HOLOMORPHIC FOCK SPACES FOR POSITIVE LINEAR TRANSFORMATIONS

HOLOMORPHIC FOCK SPACES FOR POSITIVE LINEAR TRANSFORMATIONS

R. FABEC, G. ÓLAFSSON, and A. N. SENGUPTA^∗

Abstract

SupposeAis a positive real linear transformation on a finite dimensional complex inner product spaceV. The reproducing kernel for the Fock space of square integrable holomorphic functions onV relative to the Gaussian measuredµA(z) = ^√det_πn^Ae^−ReAz,zdzis described in terms of the linear and antilinear decomposition of the linear operatorA. Moreover, ifAcommutes with a conjugation onV, then a restriction mapping to the real vectors inV is polarized to obtain a Segal-Bargmann transform, which we also study in the Gaussian-measure setting.

Introduction

The classical Segal-Bargmann transform is an integral transform which defines a unitary isomorphism of L²(Rⁿ) onto the space F(Cⁿ) of entire functions on Cⁿ which are square integrable with respect to the Gaussian measure µ=π⁻ⁿe^−z2dxdy, wheredxdystands for the Lebesgue measure onR²ⁿCⁿ, see [1], [3], [4], [5], [11], [12]. There have been several generalizations of this transform, based on the heat equation or the representation theory of Lie groups [6], [10], [13]. In particular, it was shown in [10] that the Segal-Bargmann transform is a special case of the restriction principle, i.e., construction of unitary isomorphisms based on the polarization of a restriction map. This prin- ciple was first introduced in [10], see also [9], where several examples were explained from that point of view. In short the restriction principle can be ex- plained in the following way. LetMCbe a complex manifold and letM ⊂MC

be a totally real submanifold. Let F = F(MC) be a Hilbert space of holo- morphic functions onMCsuch that the evaluation mapsFF →F (z)∈C are continuous for allz ∈ MC, i.e.,Fis areproducing kernel Hilbert space.

There exists a functionK :MC×MC →Cholomorphic in the first variable, anti-holomorphic in the second variable, and such that the following hold:

(a) K(z, w)=K(w, z)for allz, w∈MC;

∗The research of G. Ólafsson was supported by DMS-0070607, DMS-0139473 and DMS- 0402068. The research of A. Sengupta was supported by DMS-0201683. The authors would like to thank the referee for valuable comments and remarks.

Received January 31, 2004; in revised form March 20, 2005

(b) IfKw(z):=K(z, w)thenKw ∈Fand

F (w)= F, KwF, ∀F ∈F, z∈MC.

The functionKis thereproducing kernelfor the Hilbert spaceF. LetD : M →C^∗be measurable. Then the restriction mapR:F →RF :=DF|Mis injective. Assume that there is a measureµonM such thatRF ∈L²(M, µ) for allF in a dense subset ofF. Assuming thatRis closeable, Im(R)is dense inL²(M, µ), and by polarizingR^∗, we can write

R^∗=U|R^∗|

whereU :L²(M, µ)→Fis a unitary isomorphism and|R^∗| =√

RR^∗. Using the fact thatFis a reproducing kernel Hilbert space we get

Uf (z)= Uf, KzF= f, U^∗KzL² =

Mf (m)(U^∗Kz)(m) dµ(m) . ThusU is always an integral operator. We notice also that the formula for U shows that the important object in this analysis is the reproducing kernel K(z, w).

We will use the following notation through this article: Letz, w =z1w1+

· · · +znwnbe the standard inner product onCⁿand let(z, w)=Re(z, w)be the corresponding inner product onCⁿviewed as a 2n-dimensionalrealvector spaces. Notice that(x, y) = x, y = x1y1+ · · · +xnynforx, y ∈ Rⁿ. We writez²=z²1+ · · · +z_n²forz∈Cⁿ.

The reproducing kernel for the classical Fock space is given byK(z, w)= e^z,w, wherez, w∈Cⁿ. By takingD(x):=(2π)^−n/4e^−x2/2, forx ∈Rⁿ, which is closely related to the heat kernel, we arrive at the classical Segal-Bargmann transform, given as the holomorphic continuation of

Ug(x)=(2/π)^n/4e^x2/2

Rⁿg(y)e^−(x−y)2dy .

Notice thatRⁿ x→Ug(x)∈Chas a unique holomorphic extension toCⁿ. The same principle can be used to construct the Hall-transform for compact Lie groups, [6]. In [2], Driver and Hall, motivated by application to quantum Yang-Mills theory, introduced a Fock space and Segal-Bargmann transform depending on two parametersr, s >0, giving different weights to thexandy directions, wherez=x+iy∈Cⁿ(this was also studied in [13]). ThusFis now the space of holomorphic functionsF (z)onCⁿ which are square-integrable with respect to the Gaussian measuredMr,s(z)= _(πr)n/2¹(πs)^n/2 e^−xr2−ys2 dxdy. A Segal-Bargmann transform for this Fock space is given in [13] and in Theorem 3

of [7]. We show this is a very special case of a larger family of Fock spaces and associated Segal-Bargmann tranforms. Indeed, ifAis a real linear positive definite matrixAon a complex inner product spaceV, then

(0.1) dµA(z)=

√det(A)

πⁿ e^−(Az,z)|dz|

gives rise to a Fock spaceF_A. We find an expression for the reproducing kernel KA(z, w). We use the restriction principle to construct a natural generaliza- tion of the Segal-Bargmann transform for this space, with a certain natural restriction onA. We study this also in the Gaussian setting, and indicate a generalization to infinite dimensions.

We will fix the following notation for the types of bilinear pairings that we shall be using in this paper:

(i) z, wdenotes a Hermitian inner product on a complex vector spaceV, i.e., a pairing which is complex-linear inz, complex-conjugate-linear in w, andz, z>0 ifz=0. We denote byz =√

z, zthe correspond- ing norm;

(ii) (x, y)denotes an inner product onV viewed as arealvector space. The standard choice is(x, y)=Rex, y;

(iii) z·wdenotes a complex-bilinear pairing. In the standard situation we havez·w= z, w. We setz²=z·z.

1. The Fock space and the restriction principle

In this section we recall some standard facts about the classical Fock space of holomorphic functions onCⁿ. We refer to [5] for details and further informa- tion. Letµbe the measuredµ(z)=π⁻ⁿe^−z2dxdyand letFbe the classical Fock-space of holomorphic functionsF :Cⁿ →Csuch that

F²F:=

Cⁿ

|F (z)|² dµ(z) <∞.

The spaceFis a reproducing Hilbert space with inner product F, GF=

CⁿF (z)G(z) dµ(z) and reproducing kernelK(z, w)=e^z,w.

Thus

F (w)=

CⁿF (z)K(z, w) dµ(z)= F, KwF

whereKw(z)=K(z, w). The functionK(z, w)is holomorphic in the first vari- able, anti-holomorphic in the second variable, andK(z, w)=K(w, z). Notice thatK(z, z)= Kz, KzF. HenceKzF= e^z2/2. Finally the linear space of finite linear combinations

cjKzj,zj ∈Cⁿ,cj ∈C, is dense inF. An orthonor- mal system inFis given by the monomialseα(z) = z^α1¹· · ·z^α_nⁿ/√

α1!· · ·αn!, α∈Nⁿ₀. If the reference to the Fock space is clear, then we simply write·,· instead of·,·F, and similarly for the corresponding norm.

ViewRⁿ ⊂ Cⁿ as a totally real submanifold ofCⁿ. We will now recall the construction of the classical Segal-Bargmann transform using therestriction principle, see [9], [10]. For constructing a restriction map as explained in the introduction we need to choose the functionD(x). One motivation for the choice ofD is the heat kernel, but another one, more closely related to representation theory, is that the restriction map should commute with the action ofRⁿon the Fock space andL²(Rⁿ). Indeed, take

T (x)F (z)=m(x, z)F (z−x)

forFinFwherem(x, z)has properties sufficient to makex→T (x)a unitary representation ofRⁿonF. Namely, we need a multipliermsatisfying

|m(x, z)| =

dµ(z−x) dµ(z)

₂¹

=e^{(Rez,x−x2/2)}.

We takem(x, z):=e^z,x−x2/2. Set

D(x)=(2π)^−n/4m(0, x)=(2π)^−n/4e^−x2/2 and defineR:F→C^∞(Rⁿ)by

RF (x):=D(x)F (x)=(2π)^−n/4e^−x2/2F (x).

Then RT (y)F (x)=RF (x−y).

SinceFis holomorphic, the mapRis injective. Furthermore, the holomorphic polynomialsp(z)=

aαz^αare dense inFand obviouslyRp∈L²(Rⁿ). Thus, we may and will considerRas a densely defined operator fromFintoL²(Rⁿ). The Hermite functionshα(x)= (−1)^|α|

D^αe^−x2

e^x2/2are images underR of polynomials and thus are in the image of the operatorR. Hence, Im(R)is dense inL²(Rⁿ). Using continuity of the evaluation mapsF → F (z), it can be checked thatRis a closed operator. Hence,Rhas a densely-defined adjoint

R^∗:L²(Rⁿ)→F. Forz, w∈Cⁿ, recall thatz·w=

zjwj. Then, for anyg in the domain ofR^∗, we have:

R^∗g(z)= R^∗g, Kz = g, RKz =(2π)^−n/4

Rⁿg(y)e^−y2/2e^z·ydy

=(2π)^−n/4e^z2/2

Rⁿg(y)e^{−(z−y)2/2}dy

=(2π)^n/4e^z2/2g∗p(z)

wherep(z) = (2π)^−n/2e^−z2/2 is holomorphic. Applying the map R : F → C^∞(Rⁿ), we have

(1.1) RR^∗g(x)=g∗p(x) .

Sincep∈L¹(Rⁿ)andg∈L²(Rⁿ), it follows thatg∗p∈L²(Rⁿ), and so the preceding equation shows thatR^∗gis in the domain of the operatorR, and sog is in the domain of the operator compositeRR^∗. This argument also shows that RR^∗is a bounded operator, on its domain, with operator normRR^∗p1. Moreover, for everygin the domain ofR^∗, we have

R^∗g, R^∗g = RR^∗g, g ≤ RR^∗ g2.

ThusR^∗is a bounded operator. Being an adjoint, it is also closed. Therefore, the domain ofR^∗ is in fact the full spaceL²(Rⁿ). So for anyf ∈D(R), we have Rf, Rf = R^∗(Rf ), f ≤ R^∗ Rf2f2,

which implies that the operatorRis bounded. Again, being a closed, densely- defined bounded operator,Ris, in fact, defined on all ofF. In summary,

Lemma1.1. The linear operatorsR and R^∗ are everywhere defined and continuous.

Letpt(x)= (2πt)^−n/2e^−x2/2t be the heat kernel onRⁿ. Then(pt)_t>0 is a convolution semigroup andp=p1. Hence√

RR^∗=p1/2∗or RUg(x)=R^∗g(x)=p1/2∗g(x)=π^−n/2

Rⁿg(y)e^−(x−y)2dy . It follows that

Ug(x)=(2/π)^n/4e^x2/2

Rⁿg(y)e^−(x−y)2dy

forx∈Rⁿ. But the function on the right hand side is holomorphic inx. Analytic continuation gives the following classical Segal-Bargmann tranform.

Theorem1.2.The mapU :L²(Rⁿ)→Fgiven by Ug(z)=(2/π)^n/4

Rⁿg(y)exp(−y²+2z, y −z²/2) dy is a unitary isomorphism.

2. Twisted Fock spaces

LetV Cⁿbe a finite dimensional complex vector space of complex dimen- sionnand let·,·be a complex Hermitian inner-product.

We will also considerV as a real vector space with real inner product defined by(z, w)= Rez, w. Notice that(z, z)= z, zfor allz ∈Cⁿ. LetJ be the real linear transformation ofV given byJ z=iz. Note thatJ^∗ = −J =J⁻¹ and thusJ is a skew symmetric real linear transformation. Fix a real linear transformationA. ThenA=H +Kwhere

H := A+J⁻¹AJ

2 and K:= A−J⁻¹AJ

2 .

We have H J = ¹₂(AJ −J⁻¹A) = ¹₂J (J⁻¹AJ + A) = J H and KJ =

1

2(AJ +J⁻¹A)= ¹₂J (J⁻¹AJ −A)= −J K. ThusH is complex linear and Kis conjugate linear.

We now assume thatAis symmetric and positive definite relative to the real inner product(·,·).

Lemma2.1.The complex linear transformationH is self adjoint, positive with respect to the inner product·,·, and invertible.

Proof. SinceAis positive and invertible as a real linear transformation, we have(Az, z) >0 for allz=0. ButJ is real linear and skew symmetric.

Hence(J AJ⁻¹z, z) > 0 for allz =0. In particularH = ¹₂(A+J AJ⁻¹)is complex linear, symmetric with respect to the real inner product(·,·), and pos- itive. Consequently ReHiv, w = Reiv, H w. This implies ImH v, w = Imv, Hwand henceHv, w = v, H w. ThusH is complex self adjoint andH z, z>0 forz=0.

Lemma2.2.Let w ∈ V. ThenAw, w = (Aw, w)+iImKw, wand (Aw, w)=(H w, w)+(Kw, w).

Proof. The first statement follows from

Aw, w = Hw, w + Kw, w =(Hw, w)+(Kw, w)+iImKw, w

=(Aw, w)+iImKw, w.

Taking the real part in the second line gives the second claim, which also follows directly from bilinearity of(·,·).

Denote by detR(·)the determinant of a R-linear map on V Cⁿ R²ⁿ and by det(·)the determinant of a complex linear map ofV. LetµA be the measure defined bydµA(z) = π⁻ⁿ√

detRA e^−(Az,z)dxdy and letF_A be the space of holomorphic functionsF :Cⁿ →Csuch that

(2.1) F²_A:=

|F (z)|²dµA(z) <∞.

Our normalization ofµis chosen so that1A = 1. Just as in the classical case one can show thatF_Ais a reproducing kernel Hilbert space, but this will also follow from the following Lemma. We notice that all the holomorphic polynomialsp(z)are inF. To simplify the notation, we letT1=H^−1/2. ThenT1

is symmetric, positive definite and complex linear. LetcA=

detR(A^1/2T1)= (det_R(A)/det_R(H ))^1/4.

Lemma2.3. LetF :V →Cbe holomorphic. ThenF ∈F_A if and only if F ◦T¹∈F. Moreover, the map- :F_A→Fgiven by

-(F )(w):=cAexp

−KT¹w, T¹w/2

F (T¹w) is a unitary isomorphism. In particular

-^∗F (w)=-⁻¹F (w)=c⁻_A¹exp

Kw, w/2 F (√

H w) .

Proof. NoteF is holomorphic if and only ifF ◦T1is holomorphic asT1

is complex linear and invertible. Moreover, unitarity follows from -F²=π⁻ⁿ

V|-F (w)|²e^−w,wdw

=π⁻ⁿ

det_RA

V|F (w)|²e^−(Kw,w)e^{−√Hw,√H w}dw

=π⁻ⁿ detRA

V|F (w)|²e^−(Kw,w)e^−Hw,wdw

=π⁻ⁿ

det_RA

V|F (w)|²e−((H+K)w,w)dw

=π⁻ⁿ detRA

V|F (w)|²e^−(Aw,w)dw

= F²_A.

Theorem 2.4.The space F_A is a reproducing kernel Hilbert space with reproducing kernel

KA(z, w)=c_A⁻²e^12Kz,ze^Hz,we^12Kw,w. Proof. By Lemma 2.3 we get

cAexp(−KT1w, T1w/2)F (T1w)=-(F )(w)

=(-(F ), Kw)F

=(F, -^∗(Kw))FA. Hence

KA(z, w)=c⁻_A¹exp

Kw, w/2

-^∗(K^√_Hw)=c⁻_A²e^12Kz,ze^Hz,we^12Kw,w.

3. The Restriction Map

We continue to assume A > 0. We notice that Lemma 2.3 gives a unitary isomorphism-^∗U :L²(Rⁿ)→F_A, whereU is the classical Segal-Bargmann transform. But this is not the natural transform that we are looking for. AsH is positive definite there is an orthonormal basise1, . . . , enofV and positive numbersλj > 0 such thatHej = λjej. LetVR :=

Rek. Setσ(

aiei) = a¯iei. Thenσ is a conjugation withVR = {z: σz=z}. Forz ∈V we will when convenient writez¯ forσ (z). We say that a vector isrealif it belongs to VR. AsH ej = λjej withλj ∈Rit follows thatH VR ⊆VR. We note that for the complex linear mappingH that detRH =(detH )²and that detHis equal to the determinant of the real linear transformationH|VR.

Lemma3.1. Kz, w = Kw, z.

Proof. Note thatσ K is complex linear. Since J^∗ = −J, K = ¹₂(A− JAJ⁻¹)is real symmetric. Thus(Kw, z)=(w, Kz)=(Kz, w). Also note

(iKz, w)=(J Kz, w)= −(KJ z, w)= −(J z, Kw)= −(iz, Kw).

Hence ReiKz, w = −Reiz, Kw. So−ImKz, w = Imz, Kw. This gives ImKw, z =ImKz, w. HenceKz, w = Kw, z.

Lemma3.2.(σ K)^∗=Kσ.

Proof. We haveσ z, σw = w, z. Hence

σKz, w = σ w, σ²Kz = σ w, Kz = z, Kσ w.

Corollary3.3. Ifx, y∈VR, thenHx, yis real andAx, y = Ay, x. Proof. Clearly·,·is real onVR×VR. SinceH VR⊆VR, we seeH x, y is real. Next,Ax, y = Hx, y+Kx, y. The termH x, yequalsHy, x becauseHx, yis real andH is self-adjoint. On the other hand,Kx, y = Ky, xby Lemma 3.1. SoAx, y = Ay, x.

As before we would like to have a multiplierm:VR×V →C^∗such that T (x)F (z)=m(x, z)F (z−x)

is a unitary representation ofF_Athat commutes with translation onL²(VR). It turns out the multipliers we construct are co-boundaries under translation by elements ofVRonV.

Definition3.4. A functionmis aco-boundaryonV ∼=Cⁿunder transla- tion byVRif there is a nonzero complex valued functionbonV with

m(x, z)=b(z−x)b(z)⁻¹ for x∈VR and z∈V.

It is well known and easy to verify that every co-boundarymonV under translation byVRis a multiplier.

Lemma3.5.The function

m(x, z)=e^Hz,xe^Kz,x¯ e^−Ax,x/2=e^{Ax,¯z−Ax,x/2} is a co-boundary.

Proof. Defineb(z)=e^{−Hz+K¯z,¯z/2}. Then

b(z−x)b(z)⁻¹=e−H(z−x)+K(¯z−x),¯z−x/2e^{Hz+Kz,¯¯z/2}

=e^{(Hx+Kx,¯z+Hz+K¯z,x)/2}e^−Hx+Kx,x/2

=e^Hz,xe^Kz,x¯ e^−Ax,x/2

=e^{Hx,¯z+Kx,¯z}e^−Ax,x/2

=e^{Ax,¯z−Ax,x/2}

sinceA=H+K,Hx,z = z, σH x = z, H x = H z, x¯ , andKx,z =¯ σz, σ Kx = Kσ z, x = K¯ z, x¯ .

Corollary3.6.Letm(x, z)=e^{Ax,¯z−Ax,x/2}. SetTxF (z):=m(x, z)F (z

−x)forx∈VR. Thenx →Txis a representation of the abelian groupVRon F_A. It is unitary if and only ifKVR⊆VR, or equivalentlyAVR⊆VR.

Proof. Sincemis a multiplier, we haveTxTy = Tx+y. For eachTx to be unitary, we need|m(x, z)| =e(Az,x)−(Ax,x)/2. But

|m(x, z)| =e^(Hz,x)e^(Kz,x)¯ e^−(Ax,x)/2=e(Az,x)−(Ax,x)/2e^{(Kz−Kz,x)¯} . ThusTxis unitary for allxif and only if the real part of every vectorKz¯−Kz is 0. Sincez¯−zruns overiVRaszruns overV,Txis unitary for allxif and only ifK(iVR)⊂iVR, which is equivalent toK(VR)⊂VR. But sinceA=H +K andH leavesVRinvariant, this is equivalent toVRbeing invariant underA.

Remark. There is no uniqueness in the choice of a real vector space VR

such that H VR ⊆ VR and V = VR ⊕iVR. Indeed, any orthonormal basis e1, e2, . . . , en of eigenvectors for H gives such a subspace. But since Ais only real linear onV, an interesting question is when one can chooseVRwith AVR ⊆ VR, and in this case how unique is the choice ofVR? This probably depends on the degree of non complex linearity of the tranformationA.

Recall that detRH =(detH )². To simplify some calculations later on we definec:=(2π)^−n/4_detRA

detH

1/4

. We remark for further reference:

Lemma3.7. c⁻_A²c²= ^√₍₂^det_π)n/2^H andc^−1√det_πn/2^(H) =₂

π

n/4 (detH)^3/4 (detRA)^1/4. LetD(x)=c m(x,0)=c e^−Ax,x/2and defineR:F_A→C^∞(VR)by

RF (x):=D(x)F (x).

Sincemis holomorphic onV²,Dhas a holomorphic extension toV.

Lemma3.8.The restriction mapRintertwines the action ofVRonF_Aand the left regular actionLon functions onVR.

Proof. For allx, y∈VR, we have

R(TyF )(x)=c m(x,0)TyF (x)

=c m(x,0)m(y, x)F (x−y)

=c m(x,0)m(−y,−x)F (x−y)

=c m(x−y,0)F (x−y)

=LyRF (x).

4. The Generalized Segal-Bargmann Transform

As for the classical space,Rspecifies a densely defined closed operatorF_A→ L²(VR). It also has dense image inL²(VR). To see this, let{hα}_αbe the orthonor- mal basis ofL²(VR)given by the Hermite functions. Then det(A)¹⁴hα(√

Ax)

α

is an orthonormal basis ofL²(VR)which is contained in the image of the set of polynomial functions underR. It follows again thatRhas a densely defined adjoint and

R^∗h(z)= R^∗h, KA,z = h, RKA,z whereKA,z(w)=KA(w, z)=c_A⁻²e^12Kw,we^Hw,ze^12Kz,z. Thus

R^∗h(z)=c

h(x)e^−Ax,x/2KA(x, z) dx

=c⁻_A²c

h(x)e^−Ax,x/2e^12Kz,ze^z,Hxe^12Kx,xdx

=c⁻_A²c e^12Kz,z

h(x)e^−Hx,x/2e^−Kx,x/2e^z,Hxe^12Kx,xdx

=c⁻_A¹c e^12Kz,z

h(x)e^−x,Hx/2e^z,Hxdx

=c⁻_A²c e^12Kz,ze^12z,Hz¯

h(x)e^{−(z,Hz−z,Hx−x,H¯ ¯z+x,Hx)/2}dx

=c⁻_A²c e^12Kz,ze^12z,Hz¯

h(x)e^{−z−x,H(¯z− ¯x)/2}dx

forz, H x = H x,z = Hx,¯ z = x, H¯ z¯ andz, H x = z, Hx¯ . Thus we finally arrive at

(4.1) R^∗h(z)=c_A⁻²c e^12z,Hz+Kz¯ e^−12x,Hx¯ ∗h(z).

LetP :VR→VRbe positive. DefineφP(x)=√

det(P )(2π)^−n/2e^{−√P x2/2}. Fort >0, letP (t)= ¹_tP.

Lemma 4.1. Let the notation be as above. Then 0 < t → φP (t) is a convolution semigroup, i.e.,φP (t+s)=φP (t)∗φP (s).

Proof. This follows by change of parametersy=√

P x from the fact that φ^Id(t)(x)=(2πt)^−n/2e^−x2/2t is a convolution semigroup.

Define a unitary operatorW onL²(VR)by

(4.2) Wf (x)=e^iImx,Kxf (x)=e^iImx,Axf (x).

We seeW = I ifAVR ⊆ VRwhich is equivalent to the translation operators T (x)being unitary.

Lemma4.2. Lethbe in the domain of definition ofR^∗. Then RR^∗h=W(φH ∗h).

Proof. We notice first thatc⁻_A²c²=(2π)^−n/2√

detH by Lemma 3.7. From (4.1) we then have

RR^∗h(x)=c e^−12Ax,xR^∗h(x)

=c⁻_A²c²e^−12Ax,xe^12x,Hx+Kx¯ e^−12y,Hy¯ ∗h(x)

=(2π)^−n/2

det(H ) e^−12Ax,xe^12x,Axe^−12y,Hy¯ ∗h(x)

=(2π)^−n/2

det(H ) e^iImx,Ax

e^−12(y,Hy)h(x−y) dy.

=(2π)^−n/2

det(H )e^iImx,Ax e⁻

√H y2

2 h(x−y) dy

=W(φH∗h)(x).

In the last step, we used the fact thatφH ∗h∈ L²(VR), which follows from φH being inL¹(VR)andh∈L²(VR).

Arguing as in the classical case, we see that R and R^∗ are everywhere defined and continuous.

Lemma 4.1 and Lemma 4.2 leads to the following corollary:

Corollary4.3.SupposeAVR⊆VR. Then

|R^∗|h(x)=φH(1/2)∗h(x)=

√det(H ) π^n/2

VR

e^−√Hy2h(x−y) dy.

Theorem 4.4 (The Segal-Bargmann Transform). Suppose Aleaves VR

invariant. Then the operatorUA :L²(VR)→F_Adefined by UAf (z)=

2 π

_n/4

(detH)^3/4

(detRA)^1/4e^{12(Hz,¯z+z,Kz)}

e(H(z−y))·(z−y)f (y) dy is a unitary isomorphism. We call the mapUAthegeneralized Segal-Bargmann transform.

Proof. By polar decomposition, we can writeR^∗ = U|R^∗| whereU : L²(VR)→F_A is a unitary isomorphism. Taking adjoints gives|R^∗|U^∗ =R. HenceRU = |R^∗|. Thus

c m(x)Uh(x)=RUh(x)=(|R^∗|h)(x)

=

√det(H ) π^n/2

VR

e^−√Hy2h(x−y) dy.

Sincem(x)=e⁻¹²(x,Hx+x,Kx), we have using Lemma 3.7:

Uf (x)= 2

π _n/4

(detH)^3/4

(detRA)^1/4e¹²(x,Hx+x,Kx)

e(x−y,H(x−y))f (y) dy.

Holomorphicity ofUf now impliesUf =UAf. 5. The Gaussian Formulation

In infinite dimensions there is no useful notion of Lebesgue measure but Gaus- sian measure does make sense. So, with a view to extension to infinite dimen- sions, we will recast our generalized Segal-Bargmann transform using Gaus- sian measure instead of Lebesgue measure as the background measure onVR. Of course, we have already defined the Fock spaceF_Ausing Gaussian measure.

As before,V is a finite-dimensional complex vector space with Hermitian inner-product·,·, andA:V →V is areal-linear map which issymmetric, positive-definite with respect to the real inner-product (·,·) = Re·,·, i.e.

(Az, z) > 0 for allz ∈ V exceptz = 0. We assume, furthermore, that there is a real subspace VR for which V = VR +iVR, the inner-product ·,· is real-valued onVR, andA(VR)⊂VR. Denote the linear mapv→ivbyJ. As usual,Ais the sum

A=H +K

whereH =(A−J AJ )/2 is complex-linear onV andK=(A+J AJ )/2 is complex-conjugate-linear. The real subspacesVRandJ VRare(·,·)-orthogonal because for anyx, y ∈ VRwe have(x, Jy) = Rex, Jy = −Re(Jx, y), sincex, yis real, by hypothesis. SinceApreservesVRand is symmetric, it also preserves the orthogonal complementJ VR. ThusAhas the block diagonal form:

A=

R 0

0 T

=d(R, T )

Here, and henceforth, we use the notationd(X, Y )to mean the real-linear map V →V given bya → XaandJ a → J Y afor alla ∈ VR, whereX, Y are real-linear operators onVR. Note thatd(X, Y )is complex-linear if and only if X=Y and is complex-conjugate-linear if and only ifY = −X. The operator d(X, X)is the unique complex-linear mapV →V which restricts toX on VR, and we denote it:

XV =

X 0

0 X

The hypothesis thatAis symmetric and positive-definite means thatRandT are symmetric, positive definite onVR. Consequently, the real-linear operator SonVRgiven by

S=2(R⁻¹+T⁻¹)⁻¹

is also symmetric, positive-definite.

The operatorsH andKonV are given by H = 1

2(RV +TV), K=d 1

2(R−T ),1

2(T −R)

.

Using the conjugation map

σ :V →V :a+ib→a−ib for a, b∈VR

we can also writeKas

(5.1) K= 1

2(RV −TV)σ

Now consider the holomorphic functionsρT andρSonV given by ρT(z)= (detT )^1/2

(2π)^n/2 e^{−12(TVz)·z} ρS(z)= (detS)^1/2

(2π)^n/2 e^{−12(SVz)·z} wheren=dimVR. Restricted toVR, these are density functions for Gaussian probability measures.

The Segal-Bargmann transform in this setting is given by the map SA:L²(VR, ρS(x)dx)→F_A :f →SAf

where

(5.2) SAf (z)=

VR

f (x)ρT(z−x) dx=

VR

f (x)c(x, z)ρS(x) dx

withc(x, z)given, forx ∈VRandz∈V, by c(x, z)= ρT(x−z)

ρS(x) .

It is possible to take (5.2) as the starting point, withf ∈ L²(VR, ρS(x)dx) and prove that: (i) SAf (z) is well-defined, (ii) SAf is in F_A, (iii) SA is a unitary isomorphism onto F_A. However, we shall not work out everything in this approach since we have essentially proven all this in the preceding sections. Full details of a direct approach would be obtained by generalizing the procedure used in [13]. In the present discussion we shall work out only some of the properties ofSA.

Lemma5.1. Letw, z∈V. Then:

(i) The functionx→c(x, z)belongs toL²(VR, ρS(x)dx), thereby ensuring that the integral (5.2) definingSAf (z)is well-defined;

(ii) TheSA-transform ofc(·, w)isKA(·,w)¯ :

[SAc(·, w)](z)=KA(z,w)¯ and so, in particular:

KA(z, w)=

VR

ρT(x−z)ρT(x− ¯w) ρS(x) dx

(iii) The transform SA preserves inner-products on the linear span of the functionsc(·, w):

c(·, w), c(·, z)_L²_(VR,ρS(x)dx) =KA(w, z)= KA(·,w), K¯ A(·,z)¯ FA

Proof. (i) is equivalent to finiteness of

VR

|ρT(x−z)|²

ρS(x) dx, which is equivalent to positivity of the operator 2T−S. To see that 2T−Sis positive observe that

2T −S=2T[(R⁻¹+T⁻¹)−T⁻¹](R⁻¹+T⁻¹)⁻¹

=2T R⁻¹(R⁻¹+T⁻¹)⁻¹=T R⁻¹S (5.3)

=2(T⁻¹+T⁻¹RT⁻¹)⁻¹ (5.4)

and in this last lineT⁻¹ >0 (sinceT >0) and(T⁻¹RT⁻¹x, x)=(RT⁻¹x, T⁻¹x)≥0 by positivity ofR. Thus 2T−Sis positive, being twice the inverse of the positive operatorT⁻¹+T⁻¹RT⁻¹.

(ii) Recall v·wforv, w ∈ V is the symmetric complex bilinear pairing given byv·w= v,w¯ , and we writev²forv·v. We shall denote the complex- linear operatorTV which restricts toT onVRsimply byT. It is readily checked thatT continues to be symmetric in the sense thatT v·w = v·T w for all v, w∈V. We start with

a^def= [SAc(·, w)](z)

=

VR

ρT(x−w)

ρS(x) ρT(z−x) dx

=(2π)^−n/2 detT (detS)^1/2

VR

e^−12[T (x−w)·(x−w)+T (x−z)·(x−z)−Sx·x]dx

=(2π)^−n/2 detT (detS)^1/2

VR

e^{−12[(2T−S)x·x−2}T x·(w+z)+T w·w+T z·z]dx.

Recall from the proof of (i) that 2T −S > 0. For notational simplicity let L=(2T −S)^1/2andM =L⁻¹T. Then

a=(2π)^−n/2 detT (detS)^1/2

VR

e^{−12(Lx−M(w+z))2}dx e^−12[T w·w+T z·z−M(w+z)·M(w+z)]

= detT

(detS)^1/2(detL)e^−12[T w·w+T z·z−M(w+z)·M(w+z)].

To simplify the last exponent observe that by (5.4) and (5.1) we have T w·w−Mw·Mw=T w·w−T w·L⁻²T w

=T w·w−T w·(2T −S)⁻¹T w

=T w·w− 1

2T w·(T⁻¹+T⁻¹RT⁻¹)T w

=T w·w− 1

2T w·(w+T⁻¹Rw)

= 1

2(T w·w−Rw·w)

= −Kw,¯ w.¯

The same holds withzin place ofw. For the “cross term” we have Mw·Mz=T w·L⁻²T z

=T w·(2T −S)⁻¹T z

= 1

2T w·(T⁻¹+T⁻¹RT⁻¹)T z

= 1

2(T w·z+w·Rz)

=2w·Hz.

Putting everything together gives [SAc(·, w)](z)= detT

(detS)^1/2(detL)e^12Kw,¯w¯ e^Hw,¯ze^12K¯z,¯z. In Lemma 6.2 below we prove that

detT

(detS)^1/2(detL) =

det_R(A) detR(H)

₋1/2

=c⁻_A². So

[SAc(·, w)](z)=KA(w,z).¯

For (iii), we have first:

c(·, w), c(·, z)L²(ρS(x)dx) =[SAc(·, w)](¯z)=KA(¯z,w)¯ =KA(w, z).

The second equality in (iii) follows sinceKAis a reproducing kernel.

6. The evaluation map and determinant relations Recall

KA(z, w)=c⁻_A²e^{12z,Kz+12Kw,w+Hz,w} where

c_A⁻²=

det_VH det_V A

2

is a reproducing kernel forF_A. Thus f (w)= f, KA(·, w) =π⁻ⁿ

Vf (z)KA(w, z)|dz|

where|dz| =dxdysignifies integration with respect to Lebesgue measure on the real inner-product spaceV. Thus we have

Proposition6.1. For anyz∈V, the evaluation map δz:F_A→C:f →f (z) is a bounded linear functional with norm

δz =KA(z, z)^1/2=c⁻_A¹e^(Az,z).

Proof. Note

(6.1) |δzf| = |f (z)| = |f, KA(·, z)| ≤ fFAKA(z, z)^1/2 follows from the reproducing kernel property

KA(·, z)²F_A =

KA(·, z), KA(·, z)

FA =KA(z, z).

This last calculation also shows that the inequality in (6.1) is an equality if f = KA(·, z) and thereby shows thatδzis actually equal to KA(z, z)^1/2. The latter is readily checked to be equal toc_A⁻¹e^(Az,z).

We have already used the first of the following two facts aboutcA.