TOEPLITZ OPERATORS. Toeplitz studied infinite matrices with NWSE diagonals constant. f e C :


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1 TOEPLITZ OPERATORS EFTON PARK 1. Introduction to Toeplitz Operators Otto Toeplitz lived from in Goettingen, and it was pretty rough there, so he eventually went to Palestine and eventually contracted tuberculosis and died. He developed this theory when common operator theory notions were very new. Toeplitz studied infinite matrices with NWSE diagonals constant. a 0 a 1 a 2 a 3... a 1 a 0 a 1 a 2... a 2 a 1 a 0 a Modern formulation: Let S 1 be the unit circle in C. L ( 2 S 1) { = f : S 1 1 C : 0 ( ) } f e iθ 2 dθ < The set { e inθ : n Z } is an orthonormal basis with respect to the L 2 inner product. Given L (S 1 ), define M : L 2 (S 1 ) : L 2 (S 1 ) by M f = f. Easy results: (1) M = M ; (2) M +ψ = M + M ψ ; (3) M M ψ = M ψ ; (4) M is invertible iff L (S 1 ) is invertible. Hardy space: H 2 = H 2 ( S 1) = Hilbert subspace of L 2 (S 1 ) spanned by { e inθ : n 0 } = set of elements of L 2 ( S 1) that have an analytic extension to D = {z C : z < 1}. Let P : L 2 (S 1 ) H 2 be the orthogonal projection: ( ) P c n e inθ = c n e inθ. n Z For each in L (S 1 ), define the Toeplitz operator T : H 2 H 2 by the formula T = P M. Question: What is the matrix for T with respect to the orthonormal basis {e inθ : n 0}? 1 n=0
2 2 EFTON a mn = T e inθ, e imθ = P M e inθ, e imθ = M e inθ, P e imθ = M e inθ, e imθ = 1 = = (n m), ( e iθ) e inθ e imθ dθ ( e iθ) e i(n m)θ dθ so (a mn ) is the same type of matrix that Toeplitz studied. The function is called the symbol of T. Question: How do operatortheoretic properties of T relate to functiontheoretic properties of? Answer: It s complicated. Easy results: (1) T = T. (2) T +ψ = T + T ψ (3) T ψ T T ψ in general (4) T invertible implies that is invertible. In general, the converse to this last statement is false. Example 1.1. T z = T e iθ. Then T z (z n ) = z n+1 for n 0. So 1 = z 0 / range(t z ), so T z is not invertible. The operator T z is the unilateral shift operator. From now on, we only consider continuous symbols. 2. Fredholm operators and the index Let H be an infinite dimensional complex Hilbert space. Let S : H H be a linear map, and define { } Sv S = sup v : v 0. Define B (H) = {S : H H linear such that S < }. It is easy to show that B (H) is an algebra under addition and composition. Let F be the set of finite rank operators on H; i.e. the set of bounded operators on H that have finitedimensional range. The set F is an ideal in B (H). Let K = F. This is the ideal of compact operators. Alternate definition: K is compact iff the image of the unit ball in H is compact (in the norm topology).
3 TOEPLITZ OPERATORS 3 Remark: if H is separable, then K is the only topologically closed nontrivial proper ideal in B (H). The quotient B(H)/K is called the Calkin algebra, and the obvious function π : B (H) B (H) /K is a quotient map in both the algebraic and topological senses. If S B (H) is invertible modulo K (H) (i.e., π (S) is invertible in B (H) /K ), we say S is a Fredholm operator. Theorem 2.1. (Atkinson s Theorem) S is Fredholm iff (1) ker S is finite dimensional (2) ker S is finite dimensional (3) ran S is (topologically) closed. If S is Fredholm, we define its index to be index S = dim ker S dim ker S. Properties of index: (1) If {S t } is a path of Fredholm operators, then index S 0 = index S 1. [Note dim ker S t and dim ker St may change as a function of t.] In fact, the converse is also true. (2) index (S + K) = index S for all K compact. (3) index (RS) = index R + index S. Questions: Which Toeplitz operators are Fredholm? If T is Fredholm, can we compute its index from its symbol? Proposition 2.2. If and ψ are in C(S 1 ), then T T ψ T ψ is compact. Idea of proof: For all integers m and n, the operator T z nt z m T z n zm is a finite rank operator. Example: Look at T z 2T z 3 T z 2 z 3. For k 0, and Thus T z 2 z 3(zk ) = T z 1(z k ) = (T z 2T z 3)(z k ) = (T z 2T z 3 T z 2 z 3)(zk ) = whence T z 2T z 3 T z 2 z 3 has rank 2. { 0 k = 0 z k 1 k 1 { { 0 k < 3 T z 2(z k 3 ) k 3 = 0 k < 3 z k 1 k 3. { 0 k = 0, 3, 4, 5,..., z k 1 k = 1, 2 Back to idea of proof: If and ψ are Laurent polynomials, then T T ψ T ψ is finite rank, and an approximation argument then shows that T T ψ T ψ is compact for general continuous and ψ. The Toeplitz algebra is T = {T + K : C(S 1 ), K K}.
4 4 EFTON Remark: T is the universal C algebra generated by a non unitary isometry (Coburn s Theorem). Theorem 2.3. There exists a short exact sequence 0 K T C(S 1 ) 0, where the map σ : T C(S 1 ) is defined by σ(t + K) =. Remark: We also have a linear splitting ξ : C(S 1 ) T given by ξ() = T, but ξ is not multiplicative. Corollary 2.4. An element T of the Toeplitz algebra is Fredholm if and only if its symbol σ(t ) is invertible; i.e., nowhere vanishing on the circle. Theorem 2.5. If T in T is Fredholm, then Index T = (winding number of σ(t )) = 1 0 dσ(t ) σ(t ). Proof: Because of Property (2) of the index, we need only prove the result for honest Toeplitz operators T with nowhere vanishing on the circle. Next, every nonvanishing function on S 1 can be homotoped to z n for some n, so by Property (1) of the index and homotopy invariance of winding number, it is enough to prove the theorem for Toeplitz operators T z n. And because of Property (3) of the index (along with the obvious property Index(S ) = Index S for all Fredholm operators S) and the algebraic properties of winding number, it suffices to establish the result for T z. As we saw earlier, T z is injective, so dim ker T z = 0. It is easy to show that the kernel of T z = T z 1 is spanned by the constant function 1, and hence dim ker T z = 1. Therefore Index T z = 0 1 = 1 = (winding number of z). If S in B(H) is invertible, then S is also invertible, whence S is a Fredholm operator of index 0. The converse of this result is not generally true, but surprisingly it is for Toeplitz operators (but not for arbitrary elements of T ). Lemma 2.6 (F. and M. Riesz). If f H 2 is not the zero function, then f in nonzero almost everywhere. Proposition 2.7 (Coburn). Suppose C(S 1 ) is nowhere vanishing. Then either ker T = {0} or ker T = {0}. Proof: Suppose there exist f and g are nonzero functions (a.e.) such that T f = 0 and T g = 0. Then P (f) = 0 and P (g) = 0, whence f and g are in H 2. Therefore fg and fg = fg are in H 1, which implies that fg equals a real constant almost everywhere. But, note that (surprisingly!) 1 fg = so fg is zero a.e., a contradiction. ( 1 ) ( 1 f ) g = 0,
5 TOEPLITZ OPERATORS 5 Corollary 2.8. T is invertible if and only if is nowhere vanishing and Index T = 0. Warning: If T is invertible, it is not generally true that (T ) 1 = T Generalizations Easy generalization: Suppose Φ is in M(n, C(S 1 )). Then we have a matrix multiplication operator and Toeplitz operator M Φ : ( L 2 (S 1 ) ) n ( L 2 (S 1 ) ) n, MΦ (F ) = ΦF T Φ : ( H 2) n ( H 2 ) n, TΦ (F ) = P M Φ, where P : (L 2 (S 1 )) n (H 2 ) n is the obvious map. There exists a short exact sequence or 0 K M(n, C) T M(n, C) C(S 1 ) M(n, C) 0, 0 M(n, K) M(n, T ) M(n, C(S 1 )) 0. Theorem 3.1. T Φ is Fredholm if and only if Φ(z) is invertible for every z in S 1, and in this case, the index of T Φ equals the negative of the winding number of det Φ. Remark: In this generality, it is not true that Index T Φ = 0 implies that T Φ is invertible. Back to scalar (nonmatrix) case for a minute. Consider D = 1 d as an (unbounded) i dθ differential operator on L 2 (S 1 ). The functions e inθ for a complete set of eigenvectors for D: D(e inθ ) = ne inθ. Note that P is the projection onto the span of the eigenspaces associated to nonnegative eigenvalues of D; in other words, P is the positive spectral projection of D. Crazy generalization: Let X be a topological space. A complex vector space is a topological space E and a continuous surjection p : E X such that for each x in X there is a neighborhood U such that p 1 (U) = U C n for some natural number n. The set E x = p 1 (x) is called the fiber of E over x. A section of E is a continuous function s : X E with the property that s(x) E x for each x in X, or, in other words, p(s(x)) = x for all x in X. We call E a Hermitian vector bundle if each E x has a complex inner product on it and these inner products vary continuously. Now suppose X is a smooth manifold and that E is a smooth Hermitian vector bundle over X. Let D be a selfadjoint elliptic differential operator acting on sections of a smooth Hermitian complex vector bundle E over a smooth manifold X; elliptic roughly means that D differentiates in all directions.
6 6 EFTON Given in C(X), we have a multiplication operator M : L 2 (X, E) L 2 (X, E) and a Toeplitz operator T = P M, where P is the positive spectral projection of D. More generally, we can form Toeplitz operators with matrixvalued symbols. Theorem 3.2 (BaumDouglas). [Odd AtiyahSinger Index Theorem] T Φ is Fredholm if and only if Φ is in GL(n, C(M)), and in this case there is a formula for the index involving F, topological invariants of M and E, and the principal symbol of D. Remark 1: In general we only get nonzero indices here if M is odddimensional. Remark 2: The formula mentioned in the theorem can be expressed as an integral over M, with one factor in the integrand being the Chern character of Φ: ch(φ) = ( 1) k k! (2k + 1)! tr ( Φ 1 dφ ) 2k+1 n=0 Note that the integrand in the winding number formula is one term in this sum. 4. Traces, Determinants, and Toeplitz Operators Given a finite matrix with complex entries, we can compute its trace and determinant in terms of those entries, and we know that these numbers are equal to the sum and the product of the eigenvalues of that matrix. We can try to do the same thing with operators on a separable infinitedimensional Hilbert space H. Take S in B(H), let {e n } be an orthonormal basis for a Hilbert space and try to compute Sen, e n. Obvious problem: this sum may not converge. Notsoobvious problem: even if this sum converges, it might converge to something different if we change our orthonormal basis or it might not converge at all! Instead look at S Se n, e n. This (possibly infinite) sum does not depend on the choice of orthonormal basis. We say S is trace class if the sum in finite. In this case, Se n, e n converges as well and also does not depend on the choice of orthonormal basis {e n }. We can therefore define the trace of S in this case to be tr S = Se n, e n. The set of trace class operators is denoted L 1 = L 1 (H).
7 Properties of trace class operators: (1) F L 1 K; (2) L 1 is a (nonclosed) ideal in B(H); (3) tr S = sum of eigenvalues of S; TOEPLITZ OPERATORS 7 (4) If K and L are trace class, then K + L is trace class and tr(k + L) = tr K + tr L; (5) If K is trace class and S is any (bounded) operator, then tr(ks) = tr(sk). Proposition 4.1. If, ψ are smooth functions on S 1, then T T ψ T ψ T is trace class. Furthermore, the trace of this commutator is invariant under trace class perturbations: if K and L are trace class, then (T + K)(T ψ + L) (T ψ + L)(T + K) is trace class and tr ((T + K)(T ψ + L) (T ψ + L)(T + K)) = tr(t T ψ T ψ T ). Proof: We noted earlier that T T ψ T ψ T is finite rank if and ψ are Laurent polynomials; we can then do an approximation argument to establish the first statement. To prove the second statement, multiply out (T +K)(T ψ +L) (T ψ +L)(T +K) and then use Properties 4 and 5 above to get rid of all the terms except T T ψ T ψ T. The proposition suggests that we should be able to write down a formula for tr(t T ψ T ψ T ) in terms of and ψ. Theorem 4.2 (HeltonHowe). With the hypotheses above, tr(t T ψ T ψ T ) = 1 dψ = 1 i i ψ d. Note that unlike the situation in finitedimensional vector spaces, the trace of a commutator can be nonzero! What about determinants? Let (L 1 ) + = {I + S : S L 1 }. Lemma 4.3. Every element A in (L 1 ) + can be written as exp S for some S in L 1. Definition: Given A in (L 1 ) +, its determinant is defined as follows: Write A = exp(s). Then det A = e tr S. In other symbols, det(e S ) = e tr S. This agrees with the usual definition of determinant for linear maps on finitedimensional vector spaces. The elements of (L 1 ) + are called determinant class operators. Suppose that and ψ are nonvanishing smooth functions on the circle that have winding number 1. Then T and T ψ are both invertible by our earlier results, and the multiplicative commutator T T ψ ψ is in (L 1 ) + : T T ψ ψ = I + (T T ψ T ψ T ) ψ.
8 8 EFTON Question: What is det(t T ψ ψ )? One property of the determinant is that det(aba 1 B 1 ) = 1 if A and B are invertible determinant class operators (but not in general!). This fact along with a bit of computation implies that det(st S 1 ) = det(t T ψ ψ ) for any invertible S and T in T such that σ(s) = and σ(t ) = ψ. This suggests that perhaps there is formula for det(t T ψ ψ ) in terms of the symbols and ψ. Because and ψ are nowhere zero and have winding number zero, log and log ψ are defined. Write these in terms of their Fourier series: log = n= a n e inθ, log ψ = n= b n e inθ Theorem 4.4 (HeltonHowe). If and ψ are nowhere zero and have winding number zero, then ( ) det(t T ψ ψ ) = exp na n b n n= Not very pretty! But there s a much prettier formula (IMHO). Suppose that and ψ are restrictions of meromorphic functions (which we also denote and ψ) defined in a neighborhood of the closed unit disk such that neither nor ψ has zeros or poles on the unit circle. For each point z in the open unit disk D, define m if has a zero of order m at z v(, z) = m if has a pole of order m at z 0 if has neither a zero nor a pole at z, and similarly define v(ψ, z). The quantity ψ(w)v(,z) lim( 1)v(,z)v(ψ,z) w z (w) v(ψ,z) is called the tame symbol of and ψ at z and is denoted (, ψ) z. Theorem 4.5 (CareyPincus). If and ψ are nowhere zero and have winding number zero, then det(t T ψ ψ ) = (, ψ) 1 z. z D Remark: The quantity (, ψ) 1 z the RHS is well defined. equals 1 for all but finitely many z in the open unit disk, so
9 TOEPLITZ OPERATORS 9 If and ψ are nowhere zero but do not have winding number zero, there is a way to generalize the CareyPincus result. Define matrices Φ = , Ψ = ψ ψ 1 Then and R = 2T T T 1T T T 1 I 0 I T 1T T I S ψ = 2T ψ T ψ T ψ 1T ψ 0 T ψ T ψ 1 I 0 I 0 I T ψ 1T ψ 0 T ψ 1 are invertible matrices with entries in T, and therefore det(r S ψ R 1 S 1 ψ ) is defined. Furthermore, its value does not depend on our choices of invertible lifts of Φ and Ψ. Theorem 4.6. det(r S ψ R 1 S 1 ψ ) = (, ψ) 1 z z D We see that the determinant in the theorem above only depends on and ψ. But in fact, this determinant really only depends on the Steinberg symbol (different meaning of the word symbol!) {, ψ} in the algebraic Ktheory group K alg 2 (C (S 1 )). Department of Mathematics, Texas Christian University, Fort Worth, Texas 76129, USA address:
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