzj + 1 dxdy (1 it) (1 + it) Cn. =. n2. Hn. 1. Cn. dt:. 1. We evaluate the integral in c . Recall that the area of the unit sphere in Cn is 2. 24. n. = (n). The …rst integral.
(27) is Z. Cn. Z. n 1. 2. jzj + 1. dxdy. =. 1. jzj=r n Z 1. 0. 2. =. Z. (n) n. =. (n) n. =. Z. (n). 1. r2 + 1. n 1 2n 1. d (x; y) dr; r. (using polar coordinnates,). dr. s. n 1. n 1. (s. 1). s = r2 + 1; ds = 2rdr;. ds;. 1. Z. 0 n 1. 1 1. 1. n 1. d. n. 1. (1. ). (1. ). 1. n 1. d ;. =s. 1. ;d =. s. 2. ds;. 0. n. =. n 1. 0. (n) n Z. =. r2 + 1. (n). n. =0. n. 1 (n) n. =. n. =. (n + 1). For the second integral, we assume R1 0. e. xs. x. 1. dx =. ( )s. :. n. n for the moment, then. holds for Re s > 0. Put s = 1 + it, then (1 + it). ( )=. Z. 1. ixt. e. x. e. 1. x. 0. where fb denotes the Fourier transform of f , and 8 < e xx f (x) = : 0. Now put s = 1. it and replace (1. 1. dx = fb(t) ;. for x > 0; for x. 0:. by :. it). ( ). =. Z. 1. eixt e. x. 1. x. dx. 0. =. Z. 1. ixt x. e. e ( x). 1. d ( x). 0. = where. 8 < 0 g (x) = : e. Z. 0 ixt. e. e. 1. jxj. jxj. for x jxj. 1. jxj. 1. dx = gb (t) ; 0;. for x < 0:. By Plancherel formula in the form Z. 1 1. fb(t) gb (t) dt = 2. Z. 25. 1 1. f (x) g ( x) dx;. 1;. 1. Recall that.
(28) we have. Z. 1. (1 + it). ( ) (1. it). Z. ( ) dt = 2. 1. 1. x. e. 1. x. e. x. 1. x. dx;. 0. and thus ( ) ( ). Z. 1. (1. it). (1 + it). dt =. 2. 1. Z. 1. e. 2x. e. 2x n. x. +. 2. dx. 0. =. 2. Z. 1. x dx. 0. Therefore,. for. n. Z. 1. where the identity. it). i 2. 1 2 (n+. +2). it). (1 + it). dt =. 2. =. 2. 2. n. 1. n. Now for arbitrary (1. (1. =. n. 1 2 (n. (n + 1) :. (n + 1) ; ( ) ( ). +2). =. 1 + t2. =. 1 + t2. =. 1 + t2. =. 1 + t2. =. 1 + t2. =. 1 + t2. =. 1 + t2. n+2 2. n+2 2. n+2 2. n+2 2. n+2 2. n+2 2. n+2 2. (1. it). (i + t). 2. 2. (1 + it) 2. (i. t) 2. e. 2. log(i+t). e. 2. (log(i+t) log(i t)). e. 2. log( ii+tt ). e2. log(i t). log( ii+tt ). ei. i 2. ei. arctan t. ;. t)) = arctan t can be checked by noting that when t = 0, i log 2. i+t i t. =. i log 1 2. =. 0. =. arctan 0;. t=0. and the derivatives of these 2 functions i 2. n 1. 2 C, since. (1 + it). log ((i + t) = (i. (n + 1) 2. 1 i+t. ( 1) i t. =. i 2. =. 1. t=0. = =. 26. 1 1 + i i 1 1 + t2 t=0 d arctan t dt. t=0.
(29) are equal. Whence. Z. 1. (1. it). (1 + it). Z. dt =. 1. 1. n+2 2. 1 + t2. ei. arctan t. dt. 1. de…nes an entire function of . Thus by analytic continuation, Z holds for all. 1. (1. it). (1 + it). n. 2. dt =. 1. (n + 1) ( ) ( ). 2 C.. Finally, c. =. 2. =. 2 22. =. n. n2. n. 2. 2. (n + 1) n2. 2n n+1 n+ 2. n. (n + 1). n+ +2 2 2. n+ 2. n. +2 2. n. n 2. 2. 1. 2n n+1 n+ 2. n 2. 2 2n n+1. 2. =. ;. n. n+ 2. 2. and the proof is complete. Since. 1 (z). is an entire function with zeroes at z = 0; 1; 2;. = n; n + 2; n + 4;. n 2. = 0; 1; 2;. , i.e.,. , we have c = 0.. De…nition 22 If c 6= 0, then If we further de…ne. 5. , so when. =c. is called admissible. 1. ' , then. is a fundamental solution for L with source at 0.. Solutions of L f = g. We denote D0 to be the spaces of distributions and E 0 the spaces of distributions with compact support. De…nition 23 The convolution of two functions f and g on Hn is de…ned by f. g (u) =. Z. f (v) g v. 1. u dV =. Hn. Proposition 24 If we set ge (u) = g u Z. (f. Hn. Z. f uv. 1. g (v) dV:. Hn. 1. , then. g) (u) h (u) dV (u) =. Z. Hn. whenever both sides make sense. 27. f (u) (h ge) (u) dV (u) ;.
(30) Proof. We start with Z. Hn. f (u) (h ge) (u) dV (u). Z. =. f (u). Hn. Z. =. Hn. f (u). Hn. Z. =. Z. Z. h (v) ge v. 1. h (v) g u. u dV (v) dV (u). 1. v dV (v) dV (u). Hn. (f. g) (v) h (v) dV (v) ;. Hn. thus. R. Hn. (f. g) (u) h (u) dV (u) =. De…nition 25 For f 2 C01 and Lemma 26 For ' 2 C 1 and Proof. We …rst compute. R. Z. R. f (u) (h ge) (u) dV (u) whenever both sides are de…ned.. Hn. admissible, we de…ne the operator K : C01 ! C 1 by K f = f. 2 C01 , we have. Zj '. dV and. Zj '. dV. Z. R. R. L '. i T'. dV =. Z. dV. Z. =. Z. =. Z. Z. Zj'. Zj Z j ' Z j Zj '. dV =. @ i ' dV @t Z @ i ' dV @t ' ( i T ) dV. Z. =. It follows that. dV .. dV by integrating by parts.. =. Z. ' L. @ @ ' dV + iz j @zj @t Z Z @ @ = ' dV dV ' iz j @zj @t Z @ @ = ' iz j dV @zj @t Z = ' ( Zj ) dV: =. i T'. Similarly,. R. Z. dV = dV =. Z Z. 28. ' i(. '. ) T dV:. Zj. dV:. '. Z j Zj. dV;. '. Zj Z j. dV:. ..
(31) Since L =. 1 2. Pn. j=1. Z. Zj Z j + Z j Zj + i T ,. L '. dV. Z. =. Z. =. Proposition 27 For f 2 C01 and. 0. n. 1X ' @ Z j Zj + Zj Z j + i ( 2 j=1. ' L. dV:. 1. ) T A dV. admissible, we have L K f = K L f = f .. Proof. Since L is a combination of left-invariant vector …elds Zj , Z j and T , L is left-invariant, which means L ( for any u 2 Hn and. Lu ) = (L. ) Lu. 2 C 1 . Therefore, for any u 2 Hn , L K f (u). = L (f ) (u) Z = L f (v) H Z n = L f (v) Hn Z = f (v) L ( ZHn = f (v) (L ZHn = f (v) (L Hn. = f = f. L. (u). (u). = f (u) :. 29. 1. v (Lv. u dV (v) 1. Lv ) Lv ) v. 1. (u)) dV (v) 1. 1. (u)) dV (v) (u) dV (v). u dV (v).
(32) So we have L K f = f . On the other hand, for any g 2 C01 , Z. g (u) f (u) dV (u). Z. =. Z. =. Z. =. Z. =. Z. =. Z. = where we use the fact that e. L. K. K. g (u) L f (u) dV (u). g. by above result, by lemma,. (u) L f (u) dV (u) e. g (u) L f g (u) (L f. (u) dV (u) ) (u) dV (u). g (u) K L f (u) dV (u) ; u. (u) =. g (u) f (u) dV (u). 1. 1. =c. '. u. 1. =c. 1. '. u. 1. =c. 1. ' (u) =. (u).. Hence K L f = f .. Since the operator K is continuous from C01 to C 1 , we may thus consider the dual mapping K 0 : ) for all G 2 C 10 and. C 10 ! C010 given by (K 0 G; ) = (G; K. 2 C01 . Note that C 10 and C010 are. what we have just denoted by E 0 and D0 , respectively. We now embed C01 in E 0 and C 1 in D0 and denote R these embeddings by the same , where the embedding is de…ned by ( (f ) ; ) = f dV for all f 2 C01. and. 2 C 1 . We also denote the image of f by f _ ; i.e., f _ = (f ). As for g 2 C 1 , the de…nition of the. embedding. is similar.. We have for f 2 C01 (, thus f _ 2 E 0 ,) and (K 0 f _ ; ). = = = = = =. Thus the restriction of K 0 to C01 is K on C01 or C 1 is L0. 2 C01 , (f _ ; K ) Z f (K ) dV Z f( ) dV Z (f ) dV Z (K f ) dV (K. _. f) ;. by de…nition, by de…nition,. :. ; we therefore extend K to E 0 by K 0 . Likewise, the dual of L. on D0 or E 0 , respectively. We still denote K 0. Proposition 28 If F 2 E 0 and. and L0. by K and L , respectively.. is admissible, we have L K F = K L F = F .. 30.
(33) Proof. For any. 2 C01 , we have (L K F; ). =. L0 K 0 F;. =. K 0 F; L. =. (F; K. L. =. (F; ) :. ). Hence L K F = F . Similarly, K L F = F . Corollary 29 L is locally solvable: for any g 2 E 0 there is an f 2 D0 such that L f = g. Proof. For g 2 E 0 , we take f = K g; then L f = L K g = g. Corollary 30 The equation L f = 0 has no nontrivial solutions in E 0 . Proof. If there is an f 2 E 0 with f 6= 0 such that L f = 0, then f = K L f = K 0 = 0, a contradiction.. De…nition 31 A di¤ erential operator D is called hypoelliptic if, whenever F 2 D0 and DF is a C 1 function on some open set U , then F is also a C 1 function on U . Proposition 32 L is hypoelliptic if and only if Proof. If. is admissible. 2. is not admissible, i.e., c = 0, then F = ' = jzj. solution of L F = c. it. n 2. 2. 2. jzj + it. n 2+2. is a non-smooth. = 0. That is, we have L F = 0 2 C 1 but F is not C 1 .. Conversely, suppose. is admissible, F 2 D0 ; L F = G is a C 1 function on U , and u 2 U . Choose a. neighborhood V of u such that V. U . We claim that F is a C 1 function on V . Let. 2 C01 (U ) with. = 1 on V , then G 2 C01 (U ) and F 0 = K ( G) 2 C 1 (U ). It follows that L F 0 = L K ( G) = G on V . Now it su¢ ces to show that H = and L H = L H=K 0=0. (F. F 0) = L F. (F. F 0 ) 2 E 0 is a C 1 function on V . Note that H = K L H. L F0 = G. G = 0 on V . Since. is C 1 on V .. 31. is C 1 away from the origin,.
(34) 6. References. References [1] O. Calin, D.-C. Chang, P. Greiner, Geometric analysis on the Heisenberg group and its generalizations. AMS/IP Studies in Advanced Mathematics, 40. American Mathematical Society, Providence, RI; International Press, Somerville, MA, 2007. x+244 pp. ISBN: 978-0-8218-4319-2; 0-8218-4319-2 . [2] S.-C. Chen, M.-C. Shaw, Partial Di¤erential Equations in Several Complex Variables. AMS/IP Studies in Advanced Mathematics, 19. American Mathematical Society, Providence, RI; International Press, Boston, MA, 2001. xii+380 pp. ISBN: 0-8218-1062-6 (Reviewer: Harold P. Boas) 32Wxx (32-02 35A20 35N15 35S05 47N20) [3] G. B. Folland, A fundamental solution for a subelliptic operator. Bull. Amer. Math. Soc. 79 (1973), 373–376. [4] G. B. Folland, E. M. Stein, Estimates for the @ b Complex and Analysis on the Heisenberg Group. Comm. Pure Appl. Math. 27 (1974), 429–522. (Reviewer: S. G. Gindikin) 35N15 (22E30 32K15 47G05) [5] Sorin Dragomir, Giuseppe Tomassini, Di¤erential geometry and analysis on CR manifolds. Progress in Mathematics, 246. Birkhäuser Boston, Inc., Boston, MA, 2006. xvi+487 pp. ISBN: 978-0-8176-4388-1; 0-8176-4388-5 [6] Loring W. Tu, Di¤erential geometry. Connections, curvature, and characteristic classes. Graduate Texts in Mathematics, 275. Springer, Cham, 2017. xvi+346 pp. ISBN: 978-3-319-55082-4; 978-3-319-55084-8 [7] E. M. Stein, R. Shakarchi, Complex analysis. Princeton Lectures in Analysis, 2. Princeton University Press, Princeton, NJ, 2003. xviii+379 pp. ISBN: 0-691-11385-8 [8] E. M. Stein, R. Shakarchi, Functional analysis. Introduction to further topics in analysis. Princeton Lectures in Analysis, 4. Princeton University Press, Princeton, NJ, 2011. xviii+423 pp. ISBN: 978-0-69111387-6. 32.
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