MATH 668M NOTES. References

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1 MATH 668M NOTES This is a quick introduction to some aspects of several complex variables emphasizing the role of pseudocovexity and the model strongly pseudoconvex domain (Siegel domain) and its boundary, the Heisenberg group. Standard references (from which I will borrow freely) are the books by Hörmander, [3], Folland and Kohn [2], and the expository paper by Beals, Fefferman and Grossman[1].Sections 1-5 follow Hörmander s book. References [1] M. Beals, C. Fefferman, R Grossman, Strictly pseudoconvex domains in C n, Bulletin of the AMS, Vol 8, No 2, March [2] G. Folland, J. J. Kohn, The Neumann problem for the Cauchy-Riemann complex, Princeton University Press, 1972 [3] L. Hörmander, An introduction to complex analysis in several variables, North Holland [4] Kranz Function theory of several complex variables, Wiley [5] E. M. Stein Harmonic analysis, Princeton University Press 1. General properties of holomorphic functions in C n Definition 1.1. Let Ω be an open set in C n. A C 1 function f : Ω C is holomorphic (or analytic) in Ω if it satisfies the Cauchy- Riemann equations in each variable separately, or f := f dz k = 0 An expression of the form ω = ω k dz k is called a (0, 1) form. Let D be a Cartesian product of discs (called a polydisc) D = D 1 D n, and assume f is holomorphic in a neighborhood of the closure of D. Applying the one dimensional Cauchy formula n times, followed by Fubini s theorem, we have f(z) = 1 (2πi) n D 1 D n f(ζ) (ζ 1 z 1 ) (ζ n z n ) dζ 1 dζ n for all z D. Notice D 1 D n is not the boundary of D. From this formula we conclude that holomorphic functions are in fact C and we can get Taylor s formula: 1

2 2 MATH 668M NOTES Theorem 1.2. Let f be holomorphic in a neighborhood of the closure of a polydisc D centered at z. Then f(z) = 1 α f ( z)(z z)α α! zα α The convergence is uniform in D. Here α = (α 1, α n ) Z n, α! = α 1! α n!, α = α 1 αn z α z α 1 z αn As a consequence of Taylor s formula we have the principle of analytic continuation: If f and g are analytic in a connected open set Ω and α f(z) = α g(z) for all α and one fixed z Ω, then f = g in Ω. z α z α 2. Practical guide to (0, 1) and (0, 2) forms Let ω = ω k dz k be a (0, 1) form. Define ω = k<l ( ωk z l ω ) l dz l dz k If you are not familiar with forms, you can just treat dz k and dz l dz k (k < l) as abstract bases of the spaces of (0, 1) forms and (0, 2) forms. However, our calculations will be easier if we recall their conceptual meaning. Facts: ω = ω k dz k is as a linear function on the complex vector space of (0, 1) vectors (spanned by the basis z l ) defined according to dz k ( z l ) = δ kl.. It can also be regarded as a linear function on all vectors, setting dz k ( z l ) = 0. If f is C 2, (f)) = 0 Stokes theorem: If ω 1 is C 1 in a bounded domainω C with smooth boundary (Ω = {r < 0}, r smooth, r(z) 0 if r(z) = 0), then ω 1 ω 1 dz = dz dz (1) Ω Ω z where, for the purpose of integrating, dz dz = 2idxdy. We need this when Ω is an annulus. This really is just the Math 241 Green s theorem in fancy, but natural notation. Wedge products: If ω = ω k dz k and ω = ω k dz k are (0, 1) forms at z fixed, ω ω is a (0, 2) form computed according to dz k dz l = dz l dz k. It will help with calculations to assign meaning to this: For V, V (0, 1) vectors define (dz k dz l )(V, V ) = dz k (V )dz l (V ) dz l (V )dz k (V )

3 MATH 668M NOTES 3 and extend the definition to ω ω by linearity. This makes ω ω a complex-bilinear alternating function on the Cartesian product of two copies of the vector space of (0, 1) vectors satisfying (ω ω )(V, V ) = ω(v )ω (V ) ω(v )ω (V ) 3. Solving f = ω for compactly supported smooth ω Theorem 3.1. Let ω = ω 1 dz ω n dz n be a C 1, compactly supported (0, 1) form in C n satisfying the necessary condition ω = 0. Define the function ω 1 (z 1 ζ 1, z 2,, z n ) f(z 1,, z n ) = c dm(ζ 1 ) (2) ζ 1 C Then f is C 1, f = ω and, if n > 1, then f is compactly supported. In particular, f is 0 in the unbounded component of the complement of the support of ω. Also, c = 1 π. Proof. That z 1 f = ω 1 follows from the fundamental solution of the Laplace equation in two dimensions (so you already know this), or by using (1). We will do both in class. That f = ω k for k 2 follows from ω = 0. Now we know f = ω, so f is analytic outside the support of ω. If z 2 is sufficiently large, f = 0 by looking at the formula. So, by analytic continuation, f is 0 in the unbounded component of the support of ω. Remark 3.2. If ω is compactly supported in z 1 for each fixed (z 2,, z n ), formula (2) still solves f = ω. If f is known to be 0 in an open subset of a component of the complement of the support of ω, then it is 0 in that whole component. 4. Extending holomorphic functions We can already prove the following famous result. Theorem 4.1. Let Ω be open, K Ω compact such that Ω K is connected. in C n, n > 1. Then every holomorphic function in Ω K extends to a holomorphic function in Ω. Proof. Let φ C 0 (Ω), φ = 1 in K. Solve g = (1 φ)f. Then (1 φ)f g is the desired extension. Remark 4.2. If n = 1, for every open set U in C there exists a holomorphic function in U which does not extend past any boundary point of U.

4 4 MATH 668M NOTES Exercise 1 Let Ω be a smoothly bounded domain in C, and ω a (0, 1) form smooth up to the boundary, such that ω = 0 in Ω. Prove there exists a function f, smooth up to the boundary, such that f = ω in Ω. Does your proof work in C n, n 2? Let Ω = {r < 0} be a smooth domain ( r(z) 0 if r(z) = 0). We ask the question: what (smooth) functions on Ω extend as analytic functions in Ω. In one dimension, in the disc, the answer is in terms of Fourier coefficients. In higher dimensions, the answer is given by a PDE. Let V = a k (z) be a smooth complexified vector field which is tangent to the boundary, meaning ak (z) r = 0 for all z U. These are called tangent vectors fields of type (0, 1). Let L be the collection of all such vector fields. No such (non-zero) vector fields exist in one dimension, but they do exist in two and higher dimensions. If r z 1 (z) 0, let V 1 = r z 1 V n 1 = r z 1 N = r z 2 z 2 r z n 1 z 1 z n z 1 r r z r z n 2 Then V 1, V n 1 form a basis of L at z, and, together with N, form a basis for the space of all vectors of type (0, 1) at z. If f is analytic in Ω and smooth up to the boundary, then the boundary values of f satisfy V f = 0 for every V L. In other words, boundary values of analytic functions satisfy a PDE in several complex variables. This is also a sufficient condition, at least globally, and for domains without holes. Theorem 4.3. Let Ω = {r < 0} be a smoothly bounded domain. Assume the complement of Ω is connected. Let f be a smooth function in C n such that V f = 0 for everyv L. (3)

5 MATH 668M NOTES 5 Then there exists u : Ω C analytic in Ω, such that u = f on Ω. Proof. The main step is to to construct f 1 which agrees with f on Ω and f 1 vanishes to second order on Ω. Define ω to be f 1 in Ω and 0 outside. This form is C 1 and ω = 0. Then use theorem (3.1) to solve u 1 = ω, u 1 = 0 outside Ω, thus u 1 = 0 on Ω. The u in the conclusion of the theorem is u = f 1 u 1. We proceed with the details of constructing f 1. Lemma 4.4. Condition (3) means that f = c(z)r + rω in Ω, c smooth, ω a smooth (0, 1) form. Proof. Proof of the lemma: We should introduce the conceptual meaning of (0, 1) forms. For each fixed z, they act as linear functions on the complex vector space of vectors of type (0, 1) according to dz k ( z l ) = δ k l. Condition (3) means that f(v k ) = 0 at z Ω, k = 1,, n 1. r(v k ) = 0 at z Ω is automatically true. This shows that the linear functions f and r are scalar multiples of each other at each z Ω. To show that the multiple varies smoothly, also look at how they act on N. Let c(z) = f(n). Notice r(n) = 1, so f c(z)r vanishes on all V and N, thus is 0 on Ω and f c(z)r = r(z)ω in general, for the following reason: by the inverse function theorem, r can be regarded (locally) as a coordinate. We have the following elementary general result: if f(r, x 2, x n ) is smooth in an open ball and vanishes when r = 0, then there exists g smooth, f = rg. Back to the main proof: f = c(z)r + rω. Then f 0 = f cr agrees with f on the boundary, and f 0 = (f cr) = r(ω c) := rω 1 vanishes to first order on Ω. If we define this to be 0 outside Ω, this extension has bounded, but not continuous derivatives across Ω, so it needs to be refined. From 0 = (rω 1 ) = r ω 1 + rω 1 (product rule for function times form, but be aware that (ω ω = (ω) ω + ( 1) p ω (ω ) is ω is a (0, p) form ) we conclude r ω 1 = 0 on Ω. General fact: If ω, ω are a (0, 1) forms, then ω ω = 0, and if ω ω = 0 and ω 0, then ω = c(z)ω. See the lemma below for a proof. Thus ω 1 is a multiple of r for z Ω 1. c(z) is smooth on Ω (it equals ω 1 (N)), and therefore for a smooth (0, 1) form ω 2. ω 1 = c 1 (z)r + r(z)ω 2

6 6 MATH 668M NOTES Thus f 0 = r(c 1 r + rω 2 ) Define f 1 = f 0 c 1 r 2 2 Then f 1 = r 2 (ω c 1). Modify this to be 0 outside Ω, and call the resulting form ω. Lemma 4.5. If ω and ω are non-zero (0, 1) forms at a fixed point z, and ω ω = 0, then there exists λ such that ω = λω Proof. This follows from (ω ω )(V, V ) = ω(v )ω (V ) ω(v )ω (V ) If the linear functions ω, ω have the same null space, they are multiples of each other. If not, pick V 1 in the null space of ω but not the null space of ω, and V 2 not in the null space of either. Exercise 2 The local version of this result is not true. If V f = 0 on Ω for all V L and all z Ω intersected with a neighborhood of a point z 0 Ω, it does not follow there exists an analytic function in a neighborhood of z 0 intersected with Ω which agrees with f on Ω. Show this in the case Ω = {Iz n > 0}. Show there exists a smooth real function f(x n ) (depending only on x n ) which not the boundary value of an analytic function in Ω { z < 1} which is continuous up to the boundary. Show that V f = 0 for all V L if f = f(x n ). What is L in this case? Answering this question requires a Math 660 theorem. 5. The Levi form Warning: In this section V are (1, 0) vector fields, so, to be consistent, the V of the previous section should have been called V. We investigate the local version of the preceding theorem. This is our introduction to the Levi form. Definition 5.1. Let Σ = {r = 0}, r C, r 0 on Σ. The Levi form is the Hermitian form L(z)(V, V ) = 2 r(z) z i z j v i v j (4) restricted to (1, 0) vectors (of the form V = v k and V = v k tangent to the Σ: r(z) v k = 0, v k r(z) = 0. Remark 5.2. If you are familiar with exterior derivatives, L(V, V ) = r(v, V ) )

7 where MATH 668M NOTES 7 r = 2 r z j dz j dz k is a (1, 1) form. This shows that, for smooth vector fields of the type considered, r(v, V ) = dr(v, V ) = V (r(v )) V (r(v )) r([v, V ]) = ([V, V ]) = the coefficient of N N in [V, V ] Theorem 5.3. Let Σ = {r = 0}, r C, r 0 on Σ. Assume the Levi form has at lest one negative eigenvalue at a point z 0. In other words, there is a (1, 0) tangent vector V such that 2 r(z 0 ) z i z j v i v j < 0 Then there exists U a neighborhood of z 0 such that every smooth function f defined on Σ which satisfies the compatibility condition V f = 0 for all V tangent (0, 1) vector fields extends as an analytic function to {r > 0} U. Proof. After an analytic change of variables, we can assume z 0 = 0 and, r(z) = Iz n z Λ 22 z Λ n 1,n 1 z n O( z n z + z 3 ) (5) r(z) = Iz n + a ij z j z k + O( z 3 ) with Λ kk = ±1 or 0. (DETAILS: The Taylor expansion of the real C function r is r(z) = r(0) z k + r(0) z k z ( k r(0) z j z k + 2 r(0) z j z k + 2 ) 2 r(0) z j z k + O( z 3 ) 2 z j z j z j Assume r(0) z n 0. The analytic change of variables is w 1 = z 1 w n = r(0) z k r(0) z j z j z k The Jacobian matrix ( w i ) has determinant r(0) z n. By the analytic inverse function theorem (which is an immediate consequence of the real inverse function theorem, as we will prove in class), the map z w is

8 8 MATH 668M NOTES locally one-to-one, onto, with an analytic inverse. In these coordinates r(w) = r(z(w)) has Taylor expansion r(w) = 2Rw n + 2 r(0) w j w k + O( w 3 ) w j w k ( ) Computing the matrix 2 r(0) w j w k by the chain rule is easy, but not really necessary if you know/believe that the statement r( v k, v k ) < 0 for some vector v k tangent to Σ at 0 is independent of coordinates. The (1, 0) vectors tangent at 0 are spanned by w 1, w n 1, so we know that the Hermitian matrix ( ) 2 r(0) M = w j w k 1 j,k n 1 has a negative eigenvalue, and let (v 1, v n 1 ) such that 2 r(0) v j v k = 1 w j w k 1 j,k n 1 Diagonalize this quadratic form: let U non-singular, (n 1) by (n 1) be such that U MU = Λ with Λ 11 = 1 and all other Λ jj = 1, 0 or 1. and the first column of U is (v 1, v n 1 ). Now make one last linear change of variables w z such that Iz n = 2Rw n and Uz = w, so that z Λz = w Mw. CONTINUING THE MAIN PROOF: Now we have r(z) = Iz n z Λ 22 z Λ n 1,n 1 z n O( z n z + z 3 ) Let δ > 0 be small. Pick 0 < δ 1 < δ so that r(z 1, 0) < 0 if z 1 = δ 1 and 2 r z 1 z 1 (z 1, 0) < 0 if z 1 δ 1 Next, pick 0 < δ 2 < δ 1 so that r(z 1, z ) < 0 if z 1 = δ 1 and z 2 δ 2,, z n δ 2, and 2 r z 1 z 1 (z 1, z) < 0 if z 1 δ 1 and z 2 δ 2, z n δ 2. Then r(z 1, z ) = Iz n + (negative) + O( δ δ 1 δ 2 ) So r(z 1, z ) < 0 if δ 2 /2 Iz n δ 2 /3 and z 1 δ 1, z 2 δ 2, z n δ 2.

9 MATH 668M NOTES 9 As before, we can find a C extension of f such that f = 0 to second order on Σ. Define ω to be f in {r > 0} { z 1 δ 1 } { z 2 δ 2, z n δ 2 } and 0 otherwise. Thus ω is C 1 in { z 2 < δ 2, z n < δ 2 } with no restrictions on z 1, and ω = 0. For each fixed z, it is compactly supported in z 1. Let s solve f 1 = ω in { z 2 < δ 2, z n < δ 2 } using formula (2). Notice f 1 = 0 if δ 1 /2 Iz n δ 1 /3 and f 1 is analytic if { z 2 < δ 2,, z n < δ 2 } and r < 0. For each fixed (z 2, z n ), z 2 < δ 2, z n < δ 2, the set {z 1 r(z 1, z ) < 0, z 1 δ 1 } is connected (if you took Math 673, you know why) and contains a neighborhood of { z 1 = δ 1 }. Thus f 1 = 0 in this set. The desired extension is f f 1. Definition 5.4. Ω = {r < 0} is pseudoconvex at z U if 2 r z j (z)v j v k 0 (6) for all (v 1,, v n ) such that z j r(z)v j = 0 (7) (that is, the vector v j z j is tangent to the boundary). Ω is strongly (or strictly) pseudoconvex at z U if 2 r z j (z)v j v k > 0 (8) for all (v 1,, v n ) 0 which satisfy (7). Ω is a domain of holomorphy if for every point in z Ω and for every neighborhood U of z there exists a holomorphic function in Ω which does not extend to U. A slight modification of the above proof shows Ω not pseudoconvex => Ω not a domain of holomorphy. The converse is also true but much harder, so Ω pseudoconvex <=> Ω is a domain of holomorphy. Also, the proof of (5) shows that if Ω is strictly pseudoconvex at z 0 = 0, then, after an analytic change of variables Ω is given by Iz n > z z z n O( z n z + z 3 ) Exercise 3. Let Ω = {r < 0} be a smoothly bounded strongly pseudoconvex domain, and assume, for convenience, 0 Ω. Assume the following two properties:

10 10 MATH 668M NOTES 1) One can make an analytic change of variables such that, in a neighborhood of 0, Ω agrees with n {Iz n > z k 2 + O( z 3 )} k=1 This is stronger than what we proved in class, but still elementary (see page 112 of Kranz s book). In particular, Ω is locally convex. 2) If ω is a (0, 1) form which satisfies ω = 0 in the open set Ω and ω is smooth in Ω, then there exists f smooth in Ω such that f = ω in Ω. This type of result is due to of J.J. Kohn (see the book by Folland and Kohn.) Implicit in this is an assumption on the cohomology of Ω. Prove there exists an analytic function in Ω which does not extend to any neighborhood of The Siegel domain and the Heisenberg group The Siegel domain S is the model strongly pseudoconvex domain in C n+1 S = {Iz n+1 > z z n 2 } The boundary of the above domain is the Heisenberg group H = {(z, t) C n R} with group multiplication (z, t)(w, s) = (z + w, t + s + 2Iz w). The meaning of is is as follows: to each (w, s) we associate h w,s a biholomorphic map of S. First, define and then h w,s (0, 0) = (w, s + i w 2 ) H h w,s (z, z n+1 ) = (z + w, z n+1 + s + i w 2 + 2iz w) (9) Check that H preserves the defining function of S and is a group homomorphism: H w1,s 1 (H w2,s 2 (z, z n+1 )) = H (w1,s 1 )(w 2,s 2 )(z, z n+1 ). 7. The Heisenberg group as a subgroup of SU(n + 1, 1) We will show that fractional linear transformations with coefficients in SU(n + 1, 1) are biholomorphic maps of S. The right was to think of S is in homogeneous coordinates (z 0, z 1,, z n+1 ) with z 0 = 1. The relevant quadratic form in C n+2 is Q(z 0, z, z n+1 ) = z n+1z 0 z 0 z n+1 2i z 2 (10)

11 MATH 668M NOTES 11 Notice Q(α(z 0, z, z n+1 )) = α 2 Q(z 0, z, z n+1 ). The Siegel domain is obtained by intersecting the interior light cone Q > 0 with the hyperplane z 0 = 1. The quadratic form (10) has one positive and n + 1 negative eigenvalues, and the linear transformations (of determinant 1) which preserve it form the group SU(n + 1, 1). If L : C n+2 C n+2 is linear and satisfies Q(L(z 0, z, z n+1 ) = Q(z 0, z, z n+1 ), in other words L U(n + 1, 1), then ( L1 (1, z, z n+1 ) (z, z n+1 ) L 0 (1, z, z n+1 ),, L ) n+1(1, z, z n+1 ) L 0 (1, z, z n+1 ) is a biholomorphic fractional linear transformation of S. Indeed, notice L 0 (1, z, z n+1 ) 0 if (z, z n+1 ) S and Iz n+1 ( z z n 2) = Q(1, z, z n+1 ) = Q(L 0, L, L n+1 ) = L 0 2 Q(1, L L 0, L n+1 L 0 ) = L 0 2 ( IL n+1 ( L L n 2)) The Heisenberg translations (9) (written for typing ease in the case n = 1) correspond to the matrix h w,s = w 1 0 s + i w 2 2iw 1 while the quadratic form Q, in matrix notation, is (z 0, z, z n+1 ) Q(z 0, z, z n+1 ) where Q = i (11) i and h Qh = Q, thus h SU(2, 1).

12 12 MATH 668M NOTES 8. The Siegel domain is biholomorphic to the unit ball This is also seen best in homogeneous coordinates. Formula (10) can be re-written as Set Q(z 0, z, z n+1 ) = z n+1 + iz 0 2 z n+1 iz w 0 = z n+1 + iz 0 w = 2z w n+1 = z n+1 iz 0 z 2 so that the forward light cone Q > 0 intersected with w 0 = 1 is the unit ball w 2 + w 2 n+1 < 1 Now we have to work out the algebra for the right scaling factor which maps z 0 = 1 to w 0 = 1. The resulting FLT will map the Siegel domain to the unit ball. Start with (1, z, z n+1 ). In the new coordinates this is (w 0, z, w n+1 ) with has w 0 = z n+1 + i, w n+1 = z n+1 i. The map is (1, z, z n+1 ) (w 0, w, w n+1 ) = (z n+1 + i, 2z, z n+1 i) If we drop the first coordinate, ( 2z (z, z n+1 ) z n+1 + i, z ) n+1 i z n+1 + i ( 1, 2z z n+1 + i, z n+1 i z n+1 + i is a biholomorphic map of the Siegel domain to the unit ball. In particular, if n = 0 we recovered z z i which maps the upper half plane z+i to the unit ball in C. 9. The Szegö kernel This is a formula for recovering the values of an analytic function from its boundary values, somewhat similar to Cauchy s integral formula. Unlike Cauchy s formula in C, there isn t a simple formula which works for all domains. We will write down (but not prove) the formula for the Siegel domain. We will meet this formula later in the seemingly unrelated context of wave packets. Let U = {r < 0} be a smooth domain. Look at the boundary values of functions continuous on U, holomorphic in U, whose boundary values are in L 2 (U). The closure of this space in L 2 (U) is, by definition, H 2 (U). (This is the Hardy space, not Sobolev space). )

13 MATH 668M NOTES 13 Definition 9.1. The Szegö kernel S(z, w) is defined by the following properties: S(z, w) is originally defined for z U, w U, and, for fixed w, is holomorphic in z U. For fixed z, it is in L 2 (U) in w. for fixed z, S(z, w) extends as an antiholomorphic function in w U, and S(z, w) = S(w, z) If f is holomorphic in U and continuous up to the boundary, (and, in the unbounded case, the boundary values of f are in L 2 (U), then f(z) = S(z, w)f(w) d surface measure(w) U If the surface measure is replaced with another smooth measure on U, the Szegö kernel changes, but still satisfies the above properties. In strongly pseudoconves domains, S(z, w) is singular when z = w U, but otherwise smooth. There is a famous formula, originally due to Fefferman, describing the singularities of S(z, w) in a strongly pseudoconvex domain U in C n given by U = {r < 0} S(z, w) = φ 1(z, w) r n (z, w) + φ 2(z, w) log r(z, w) (12) where φ 1, φ 2 are smooth up to the boundary and r(z, w) is made up out of r(z) by the requirement r(z, z) = r(z), z r(z, w) vanishes to infinite order if z = w and w r(z, w) vanishes to infinite order if z = w. In the case of the Siegel domain, φ 1 will turn out to be a constant and φ 2 will be 0. The following theorem is taken from Stein s book. (I think it is due to Folland and Stein). Theorem 9.2. Let r(z, z n+1 ; w, w n+1 ) = z n+1 w n+1 z w. Then the 2i Szegö kernel for the Siegel domain is given by S(z, z n+1 ; w, w n+1 ) = c r(z, z n+1 ; w, w n+1 ) n+1 Remark 9.3. Parametrize z n+1 = t + i z 2 + iɛ, w n+1 = s + i w 2, and notice r(z, z n+1 ; w, w n+1 ) = 1 ( ( t s 2Iwz + i z w 2 + ɛ )) 2i thus, if we define S ɛ (z, t) = c (t + i ( z 2 + ɛ)) n+1

14 14 MATH 668M NOTES then S(z, z n+1 ; w, w n+1 )f(w, w n+1 )dm(w)ds = S H F (w, s)s ɛ ( (w, s) 1 (z, t) ) dm(w)dt where F (w, s) = f(w, s+i w 2 ). This is a convolution on the Heisenberg group. In general, the convolution of two functions on a Lie group G is defined by f g(x) = f(y)g(y 1 x)dm(x) where m is the left-invariant Haar measure. Also, G lim S ɛ(z, t) = S(z, t) ɛ 0+ exists in the sense of distribution theory. 10. Left invariant vector fields on the Heisenberg group A vector V based at x 0 in R n (or in the tangent space at x 0 on a manifold parametrized by x) can be regarded as the tangent vector to a curve x(t) with x(0) = x 0. V = x (0) = dx k dt (0) x k If φ : R n R m is smooth and φ(x 0 ) = y 0, (Dφ)(x 0 )(V ) = d dt t=0φ(x(t)) = d dt (φ k x)(0) y k defines a vector at y 0, which is independent on the choice of x(t), as long as x(0) = x 0 and x (0) = V. Dφ agrees with the usual matrix from vector calculus. If V is a vector field and φ : R n R n is one-to-one and onto, W (y 0 ) = Dφ(x 0 )(V (x 0 )) defines a new vector field. If R n has a group structure (or on a Lie group in general) a vector field V is called left invariant if for each g and φ g (x) = gx (group multiplication) it is true that Dφ g (x)(v (x)) = V (gx) for all x. There is an obvious recipe to construct left invariant vector fields: start with vectors based at the group identity (0 in the case of the Heisenberg group) and push them to g using φ g. The resulting vector field is easily shown to be left invariant. We do this, for simplicity, for the 3 dimensional Heisenberg group.

15 MATH 668M NOTES 15 Let g = (x, y, t) (z = x + iy). The curve (u, 0, 0) gets pushed to (x, y, t)(u, 0, 0) = (x + u, y, t + 2uy) and the tangent vector at the origin gets pushed to + 2y. x x t By the same procedure, In general, we obtain X k = + 2y k x k t Y k = 2x k y k t T = t which satisfy the commutation relations [X k, Y j ] = 4δ jk t These vector fields lead to two famous results in linear PDEs. The linear operator n ( ) j=1 X 2 j + Yj 2 is not elliptic because, at each point, the vectors only span 2n directions in R 2n+1. However, [X j, Y j ] recover the missing direction and we have the subelliptic estimate n C φ 2 L + ( ) 2 Xj φ 2 + Y j φ 2 L c φ 2 2 H 1/2 j=1 and all the existence and regularity theorems go through. This is an example of Hörmander s sums of squares of vector fields theorem. It would be easy to prove if we had access to pseudodifferential operators (but we don t). The second result is Lewy s example of a linear PDE with no solutions. In R 3 consider Z = 1 2 X iy = z + iz t Theorem There exists a continuous, compactly supported function f such that the equation Zu = f has no solutions (in the sense of distribution theory) in any neighborhood of 0. This is proved in Stein s book.

16 16 MATH 668M NOTES Notes on Folland s book 11. Operators on L 2 (R n ) This is background for Theorem Let H be a (separable) Hilbert space, T : H H linear. T : H H is bounded if M such that T f M f for all f H. The infimum of such M is the operator norm T op. T is compact if f i 1 implies T f i has a convergent subsequence. If H = L 2 (R n ) and T f = K(x, y)f(y)dy, T is Hilbert-Schmidt if K L 2 (R 2n ), and T HS = K L 2 (R 2n ). Proposition Compact operators form a closed subspace of bounded operators (with the norm). Proof. Let T k T in the operator norm, T k compact. Letf i be a bounded sequence. Each T k (f i ) (k fixed) has a convergent subsequence. By the diagonalization process find a subsequence (also called f i ) which works for all f k. The family T k T is equicontinuous, so the usual proof shows that T f i is Cauchy. Proposition Hilbert-Schmidt operators are bounded, and T op T HS Proof. This is best seen by duality: Look at (T f)g, apply Fubini and Cauchy-Schwarz. Proposition Hilbert-Schmidt operators are compact. Proof. One can find a sequence of simple functions (each of the form finite ci χ A (x)χ B (y) which converge to K in L 2 (R 2n ). The corresponding operators T k converge to T in the Hilbert-Schmidt norm, thus in the operator norm. But all T k are compact (they have finite dimensional ranges). Schwartz functions S are defined by requiring that the semi-norms N (α,β) (φ) := sup x α β φ(x) x are finite for each multiindex α, β. φ i φ in S if N (α,β) (φ i φ) 0 for all (α, β). One can define a metric compatible with this notion of convergence. The space of tempered distributions S is the set of all bounded linear functionals L C.

17 MATH 668M NOTES 17 All L p functions act as tempered distributions by (u, φ) = uφ The Fourier transform maps S to S (it basically interchanges x α and α ). Any tempered distribution u has a well defined Fourier transform which is also a tempered distribution defined by so it agrees with valid for nice functions u and φ. (û, φ) = (u, ˆφ) ûφ = u ˆφ