Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 (Analysis of Operators)

A bounded operator $T:H\to H$ is a Fredholm operator when its range is closed and both $\ker T$ and $\mathrm{ran}(T{)}^{\perp }=\ker T^{\ast }$ have finite dimension. Its Fredholm index isFor index zero, choose an isomorphism $J:\ker T\to \ker T^{\ast }$ and extend it by zero on $(\ker T{)}^{\perp }$. The restricted map $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ is a bounded isomorphism, by the bounded inverse theorem. The finite-rank operator $J$ fills exactly the missing kernel-to-cokernel block. Thus $T+J$ is invertible and $T=(T+J)-J$ is an invertible operator plus a compact operator.

Conversely suppose $T=A+K$ with $A$ invertible and $K$ compact. The Atkinson theorem proved in part (b) makes every $A+tK$, $0\le t\le 1$, Fredholm: its quotient image is always $[A]$. The block argument in part (c) proves local constancy of the Fredholm index, so its index is constant along this interval and equals $\mathrm{ind}A=0$. This invokes the actual parametrix and block proofs below, not an unproved perturbation assertion. The claimed equivalence holds.

For a Fredholm operator $T$, define $B$ to invert $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ and to vanish on the range complement. The bounded inverse theorem makes $B$ bounded, andThe errors are finite rank, so $T$ is invertible in the Calkin algebra.

Conversely suppose $BT=I-K$ and $TB=I-L$ for bounded $B$ and compact $K,L$. On $\ker T$, $K$ is the identity. An infinite-dimensional closed kernel would contain an orthonormal sequence, whose images under $K$ have no convergent subsequence, contradicting compactness. Thus the kernel is finite-dimensional.

There is a positive lower bound for $T$ on $(\ker T{)}^{\perp }$. Otherwise choose unit vectors $x_j$ there with $T{x}_j\to 0$. Since $x_j=BT{x}_j+K{x}_j$, compactness gives a norm-convergent subsequence with unit limit $x$. The limit lies in the kernel complement and has $Tx=0$, a contradiction. This lower bound proves closed range: if $T{x}_j$ converges, remove each kernel component; the remaining $x_j$ form a Cauchy sequence and converge to a preimage of the limit.

Finally adjointing $TB=I-L$ gives $B^{\ast }{T}^{\ast }=I-L^{\ast }$. The same kernel argument makes $\ker T^{\ast }$ finite-dimensional. Therefore $T$ is Fredholm. $\boxed{T\text{ is Fredholm}⟺[T]\text{ is invertible modulo compacts}}$, the Atkinson theorem. The finite-dimensional case also satisfies the parametrix formulation directly, though its quotient algebra is the zero algebra.

Fix a Fredholm operator $T$. Split its domain as $X\oplus N=(\ker T{)}^{\perp }\oplus \ker T$ and its target as $Y\oplus M=\mathrm{ran}T\oplus \ker T^{\ast }$. For a sufficiently small norm perturbation $R$, writewhere $A:X\to Y$ remains invertible by the Neumann series. Invertible triangular row and column operations reduce this matrix toThe second block maps the finite-dimensional space $N$ into $M$. Hence the perturbed operator has closed range, finite kernel and cokernel, andsince the rank of that finite block cancels from the difference. Thus the Fredholm set is open and its index is locally constant, hence norm continuous.

If $S,T$ are Fredholm, their product is Fredholm by multiplying their invertible images in the Calkin algebra. For index additivity use the following finite-dimensional exact sequence:Here $q$ is reduction modulo $\mathrm{ran}T$, the induced $S$-map sends $[x]$ to $[Sx]$, and the last map is the quotient by the larger range $\mathrm{ran}S$. These maps are well-defined. For example, $\ker q=\ker S\cap \mathrm{ran}T$ is the image of the preceding map. If $Sx\in \mathrm{ran}ST$, subtracting a suitable $Ty$ makes $x-Ty\in \ker S$, proving exactness at $\mathrm{coker}T$. The other positions follow directly from the definitions. Alternating dimensions in this exact sequence gives

On the Hardy space of the circle $H^2(\mathbb{T})\subset L^2(\mathbb{T})$, let $P_{+}$ be the projection onto nonnegative Fourier modes and $T_f=P_{+}{M}_f{|}_{H^2}$. The Toeplitz index theorem for continuous symbols states: for continuous $f$, $T_f$ is Fredholm exactly when $f$ never vanishes, and thenwith counterclockwise winding positive.

First, $T_f{T}_g-T_{fg}=-P_{+}{M}_f(I-P_{+})M_g{P}_{+}$ is compact. For Laurent-polynomial symbols the negative-mode crossing involves only finitely many modes and is finite rank. Uniform approximation and $\Vert T_f\Vert \le \Vert f{\Vert }_{\infty }$ extend this to continuous symbols. If $f$ is nonvanishing, $T_{1/f}$ is therefore a two-sided parametrix, and part (b) gives the Fredholm property.

Writing $n=\mathrm{wind}(f,0)$, a continuous argument along the parametrized circle increases by $2\pi n$. Removing this increase gives a periodic continuous logarithm $h$ of $z^{-n}f(z)$, so $f=z^n{e}^h$. The symbols $z^n{e}^{th}$ give a continuous Fredholm path from $z^n$ to $f$. Part (c) makes their indices equal. For $n\ge 0$, $T_{z^n}$ is the $n$th unilateral shift, with zero kernel and $n$-dimensional cokernel. For negative $n$ it is the corresponding backward shift, with kernel dimension $-n$ and zero cokernel. In both cases the index is $-n$.

For necessity, if $f(\zeta )=0$ use the normalized Hardy kernelsThey have norm one and converge weakly to zero, as their inner products with each Fourier basis vector tend to zero. Their squared moduli are Poisson kernels concentrating at $\zeta$, so continuity gives $\Vert f{k}_{r,\zeta }{\Vert }_2\to 0$ and hence $\Vert T_f{k}_{r,\zeta }{\Vert }_2\to 0$. A Fredholm parametrix would imply $k_{r,\zeta }=B{T}_f{k}_{r,\zeta }+K{k}_{r,\zeta }\to 0$ in norm, because a compact operator takes this weakly null bounded family to zero in norm. This contradicts unit norm, proving the converse.

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

Pass to $B={\overline{A}}^{\Vert \cdot \Vert }$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ is a star-algebra, every closed invariant subspace reduces it.

Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element $b$; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection $e\in B$ by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb{C}p$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb{C}v$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in \mathbb{C}v$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ is irreducible.

Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts $H$.

Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\Vert bv\Vert >0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.

Use ${\mathbb{T}}^n={\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$ with normalized Haar measure and Fourier basis $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$. For any real $s$, the periodic Sobolev space isWeighted Fourier coefficients identify it isometrically with ${\ell }^2$, so it is a Hilbert space and complete. This definition includes negative orders and distributions. For nonnegative integer $s$ the norm is equivalent to the sum of squared L2 norms of weak derivatives through order $s$.

Write $⟨k⟩=(1+|k{|}^2{)}^{1/2}$. There are $O(2^{jn})$ lattice points in the dyadic shell $2^j\le |m|<2^{j+1}$, while each contributes at most a constant times $2^{-2js}$. The geometric series ${\sum }_j{2}^{j(n-2s)}$ converges when $2s>n$, proving ${\sum }_m⟨m{⟩}^{-2s}<\infty$. For $|\alpha |\le k$ and $u\in H_{s+k}$, the Cauchy-Schwarz inequality yieldsThus every derivative Fourier series through order $k$ converges absolutely and uniformly. Starting with the continuous sum for $u$, uniform convergence of successive derivative series justifies differentiation, so these are its continuous classical derivatives. $\boxed{H_{s+k}({\mathbb{T}}^n)\hookrightarrow C^k({\mathbb{T}}^n),s>n/2}$, with a continuous norm bound.

For every real $s$, the elementary weight estimatefollows from $⟨x+y⟩\le \sqrt{2}⟨x⟩⟨y⟩$, using the reverse rearrangement for negative $s$. If $a$ is smooth its Fourier coefficients decay faster than every power, so $c_j=⟨j{⟩}^{|s|}|\hat{a}(j)|$ is summable. The Fourier convolution formula and Young convolution inequality giveInitially this holds for trigonometric polynomials, and density extends it to every $u\in H_s$, agreeing with distributional multiplication. Hence smooth multiplication preserves every order.

If $s>n/2$, the stronger estimate $⟨m{⟩}^s\le C_s(⟨l{⟩}^s+⟨m-l{⟩}^s)$ givesusing the summability estimate from part (b) with $k=0$. Approximation extends the product, and the continuous embedding identifies it with the pointwise product. Completeness from part (a), this estimate and the constant unit make $H_s$ a commutative Sobolev algebra for $s>n/2$. Rescaling the norm by $C_s^{\prime }$ if necessary gives an equivalent strictly submultiplicative Banach-algebra norm.

Start with $u\in H_s$ for some real $s$. The equation becomesSmooth $f$ belongs to every periodic Sobolev space, and multiplication by the smooth coefficient preserves $H_s$ by part (c), so the right side lies in $H_s$. The Fourier multiplier $(1+\Delta {)}^{-1}$ has coefficient $(1+|m{|}^2{)}^{-1}$ and maps $H_s$ isometrically to $H_{s+2}$. Thus $u\in H_{s+2}$. Repeat this argument to get $u\in H_{s+2j}$ for every nonnegative integer $j$. For any derivative order $k$, choose $j$ with $s+2j>k+n/2$ and apply part (b). Therefore $\boxed{u\in C^{\infty }({\mathbb{T}}^n)}$. This is elliptic bootstrapping directly from the Fourier multiplier, with no assumption that the solution initially has nonnegative Sobolev order.

The finite-dimensional Krein-Milman theorem says that a nonempty compact convex set $K$ is the convex hull of its extreme points. A point is extreme if a proper convex combination of two points of $K$ equals it only when both points equal it.

Prove the theorem by induction on the dimension of the affine hull. Dimension zero is immediate. In positive dimension work inside that affine hull, where $K$ has interior. Every boundary point $x$ lies on a supporting hyperplane. To see this, choose exterior points $y_j\to x$ and nearest points $p_j\in K$. Convexity and the minimum-distance condition give $(y_j-p_j)\cdot (z-p_j)\le 0$ for $z\in K$. A subsequence of the normalized nonzero vectors $y_j-p_j$ converges to a unit vector $v$, while $p_j\to x$, giving $v\cdot (z-x)\le 0$. The resulting face $F=\{z\in K:v\cdot (z-x)=0\}$ has smaller affine dimension. Its extreme points are extreme in $K$, since equality in the supporting inequality forces both endpoints of any decomposition to lie in $F$. Induction expresses $x$ as a convex combination of extreme points of $K$.

Every interior point is a convex combination of the two endpoints of a line segment through it maximal in $K$; compactness gives the endpoints, and both are boundary points. Combining their convex decompositions completes the proof. In the generalization requested, a nonempty weak-star compact convex subset $K$ of the dual unit ball of a separable Banach space satisfiesThe closure is now essential; separability makes this compact weak-star setting metrizable but the general Krein-Milman theorem holds more broadly in locally convex spaces.

For a bounded self-adjoint $T$, nonnegative spectrum is equivalent to $⟨Tx,x⟩\ge 0$. Here is a useful elementary justification of the nontrivial direction. Let $a={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $S=T-aI$, so $S$ has nonnegative quadratic form. Cauchy-Schwarz for that positive form gives $\Vert Sx{\Vert }^2\le \Vert S\Vert ⟨Sx,x⟩$. Taking unit vectors approaching the infimum makes $S{x}_j\to 0$, hence $a\in \sigma (T)$. Thus nonnegative spectrum forces $a\ge 0$. The reverse direction follows from the lower bound and adjoint argument for $T-\lambda$ when $\lambda <0$, together with reality of the self-adjoint spectrum.

If $T=0$ its positive square root is zero. Otherwise set $M=\Vert T\Vert$ and $C=I-T/M$, so $0\le C\le I$ and $\Vert C\Vert \le 1$. Write the scalar binomial seriesThe last equality follows by taking $z\uparrow 1$ in this positive-coefficient series. Consequentlyconverges absolutely in operator norm. Each partial sum is positive, because $0\le C^n\le I$ and the sum of its coefficients is at most one. The limit is positive and self-adjoint. Multiplying the absolutely convergent series gives $B^2=M(I-C)=T$.

For uniqueness let $R$ be another positive square root. It commutes with $T=R^2$, hence with the polynomial-norm-limit $B$. Thus $(R-B)(R+B)=0$. The kernel of $R+B$ is $\ker R\cap \ker B$: zero quadratic form for a positive operator implies it annihilates the vector, by positive-form Cauchy-Schwarz. The difference $R-B$ vanishes there and on $\mathrm{ran}(R+B)$; the latter is dense in the kernel complement because $R+B$ is self-adjoint. Therefore $R=B$. The positive square root of an operator exists and is unique.

Let $X$ be the character space of an algebra, consisting of nonzero multiplicative complex linear functionals on $A$. We construct the required Gelfand transform and prove its properties.

Characters exist. A maximal proper ideal exists by the axiom of choice and is closed: if it were dense it would contain an element within distance one of the identity, hence an invertible element, contradicting properness. The quotient is a complex unital Banach division algebra, and the Gelfand-Mazur theorem identifies it with $\mathbb{C}$, producing a character. The same argument applied to the proper ideal generated by $a-\lambda$ provesEvery character has norm one, since its value belongs to the spectrum and is bounded by $\Vert a\Vert$, while $\chi (1)=1$. The character equations are weak-star closed in the dual unit ball. Thus Banach-Alaoglu theorem makes $X$ compact Hausdorff; separability of $A$ makes it metrizable.

The algebra is inverse-closed in $B(H)$. If $a\in A$ is invertible there, $a^{\ast }a\ge cI$ for some $c>0$. For $M=\Vert a^{\ast }a\Vert$, the Neumann series for $(a^{\ast }a{)}^{-1}=M^{-1}[I-(I-a^{\ast }a/M){]}^{-1}$ converges in $A$, since the positive contraction in brackets has norm at most $1-c/M<1$. Therefore $a^{-1}=(a^{\ast }a{)}^{-1}{a}^{\ast }\in A$. This proves spectral permanence for C-star algebras in the present concrete setting. In particular a self-adjoint element has real spectrum in $A$, so characters take real values on it. Decomposing $a$ into self-adjoint real and imaginary parts gives $\chi (a^{\ast })=\overline{\chi (a)}$.

Define $\hat{a}(\chi )=\chi (a)$. This is a unital star-homomorphism to $C(X)$. Every element of the abelian star-algebra is normal, and the C-star identity gives $\Vert a^{2^j}\Vert =\Vert a{\Vert }^{2^j}$. The spectral radius formula therefore gives $r_A(a)=\Vert a\Vert$. The displayed spectrum identity now yields $\Vert \hat{a}{\Vert }_{\infty }=\Vert a\Vert$, proving isometry and injectivity. Its image is closed because $A$ is complete, contains the constants, is closed under conjugation, and separates distinct characters. The Stone-Weierstrass theorem makes that image dense in $C(X)$, hence equal to it. $\boxed{A\cong C(X)}$ isometrically as a star-algebra.

Apply part (c) to the commutative unital C-star algebra $A=C^{\ast }(I,T)$, commutative because $T$ is normal. A character is determined by $z=\chi (T)$, since $T,T^{\ast }$ generate the algebra and $\chi (T^{\ast })=\overline{z}$. The spectrum identity in part (c) shows that every $z\in {\sigma }_A(T)$ occurs. Thus $X\to {\sigma }_A(T)$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Spectral permanence identifies this spectrum with the operator spectrum.

For any complex polynomial $p$ in two commuting variables, evaluation gives $\chi (p(T,T^{\ast }))=p(z,\overline{z})$. Again using the spectrum identity and spectral permanence,This proves the requested spectral mapping theorem, including constants, and explains why conjugation in the second coordinate is required.

The Weyl-von Neumann theorem states that on a separable Hilbert space a self-adjoint operator is a diagonal self-adjoint operator plus a compact self-adjoint operator. For a bounded operator $T$, the perturbation can in fact be chosen Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. The separable-space formulation is also recorded in Kadison's Corollary B.

Here is a constructive proof. The bounded self-adjoint spectral theorem decomposes a separable $H$ into countably many cyclic reducing spaces, on each of which $T$ is multiplication by $x$ in $L^2(\mu )$ for a finite measure supported in a bounded real interval. First work on one such space. Partition the interval into $N$ intervals of length at most $\delta$, and repeatedly bisect them. Let $E_j$ be the finite-dimensional space of functions constant on the level-$j$ intervals. These spaces are nested and their union is dense, since dyadic step functions approximate continuous functions and continuous functions are dense in $L^2(\mu )$.

Choose normalized interval indicators as a basis of $E_0$, and choose orthonormal bases of $E_j\ominus E_{j-1}$ within each parent interval. This is possible because the refinement splits into orthogonal parent blocks. Each new basis vector at level $j\ge 1$ is supported in an interval of length at most $\delta 2^{-(j-1)}$, and there are at most $N{2}^{j-1}$ such vectors. For every basis vector $e$ choose a real ${\lambda }_e$ in its supporting interval and define the diagonal operator $De={\lambda }_{e}e$. ThensoOne may take $N=O({\delta }^{-1})$, so this bound tends to zero as $\delta \to 0$. The error is Hilbert-Schmidt, hence compact; it is self-adjoint because both $T$ and $D$ are.

On the countably many cyclic summands choose error bounds ${\epsilon }_j$ with ${\sum }_j{\epsilon }_j^2<{\epsilon }^2$. Combining their orthonormal bases and diagonal operators gives $T=D+K$ with $\Vert K{\Vert }_{\mathrm{HS}}<\epsilon$, proving the assertion. For an unbounded self-adjoint operator, first split by spectral intervals $[j,j+1)$, apply the bounded construction on each, and choose summable error bounds. The resulting total $K$ is bounded compact, and $D=T-K$ has the same domain as $T$; the diagonal domain condition follows from the boundedness of $K$.

Separability matters. Take an uncountable orthogonal sum of copies of multiplication by $x$ on $L^2[0,1]$. The range closure of a compact operator is separable, and a separable subspace of this direct sum is supported in only countably many summands: choose a countable dense set and unite its countable supports. A self-adjoint compact error therefore vanishes on every remaining summand, each of which reduces the proposed diagonal operator $D=T-K$. The restriction of a diagonal self-adjoint operator to a reducing subspace has pure point spectrum, whereas multiplication by $x$ on $L^2[0,1]$ has no eigenvectors. This is a contradiction, qualifying the theorem if the unspoken separability convention is dropped.

For the Fredholm convention, defineThe equality follows from the Atkinson theorem. The quotient is a unital complex Banach algebra when $H$ is infinite-dimensional. Its invertible set is open by the Neumann series, so the spectrum is closed, and it lies in the disk $|\lambda |\le \Vert [T]\Vert$.

For nonemptiness, suppose the quotient resolvent $R(\lambda )=(\lambda 1-[T]{)}^{-1}$ existed for every complex $\lambda$. It is analytic everywhere, and outside a sufficiently large disk its Neumann expansion gives $\Vert R(\lambda )\Vert \le 1/(|\lambda |-\Vert [T]\Vert )$. Continuity bounds it on the remaining compact disk. Every continuous linear functional applied to $R$ is therefore a bounded entire scalar function, constant by Liouville's theorem and zero because it tends to zero at infinity. The Hahn-Banach theorem, or separation by the dual, forces $R(\lambda )=0$, contradicting its inverse identity. The essential spectrum is nonempty and closed.

Infinite dimension is necessary: every operator on a finite-dimensional Hilbert space is Fredholm, so its essential spectrum in this convention is empty. The proof works on nonseparable infinite-dimensional spaces as well; it needs neither self-adjointness nor a sequence basis.

Interpret limit points as subsequential limits, including a value repeated infinitely often. Let $L$ be the common nonempty compact cluster set. Boundedness implies $\mathrm{dist}(a_n,L)\to 0$ and likewise for $b_n$: otherwise a subsequence staying a fixed distance away would have a convergent further subsequence whose limit both belongs to $L$ and stays away from it.

Construct a matching of indices in alternating steps. At step $2j-1$ choose the least unused index $m$ in the first sequence. Choose a nearest point $\ell \in L$ to $a_m$ and an unused index $n$ in the second sequence with $|b_n-\ell |<1/j$. Such indices exist infinitely often because $\ell$ is a subsequential limit. Pair $m$ with $n$; their values differ by at most $\mathrm{dist}(a_m,L)+1/j$. At step $2j$ choose the least unused second-sequence index and match it to an unused first-sequence index within $1/j$ of a nearest point of $L$. This gives the analogous bound with $b_n$.

Every index is used exactly once: no step reuses an index, and by step $2j$ at least the first $j$ indices in each sequence have been used. Hence the produced lists are permutations satisfyingThe indices actively chosen as least unused tend to infinity on their respective odd and even steps. Their distances to $L$ therefore tend to zero, and $1/j\to 0$. The two matching estimates prove $\boxed{|a_{m_k}-b_{n_k}|\to 0}$. The cluster-set convention is important here; accumulation points of the set of distinct values would ignore infinite repetitions.

Work with bounded self-adjoint operators on separable infinite-dimensional Hilbert spaces, the setting in which the classification holds. If $UT{U}^{\ast }-S$ is compact for a unitary $U$, their images in the Calkin algebra have identical spectra, by compact perturbation invariance and unitary conjugation. Thus their essential spectra coincide.

Conversely apply the Weyl-von Neumann theorem to write $T=D_a+K_a$ and $S=D_b+K_b$ in orthonormal bases, with bounded real diagonal sequences and compact errors. The essential spectrum of a diagonal operator is exactly its sequence cluster set. Indeed outside that cluster set all but finitely many entries stay a positive distance from $\lambda$, so the diagonal has an inverse on a finite-codimensional subspace and is Fredholm. At a cluster point, infinitely many distinct basis vectors have $(D-\lambda )e_j\to 0$; these are weakly null unit vectors, contradicting any Fredholm parametrix. This also handles a value with infinite multiplicity.

The two diagonal sequences therefore have the same cluster set. Part (c) provides permutations with $a_{m_k}-b_{n_k}\to 0$. Send the $m_k$th basis vector for $T$ to the $n_k$th basis vector for $S$. Under this unitary, the diagonal difference has entries tending to zero and is compact, by finite-rank truncation. Adding the two conjugated compact errors gives $\boxed{UT{U}^{\ast }-S\text{ compact}}$.

Without separability, equality of essential spectra need not suffice, even on the same Hilbert space. Let $H$ have uncountable dimension and let $T,S$ be projections with infinite-dimensional kernels, but with range dimensions respectively countable and uncountable. Both essential spectra are $\{0,1\}$. After any unitary conjugation the range of $T$ is still separable. If its difference from $S$ were compact, the range of $S$ would lie in the sum of that separable range and the separable closure of the compact error's range, which is impossible. Finite-dimensional spaces also need a separate dimension convention, since essential spectra are then empty. These are genuine qualifications to a statement made without the standard separable infinite-dimensional setting.

The irreducible finite-dimensional unitary representations of $SU(2)$ are $E_n={\mathrm{Sym}}^n({\mathbb{C}}^2)$, one for each $n\ge 0$, with dimension $n+1$. Equivalently they are spin-$j$ representations with $j=n/2$. The defining representation has $n=1$, and $n=0$ is trivial. Every finite-dimensional unitary representation is an orthogonal direct sum of these.

Choose self-adjoint angular-momentum generators $J_x,J_y,J_z$ with $[J_x,J_y]=i{J}_z$ cyclically. The actual SU(2) Lie algebra generators are $L_a=-i{J}_a$, so $[L_x,L_y]=L_z$. On an orthonormal weight basis $v_m$, $m=-j,-j+1,\dots ,j$, the complexified action isThe Casimir operator in this normalization is $C=-{\sum }_a{L}_a^2={\sum }_a{J}_a^2$, acting byOther Lie-algebra metric normalizations rescale this scalar. On the diagonal subgroup $\mathrm{diag}(e^{it},e^{-it})$, the weights of $E_n$ are $n,n-2,\dots ,-n$, all with multiplicity one.

The kernel and cokernel of an equivariant Fredholm operator are finite-dimensional representations. Define its equivariant Fredholm index in the representation ring of a compact group byFor $G=SU(2)$ write $H={\widehat{⨁}}_n{E}_n\otimes {\mathbb{C}}^{m_n}$ and $H^{\prime }={\widehat{⨁}}_n{E}_n\otimes {\mathbb{C}}^{m_n^{\prime }}$, with finite multiplicities. By Schur lemma an intertwiner has blocks $I_{E_n}\otimes s_n$, where $s_n:{\mathbb{C}}^{m_n}\to {\mathbb{C}}^{m_n^{\prime }}$. If its rank is $r_n$, the kernel and cokernel multiplicities are $m_n-r_n$ and $m_n^{\prime }-r_n$. ConsequentlyOnly finitely many coefficients can be nonzero, because the global kernel and cokernel are finite-dimensional. Closed range is part of the Fredholm hypothesis and supplies the uniform lower bound on the remaining blocks; multiplicity matching alone would not establish that hypothesis. The ordinary index is obtained by taking dimensions, $\mathrm{ind}S={\sum }_n(n+1)(m_n-m_n^{\prime })$. This calculation applies to any compact group with its finite-dimensional irreducibles in place of the $E_n$.

Use the round unit sphere with its complex orientation and unique spin structure. Its chiral spinor bundles are $S^{+}=\mathcal{O}(-1)$ and $S^{-}={\Lambda }^{0,1}\otimes \mathcal{O}(-1)$, the latter a smooth line bundle of degree $+1$. The Dirac operator isHere Clifford multiplication has $c(v{)}^2=-|v{|}^2$, and $D$ acts self-adjointly on the L2 spinors with first-order Sobolev domain. The geometric spin connection is invariant under the lifted $SU(2)$ action, hence $D$ is equivariant.

We justify the spectral properties needed for its phase. The Peter-Weyl theorem and Frobenius reciprocity for compact groups give the decomposition of sections of an isotropy-weight-$q$ line bundle: $E_n$ occurs once exactly when $n\ge |q|$ and $n\equiv q(\mathrm{mod}2)$, because its torus weights are $n,n-2,\dots ,-n$. The spinor isotropy weights are $+1,-1$, so each chirality contains $E_1,E_3,E_5,\dots$, each once. The horizontal connection Laplacian is the total Casimir minus the vertical generator square, ${\nabla }^{\ast }\nabla =C-1/4$ on these weight-one bundles. Expanding the Dirac square in a normal orthonormal frame gives the diagonal connection-Laplacian terms and the off-diagonal curvature contraction. The Clifford relations contract spin curvature to $\mathrm{Scal}/4$, the Lichnerowicz formula for a twisted Dirac operator with no twisting bundle. Since the scalar curvature here is two,Thus there is no kernel, the eigenvalues of $D$ are $\pm 1,\pm 2,\dots$ with finite multiplicities, and the resolvent is compact by spectral truncation. These round-sphere spectral facts agree with the explicit calculation in Abrikosov's paper.

Define the chiral phase of the round-sphere Dirac operatorIt is unitary because both chiral kernels vanish; equivariance follows from that of $D$. The full phase is $F=D|D{|}^{-1}=\left(\begin{array}{cc}0& V^{\ast }\\ V& 0\end{array}\right)$.

For a smooth scalar $f$, the Leibniz rule gives $[D,M_f]=c(df)$, a bounded bundle endomorphism. The spectral integral for the sign operator gives, in the strong topology,Write $R_{\pm }=(D\mp it{)}^{-1}$. Since $2D(D^2+t^2{)}^{-1}=R_{+}+R_{-}$ and $[R_{\pm },M_f]=-R_{\pm }[D,M_f]R_{\pm }$, the commutator integral is norm convergent: its integrands are compact and bounded by $\Vert [D,M_f]\Vert /(1+t^2)$. A norm limit of their integrals is compact. Thus $[F,M_f]$ is compact, and its off-diagonal blocks say $V{M}_f^{+}-M_f^{-}V$ is compact. Uniform approximation of continuous $f$ by smooth functions extends this to all $f\in C(S^2)$, using the bound $2\Vert f-g{\Vert }_{\infty }$ for the change of commutator.

The chiral phase is an equivariant unitary which intertwines the two multiplication representations modulo compacts. This specifies the grading implicit in the subsequent index question; treating the full self-adjoint phase as an endomorphism would yield a different, zero, compressed index.

Amplify the chiral phase by the identity on the auxiliary ${\mathbb{C}}^2$, and let $P$ act by multiplication on each chirality. The displayed projection has image line $E_z=\mathbb{C}(1,\overline{z})$. On the chart $w=1/z$ near infinity it has frame $(\overline{w},1)$. On the equator the normalized second frame is $z$ times the first, so its clutching function has winding $+1$. Equivalently it is the conjugate of the tautological degree-$-1$ line. Thus $E$ has degree $+1$ with the complex orientation fixed in part (c).

It is equivariant for the conjugate defining action on the auxiliary ${\mathbb{C}}^2$; this representation is equivalent to the defining representation of $SU(2)$. At $z=0$ its isotropy character under $\mathrm{diag}(e^{it},e^{-it})$ is $e^{-it}$. Tensoring with the positive spinor weight $+1$ gives weight zero on $S^{+}\otimes E$, whereas tensoring with the negative weight $-1$ gives weight $-2$ on $S^{-}\otimes E$.

The compressed map isIts parametrix is $P{V}^{\ast }P$: the two products equal the respective identity $P$ modulo compacts, since every matrix entry of $P$ is a continuous function and the phase commutators in part (c) are compact. Hence this is an equivariant Fredholm operator.

Use the precisely stated homogeneous-bundle multiplicity rule from part (c). The domain, of weight zero, decomposes as $E_0\oplus E_2\oplus E_4\oplus \cdots$. The target, of weight $-2$, decomposes as $E_2\oplus E_4\oplus \cdots$. Part (b) subtracts these multiplicities, leaving only the trivial representation. ThereforeThis computes the index from the actual projection and representation weights without invoking an index theorem that merely supplies the requested number. Reversing the chiral direction gives $-1$. If $V$ instead denotes the full self-adjoint phase on $H_{+}\oplus H_{-}$, its compression is self-adjoint Fredholm and has index zero; the nonzero answer uses the standard chiral interpretation made explicit above.

Fix a separable infinite-dimensional complex Hilbert space $\mathcal{H}$ and letwith its operator-norm topology. The odd topological K-theory group of a compact metric space is defined by continuous-map homotopy classesHomotopies use the same norm topology. Its finite-dimensional stable realization and group operation are established in the following parts, so this definition does not yet assume that homotopy classes form a group.

Let $u:X\to U_{\mathcal{K}}$ be continuous. The family $\{u(x)-I:x\in X\}$ is norm compact in the compact-operator ideal. For any $\epsilon >0$ there is one finite-rank orthogonal projection $P$ withTo prove it, take a finite $\epsilon$-net of compact operators, approximate each by a finite-rank operator, and include both their ranges and adjoint ranges in $P\mathcal{H}$. Contractivity of compression transfers the estimate to the whole family.

Take $\epsilon <1$ and set $g(x)=I+P(u(x)-I)P$. It is invertible because it is within distance one of the unitary $u(x)$. It has the form $g_V(x)\oplus I$ on $V\oplus V^{\perp }$, where $V=P\mathcal{H}$ is finite-dimensional and $g_V:X\to GL(V)$ is continuous. The straight segment from $u$ to $g$ stays invertible; taking its unitary polar factors supplies a homotopy in $U_{\mathcal{K}}$, since inverse square root depends continuously on the positive invertible factors and preserves identity modulo compacts.