Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 (Analysis of Operators)

Use the complex inner product convention linear in its first argument. Give the exterior algebra its induced inner product, with an orthonormal wedge basis. Exterior multiplication is $e(v)\eta =v\wedge \eta$. Its adjoint is contraction:Applying this formula to $a\wedge \eta$ separates its first term from the remaining terms, giving $e(b{)}^{\ast }e(a)\eta =(a,b)\eta -e(a)e(b{)}^{\ast }\eta$. Hence the canonical anticommutation relations areFor irreducibility, choose an orthonormal basis $u_1,\dots ,u_d$. The occupancy projections are $N_j=e(u_j)e(u_j{)}^{\ast }$, and $P_0={\prod }_j(I-N_j)$ projects onto the vacuum $1\in {\Lambda }^{0}V$. For an increasing index set $J$, put $E_J=e(u_{j_1})\cdots e(u_{j_r})$. Then $E_I{P}_0{E}_J^{\ast }$ maps the wedge basis vector $u_J$ to $u_I$ and kills every other wedge basis vector. Thus these products are all matrix units and the generated star-algebra is $\mathrm{End}(\Lambda V)$. Any common invariant subspace for the creation and annihilation operators is therefore either zero or all of $W$.

The commutant is $A^{\prime }=\{T\in \mathrm{End}V:Ta=aT\text{ for every }a\in A\}$. Clearly $A\subseteq A^{\prime \prime }$. For the reverse inclusion, first note that $A\xi$ is an invariant subspace for the unital star-algebra $A$, and its orthogonal complement is invariant too. Its projection belongs to $A^{\prime }$. Thus if $B\in A^{\prime \prime }$, $B\xi \in A\xi$.

To make the same algebra element work for all vectors, amplify to $V^d$, where $d=\dim V$. The diagonal action $a^{(d)}$ has commutant consisting of block matrices whose entries belong to $A^{\prime }$. Therefore $B^{(d)}$ commutes with this amplified commutant. For a basis ${\xi }_1,\dots ,{\xi }_d$, the subspace $\{(a{\xi }_1,\dots ,a{\xi }_d):a\in A\}$ is invariant and reducing for the amplified algebra. Its projection lies in that commutant. The preceding argument gives some $a\in A$ with $(B{\xi }_1,\dots ,B{\xi }_d)=(a{\xi }_1,\dots ,a{\xi }_d)$. Since these vectors form a basis, $B=a$. HenceThis finite-dimensional double commutant theorem for star-algebras uses the star hypothesis to make the orbit subspaces reducing.

A strongly continuous unitary representation of the circle group decomposes as $H={\widehat{⨁}}_{m\in \mathbb{Z}}{H}_m$, with $U_z\xi =z^m\xi$ on $H_m$. Its positive energy unitary representation of the circle condition is that the occurring integers are bounded below. Equivalently $U_{e^{it}}=e^{itL}$ with a self-adjoint generator bounded below. An integer character can shift the lowest weight to zero.

Let $M=S^{\prime \prime }$. Because $S=S^{\ast }$, this is a Von Neumann algebra, and rotation invariance makes $U_z{M}^{\prime }{U}_z^{\ast }=M^{\prime }$. Joint irreducibility implies $M^{\prime }\cap \{U_z{\}}^{\prime }=\mathbb{C}I$: a nonscalar self-adjoint element of that joint commutant would have a nontrivial invariant spectral projection gives a reducing subspace.

For $T\in M^{\prime }$, take its strong Fourier coefficients $T_k={\int }_{\mathbb{T}}{z}^{-k}{U}_{z}T{U}_z^{\ast }dz$. They lie in $M^{\prime }$ and satisfy $U_z{T}_k{U}_z^{\ast }=z^k{T}_k$. Both $T_k^{\ast }{T}_k$ and $T_k{T}_k^{\ast }$ are rotation-invariant elements of $M^{\prime }$, hence scalar. If $T_k\ne 0$, both scalars equal $\Vert T_k{\Vert }^2$, so $T_k/\Vert T_k\Vert$ is a unitary. It shifts energy by $k$. For $k\ne 0$, applying it or its adjoint to a nonzero lowest-energy vector produces a vector below the lowest energy, a contradiction. Thus $T_k=0$ for every $k\ne 0$, while $T_0$ is scalar.

Fejér sum averaging of the strongly continuous orbit $U_{z}T{U}_z^{\ast }$ converges strongly to $T$. Its Fourier coefficients have just shown every such average is $T_0$, so $T=T_0$ is scalar. In particular every bounded operator commuting with $S$ commutes with every $U_z$. In fact the hypotheses force the stronger conclusion $S^{\prime }=\mathbb{C}I$.

Write $X=I+E$. Use a submultiplicative Sobolev algebra norm for which the Fourier projections are contractive, and its compatible matrix/column norms. The numeric hypothesis $\Vert E\Vert <1$ is understood in this algebra norm; the unscaled Fourier Sobolev norm need not have multiplication constant one.

Let $A_{>0}$ and $A_{<0}$ denote the strictly positive and strictly negative Fourier subspaces. The equation $y_{+}=-P_{>0}(E+E{y}_{+})$ is a contraction on $M_n(A_{>0})$, so it has a unique solution. Put $Y_{+}=I+y_{+}$ and $Z_{-}=X{Y}_{+}$. The equation says $Z_{-}\in M_n(A_{-})$, and $Y_{+}(0)=I$. Similarly the contraction $v_{-}=-P_{<0}(E+v_{-}E)$ gives $V_{-}=I+v_{-}$ with $Z_{+}=V_{-}X\in M_n(A_{+})$. No invertibility of these solutions has yet been assumed.

Now $V_{-}{Z}_{-}=Z_{+}{Y}_{+}$ belongs to both Fourier subalgebras, hence is a constant matrix $C$. Taking constant coefficients on either side shows $C=Z_{-}(0)=Z_{+}(0)$. We claim $C$ is invertible. If $Cv=0$, then the Hardy compression $T_X=P_{\ge 0}X$ annihilates $Y_{+}v$, because $P_{\ge 0}{Z}_{-}v=Cv=0$. But $T_X=I+P_{\ge 0}E$ on the Hardy Sobolev space has inverse given by a Neumann series, since the perturbation has norm less than one. Thus $Y_{+}v=0$, and its constant coefficient is $v$, forcing $v=0$. Finite-dimensionality proves the claim.

Consequently $C^{-1}{Z}_{+}$ is a left inverse of $Y_{+}$, and $C^{-1}{V}_{-}$ is a left inverse of $Z_{-}$. Matrix entries are functions in a commutative algebra, so pointwise finite-dimensional inversion makes these left inverses two-sided inverses in their respective Fourier subalgebras. SetThen $X=X_{-}{X}_{+}$, both factors are invertible in the stated algebras, and $X_{+}(0)=I$. If two normalized factorizations exist, ${\tilde{X}}_{-}^{-1}{X}_{-}={\tilde{X}}_{+}{X}_{+}^{-1}$ lies in $A_{-}\cap A_{+}$ and is constant; evaluating the analytic side at zero makes it $I$. This proves uniqueness. The argument is near-identity Sobolev loop factorization and, crucially, proves invertibility rather than treating the contraction solution itself as automatically invertible.

Give $G(H)$ its natural trace-class topology: differentiability means $F(t+h)=F(t)+h\dot{F}(t)+o_1(h)$, with remainder small in trace norm. The Fredholm determinant is the continuous extension of finite-rank determinants; it is multiplicative by finite-rank approximation and satisfiesnear zero. For example its exterior-power series bounds the remainder by $e^{\Vert K{\Vert }_1}-1-\Vert K{\Vert }_1$, proving that estimate directly.

Factor $F(t+h)=F(t)[I+hF(t{)}^{-1}\dot{F}(t)+o_1(h)]$. Multiplicativity and the preceding expansion giveThe determinant is nonzero for invertible $F$, so division yields the required logarithmic derivative. Differentiability only in the bounded-operator norm would not justify this trace formula; the specified topology is essential.

Let $C=[A,B]$ be trace class and set $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$. The integral identity $e^{B}A{e}^{-B}-A={\int }_0^1{e}^{sB}[B,A]e^{-sB}ds$ is trace-norm convergent. It shows that this difference is trace class. The analogous Duhamel identity then makes $F(t)-I$ trace class and differentiable in that norm. Hence $F(t)\in G(H)$.

Compute the right logarithmic derivative without taking separate traces of $A$:Trace invariance under bounded invertible similarity and the integral identity give $\mathrm{Tr}(\dot{F}{F}^{-1})=-{\int }_0^1\mathrm{Tr}[B,A]ds=\mathrm{Tr}[A,B]$. Cyclicity with a trace-class factor makes this equal to the trace in part (b). As $F(0)=I$, integrating the scalar differential equation givesThis Fredholm determinant of an exponential commutator can be nontrivial in infinite dimension: it is not legitimate to set the trace of a trace-class commutator to zero when $A,B$ themselves need not be trace class.

Split $f=f_{-}+a_0+f_{+}$ into strictly negative and strictly positive Fourier parts, and let $A=T(f_{-})$, $B=T(f_{+})$. The Toeplitz multiplication rule is $T(g)T(h)=T(gh)$ when $g$ is coanalytic or $h$ is analytic. It follows directly because the discarded negative Fourier part cannot return to the Hardy space in those cases. Thus exponentiation within either subalgebra is exact, andThe constant terms cancel. If $S=T(z)$ is the unilateral shift, $A={\sum }_{m>0}{a}_{-m}(S^{\ast }{)}^m$ and $B={\sum }_{n>0}{a}_n{S}^n$. The commutator $[(S^{\ast }{)}^m,S^n]$ is finite rank, with trace $n$ when $m=n$ and zero otherwise. For equal powers it is the projection onto the first $n$ basis vectors; for unequal powers its matrix has no diagonal entries. Its trace norm is bounded by $2\min (m,n)$. Smoothness of $f$ makes the weighted double coefficient sum finite, so the commutator series converges in trace norm andApplying part (c) proves the Toeplitz exponential determinant identityIn particular the product is an invertible identity-plus-trace-class operator, so the determinant in the formula is well-defined.

A unitary family with the required Weyl relations isTranslations preserve the $L^2$ norm and the multiplier has modulus one. Multiplication of two such operators leaves the phase $x_1{y}_2-y_1{x}_2$ after extracting $W(x_1+x_2,y_1+y_2)$, proving the relation. Translation and modulation are strongly continuous, first on compactly supported smooth functions by direct estimates, then on all of $L^2$ by density and unitarity.

If a bounded operator $A$ commutes with all $W$, it commutes with translations and every modulation $e^{2iyt}$. The spectral theorem for the multiplication coordinate therefore makes it commute with all bounded multiplication operators and interval projections. One can see its form directly: on a finite interval $I$, set $h_I=A{1}_I$. For bounded $\varphi$ supported in $I$, commutation gives $A\varphi =\varphi h_I$. Testing characteristic functions shows $|h_I|\le \Vert A\Vert$ almost everywhere. On increasing intervals these functions agree on overlaps and yield a single $h\in L^{\infty }$ with $A=M_h$.

Commutation with translations says $h(t+x)=h(t)$ almost everywhere for every $x$. Convolving with a smooth approximate identity produces smooth functions invariant under every translation, hence constants. Passing to the local $L^1$ limit makes $h$ constant almost everywhere. Therefore $A$ is a scalar operator, and the unitary family acts irreducibly.

Let $\tilde{W}$ be the given regular irreducible family. Write $a=(x,y)$ and $a\wedge b=x_1{y}_2-y_1{x}_2$. Define the strongly integrated operatorThe Gaussian is integrable and the operators are unitary, so the integral defines a bounded operator on each vector. The Weyl relation and $\tilde{W}(a{)}^{\ast }=\tilde{W}(-a)$ give $P^{\ast }=P$. For its square, the coefficient at $\tilde{W}(c)$ isby the two-dimensional Gaussian integral. Thus $P^2=P$. The same Gaussian integration, with an inserted Weyl operator, givesThis is the Gaussian projection in a Weyl representation.

It is nonzero. Otherwise all its conjugates would vanish; the Weyl relation makes their matrix coefficients the Fourier transforms of $e^{-|a{|}^2/2}(\tilde{W}(a)\xi ,\eta )$, with the dual variable $2b\wedge a$. Uniqueness of the Fourier transform of an integrable function would make each such function zero. Strong continuity then makes its value at zero vanish, contradicting $(\xi ,\xi )>0$ for a nonzero vector.

Choose a unit vector $v\in PH$. Irreducibility makes the span of $\tilde{W}(a)v$ dense. If $w\in PH$ is perpendicular to $v$, the boxed identity makes it perpendicular to this entire span, so $w=0$. Hence $P$ has rank one, and its identity gives $(\tilde{W}(a)v,v)=e^{-|a{|}^2/2}$.

In the concrete representation of part (a), the Gaussian $v_0(t)=(2/\pi {)}^{1/4}{e}^{-t^2}$ has precisely this expectation. With the linear-in-first inner-product convention, both orbit families have Gram matrixTherefore the map $\sum c_{j}W(a_j)v_0\mapsto \sum c_j\tilde{W}(a_j)v$ is well-defined and isometric. Its source and target spans are dense, so it extends to a unitary operator intertwiner. This proves the unitary equivalence, the Stone-von Neumann theorem in the normalization of this question, rather than invoking that theorem as the proof.

The real matrix $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with determinant one preserves $u\wedge v$. Hence $W_g(u)=W(gu)$ obeys the same Weyl relations. It is strongly continuous and irreducible because $g$ is invertible and its arguments range over the whole plane. Part (b) gives a unitary $V$ intertwining $W$ and $W_g$, so $VW(u)V^{\ast }=W(gu)$. If $V_1,V_2$ both work, $V_2^{\ast }{V}_1$ commutes with all $W$ and is scalar by part (a); its scalar has modulus one. Thus the implementer is unique up to phase.

For the displayed quarter-turn, choose the scaled inverse Fourier transformThe formula starts on Schwartz functions and extends unitarily to $L^2$ by Plancherel and a change of scale. It conjugates $T_{x}f(t)=f(t+x)$ to multiplication by $e^{-2ixs}$, and multiplication by $e^{2iyt}$ to $T_y$. Keeping the Weyl phase therefore gives $VW(x,y)V^{\ast }=W(y,-x)$, as required. With the opposite Fourier-kernel sign one obtains the opposite quarter-turn, so that convention must be checked.

One form of the Jacobi triple product is, for $0<|q|<1$ and $z\ne 0$,Both sides converge locally uniformly on the punctured $z$ plane. Here is a finite-product proof which also determines the multiplicative constant.

Put $Q=q^2$ and $(Q;Q{)}_m={\prod }_{j=1}^m(1-Q^j)$, with value one at $m=0$. Define the Gaussian binomial coefficient ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q=(Q;Q{)}_m/[(Q;Q{)}_k(Q;Q{)}_{m-k}]$. The recurrence ${\left[\genfrac{}{}{0ex}{}{m}{k}\right]}_Q={\left[\genfrac{}{}{0ex}{}{m-1}{k}\right]}_Q+Q^{m-k}{\left[\genfrac{}{}{0ex}{}{m-1}{k-1}\right]}_Q$ proves by induction thatRewrite the two finite factors asThe finite identity, with $n=k-N$, makes this ${\sum }_{n=-N}^N{z}^n{q}^{n^2}{\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q$. Multiply by $(Q;Q{)}_N$. For each fixed integer $n$, the resulting coefficient tends to one, because $(Q;Q{)}_m$ tends to a nonzero limit and ${\left[\genfrac{}{}{0ex}{}{2N}{N+n}\right]}_Q\to 1/(Q;Q{)}_{\infty }$.

All finite $(Q;Q{)}_m$ are uniformly bounded above and away from zero. Thus these coefficients have one uniform bound, and ${\sum }_{n\in \mathbb{Z}}|q{|}^{n^2}|z{|}^n$ is uniformly summable on compact annuli. Dominated convergence passes the limit through the finite sums and proves the boxed formula. At $q=0$ the formula follows by its limiting value one. The frequently used alternative form isIt follows by the substitution $q^2=t$, $z=u/q$, and is the version used for the charge-counting identity in Question 8.

Write $p:U(H)\to PU(H)$ for the scalar quotient. Local strongly continuous sections exist: near a fixed projective unitary, fix its phase by requiring one nonzero matrix coefficient, taken relative to a fixed representative, to be positive real. Thus the pullback $E=\{(z,V):p(V)=\rho (z)\}$ is a locally trivial central extension of the circle by the circle. It is compact, since finitely many closed arcs inside such trivializing neighbourhoods cover the compact base and each pullback is a product with the compact fibre.

Any two lifts commute up to a scalar. Their commutator defines a continuous alternating bicharacter $c(z,w)$ on the base: the scalar commutator is independent of phase choices and multiplicative in each argument. If $z$ is an irrational rotation, its powers are dense and $c(z,z^m)=1$ for every integer $m$. Continuity makes $c(z,w)=1$ for every $w$. Such $z$ are themselves dense, so all the lifts commute, and $E$ is abelian.

The tautological representation $R(z,V)=V$ is strongly continuous. Choose a unit vector $\xi$ and average its rank-one orthogonal projection over Haar measure on $E$:The integrand is trace-norm continuous, so $K$ is positive compact with trace one, and commutes with $R(E)$. A positive eigenvalue has a nonzero finite-dimensional eigenspace invariant under $E$. Commuting unitary matrices on that space have a common unit eigenvector $\eta$. Therefore $R(e)\eta =\chi (e)\eta$ for a continuous character $\chi$ of $E$.

On the central fibre, $R(1,\lambda I)=\lambda I$, so $\chi (1,\lambda I)=\lambda$. Consequentlyis independent of that lift. It is a homomorphism and strongly continuous, because the continuous map on $E$ descends through its quotient to the base. It satisfies $p(U_z)=\rho (z)$. This proves circle projective representations lift to unitary representations using an actual central character, rather than assuming a global choice of phases.

First, all lifts of the real one-parameter projective group commute. Its scalar commutator $c(s,t)$ is again a continuous alternating bicharacter. For fixed nonzero $t$ and an integer $n>0$, a lift of $t$ is a phase multiple of the $n$th power of a lift of $t/n$; hence $c(t,t/n)=1$ and $c(t,mt/n)=1$ for every integer $m$. Rational multiples of $t$ are dense in $\mathbb{R}$, so continuity makes $c(t,s)=1$ for every $s$.

Choose a unitary $U$ representing $\rho (1)$ and the allowed self-adjoint logarithm $A\in U^{\prime \prime }$ with $U=e^{iA}$. Every lift of $\rho (t)$ commutes with $U$, hence with $A$ by the double commutant theorem for star-algebras condition. Thereforeis a continuous projective homomorphism, with $\tilde{\rho }(1)=1$. It is periodic of period one and descends to $\mathbb{R}/\mathbb{Z}$. Part (b) lifts it to a continuous unitary homomorphism $L$ of that circle. Each $L(t)$ is a lift of $\tilde{\rho }(t)$ and commutes with $A$, since a representative has the form of an original lift times $e^{-itA}$. Thusis a strongly continuous unitary homomorphism with projective class $\rho (t)$. This gives the required lift of the whole real group, not just unrelated unitary representatives at each time.

The Fredholm index measures the finite-dimensional obstruction to inverting an operator. For a bounded operator $T:H_1\to H_2$ between Hilbert spaces, the Fredholm conditions are that its range of a bounded linear operator is closed and its kernel and cokernel are finite-dimensional. The orthogonal complement of the range is $\ker T^{\ast }$, soAn invertible linear map has Fredholm index zero; the converse needs an additional finite-dimensional correction, since an operator can have equally large kernel and cokernel.

First construct an inverse up to finite-rank operators. On $(\ker T{)}^{\perp }$, the restriction of $T$ is a bounded bijection onto its closed range, with bounded inverse by the bounded inverse theorem. Extend that inverse by zero on $(\mathrm{ran}T{)}^{\perp }$, obtaining $S:H_2\to H_1$. ThenConversely, suppose $ST-I$ and $TS-I$ are compact operators. On $\ker T$, the first identity expresses the identity as a compact operator, so this kernel is finite-dimensional. If $T$ were not bounded below on $(\ker T{)}^{\perp }$, there would be unit vectors $x_j$ in that complement with $T{x}_j\to 0$. Write $ST=I+K$. Compactness of $K$ gives a subsequence for which $K{x}_j$ converges, hence $x_j=ST{x}_j-K{x}_j$ converges to a unit vector $x$ in $(\ker T{)}^{\perp }$ with $Tx=0$, a contradiction. This lower bound makes the range closed. Applying the analogous argument to $T^{\ast }$, using $S^{\ast }{T}^{\ast }=I+K^{\prime \ast }$, proves that its kernel is finite-dimensional. This proves the Atkinson theorem: a bounded operator is Fredholm exactly when it has an inverse modulo compact operators. For operators on one Hilbert space this says that its image in the Calkin algebra is invertible.

The Fredholm index is stable. Decompose the domain as $N\oplus N^{\perp }$, where $N=\ker T$, and the target as $C\oplus R$, where $C=\ker T^{\ast }$ and $R=\mathrm{ran}T$. A sufficiently small norm perturbation has block matrixMultiplying by invertible triangular block matrices reduces this matrix to the direct sum of $D$ and the Schur complement $a-b{D}^{-1}c:N\to C$. The latter's kernel dimension minus cokernel dimension is $\dim N-\dim C$, whatever its rank. Therefore the Fredholm set is open and the index is locally constant. If $K$ is compact, the same parametrix works modulo compact operators for every $T+tK$, $0\le t\le 1$. Local constancy along this path gives

For Fredholm operators $A:H_1\to H_2$ and $B:H_2\to H_3$, multiplication of parametrices shows that $BA$ is Fredholm. Its index is additive. Here is a direct proof, rather than an assertion of additivity. There is an exact sequence of finite-dimensional spacesThe map from $\ker B$ takes a vector to its class modulo $\mathrm{ran}A$; the next takes the class of $y$ to that of $By$; the last is the quotient map. For example, a class $y+\mathrm{ran}A$ lies in the kernel of the middle map exactly when $By=BAx$ for some $x$, so $y-Ax\in \ker B$. This verifies the only less immediate exactness assertion; the others follow directly from their definitions. Alternating the dimensions givesAlso $\mathrm{ind}{T}^{\ast }=-\mathrm{ind}T$, and indices add under direct sums. For $I+K$ with $K$ compact, the index is zero, so injectivity is equivalent to surjectivity. More generally $Tx=f$ is solvable exactly when $f$ is orthogonal to $\ker T^{\ast }$: these are the finitely many compatibility conditions in the Fredholm alternative.

Toeplitz operators exhibit how this analytic index detects a topological winding. Let $H^2\subset L^2(S^1)$ be the Hardy space of the circle, spanned by $1,z,z^2,\dots$, and let $P$ be its orthogonal projection. For a continuous symbol $g$, define $T_g=P{M}_g{|}_{H^2}$, where $M_g$ is multiplication by $g$. Then $\Vert T_g\Vert \le \Vert g{\Vert }_{\infty }$. The product defect isFor trigonometric polynomials the crossing of negative and nonnegative Fourier modes has finite rank. Uniform approximation of continuous symbols by trigonometric polynomials, together with the displayed norm estimate, therefore makes this defect a compact operator for all continuous $f,g$. If $g$ is nowhere zero, $1/g$ is continuous and $T_{1/g}$ is a two-sided parametrix modulo compact operators. Thus $T_g$ is Fredholm.

The converse is instructive. If $g(\zeta )=0$ at $\zeta \in S^1$, consider the normalized Hardy kernelsTheir norms are one, their individual Fourier coefficients tend to zero as $r\uparrow 1$, and hence they converge weakly to zero. Their squared moduli are the Poisson kernel, so continuity of $g$ givesA Fredholm operator is bounded below off its finite-dimensional kernel. The projection of these weakly vanishing vectors onto that kernel tends to zero, contradicting this lower bound. Thus a continuous-symbol Toeplitz operator is Fredholm exactly when its symbol never vanishes. Applying the same argument to $g-\lambda$ gives its essential spectrum as $g(S^1)$, with essential spectrum defined by failure of the Fredholm property.

For nonvanishing $g$, let $m=\mathrm{wind}(g,0)$. Lifting a continuous argument of $g(e^{it})$ on $0\le t\le 2\pi$ gives an argument change $2\pi m$. After division by $z^m$, its argument has matching endpoints, so there is a continuous function $h:S^1\to \mathbb{C}$ with $g(z)=z^m{e}^{h(z)}$. The nonvanishing-symbol homotopy $g_t=z^m{e}^{th}$ reduces the index to that of $T_{z^m}$. For $m\ge 0$ this is the $m$th power of the unilateral shift operator, whose kernel is zero and whose range has codimension $m$. For $m<0$ it is the $(-m)$th power of the adjoint shift, whose kernel has dimension $-m$ and whose range is all of $H^2$. ConsequentlyThe minus sign comes from counting missing nonnegative Fourier modes. This is a concrete bridge between the Fredholm index, stability under compact perturbations, and the winding number of a loop.

On the torus ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$, choose normalized Haar measure, so that $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$, is an orthonormal basis of $L^2$. A periodic Sobolev space is defined, for any real $s$, byFor negative $s$, the coefficients describe distributions rather than necessarily functions. For a nonnegative integer $s$, the weight is comparable to ${\sum }_{|\alpha |\le s}|k^{\alpha }{|}^2$, so this norm is equivalent to the sum of the squared $L^2$ norms of all weak derivatives through order $s$. Fourier truncation proves that smooth trigonometric polynomials are dense. The $L^2$ pairing identifies the continuous dual of $H^s$ with $H^{-s}$, giving Sobolev duality.

Two Fourier estimates explain the usefulness of these spaces. If $s>m+n/2$, the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding $H^s\hookrightarrow C^m$. If $\delta >0$, a bounded subset of $H^{s+\delta }$ has uniformly small high-frequency tails in $H^s$:Finite-dimensional Fourier truncation therefore proves the compact embedding $H^{s+\delta }\hookrightarrow H^s$, the periodic version of Rellich-Kondrachov compactness theorem. In addition, for $s>n/2$ multiplication is continuous $H^s\times H^s\to H^s$, making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of $⟨k{⟩}^s|\hat{u}(k)|$ with $|\hat{v}(k)|$ and of $|\hat{u}(k)|$ with $⟨k{⟩}^s|\hat{v}(k)|$; Cauchy-Schwarz puts unweighted coefficients in ${\ell }^1$, and the ${\ell }^1\ast {\ell }^2\to {\ell }^2$ case of Young's convolution inequality finishes the estimate.

For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, $a_{ij}=a_{ji}$, real $V$, and uniform ellipticityTake the self-adjoint realization in $L^2$ with domain $H^2$. Its construction and spectral properties can be seen from the associated form on $H^1$,For a sufficiently large constant $C$, $b_C=b+C(\cdot ,\cdot )$ is a positive coercive form: $b_C(u,u)\ge c_1\Vert u{\Vert }_{H^1}^2$, and $L+C\ge I$. The Lax-Milgram theorem supplies the unique weak solution of $(L+C)u=f$ for $f\in H^{-1}$; the form is Hermitian, which gives symmetry of its inverse on $L^2$.

The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol $\sum a_{ij}{\xi }_i{\xi }_j$ dominates $c|\xi {|}^2$, and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with $L$ have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak $H^1$ solution, the same localization with difference quotients first bounds its second weak derivatives for $f\in L^2$; differentiating and repeating then gives $f\in H^s\Rightarrow u\in H^{s+2}$ for nonnegative integer $s$. Interpolation and duality extend the scale to real orders. In particular $(L+C{)}^{-1}:L^2\to H^2$ is bounded.

Let $R=(L+C{)}^{-1}$. Compactness of $H^2\hookrightarrow L^2$ makes $R$ a compact operator. Symmetry of the form makes $R$ self-adjoint, and positivity of $L+C$ makes $R$ positive and injective. Its range is dense because $\ker R^{\ast }=\ker R=0$. The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis ${\varphi }_j$ withElliptic regularity bootstraps every ${\varphi }_j$ to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.

Writing $u_j=(u,{\varphi }_j)$, every $u\in L^2$ has $u={\sum }_j{u}_j{\varphi }_j$ in $L^2$, with $\Vert u{\Vert }_2^2={\sum }_j|u_j{|}^2$. Iterating the elliptic estimate shows that the graph norm of $(L+C{)}^m$ is equivalent to the $H^{2m}$ norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and $s$. In particular smooth functions have coefficients decaying faster than every power of ${\lambda }_j+C$, and their expansions converge in every Sobolev norm and hence in every $C^m$ norm.

This gives a practical way to solve elliptic equations. The equation $Lu=f$ is solvable if and only if $f_j=0$ whenever ${\lambda }_j=0$; then one may take $u_j=f_j/{\lambda }_j$ on the other eigenspaces and add an arbitrary vector in $\ker L$. The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation ${\partial }_{t}u+Lu=0$ isFor every $t>0$ the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.

For the Laplacian $-\Delta$ the eigenfunctions are precisely $e^{ik\cdot x}$ with eigenvalues $|k{|}^2$. The number with $|k{|}^2\le R$ is the number of lattice points in a radius-$\sqrt{R}$ ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of $\Vert \nabla u{\Vert }_2^2$ plus constants, together with the Courant–Fischer min-max principle, imply ${\lambda }_j+C\asymp j^{2/n}$. Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on ${\mathbb{R}}^n$, translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.

Use the unitary normalization of the Fourier transform,The Schwartz space $\mathcal{S}({\mathbb{R}}^n)$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^{\alpha }{\partial }^{\beta }f(x)|$ is finite. Such functions and all their derivatives are integrable. Differentiating under the integral and integrating by parts giveConsequentlyExpand the derivative by the product rule and insert the integrable weight $(1+|x|{)}^{-n-1}$; the right side is bounded by finitely many Schwartz seminorms of $f$. This proves continuity $\mathcal{F}:\mathcal{S}\to \mathcal{S}$. The identical argument with the opposite sign proves continuity of $\mathcal{G}$.