The Fredholm determinant Jordan Bell [email protected] Department of Mathematics, University of Toronto May 15, 2014 1 Introduction By N we mean the set of positive integers. In this note we write inner products as conjugate linear in the first variable, following the notation of Reed and Simon. The purpose of this note is to make sense of det(I + A) for bounded trace class operators on a Hilbert space. This definition is consistent with the definition of the determinant for a finite dimensional Hilbert space. 2 Singular value decomposition Let H be a Hilbert space with an inner product that is conjugate linear in the first variable. We do not presume unless we say so that H is separable.1 We denote by B(H) the set of bounded linear operators H ! H. For any A 2 B(H), A∗A is positive and one proves that it has a unique positive square root jAj 2 B(H). We call jAj the absolute value of A. We say that U 2 B(H) is a partial isometry if there is a closed subspace X of H such that the restriction of U to X is an isometry X ! U(X) and ker U = X?. One proves that for any A 2 B(H), there is a unique partial isometry U satisfying both ker U = ker A and A = UjAj, and A = UjAj is called the polar decomposition of A. Some useful identities that the polar decomposition satisfies are U ∗UjAj = jAj;U ∗A = jAj;UU ∗A = A: If A 2 B(H) is compact and self-adjoint, the spectral theorem tells us that there is an orthonormal set fen : n 2 Ng in H and λn 2 R, jλ1j ≥ jλ2j ≥ · · · , such that X Ax = λn hen; xi en; x 2 H: n2N 1Often one assumes separability not because statements are false for nonseparable Hilbert spaces but because it is notationally easier to talk about separable Hilbert spaces, and this indolence hides where separability matters. Moreover, whether or not H is separable, it does not take long to prove that the image of a compact linear operator is separable. 1 If A 2 B(H) is compact, then jAj is compact (the compact operators are an ideal and jAj = U ∗A), so the spectral theorem tells us that there is an orthonormal set fen : n 2 Ng in H and λn 2 R, λ1 ≥ λ2 ≥ · · · ≥ 0, such that X jAjx = λn hen; xi en; x 2 H: n2N Write σ(A) = λn(jAj). σn(A) are called the singular values of A. Using the spectral theorem and the polar decomposition, one proves that for any compact A 2 B(H), there are orthonormal sets fen : n 2 Ng and ffn : n 2 Ng in H such that X Ax = σn(A) hfn; xi en; x 2 H: n2N This is called the singular value decomposition of A. 3 Trace class operators We denote by B1(H) the set of those A 2 B(H) that are compact and such that X kAk1 = σn(A) < 1: n2N Elements of B1(H) are called trace class operators. It can be proved that B1(H) with the norm k·k1 is a Banach space. Let E be an orthonormal basis for H. We define tr : B1(H) ! C by X tr A = he; Aei ;A 2 B1(H): e2E One proves that the value of this sum is the same for any orthonormal basis of H, and that tr is a bounded linear operator. ∗ ∗ ∗ It is a fact that if A 2 B1(H) then A 2 B1(H), tr A = tr A, and kA k1 = kAk1. Another fact is that if A 2 B1(H) and B 2 B(H), then AB; BA 2 B1(H), tr (AB) = tr (BA), jtr (BA)j ≤ kBk kAk1, and kABk1 ≤ kAk1 kBk, kBAk1 ≤ kBk kAk1. Finally, if A 2 B1(H) then kAk ≤ kAk1. 4 Logarithms We say that A 2 B(H) is invertible if A−1 2 B(H). If A 2 B(H) and kAk < 1, −1 P1 n one checks that I −A is invertible: (I −A) = n=0 A . Furthermore, suppose that A 2 B1(H) and kAk < 1. Then, 1 X An log(I − A) = − ; n n=1 2 hence 1 X tr (An) tr log(I − A) = − ; n n=1 and 1 Y tr (An) exp tr log(I − A) = exp − : (1) n n=1 Because det exp B = exp tr B for any B 2 B(RN ), if A 2 B(RN ) satisfies kAk < 1 then det(I −A) = exp tr log(I −A). The above expression makes sense for any A 2 B1(H) with kAk < 1, so it makes sense to define det(I − A) = exp tr log(I − A) in this case, for which we have the explicit formula (1). We have not shown that det has the properties we might expect it to have, but at least in the case H = RN it is equal to the ordinary determinant, which is the least we could demand of a function which we denote by det. 5 Tensor products Suppose that H is a Hilbert space over K with inner product h·; ·i that is conjugate linear in the first variable. For each n 2 N and x1; : : : ; xn 2 H, let n x1 ⊗ · · · ⊗ xn be the multilinear function on H defined by n Y n (x1 ⊗ · · · ⊗ xn)(y1; : : : ; yn) = hxk; yki ; (y1; : : : ; yn) 2 H : k=1 The set of all finite linear combinations of such multilinear functions is a vector space for which there is a unique inner product that satisfies n Y hx1 ⊗ · · · ⊗ xn; y1 ⊗ · · · ⊗ yni = hxk; yki ; x1; : : : ; xn; y1; : : : ; yn 2 H: k=1 We denote by Nn H the completion of this vector space using this inner prod- uct.2 Thus, Nn H is a Hilbert space. We call this a tensor product, but it does not have the universal property of tensor products, so is not a categorical tensor product.3 n If A 2 B(H), one proves that there is a unique T 2 B(N H) that satisfies Nn T (x1 ⊗ · · · ⊗ xn) = Ax1 ⊗ · · · ⊗ Axn; x1 ⊗ · · · ⊗ xn 2 H; Nn Nn and we write T = A. For A; B 2 B(H) and x1 ⊗ · · · ⊗ xn 2 H, it is straightforward to check that Nn Nn Nn ( (AB))(x1 ⊗ · · · ⊗ xn) = ( A)( B)(x1 ⊗ · · · ⊗ xn); from which it follows that Nn(AB) = Nn A Nn B. 2See Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, volume I: Functional Analysis, revised and enlarged ed., p. 50. 3Paul Garrett, http://www.math.umn.edu/~garrett/m/v/nonexistence_tensors.pdf 3 6 Exterior powers Let Sn denote the group of permutations on n symbols and let sgn (π) denote the Nn sign of the permutation π. For x1; : : : ; xn 2 H, we define x1 ^ · · · ^ xn 2 H by −1=2 X x1 ^ · · · ^ xn = (n!) sgn (π)xπ(1) ⊗ · · · ⊗ xπ(n): π2Sn We define Vn H to be the closure in Nn H of the set of all finite linear combi- Nn 4 Vn nations of elements of H of the form x1 ^ · · · ^ xn. Thus, H is a Hilbert space. Vn 5 For x1 ^ · · · ^ xn; y1 ^ · · · ^ yn 2 H, one proves that hx1 ^ · · · ^ xn; y1 ^ · · · ^ yni = det(hxi; yji); where X det(ai;j) = sgn (π)a1,π(1) ··· an,π(n): π2Sn In particular, if fx1; : : : ; xng is an orthonormal set in H, then kx1 ^ · · · ^ xnk = 1. Vn For A 2 B(H) and x1 ^ · · · ^ xn 2 H, it is apparent that Nn ( A)x1 ^ · · · ^ xn = Ax1 ^ · · · ^ Axn: It follows that Nn A sends an element of Vn H to an element Vn H, and hence n n n that the restriction of N A to V H belongs to B(V H). We denote this restriction by Vn A. Because Nn(AB) = Nn A Nn B, we also have Vn(AB) = Vn A Vn B. 7 Finite dimensional Hilbert spaces Suppose that H is an n-dimensional Hilbert space and that A 2 B(H). If H has dimension n and fe1; : : : ; eng is an orthonormal basis for H, one proves that fei1 ^ · · · ^ eik : 1 ≤ i1 < ··· < ik ≤ ng Vk Vk n is an orthonormal basis for H, and hence that H has dimension k . So n n n V H has dimension 1, and as V A 2 B(V H), there is some scalar α such that (Vn A)v = αv for all v 2 Vn H. (A linear map from a one dimensional vector space to itself is multiplication by a scalar.) On the one hand, Vn he1 ^ · · · ^ en; ( A)(e1 ^ · · · ^ en)i = he1 ^ · · · ^ en; αe1 ^ · · · ^ eni = α he1 ^ · · · ^ en; e1 ^ · · · ^ eni = α: 4cf. Paul Garrett, http://www.math.umn.edu/~garrett/m/algebra/notes/28.pdf 5Michael Reed and Barry Simon, Methods of Modern Mathematical Physics, volume IV: Analysis of Operators, p. 321. 4 On the other hand, Vn he1 ^ · · · ^ en; ( A)(e1 ^ · · · ^ en)i = he1 ^ · · · ^ en; Ae1 ^ · · · ^ Aeni = det A: Thus, Vn A is the map v 7! det(A)v.6 Still taking H to be n-dimensional, take A 2 B(H), and let e1; : : : ; en be an orthonormal basis for H such that Vk = spanfe1; : : : ; ekg and such that A(Vk) = Vk for k = 1; : : : ; n; such a basis is obtained using the Schur decomposition of A. The equality Vk X D Vk E tr ( A) = ei1 ^ · · · ^ eik ; ( A)(ei1 ^ · · · ^ eik ) 1≤i1<···<ik≤n X = hei1 ^ · · · ^ eik ; Aei1 ^ · · · ^ Aeik i 1≤i1<···ik≤n X = det( eii ; Aeij ) 1≤i1<···<ik≤n X = λi1 ··· λik 1≤i1<···<ik≤n and the equality det(I + A) = he1 ^ · · · ^ en; (I + A)e1 ^ · · · ^ (I + A)eni = det(hei; (I + A)eji) n Y = (1 + λj) j=1 together give n X j det(I + A) = tr (V A): j=0 If H is infinite dimensional we define det(I +A) for A 2 B1(H) following the above formula.