Determinant lines and determinant line bundles

Transcription

1 CHAPTER Determinant lines and determinant line bundles This appendix is an exposition of G. Segal s work sketched in [?] on determinant line bundles over the moduli spaces of Riemann surfaces with parametrized boundaries, especially in the genus-zero case. It requires some basic knowledge of complex analysis, functional analysis, differential geometry and geometric analysis, and is written in a selfcontained way assuming this knowledge. We supply in this appendix all the necessary definitions, results and proofs needed in this work. After the present work, including a version of this appendix, was finished, the author obtained a copy of [?] from D. Freed and a copy of [?] from P. Deligne. The second version [?] of Segal s manuscript on conformal field theory includes an appendix on determinant lines and determinant line bundles and contains more details than [?]. But the methods used in this appendix are different from those in [?] and many details omitted in [?] and [?] are given in the present appendix. Though a large part of the results and proofs in Sections D.3 D.6 follow the ideas in [?], the proofs and even the statements of some results (for example, Theorem.4.4 and Proposition.4.8) are given, to our knowledge, for the first time in the present appendix, which can be read as a separate paper on determinant lines and determinant line bundles. The letter [?] contains a discussion of one-dimensional modular functors and emphasizes that determinant lines are Z 2 -graded lines and that there are some subtle compatibility conditions. We incorporate this point of view into our treatment, but we also note that in the case in which we are mostly interested in this monograph, that is, the case of genus-zero surfaces with at least one outgoing (positively oriented) boundary component, the issues related to the Z 2 -gradings of the determinant lines in fact become trivial (see Remark?? and Section D.6). Also, this appendix incorporates answers to a number of questions raised by Deligne and an alternative argument in the proof of Proposition.4.8 supplied by him. Section D. is a summary of the definitions and some properties of certain classes of linear operators on Hilbert spaces. Section D.2

2 2. DETERMINANT LINES AND DETERMINANT LINE BUNDLES is a discussion on determinant lines of Fredholm operators. In Section D.3 we define the determinant line over a Riemann surface with parametrized boundary and the determinant line bundles over the moduli spaces of Riemann surfaces with parametrized boundaries. In Section D.4 we construct the canonical isomorphisms between determinant lines corresponding to the sewing of Riemann surfaces with parametrized boundaries. We also construct the canonical connections on the determinant line bundles over moduli spaces of genus-zero Riemann surfaces with parametrized boundaries in Section D.4. In Section D.5, we show that these determinant lines together with the canonical isomorphisms constructed in Section D.4 give the basic central extension of the diffeomorphism group of S in a natural way. In Section D.6 we give a definition of a variant of one-dimensional genus-zero modular functor and prove that any such one-dimensional genus-zero modular functor is a power of the determinant line bundle over the moduli space of genus-zero Riemann surfaces with parametrized boundaries... Some classes of bounded linear operators We give the definitions and list the basic properties of Fredholm, compact, finite rank, trace-class and Hilbert-Schmidt operators in this section. The reader can find the details in [?], [?] and [?]. For convenience, we shall always assume in this appendix that the Hilbert spaces we consider are separable and over C. We shall always use I V to denote the identity operator on a vector space V. Definition... Let H and H 2 be Hilbert spaces. A bounded linear operator F : H H 2 is called a Fredholm operator if both KerF and CokerF are finite-dimensional. The index of F is the number IndexF = dim KerF dim CokerF. A bounded linear operator R : H H 2 is called a finite-rank operator if the image of R is finite-dimensional. A bounded linear operator C : H H 2 is called a compact operator if it maps bounded subsets of H to precompact subsets of H 2, that is, to subsets of H 2 whose closure is compact. A bounded linear operator T : H H 2 is called a traceclass operator if there exists a sequence {λ i } i Z+ of complex numbers and orthonormal families {φ i } i Z+ in H and {ψ i } i Z+ in H 2 such that i Z + λ i < and for any h H, T (h) = i Z + λ i (φ i, h) ψ i,

3 .. SOME CLASSES OF BOUNDED LINEAR OPERATORS 3 where (, ) is the inner product of H. The trace of T is the number tr(t ) = i Z + λ i. A bounded linear operator T : H H 2 is called a Hilbert-Schmidt operator if for any complete orthonormal basis {e i } i Z+ of H, the series i Z + T e i 2 converges. Proposition..2. The operators defined above have the following properties: () Finite-rank operators are trace-class operators. Trace-class operators and Hilbert-Schmidt operators are compact operators. (2) Let C be a compact operator (respectively, trace-class or Hilbert- Schmidt operator). For any bounded linear operators P : H 0 H and P 2 : H 2 H 3, the products CP and P 2 C are still compact operators (trace-class operators or Hilbert- Schmidt operators). (3) The adjoint operator of a compact operator is also compact. (4) The space of all compact operators from H to H 2 is closed in the space of all bounded linear operators from H to H 2. The space of compact operators are the closure of the space of finite-rank operators. (5) The sum of a Fredholm operator F and a compact operator C (in particular, a finite rank operator or a trace-class operator) is a Fredholm operator. The index of F + C is equal to the index of C. (6) Products of Hilbert-Schmidt operators are trace-class operators. Let F : H H 2 be a Fredholm operator. A quasi-inverse (or parametrix) of F is a bounded linear operator G : H 2 H such that GF is equal to the identity of H plus a compact operator on H and F G is equal to the identity of H 2 plus a compact operator on H 2. The following classical theorem is due to Atkinson whose proof can be found in the references listed in the beginning of this section: Theorem..3 (F. W. Atkinson). A bounded linear operator F : H H 2 is Fredholm if and only if it has a quasi-inverse. Remark..4. From Theorem..3, we see that any quasi-inverse G of a Fredholm operator F is also a Fredholm operator. Remark..5. In [?], a canonical quasi-inverse G of F was constructed explicitly for each choice of closed subspaces X 0 and Y 0 of H

4 4. DETERMINANT LINES AND DETERMINANT LINE BUNDLES and H 2, respectively, such that and H = X 0 KerF H 2 = Y 0 ImF. This quasi-inverse has the additional property that F G I H2 and GF I H are finite-rank operators. Since in our case both H and H 2 are Hilbert spaces, we shall canonically choose X 0 and Y 0 to be the orthogonal complements of KerF and ImF, respectively, in H and H 2. Thus we obtain a canonical quasi-inverse G of F such that F G I H2 and GF I H are finite-rank operators. Let H be a Hilbert space. We denote the space of all trace-class operators on H by I and we define a norm on I as follows: For any T = i Z + λ i (φ i, )ψ i I, T = i Z + λ i. Then I equipped with this norm is a Banach space. The proof of the following result due to Simon can be found in [?]: Theorem..6 (B. Simon). There is a unique continuous function det : I H + I C with the following properties: () For a complex variable z and T I, det(i H +zt ) is an entire function satisfying (2) For any T, T 2 I, det(i H + zt ) e z T. det(i H + T + T 2 + T T 2 ) = det(i H + T ) det(i H + T 2 ). (3) For any T I, det(i H + T ) 0 if and only if I H + T is invertible. In this case, (I H + T ) I H is of trace class and det(i H + T ) = (det(i H + T )). (4) Let T be a trace-class operator and λ n (T ), n Z +, all the eigenvalues of T counting multiplicities. Then det(i H + T ) = n Z + ( + λ n (T )).

5 .2. DETERMINANT LINES 5 When a bounded linear operator on a Hilbert space H is of the form I H + T where T is of trace class, we say that the operator has a determinant. If H is finite-dimensional, any operator F on H has a determinant in the sense of the definition above and its determinant is equal to the determinant of an operator on a finite-dimensional space in the usual sense..2. Determinant lines In this section, we develop the theory of determinant lines of Fredholm operators. Many of the results are parallel to those for indices of Fredholm operators (see, for example, [?]). But for determinant lines, we need to construct canonical isomorphisms; proving the existence of isomorphisms is not enough. Thus many results here are harder and more subtle than the corresponding results for indices. A Z 2 -graded line or simply graded line is a one-dimensional vector space with an element of Z 2 called the degree. The degree of a graded line L is denoted by deg(l). Given any element v L, the degree of v is defined to be the degree of L and is denoted by deg(v). Isomorphisms of graded lines are defined in the obvious way. For an n-dimensional vector space V, we define the determinant line DetV of V to be the graded line n V of the highest exterior power of V with degree n modulo 2Z. It is clear that we can think of Det as a functor from the category of finite-dimensional vector spaces with isomorphisms as morphisms to the category of graded lines with isomorphisms as morphisms. Note our choice of notations in this appendix: Det will always denote a determinant line of a finite-dimensional space, a Fredholm operator or a Riemann surface with analytically parametrized boundary, while det will always denote the determinant (not the determinant line) of an operator having a determinant. Given two vector spaces of the same finite dimension, they are always isomorphic, but there is no canonical isomorphism. When the two vector spaces are the images of an object in a category under two naturally isomorphic functors, the spaces are canonically isomorphic. In this appendix, we shall write canonical isomorphisms as. Note that when we say that two vector spaces are canonically isomorphic, we have already chosen a natural isomorphism understood from the context. The same conventions also apply to graded lines and so on. The category of graded lines has a tensor product operation: Given any two graded lines L and L 2, the one-dimensional vector space L L 2 together with the degree deg(l L 2 ) = deg(l ) + deg(l 2 )

6 6. DETERMINANT LINES AND DETERMINANT LINE BUNDLES is a graded line. This tensor product is a functor from the category of ordered pairs of graded lines to the category of graded lines. There is a natural isomorphism from this functor to itself called the symmetry of graded lines defined as follows: Let L and L 2 be two graded lines. For any v L and v 2 L 2, we define an isomorphism from L L 2 to L 2 L by v v 2 ( ) deg(v ) deg(v 2 ) v 2 v 2. Clearly these isomorphisms give a natural isomorphism. (In fact, graded lines with the tensor product and this symmetry form a symmetric tensor category.) Thus L L 2 L 2 L. In this appendix, when we discuss canonical isomorphisms, we shall not indicate explicitly what the functors and the natural isomorphisms are, except in the proof of the following basic lemma: Lemma.2.. Let U be a finite-dimensional vector space and V a subspace of U. Then DetU Det(U/V ) DetV. Proof. We view (U, V ) DetU and (U, V ) Det(U/V ) DetV as functors from the category of pairs of finite-dimensional vector spaces and their subspaces to the category of graded lines. To show that DetU Det(U/V ) DetV is to construct a natural isomorphism between these two functors. Assume that the dimensions of U and V are m and n, respectively. Then n m. Let v v n be a nonzero element of DetV. Any nonzero element of DetU can be written as u u n m v v m where u,..., u m n are elements of U not in V. Thus we have an element ((u + V ) (u n m + V )) v v n DetU/V DetV. Whenever there are elements ũ,..., ũ m n U and ṽ,..., ṽ n V such that ũ ũ n m ṽ ṽ m = u u n m v v m, we have ũ i = m n k= a iku k + ˆv i, i =,..., m n, and ṽ j = n l= b jlv l where ˆv i are elements of V and (a ik ) and (b jl ) are matrices such that det(a ik ) det(b jl ) = (δ ij ). Thus ((ũ + V ) (ũ n m + V )) ṽ ṽ n = ((u + V ) (u n m + V )) v v n, and we obtain a linear map from DetU to DetU/V DetV. It is clear that this map is an isomorphism and in fact natural.

7 .2. DETERMINANT LINES 7 Let H and H 2 be Hilbert spaces and F : H H 2 a Fredholm operator. Following Quillen [?], we associate to F a one-dimensional vector space DetF = Det(KerF ) Det(CokerF ) where on the right-hand side Det is the functor discussed above. There is a natural Z 2 -grading of DetF given by the index of F modulo 2Z. With this grading, DetF becomes a graded line. This graded line DetF is called the determinant line of the Fredholm operator F. Let F : H H 2 and F 2 : H 3 H 4 be Fredholm operators. An equivalence from F to F 2 is a pair f : H H 3 and f 2 : H 2 H 4 of bounded linear isomorphisms such that the diagram F H H2 f f 2 H 3 F2 H 4 commutes. It is clear that the restriction of f to KerF gives a linear isomorphism from KerF to KerF 2 and f 2 induces a linear isomorphism from CokerF to CokerF 2. Thus the pair (f, f 2 ) induces a linear isomorphism Det f,f 2 : DetF DetF 2. Thus Det is a functor from the category of Fredholm operators with equivalences as morphisms to the category of one-dimensional vector spaces with isomorphisms as morphisms. Note that linear operators between finite-dimensional spaces are Fredholm operators. Thus they have determinant lines. Proposition.2.2. Let H and H 2 be finite-dimensional vector spaces and F : H H 2 a linear operator. Then there exists a canonical isomorphism from DetF to (DetH ) DetH 2 such that in the case that H = H 2 = H and F is invertible, the canonical isomorphism from C = DetF to C (DetH) DetH is given by multiplying (det F ) = det F to elements of C. Proof. Since H /KerF ImF, we have or equivalently, By Lemma.2., we have Det(H /KerF ) DetImF, Det(H /KerF ) (DetImF ) C. Det(H /KerF ) DetH (DetKerF )

8 8. DETERMINANT LINES AND DETERMINANT LINE BUNDLES and Thus (DetImF ) DetH 2 DetCokerF. C Det(H /KerF ) (DetImF ) DetH (DetKerF ) (DetH 2 ) DetCokerF DetH (DetH 2 ) DetF, or equivalently, DetF (DetH ) DetH 2. In the case that H = H 2 = H and F is invertible, we have KerF = 0 and ImF = H. Thus the canonical isomorphism from H /KerF = H to ImF = H is F itself and the canonical isomorphism from Det(H /KerF ) = DetH to DetF = DetH is given by multiplying det F to elements of DetH. From the construction above, we see that the canonical isomorphism from C = DetF to C (DetH) DetH is indeed given by multiplying (det F ) = det F to elements of C. Proposition.2.3 (Snake lemma). Consider the following diagram: 0 H H 2 H 3 0 () F F 2 F 3 0 H 4 H 5 H 6 0 where H i, i =,..., 6, are Hilbert spaces, F, F 2, F 3 are bounded linear operator, and the horizontal lines are exact. If two of the operators F, F 2, F 3 are Fredholm, then the other is also Fredholm and there is a canonical isomorphism from DetF DetF 3 to DetF 2. (2) Proof. The diagram () gives an exact sequence 0 KerF KerF 2 i KerF 3 k k CokerF j CokerF 2 CokerF 3 0. Assume that F and F 3 are Fredholm. From the exact sequence (2), we obtain (3) Imi KerF 2 /Keri, Keri KerF, Imi KerF 3. Since F and F 3 are Fredholm, both KerF and KerF 3 are finite-dimensional. Thus by (??) (3), KerF 2 is also finite-dimensional. Similarly, we can

9 .2. DETERMINANT LINES 9 show that CokerF 2 is finite-dimensional and consequently F 2 is Fredholm. If F, F 2 or F 2, F 3 are Fredholm, we can show similarly that F 3 or F, respectively, is Fredholm. The dual of (2) 0 (KerF ) (KerF 2 ) i (4) i (KerF 3 ) k (CokerF ) j j (CokerF 2 ) (CokerF 3 ) 0 is also exact. From the exact sequences (2) and (4), we see that (5) (6) (7) (8) (KerF ) (KerF 2 ) /Imi, CokerF 3 CokerF 2 /Imj, Imi (KerF 3 ) /Imk, Imj CokerF /Imk. By Lemma.2. and (5) (8), we have (9) DetF DetF 3 = Det(KerF ) DetCokerF Det(KerF 3 ) DetCokerF 3 Det((KerF 2 ) /Imi ) DetCokerF Det(KerF 3 ) Det(CokerF 2 /Imj) Det((KerF 2 ) /Imi ) (DetImj DetImk) (DetImi DetImk ) Det(CokerF 2 /Imj) Det((KerF 2 ) /Imi ) DetImi Det(CokerF 2 /Imj) DetImj DetImk DetImk Det(KerF 2 ) DetCokerF 2 DetImk DetImk DetF 2 DetImk DetImk. It is easy to see that Imk is in fact canonically isomorphic to (Imk). The canonical paring between Imk and (Imk) induces a nondegenerate pairing between DetImk and Det(Imk) which can be regarded as a canonical isomorphism from DetImk Det(Imk) to C. Thus (0) DetImk DetImk C. The second conclusion follows from (9) and (0). Corollary.2.4. Let F : H H 2 and F 2 : H 2 H 3 be Fredholm operators. Then Det(F 2 F ) is canonically isomorphic to DetF 2 DetF.

10 0. DETERMINANT LINES AND DETERMINANT LINE BUNDLES Proof. Consider the diagram 0 H i H H 2 F F 2 F I H2 F 2 j 0 H 2 H 3 H 2 where i, j, p, q are defined by i(u) = (u, F u), j(v) = (F 2 v, v), p(u, v) = F (u) v, q(w, v) = w F 2 (v) p H 2 0 q H 3 0 for all u H, v H 2 and w H 3. This diagram is commutative and the horizontal sequences are exact. Applying Proposition.2.3, we see that Det(F 2 F I H2 ) is canonically isomorphic to DetF 2 DetF 2. Since Det(F 2 F ) is canonically isomorphic to Det(F 2 F I H2 ), it is also canonically isomorphic to DetF 2 DetF 2. Corollary.2.5. Let R : H H be a finite-rank operator. Then there is a canonical isomorphism from Det(I H + R) to C such that in the case that I H +R is invertible, this canonical isomorphism from C = Det(I H +R) to C is given by multiplying the determinant det(i H +R) to elements of C. Proof. Let h be any finite-dimensional vector space containing ImR and (I H + R) h (the restriction of I H + R to h). Consider the diagram 0 h H H/h 0 (I H +R) h I H +R I H/h 0 h H H/h 0. The commutativity of the diagram and the exactness of the rows is clear. By Proposition.2.3, Det(I H + R) is canonically isomorphic to Det(I H + R) h DetI H/h. Since h is finite-dimensional, (I H + R) h is an operator on a finite-dimensional space. By Proposition.2.2, Det(I H + R) h C. Since DetI H/h = C, we obtain a canonical isomorphism from Det(I H + R) to C by the snake lemma. In the case that I H + R is invertible, by the construction above and Proposition.2.2, the canonical isomorphism from C = Det(I H + R) h to C is given by multiplying the determinant det(i H +R) h to elements of C. Let λ n, n =,..., N, be the eigenvalues of R. Since R is of finite rank, N must be finite. Since h contains ImR, λ n, n =,..., N, are

11 .2. DETERMINANT LINES also the eigenvalues of R h (the restriction of R to h). Thus by Theorem..6 det(i H + R) h = (det(i h + R h )) ( N = ( + λ n ) n= ) = (det(i H + R)) = det(i H + R), proving the corollary. We also have the following associativity: Proposition.2.6. Let F : H H 2, F 2 : H 2 H 3, F 3 : H 3 H 4 be Fredholm operators. Then the diagram () Det(F 3 F 2 F ) DetF 3 Det(F 2 F ) Det(F 3 F 2 ) DetF DetF 3 DetF 2 DetF obtained using the canonical isomorphisms constructed in Corollary.2.4 is commutative. Proof. We prove this result by writing down the canonical isomorphisms in the diagram () explicitly. First we consider the canonical isomorphism from Det(F 2 F ) to DetF 2 DetF which induces the right vertical arrow in (). In fact, the construction below is a direct proof of Corollary.2.4. As in the proof of the snake lemma, we have the following exact sequence: 0 KerF Ker(F 2 F I H2 ) KerF 2 CokerF Coker(F 2 F I H2 ) CokerF 2 0. Since Ker(F 2 F I H2 ) Ker(F 2 F ) and Coker(F 2 F I H2 ) Coker(F 2 F ), this exact sequence gives us the exact sequence 0 KerF Ker(F 2 F ) i k KerF2 k j CokerF Coker(F2 F ) CokerF 2 0

12 2. DETERMINANT LINES AND DETERMINANT LINE BUNDLES and its dual 0 (KerF ) (Ker(F 2 F )) i i (KerF 2 ) k (CokerF ) j j (Coker(F 2 F )) (CokerF 2 ) 0. Thus similar to (5) (8), we have (KerF ) (Ker(F 2 F )) /Imi, CokerF 2 Coker(F 2 F )/Imj, Imi (KerF 2 ) /Imk, Imj CokerF /Imk. From these formulas and the fact that DetImk (DetImk ), we have (2) (3) (4) (5) Det(KerF ) DetImi Det(Ker(F 2 F )), DetCokerF 2 DetImj DetCoker(F 2 F ), DetImi (DetImk ) Det(KerF 2 ), DetImj DetImk DetCokerF. Let u, u 2, u 3, u 4, u 5 be basis (nonzero vectors) of Det(KerF ), DetImi, DetCokerF 2, DetImj, DetImk, respectively, and u 5 (DetImk ) the dual basis of u 5. We identify the right-hand sides of (2) (5) with their left-hand sides, respectively. Then under these identifications, u u 2, u 3 u 4, u 2 u 5 and u 4 u 5 are basis of Det(Ker(F 2 F )), DetCoker(F 2 F ), Det(KerF 2 ) and DetCokerF, respectively. Thus u u 2 u 3 u 4 and u 2 u 5 u 3 u u 4 u 5 are basis of Det(F 2 F ) and DetF 2 DetF, respectively, and the canonical isomorphism from Det(F 2 F ) to DetF 2 DetF is the unique linear map given by u u 2 u 3 u 4 u 2 u 5 u 3 u u 4 u 5. Repeat the above discussion with F and F 2 replaced by F 2 F and F 3, respectively, and denote the maps i, j, k in this case by i 2, j 2, k 2, respectively. Let u 6, u 7, u 8, u 9 be basis of DetImi 2, DetImj 2, DetCokerF 3, DetImk 2, respectively, and u 9 (DetImk 2 ) the dual basis of u 9. Then u u 2 u 6, u 7 u 8, u 6 u 9 are basis of Det(Ker(F 3 F 2 F )), DetCoker(F 3 F 2 F ), Det(Kerf 3 ), respectively. Thus u u 2 u 6 u 7 u 8 and u 6 u 9 u 8 u u 2 u 3 u 4 are basis of Det(F 3 F 2 F ) and DetF 3 Det(F 2 F ), respectively. Note that we have DetCokerF 2 DetImj DetCoker(F 2 F ) DetImj 2 DetImk 2

13 .2. DETERMINANT LINES 3 and the resulting canonical isomorphism from DetCokerF 2 DetImj to DetImj 2 DetImk 2 is determined by u 3 u 4 u 7 u 9. Since we identify both DetCokerF 2 DetImj and DetImj 2 DetImk 2 with DetCoker(F 2 F ), u 3 u 4 and u 7 u 9 are the same in DetCoker(F 2 F ). Thus u 6 u 9 u 8 u u 2 u 3 u 4 and u 6 u 9 u 8 u u 2 u 7 u 9 are the same in DetF 3 Det(F 2 F ). By construction, the canonical isomorphism from Det(F 3 F 2 F ) to DetF 3 Det(F 2 F ) is given by u u 2 u 6 u 7 u 8 u 6 u 9 u 8 u u 2 u 7 u 9, or equivalently, given by (6) u u 2 u 6 u 7 u 8 u 6 u 9 u 8 u u 2 u 3 u 4. From the discussions above, we see that the upper horizontal arrow in () is given by (6) and the right vertical arrow in () is given by u 6 u 9 u 8 u u 2 u 3 u 4 u 6 u 9 u 8 u 2 u 5 u 3 u u 4 u 5. Thus the composition of the upper horizontal arrow and the right vertical arrow in () is given by (7) u u 2 u 6 u 7 u 8 u 6 u 9 u 8 u 2 u 5 u 3 u u 4 u 5. One can prove using the same method as above to show that the composition of the left vertical arrow and the lower horizontal arrow in () is also given by (7). Thus the diagram () is commutative. Remark.2.7. Consider the following two categories: The objects of both categories are Hilbert spaces. The morphisms between two Hilbert spaces in the first category are the Fredholm operators from the domain to the codomain, while the morphisms in the second category are the determinant lines of the Fredholm operators from the domain to the codomain. Then the above proposition shows that we can interpret F DetF as a functor from the first category to the second category. Remark.2.8. In the discussions above, if we take all the spaces to be vector spaces and all the maps to be linear maps, the conclusions are still true. So the results above are all algebraic in nature. But we do need Hilbert space structures below to obtain deeper results. Proposition.2.9. Let T : H H be a trace-class operator. Then there exists a canonical isomorphism from Det(I H + T ) to C such that when I H + T is invertible, the canonical isomorphism from

14 4. DETERMINANT LINES AND DETERMINANT LINE BUNDLES C = Det(I H +T ) to C is given by multiplying det(i H +T ) to elements of C. Proof. Let R : H H be any finite-rank operator such that T R <. Then the partial sum of the series k N ( )k (T R) k is a Cauchy sequence in the Banach space of bounded linear operators on H. So this series has a limit. Denote this limit by (I H + (T R)). Then it is indeed the inverse of I H + (T R). We have I H + T = (I H + (T R))(I H + (I H + (T R)) R). Since I H +(T R) is invertible, by definition Det(I H +(T R)) = C. Since R is of finite rank, (I H + (T R)) R is also of finite rank. By Corollary.2.5, By Corollary.2.4, Det(I H + (I H + (T R)) R) C. Det(I H + T ) = Det((I H + (T R))(I H + (I H + (T R)) R)) (8) Det(I H + (T R)) Det(I H + (I H + (T R)) R) C. So we obtain a map from the set V of all finite-rank operators R on H satisfying T R < to the space of isomorphisms from Det(I H + T ) to C. We want to show that when R goes to T, the image of R under this map has a limit. We prove this existence by showing that when two finite-rank operators R and R 2 are very close to T, their images under the map above are very close to each other. Given any finite-rank operators R and R 2 satisfying T R < and T R 2 <, respectively, we want to compare the isomorphisms from Det(I H +T ) to C obtained using R and R 2. Since we can identify both Det(I H + (T R )) and Det(I H + (T R 2 )) with C, we need only compare the canonical isomorphisms from Det(I H + (I H + (T R )) R ) and from Det(I H + (I H + (T R 2 )) R 2 ) to C. Write I H + (I H + (T R )) R = (I H + (I H + (T R )) (R R 2 ))(I H + (I H + (T R 2 )) R 2 ). Then by Proposition.2.4, we have Det(I H + (I H + (T R )) R ) Det(I H + (I H + (T R )) (R R 2 )) Det(I H + (I H + (T R 2 )) R 2 ).

15 .2. DETERMINANT LINES 5 When T R and T R 2 are small, R 2 R is small. When R 2 R is sufficiently small, is invertible and I H + (I H + (T R )) (R R 2 ) Det(I H + (I H + (T R )) (R R 2 )) = C. Thus we have a canonical isomorphism from to V 2 = Det(I H + (I H + (T R 2 )) R 2 ) V = Det(I H + (I H + (T R )) R ) and the composition of this isomorphism with the canonical isomorphism from V to C is equal to the canonical isomorphism from V 2 to C multiplied by det 2 = det(i H + (I H + (T R )) (R R 2 )). By Proposition.2.6, (8) and the discussion above, the isomorphisms from Det(I H + T ) to C obtained using (8) with R replaced by R and R 2 differ from each other by the factor det 2. Since det(i H + T ) exists, det 2 is very small if both R and R 2 are very close to T. Since the space of isomorphisms from Det(I H + T ) to C is complete, there is an isomorphism from Det(I H + T ) to C such that when R goes to T, the isomorphism from Det(I H + T ) to C obtained (8) is convergent to this isomorphism. This isomorphism is the canonical isomorphism from Det(I H + T ) to C. Now assume that I H + T is invertible. Then there is a complete orthonormal basis {ψ n } n Z+ of H such that T = n Z + λ n (ψ n, )ψ n, where (, ) is the inner product in H and the equality means that the right-hand side is convergent strongly to the left-hand side. We choose n R n = λ k (ψ k, )ψ k, n Z +. k= Then (T R n ) ImRn = 0. Note that when n is sufficiently large, I H + (T R n ) is invertible and we have (I H + (T R n )) = m N( ) m (T R n ) m.

16 6. DETERMINANT LINES AND DETERMINANT LINE BUNDLES Thus (I H + (T R n )) ImRn = ( ) m (T R n ) m ImRn m N = I ImRn. So for sufficiently large n, we have Since I H + T is invertible, (I H + (T R n )) R n = I ImRn R n = R n. I H + R n = I H + (I H + (T R n )) R n = (I H + (T R n )) (I H + T ) is also invertible. By Proposition.2.5, we know that the canonical isomorphism from C = Det(I H + (I H + (T R n )) R n ) to C is given by multiplying the elements of C by ( n det(i H + R n ) = ( + λ k )). On the other hand, the canonical isomorphism from C = Det(I H + T ) to C = Det(I H + (I H + (T R n )) R n ) in this case is the trivial map. So we see that the canonical isomorphism from C = Det(I H + T ) to C given by R n is also given by multiplying the elements of C by det(i H + R n ). Thus the limit of these isomorphisms is given by multiplying the elements of C by ( + λ k ) = (det(i H + T )) = det(i H + T ). k Z + Proposition.2.0. Let F : H H 2 be a Fredholm operator and G a quasi-inverse of F such that F G = I H2 + T where T is of trace class. Then DetF (DetG) such that in the case that both F and G are invertible, the canonical isomorphism from DetF = C to (DetG) = C is given by multiplying det(f G) = (det(i H2 + T )) to elements of C. k=

17 .2. DETERMINANT LINES 7 Proof. By Proposition.2.9, Det(I H2 + T ) C. By Proposition.2.4, DetF G DetF DetG. Thus DetF DetG C, or equivalently, DetF (DetG). If both F and G are invertible, I H2 + T = F G is also invertible. By Proposition.2.9, the canonical isomorphism from to C is given by multiplying DetF G = Det(I H2 + T ) = C det(f G) = (det(i H2 + T )) to the elements of C. Thus the induced canonical isomorphism from DetF = C to (DetG) = C is also given by multiplying to elements of C. det(f G) = (det(i H2 + T )) Corollary.2.. Let F : H H 2 be a Fredholm operator and T : H H 2 a trace-class operator. Then there is a canonical isomorphism from Det(F + T ) to DetF such that in the case that both F and F + T are invertible, (F + T )F has a determinant and the canonical isomorphism from Det(F + T ) = C to DetF = C is given by multiplying (det((f + T )F )) to elements of C. Moreover, if F : H H 2 is a Fredholm operator and T, T : H H 2 are traceclass operators, then the diagram Det(F + T + T ) Det(F + T ) (9) is commutative. Det(F + T ) DetF Proof. Let G be the canonical quasi-inverse of F in Remark..5. Then G is also a quasi-inverse of F + T such that (F + T )G I H2 and G(F + T ) I H are trace-class operators. By Proposition.2.0, Det(F + T ) (DetG) DetF. If both F and F + T are invertible, (F + T )F = I H + T F and the operator T F is of trace class. So (F + T )F has a determinant. In this case, the canonical quasi-inverse is G = F. Thus by Proposition.2.0, the canonical isomorphism from Det(F + T ) = C to (DetG) is given by multiplying (det((f + T )F )) to elements of C and the canonical isomorphism from (DetG) = C to DetF = C is the identity operator on C. The conclusion we need follows immediately.

18 8. DETERMINANT LINES AND DETERMINANT LINE BUNDLES To prove (9), we need only prove that the diagram (20) Det(F + T + T )G C Det(F + T )G C C Det(F + T )G C DetF G is commutative. But both paths in (20) are equal to the composition of the canonical isomorphism from Det(F + T + T )G to C and the canonical isomorphism from C to DetF G. So (20) is commutative. Proposition.2.2. Let H i, i =,..., 5, be Hilbert spaces, F : H H 4 and F 2 : H 2 H 5 Fredholm operators and T : H H 2 and T 2 : H 2 H 5 trace-class operators. If the diagrams (2) 0 H H 2 H 3 0 F F 2 I H3 0 H 4 H 5 H 3 0 and (22) 0 H H 2 H 3 0 F +T F 2 +T 2 I H3 0 H 4 H 5 H 3 0 have same exact rows and are commutative, then the diagram (23) Det(F 2 + T 2 ) Det(F + T ) DetF 2 DetF is commutative. Proof. Since (2) and (22) have exact rows and are commutative, we can embed H and H 4 into H 2 and H 5, respectively and after the

19 embedding,.3. DETERMINANT LINES OVER RIEMANN SURFACES 9 F = F 2 H, F + T = (F 2 + T 2 ) H, KerF = KerF 2, Ker(F + T ) = Ker(F 2 + T 2 ), ImF = ImF 2 H 4, Im(F + T ) = Im(F 2 + T 2 ) H 4. In particular, we can identify CokerF and Coker(F +T ) with CokerF 2 and Coker(F 2 + T 2 ), respectively. Thus after the embedding and the identification above, (24) (25) DetF = DetF 2, Det(F + T ) = Det(F 2 + T 2 ). To prove the commutativity of (23), we need only prove that the canonical isomorphisms from (25) to (24) obtained from Det(F +T ) DetF and from Det(F 2 + T 2 ) DetF 2 are the same. But from the proofs of Propositions.2.5,.2.9,.2.0 and Corollary.2., it is clear that these two canonical isomorphisms are equal..3. Determinant lines over Riemann surfaces with parametrized boundaries We construct the determinant lines over Riemann surfaces with parametrized boundaries and discuss their basic properties in this section. In this section, a Riemann surface means a one-dimensional compact connected complex manifold and a Riemann surface with boundary means a one-dimensional compact connected complex manifold with boundary. In the next section we shall also consider disjoint unions of connected Riemann surfaces. It is clear that the boundary of a one-dimensional compact connected complex manifold with boundary is a one-dimensional compact real manifold and thus is the union of its connected components, each of which is diffeomorphic to the circle. If the boundary has infinitely many connected components, any infinite sequence whose components belong to different connected components of the boundary must has a convergent subsequence since the complex manifold is compact, contradicting the fact that the boundary is a manifold. Thus there can only be finitely many connected components.

20 20. DETERMINANT LINES AND DETERMINANT LINE BUNDLES An orientation of a boundary component of a Riemann surface with boundary is said to be positive (negative) if the orientation is the same as (opposite to) the orientation induced from the orientation of the Riemann surface with boundary determined by its complex structure. A bijective map from S to a boundary component of a Riemann surface with boundary is called an analytic parametrization of the boundary component if there exists a real number r such that the map can be extended to an analytic map from when r < or from A r = {z C r z } A r = {z C z r} when r > to the Riemann surface with boundary. An analytic parametrization of a boundary component of a Riemann surface with boundary determine an orientation of the boundary component: It is the push-forward of the orientation on S induced from the orientation of A r. In the case r <, the orientation determined by the parametrization is positive and in the case r >, the orientation determined by the parametrization is negative. A Riemann surface with parametrized boundary is a Riemann surface with boundary whose boundary components are equipped with analytic parametrizations and both the sets of positively and negatively oriented boundary components are ordered. By the discussion above, we see that the boundary components of any Riemann surface with analytically parametrized boundary components have orientations determined by the parametrizations. Let Σ be a Riemann surface with parametrized boundary. We have the Cauchy-Riemann operator from the space Ω 0 (Σ) = Ω 0,0 (Σ) of smooth functions on the surface to the space Ω 0, (Σ) of (0, )-forms on the surface. The boundary of Σ can be decomposed as Σ = k i=c ɛ i i where for any i, i k, C ɛ i i is a connected component of Σ and ɛ i = ± indicates the orientation of the component. By definition, for any i, i k, C ɛ i i is parametrized analytically by a map from the circle S to C ɛ i i. Any smooth function on Cɛ i i can be decomposed as the sum of two smooth functions, one of which, as a function on S, has a Fourier expansion of the form n N a ne 2πnθi (θ is the usual parametrization of the circle by angles) and the other of which, as a function on S, has a Fourier expansion of the form n Z + a n e 2πnθi. If ɛ i = + (ɛ i = ), that is, this component is positively (negatively) oriented, we denote by Ω 0 +(C ɛ i i ) the space of all smooth functions on which, as functions on S, have Fourier expansions of the form C ɛ i i

21 .3. DETERMINANT LINES OVER RIEMANN SURFACES 2 n N a ne 2πnθi ( n Z + a n e 2πnθi ) and by Ω 0 (C ɛ i i ) the space of smooth functions on C ɛ i i which, as functions on S, have Fourier expansions of the form n Z + a n e 2πnθi ( n N a ne 2πnθi ). Thus the space Ω 0 ( Σ) of all smooth functions on Σ can be decomposed as k i=(ω 0 +(C ɛ i i ) Ω 0 (C ɛ i i )). Let Ω 0 ±( Σ) = k i=ω 0 ±ɛ i (C ɛ i i ) Ω0 ( Σ) and let pr be the composition of the restriction from Ω 0 (Σ) to Ω 0 ( Σ) and the projection from Ω 0 ( Σ) to Ω 0 +( Σ). We have an operator pr : Ω 0 (Σ) Ω 0, (Σ) Ω 0 +( Σ). We now want to show that pr can be regarded as a Fredholm operator. Since both Ω 0 (Σ) and Ω 0, (Σ) Ω 0 +( Σ) are not Hilbert spaces, we have to extend these spaces and the operator in order to discuss whether we have a Fredholm operator. We shall use results in the theory of elliptic boundary problems. Let Σ be the interior of Σ, H(s) (Σ ), s R, the Sobolev spaces on Σ defined in [?] and H (s) ( Σ), s R, the Sobolev spaces on Σ. For convenience, we shall denote H (s) (Σ ) and H (s) (Σ ) Ω 0, (Σ) simply as H (s) (Σ) and Ω 0, (s) (Σ), respectively. Let R : Ω 0 (Σ) Ω 0 ( Σ) be the operator taking a function on Ω 0 (Σ) to its restriction on Ω 0 ( Σ). Then we have Proposition.3.. The operator R can be extended uniquely to a Fredholm operator from H (s) (Σ) to (26) Ω 0, (s ) (Σ) H (s 2 )( Σ) for any s. The kernel of this Fredholm operator is in Ω 0 (Σ) and the range is the orthogonal space of a finite-dimensional subspace of Ω 0, (Σ) Ω 0 ( Σ) in (26). In particular, the kernel of this extension is equal to the kernel of R and the cokernel of this extension is canonically isomorphic to the cokernel of R. Proof. It easy to verify that the Dirichlet boundary problem f = g, R(f) = f Σ = h, g Ω 0, (Σ), h Ω 0 ( Σ) is elliptic (see, for example, [?] for the definition of elliptic boundary problem). Thus by Theorems and in [?], R can be extended to a Fredholm operator from H (s) (Σ) to (26) for any s having the properties stated in the proposition.

22 22. DETERMINANT LINES AND DETERMINANT LINE BUNDLES For any s R +, let H + (s )( Σ) and H (s )( Σ) be the completion 2 2 of Ω 0 +( Σ) and Ω 0 ( Σ), respectively, in H (s )( Σ). 2 Corollary.3.2. The operator pr can be extended uniquely to a Fredholm operator from H (s) (Σ) to (27) Ω 0, (s ) (Σ) H+ (s 2 )( Σ) for any s. The kernel of this Fredholm operator is 0 if Σ has at least one positively oriented boundary component and is the space of constant functions if Σ has no positively oriented boundary components. The range is the orthogonal space of a finite-dimensional subspace of Ω 0, (Σ) Ω 0 +( Σ) in (27). In particular, the kernel of this extension is equal to the kernel of pr and the cokernel of this extension is canonically isomorphic to the cokernel of pr. Proof. The extension of pr is defined to be the composition of the projection from H (s )( Σ) to H+ 2 (s )( Σ) and the extension of 2 R. Let f be an element of the kernel of this extension. Then f can be extended to an analytic function on a Riemann surface without boundary. Thus f must be a constant. If Σ has at least one positively oriented boundary component, this constant must be 0. If Σ has no positively oriented boundary components, any constant function is an element of the kernel. So the kernel of this extension is 0 if Σ has at least one positively oriented boundary component and is the space of constant functions if Σ has no positively oriented boundary components. By the definition, the dimension of the cokernel of this extension is less than or equal to the dimension of the cokernel of the extension of R. So this extension is a Fredholm operator. Since the range of the extension of R is the orthogonal space of a finite-dimensional subspace of Ω 0, (Σ) Ω 0 ( Σ) in (26), the range of this extension is the orthogonal space of a finite-dimensional subspace of Ω 0, (Σ) Ω 0 +( Σ) in (27). Thus the kernel of this extension is equal to the kernel of pr and the cokernel of this extension is isomorphic to the cokernel of pr. We shall use the same notation pr to denote its extension to H (s) (Σ) for any s. The determinant line Det( pr) = Det(Ker( pr)) Det(Coker( pr)) of pr is well defined. Let Hol(Σ) be the space of all analytic functions on Σ and π Σ the restriction of pr to Hol(Σ). We have:

23 .3. DETERMINANT LINES OVER RIEMANN SURFACES 23 Proposition.3.3. The operator π Σ : Hol(Σ) Ω 0 +( Σ) can be extended to a Fredholm operator from the completion Hol (s) (Σ) of Hol(Σ) in H (s) (Σ) to H + (s )( Σ) for any s. The kernel of this 2 extension is in Hol(Σ) and the range of this extension is the orthogonal complement of a finite-dimensional subspace of Ω 0 +( Σ) in H + (s )( Σ). 2 In particular, the kernel of of this extension is equal to the kernel of π Σ and the cokernel of this extension is canonically isomorphic to the cokernel of π Σ. Moreover, there is a canonical isomorphism from the determinant line Det( pr) to Detπ Σ. Proof. The commutative diagram 0 Hol(Σ) Ω 0 (Σ) π Σ pr Ω 0, (Σ) 0 I Ω 0, (Σ) 0 Ω 0 +( Σ) Ω 0, (Σ) Ω 0 +( Σ) Ω 0, (Σ) 0 (28) induces the commutative diagram 0 Hol (s) (Σ) H (s) (Σ) π Σ pr 0 H + (s 2 (29) Ω 0, (s )(Σ) 0 I 0, Ω (s ) (Σ) )( Σ) Ω0, (s ) (Σ) H+ (s )( Σ) Ω0, (s )(Σ) 0 2 where we use the same notations to denote operators and their extensions. Since the rows of the diagram (28) are exact, the rows of (29) is also exact. By the diagram (29), the kernel of π Σ is in the kernel of pr and thus is in Hol(Σ) and is finite-dimensional. Let V be the image of the finite-dimensional subspace orthogonal to the range of pr under the projection from (27) to H + (s 2 )( Σ). Then V is also finite-dimensional. By the diagram (29), we see that the range of π Σ is equal to the range of the composition of the projection above and pr. Thus the range of π Σ is the orthogonal space of V in H + (s 2 )( Σ). So the cokernel of π Σ is finite-dimensional and π Σ is a Fredholm operator. Since the finite-dimensional subspace orthogonal to the range of pr is in Ω 0, (Σ)) Ω 0 +( Σ), V is in Ω 0 +( Σ). Thus the cokernel of π Σ is canonically isomorphic to the cokernel of its extension. Finally

24 24. DETERMINANT LINES AND DETERMINANT LINE BUNDLES by (29) and Proposition.2.3, there is a canonical isomorphism from the determinant line Det( pr) to Detπ Σ. We shall call Detπ Σ the determinant line over Σ and denote it Det Σ. It is obvious that Σ Det Σ gives a functor from the category of Riemann surfaces with parametrized boundaries to the category of one-dimensional vector spaces with isomorphisms as morphisms..4. Canonical isomorphisms associated to sewing and determinant line bundles over moduli spaces Let Σ and Σ 2 be two Riemann surfaces with parametrized boundaries. Let S be a subset of the set of the boundary components of Σ and S 2 a subset of the set of the boundary components of Σ 2. If the number of negatively (positively) oriented boundary components in S is equal to the number of positively (negatively) oriented boundary components in S 2, we say that S and S 2 are compatible for sewing. In this case, we define Σ #Σ 2 to be the Riemann surface with parametrized boundary obtained by identifying the negatively and positively oriented boundary components in S with the positively and negatively oriented boundary components of Σ 2, respectively, using the given orderings of the boundary components to determine how to match the components to be sewn and using the given parametrizations of the boundary components. Note that the operation # depends on S and S 2. For simplicity, we shall not include this dependence explicitly in the notation #. The spaces Ω 0 ±(S ), Ω 0 ±(S 2 ), H ± (s 2 )(S ) and H ± (s 2 )(S 2) for s R + are defined in the obvious way. One of the most important properties of the determinant lines defined in the preceding section is that for any Σ and Σ 2 as above, there is a canonical isomorphism l Σ,Σ 2 : Det Σ Det Σ2 Det Σ #Σ 2 satisfying a natural associativity property. The first half of this section is devoted to the construction of l Σ,Σ 2 and to the proof of the associativity property. Let S be the curve on Σ #Σ 2 corresponding to S or S 2 and Σ Σ 2 the disjoint union of Σ and Σ 2. Then Σ Σ 2 is a compact Riemann surface with parametrized boundary and with two ordered components. We define Σ,Σ 2 : Hol(Σ Σ 2 ) Ω 0 (S) by Σ,Σ 2 (f) = f S f S2

25 and by.4. CANONICAL ISOMORPHISMS AND DETERMINANT LINE BUNDLES 25 π Σ,Σ 2 : Hol(Σ Σ 2 ) Ω 0 +( (Σ #Σ 2 )) Ω 0 (S) π Σ,Σ 2 (f) = ((f (Σ #Σ 2 )) +, Σ,Σ 2 (f)) where (f (Σ #Σ 2 )) + denotes the projection of f (Σ #Σ 2 ) Ω 0 ( (Σ #Σ 2 )) to Ω 0 +( (Σ #Σ 2 )). Then we have a commutative diagram Σ,Σ 0 Hol(Σ #Σ 2 ) Hol(Σ Σ 2 ) 2 Ω 0 (S) 0 π Σ #Σ 2 π Σ,Σ 2 I Ω 0 (S) 0 Ω 0 +( (Σ #Σ 2 )) Ω 0 +( (Σ #Σ 2 )) Ω 0 (S) Ω 0 (S) 0 (30) Lemma.4.. The rows of the diagram (30) are exact. Proof. The second row of (30) is exact by definition. We prove that the first row is also exact. The embedding of Hol(Σ #Σ 2 ) in Hol(Σ Σ 2 ) is by definition injective. The subspace Hol(Σ #Σ 2 ) of Hol(Σ Σ 2 ) is by definition in the kernel of Σ,Σ 2. Let f be a function in the kernel of Σ,Σ 2. Then f is continuous on Σ #Σ 2 and is analytic on (Σ #Σ 2 ) \ S. Using the Cauchy theorem, we see that f is also analytic on S. Thus f is in Hol(Σ #Σ 2 ). This proves that the first row is exact at Hol(Σ Σ 2 ). To show that the first row is exact, we need to show that Σ,Σ 2 is surjective. Let Σ #Σ 2 be the compact Riemann surface without boundary obtained from Σ #Σ 2 by sewing disks along its boundary components using their analytic parametrizations. Since when the genus of Σ #Σ 2, or equivalently, the genus of Σ #Σ 2 is zero, the proof is simpler. In this work, we need only the genus-zero case and we only prove the surjectivity of Σ,Σ 2 in this case. In this genus-zero case, by the uniformization theorem (Theorem??), we can always find a conformal equivalence from Σ #Σ 2 to Ĉ such that S is mapped to a smooth simple Jordan curve and such that the interiors of Σ and Σ 2 are mapped to the exterior and to the interior of the curve, respectively. Thus we need only prove the surjectivity of Σ,Σ 2 when S is a smooth simple Jordan curve on the complex plane and Σ and Σ 2 are the unions of S with its exterior and with its interior, respectively. In this case, for any f Ω 0 (S), we define functions f(ξ) g (z) = ξ z dξ S

26 26. DETERMINANT LINES AND DETERMINANT LINE BUNDLES when z is in the exterior of S and g 2 (z) = S f(ξ) ξ z dξ when z is in the interior of S. Obviously g and g 2 are analytic in the exterior and interior of S, respectively. Let z 0 S and z in the interior of S. Note that if we define the value of f(ξ) f(z 0) ξ z 0 at ξ = z 0 to be f (z 0 ), f(ξ) f(z 0) ξ z 0 becomes a smooth function on S. We need the following result: Lemma.4.2. When z approaches z 0 from the exterior of S along the normal line of S, g (z) converges to f(ξ) f(z 0 ) dξ ξ z 0 S uniformly with respect to z 0. When z approaches z 0 from the interior of S along the normal line of S, g 2 (z) converges to f(ξ) f(z 0 ) f(z 0 ) + dξ ξ z 0 uniformly with respect to z 0. S The proof of this lemma is elementary and can be found in Chapter 4, Section 3 of [?]. Since S is smooth, f is smooth and the convergence given in Lemma.4.2 is uniform with respect to z 0 S, we conclude that lim z z0 g (z) and lim z z0 g 2 (z) exist and are equal to f(ξ) f(z 0 ) dξ ξ z 0 and S f(z 0 ) + S f(ξ) f(z 0 ) dξ, ξ z 0 respectively. Using these limits, we define the boundary values of g and g 2. The functions g and g 2 are analytic in the exterior and interior of S, respectively, and are continuous on the unions of S with its exterior and with its interior, respectively, and satisfy the relation for z 0 S. Since d f(ξ) f(z 0 ) dξ = dz 0 ξ z 0 S g 2 (z 0 ) g (z 0 ) = f(z 0 ) S ( f(ξ) f(z0 ) f ) (z 0 ) dξ (ξ z 0 ) 2 ξ z 0

27 .4. CANONICAL ISOMORPHISMS AND DETERMINANT LINE BUNDLES 27 and f(ξ) f(z 0 ) (ξ z 0 ) 2 f (z 0 ) ξ z 0 is smooth if we define its value at ξ = z 0 to be f (z 0 ), we see that g and g 2 are differentiable on S. Similarly, we can show that the higher derivatives of g and g 2 on S exist and thus g S and g 2 S are in Ω 0 (S), proving the surjectivity of Σ,Σ 2 in this case. We extend the diagram (30) to the following commutative diagram: 0 Hol (s) (Σ #Σ 2 ) Hol (s) (Σ Σ 2 ) π Σ #Σ 2 π Σ,Σ 2 Σ,Σ 2 0 H + (s )( (Σ #Σ 2 )) H + (s )( (Σ #Σ 2 )) H (s )(S) (3) Σ,Σ 2 H (s )(S) 0 2 I H(s 2 ) (S) H (s )(S) 0. 2 Since (30) has exact rows, (3) also has exact rows. By Proposition.2.3, there exists a canonical isomorphism from Det π Σ,Σ 2 to Det Σ #Σ 2. (In fact, by Remark.2.8, to conclude Det π Σ,Σ 2 Det Σ #Σ 2, (30) is enough.) On the other hand, we note that the construction in the preceding section also works for Riemann surfaces that are not connected, and from the construction we see that Det Σ Det Σ2 = Detπ Σ Detπ Σ2 is canonically isomorphic to Det Σ Σ 2 = Detπ Σ Σ 2 where Σ Σ 2 is the disjoint union of Σ and Σ 2 and π Σ Σ 2 is the corresponding Fredholm operator from Hol (s) (Σ Σ 2 ) to H + (s )( (Σ Σ 2 )). Since 2 H + (s 2 )( (Σ Σ 2 )) = H + (s 2 )( (Σ #Σ 2 )) H (s 2 )(S ) H + (s 2 )(S 2) and H (s 2 )(S ) H + (s 2 )(S 2) can be identified with H (s 2 )(S), we see that π Σ Σ 2 can also be viewed as a Fredholm operator from Hol (s) (Σ Σ 2 ) to and the difference π Σ Σ 2 π Σ,Σ 2 H + (s 2 )( (Σ #Σ 2 )) H (s 2 )(S) is given by (π Σ Σ 2 π Σ,Σ 2 )(f) = (f S ) (f S2 ) +

28 28. DETERMINANT LINES AND DETERMINANT LINE BUNDLES for f Hol (s) (Σ Σ 2 ). Lemma.4.3. The operator π Σ Σ 2 π Σ,Σ 2 is a trace-class operator. Proof. For simplicity, we only prove the case that S contains only one negatively oriented boundary components and contains no positively oriented boundary components. Thus S 2 contains only one positively oriented boundary component and contains no negatively oriented boundary components. We need only show that the operators given by f (f S ) and f (f S2 ) + are of trace class. We show below that the first operator is of trace class. The proof that the second operator is of trace class is similar. Since S is analytically parametrized, the parametrization of S as a map from the circle S = {z C z = } to S can be extended to an analytic injective map from a closed annulus A r to Σ for some r >. Let A be the image of A r under this analytic map. Consider the restriction map P A : Hol (s) (Σ Σ 2 ) Hol (s) (A) f f A. Let s Σ Σ 2 and s A be the norms in Hol (s)(σ Σ 2 ) and Hol (s) (A), respectively. Then by definition f A s A f s Σ Σ 2 for any f Hol (s) (Σ Σ 2 ). Thus the restriction map P A is continuous and we need only show that the map f (f S ) from Hol (s) (A) to H (s )(S ) is of trace class. Since A is analytically isomorphic to the 2 annulus A r and S corresponds to the unit circle S = {z C z = }, we need only show that the map P : Hol (s) (A r ) H (s 2 )(S ) f (f S ) is of trace class. For simplicity, we only prove this fact in the case s =. The other cases can be discussed similarly. In fact, since determinant lines are independent of s, this case is enough for our purpose. Let f be an analytic function on A r. Then f(z) = n Z a nz n when z r. Thus for any f Hol () (A r ), we have f = n Z a nz n in Hol () (A r ).