MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� PROBLEM SET 10 (AND LAST) FOR 18.102, SPRING 2009 DUE 11AM TUESDAY 5 MAY. RICHARD MELROSE By now you should have become reasonably comfortable with a separable Hilbert space such as l2. However, it is worthwhile checking once again that it is rather large – if you like, let me try to make you uncomfortable for one last time. An important result in this direction is Kuiper’s theorem, which I will not ask you to prove1 . However, I want you to go through the closely related result sometimes known as Eilenberg’s swindle. Perhaps you will appreciate the little bit of trickery. First some preliminary results. Note that everything below is a closed curve in the x ∈ [0, 1] variable – you might want to identify this with a circle instead, I just did it the primitive way. Problem P10.1 Let H be a separable, infinite dimensional Hilbert space. Show that the direct sum of two copies of H is a Hilbert space with the norm 2(P10.1) H ⊕ H � (u1, u2) �−→ (�u1�H 2 + �u2�H 2 ) 1 either by constructing an isometric isomorphism (P10.2) T : H −→ H ⊕ H, 1-1 and onto, �u�H = �T u�H⊕H or otherwise. In any case, construct a map as in (P10.2). Problem P10.2 One can repeat the preceding construction any finite number of times. Show that it can be done ‘countably often’ in the sense that if H is a separable, infinite dimensional, Hilbert space then (P10.3) l2(H) = {u : N −→ H; �u�l2 2(H) = �ui�H 2 < ∞}i has a Hilbert space structure and construct an explicit isometric isomorphism from l2(H) to H. 1Kuiper’s theorem says that for any (norm) continuous map, say from any compact metric space, g : M −→ GL(H) with values in the invertible operators on a separable infinite-dimensional Hilbert space there exists a continuous map, an homotopy, h : M × [0, 1] −→ GL(H) such that h(m, 0) = g(m) and h(m, 1) = IdH for all m ∈ M. 12 RICHARD MELROSE Problem P10.3 Recall, or perhaps learn about, the winding number of a closed curve with values in C∗ = C\{0}. We take as given the following fact:2 If Q = [0, 1]N and f : Q −→ C∗ is continuous then for each choice of b ∈ C satisfying exp(2πib) = f(0), there exists a unique continuous function F : Q −→ C satisfying (P10.4) exp(2πiF (q)) = f(q), ∀ q ∈ Q and F (0) = b. Of course, you are free to change b to b + n for any n ∈ Z but then F changes to F + n, just shifting by the same integer. (1) Now, suppose c : [0, 1] −→ C∗ is a closed curve – meaning it is continuous and c(1) = c(0). Let C : [0, 1] −→ C be a choice of F for N = 1 and f = c. Show that the winding number of the closed curve c may be defined unambiguously as (P10.5) wn(c) = F (1) − F (0) ∈ Z. (2) Show that wn(c) is constant under homotopy. That is if ci : [0, 1] −→ C∗, i = 1, 2, are two closed curves so ci(1) = ci(0), i = 1, 2, which are homotopic through closed curves in the sense that there exists f : [0, 1]2 −→ C∗ continuous and such that f (0, x) = c1(x), f(1, x) = c2(x) for all x ∈ [0, 1] and f(y, 0) = f (y, 1) for all y ∈ [0, 1], then wn(c1) = wn(c2). (3) Consider the closed curve Ln : [0, 1] � x �−→ e2πix Idn×n of n × n matrices. Using the standard properties of the determinant, show that this curve is not homotopic to the identity through closed curves in the sense that there does not exist a continuous map G : [0, 1]2 −→ GL(n), with values in the invertible n × n matrices, such that G(0, x) = Ln(x), G(1, x) ≡ Idn×n for all x ∈ [0, 1], G(y, 0) = G(y, 1) for all y ∈ [0, 1]. Problem P10.4 Consider the closed curve corresponding to Ln above in the case of a separable but now infinite dimensional Hilbert space: (P10.6) L : [0, 1] � x �−→ e 2πix IdH ∈ GL(H) ⊂ B(H) taking values in the invertible operators on H. Show that after identifying H with H ⊕ H as above, there is a continuous map (P10.7) M : [0, 1]2 −→ GL(H ⊕ H) with values in the invertible operators and satisfying (P10.8) M(0, x) = L(x), M(1, x)(u1, u2) = (e 4πix u1, u2), M(y, 0) = M(y, 1), ∀ x, y ∈ [0, 1]. Hint: So, think of H ⊕ H as being 2-vectors (u1, u2) with entries in H. This allows one to think of ‘rotation’ between the two factors. Indeed, show that (P10.9) U (y)(u1, u2) = (cos(πy/2)u1 +sin(πy/2)u2, − sin(πy/2)u1 +cos(πy/2)u2) defines a continuous map [0, 1] � y �−→ U (y) ∈ GL(H ⊕ H) such that U(0) = Id, U(1)(u1, u2) = (u2, −u1). Now, consider the 2-parameter family of maps (P10.10) U−1(y)V2(x)U(y)V1(x) 2Of course, you are free to give a proof – it is not hard.3 PROBLEMS 10 where V1(x) and V2(x) are defined on H ⊕ H as multiplication by exp(2πix) on the first and the second component respectively, leaving the other fixed. Problem P10.5 Using a rotation similar to the one in the preceeding problem (or otherwise) show that there is a continuous map (P10.11) G : [0, 1]2 −→ GL(H ⊕ H) such that (P10.12) G(0, x)(u1, u2) = (e 2πix u1, e−2πix u2), G(1, x)(u1, u2) = (u1, u2), G(y, 0) = G(y, 1) ∀ x, y ∈ [0, 1]. Problem P10.6 Now, think about combining the various constructions above in the following way. Show that on l2(H) there is an homotopy like (P10.11), G˜: [0, 1]2 −→ GL(l2(H)), (very like in fact) such that (P10.13) G˜(0, x) {uk}∞= � exp((−1)k2πix)uk �∞ ,k=1 k=1 G˜(1, x) = Id, G˜(y, 0) = G˜(y, 1) ∀ x, y ∈ [0, 1]. Problem P10.7: Eilenberg’s swindle For an infinite dimenisonal separable Hilbert space, construct an homotopy – meaning a continuous map G : [0, 1]2 −→ GL(H) – with G(0, x) = L(x) in (P10.6) and G(1, x) = Id and of course G(y, 0) = G(y, 1) for all x, y ∈ [0, 1]. Hint: Just put things together – of course you can rescale the interval at the end to make it all happen over [0, 1]. First ‘divide H into 2 copies of itself’ and deform from L to M (1, x) in (P10.8). Now, ‘divide the second H up into l2(H)’ and apply an argument just like the preceding problem to turn the identity on this factor into alternating terms multiplying by exp(±4πix) – starting with −. Now, you are on H ⊕ l2(H), ‘renumbering’ allows you to