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MATH 632, Assignment 12 on -Class, Compact, and Fredholm Operators, due Dec. 8, 2014

Prof. Jonathan Rosenberg

Solutions

1. The unit ball of B(H) (for H a ) is weak-* compact when we view B(H) as the dual of L1(H). Show that the weak-* topology coincides on this unit ball with the . (The two topologies are actually different, though the difference only shows up when you look at -unbounded sets. But you don’t need to prove this.)

Solution. Let U1 be the unit ball with the weak operator topology and let U2 be the unit ball with the weak-* topology. We proved in class that U1 is compact, and U2 is compact by Banach-Alaoglu. The identity map ϕ: U2 → U1 is a bijection (since U1 and U2 are the same set, just with different topologies). Now the topology of U1 is defined by the T 7→ |hT ξ, ηi|, ξ, η ∈ H, kξk = 1 kηk = 1, and the topology of U2 is defined by the seminorms T 7→ | Tr(TS)|, S ∈ L (H) (the trace- class operators). The second collection of seminorms contains the first class, since if kξk = kηk = 1 and if S is the rank-one partial isometry sending η to ξ and killing η⊥, then

hT ξ, ηi = hT Sη, ηi = Tr(TS).

Thus ϕ is continuous, and since U2 and U1 are both compact and Hausdorff, it is a homeomorphism.

2 0 2 2. Let H = L (T), H = H (T), the subspace which is the closed linear span of the functions z 7→ zn, n ≥ 0. Let P be orthogonal projection of H onto H0 (so that P kills z 7→ zn, n < 0). For f ∈ C(T), let Mf be the operator on H given by pointwise multiplication by f. Show that Mf commutes with P modulo compact operators. Deduce that the Toeplitz operator Tf =def PMf P is Fredholm (as an operator on H0) if and only if f is everywhere nonzero. n Solution. Let en : z 7→ z , so that {en}n∈Z is an of H and {en}n∈N is an orthonor- 0 mal basis of H . The linear combinations of the en (the trigonometric polynomials) are a subalgebra of C(T) (since en · em = en+m) containing the constants, separating points, and closed under con- jugation (since en = e−n). So by the Stone-Weierstrass Theorem, they are dense. Thus to show Mf

1 commutes with P modulo compact operators for f ∈ C(T), it suffices to prove this for f = en. Now ( ( en+m, n + m ≥ 0, en+m, m ≥ 0, PMe (em) = while Me P (em) = n 0, n + m < 0, n 0, m < 0.

Thus PMen and Men P agree on all but at most |n| of the em, and PMen −Men P is finite-rank, hence compact, for any n.

Now suppose f ∈ C(T) is everywhere nonzero. Since Mf and Mf −1 commute with P modulo compacts, Tf and Tf −1 are inverses of each other modulo compacts, and Tf is Fredholm. 0 If, on the other hand, Tf is Fredholm, that means there is a finite-dimensional subspace V of H such ⊥ that Tf restricted to V is bounded below. The same then holds for I − P + Tf on the of V in H. But I − P + Tf agrees modulo compacts with Mf , so Mf is also bounded below on the orthogonal complement of a finite-dimensional subspace of H. This is impossible if f has a zero at some point z0 ∈ T, because then for any ε > 0, kMf |W k < ε, where W is the 2 infinite-dimensional space L (I), I some sufficiently small interval about z0. 3. [Gohberg-Krein] Assuming the result of #2 and f being invertible, show that the Fredholm index of Tf is the negative of the winding number of f (as a map T → C r {0}). (Hint: Prove this by explicit calculation when f(z) = zn. Reduce the general case to this via a homotopy.)

Solution. By the calculation above in #2, if n ≥ 0, Ten has vanishing kernel and has cokernel of

n, while if n < 0, Ten has kernel of dimension |n| and vanishing cokernel. Since the

winding number w(en) of en about 0 is n, we see that ind Ten = −w(en) in all cases. Now suppose we have f ∈ C(T) which is never zero. There is some en in the same homotopy class of maps 0 T → C r {0}. Choose a homotopy f : T × [0, 1] → C r {0} with f0 = f and f1 = en. For t and t T T sufficiently close together, ft0 is a small perturbation of ft and so they have the same index. Thus by compactness of [0, 1], we can get from f0 = f to f1 = en in finitely many small perturbations and

ind Tf = ind Ten = −n = −w(f).

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