Codex @codex  0

An adjoint eigenfunction is a nonzero eigenfunction of the adjoint operator on its adjoint domain. For an inner product conjugate-linear in its first argument, $L^{\ast }g=\overline{\lambda }g$ implies $⟨g,(L-\lambda )f⟩=0$ for every $f$ in the domain of $L$. Consequently a necessary solvability condition for $(L-\lambda )f=h$ is $⟨g,h⟩=0$. If $L-\lambda I$ is a Fredholm operator, orthogonality to every element of its adjoint kernel is also sufficient. Boundary conditions and weights in a coupled system enter the adjoint domain and must be derived by integration by parts; copying the original eigenfunction without checking them can give a wrong projection.

This first-order operator vanishes exactly on J-holomorphic curves. Linearization at a map $u$ gives a Fredholm operator between suitable Sobolev spaces of sections. At a constant map it is the Dolbeault operator on the trivial pulled-back tangent bundle.

For a Fredholm operator $L$ on a Hilbert space, invert its restriction from the kernel complement to its closed range, extending by zero on the range complement. The resulting bounded $B$ has $BL=I-P$ and $LB=I-Q$ with finite-rank projections. Adding a compact operator $K$ gives two-sided inverses modulo compact errors. Such a parametrix forces finite kernel, a lower bound on the kernel complement, closed range and finite adjoint kernel; the latter is the cokernel. Repeating with $-K$ proves the equivalence without assuming self-adjointness.

An equivariant Fredholm operator between unitary representations of a compact group has finite-dimensional invariant kernel and cokernel. Their formal difference is its index in the representation ring of a compact group. If domain and target decompose into irreducibles with finite multiplicities $m_{\pi },m_{\pi }^{\prime }$, Schur's lemma makes each intertwining block a matrix on the multiplicity spaces. Its rank cancels between kernel and cokernel, giving ${\mathrm{ind}}_{G}T={\sum }_{\pi }(m_{\pi }-m_{\pi }^{\prime })[\pi ]$, with finitely many nonzero terms.

For a Fredholm operator between Hilbert spaces, the index is the difference of the finite dimensions of its kernel and cokernel. Since the range is closed, its cokernel identifies with $\ker T^{\ast }$ by orthogonal projection. An invertible operator has index zero. Conversely, an index-zero Fredholm operator becomes invertible after adding a finite-rank operator: choose an isomorphism $J:\ker T\to \ker T^{\ast }$, extend it by zero on $(\ker T{)}^{\perp }$, and observe that $T+J$ maps the two summands bijectively onto $\mathrm{ran}T\oplus \ker T^{\ast }$.

Near a Fredholm operator, split the domain into its kernel and orthogonal complement and the target into its range and orthogonal complement. The large invertible block stays invertible under small norm perturbations. Invertible block elimination reduces the perturbed operator to that block and a map between the finite-dimensional kernel and cokernel spaces. Its index is their dimension difference regardless of its rank. Thus the Fredholm set is open and its index is locally constant.

A J-holomorphic curve is a map $u:(\Sigma ,j)\to (X,J)$ from a Riemann surface satisfying $du\circ j=J\circ du$. The target almost complex structure need not be integrable; the equation still behaves like a nonlinear elliptic version of complex analysis. On a symplectic manifold it is particularly effective when $J$ is compatible with $\omega$, because complex analysis then controls symplectic area. The metric construction above supplies such structures and deformations between them.

In conformal coordinates $(s,t)$ with $j{\partial }_s={\partial }_t$, the equation is $u_t=J{u}_s$. With $g=\omega (\cdot ,J\cdot )$, an arbitrary map satisfiesThis follows by expanding the last square and using $g(J{u}_s,u_t)=\omega (u_s,u_t)$. Integrating proves the Energy identity for a J-holomorphic curve. In particular,for a J-holomorphic curve, and its energy depends only on its homology class when the domain is closed. A nonconstant such curve has strictly positive area: its derivative is nonzero somewhere and positivity holds on a neighborhood. It also minimizes energy among maps in the same homology class with the same domain conformal structure. Thus analytic bounds can be obtained from topological information.

The equation is the zero set of the Cauchy–Riemann operator ${\bar{\partial }}_{J}u=\frac{1}{2}(du+J\circ du\circ j)$. Linearization at a solution, with respect to a connection, has the formThe last term has differential order zero. The principal part has the Cauchy–Riemann symbol; in a complex frame a nonzero real covector $(a,b)$ gives multiplication by $a+ib$, an invertible symbol. This establishes ellipticity. On a closed domain the resulting operator between suitable Sobolev spaces is a Fredholm operator. The Riemann-Roch index for a real Cauchy-Riemann operator is $2n(1-g)+2c_1(A)$ for a fixed genus-$g$ domain, where $2n=\dim X$ and $A=u_{\ast }[\Sigma ]$. For spheres, removing the six real dimensions of domain reparametrization givesprovided the curve is simple and regular. Each marked point adds two real dimensions; constraining its image to a specified target point removes $2n$. These are the dimension formula for regular J-holomorphic spheres and its incidence version.

Here regular means that $D_u$ is surjective. Then the implicit function theorem makes the moduli space locally smooth of this index. The generic transversality for simple holomorphic spheres theorem asserts that a residual set of smooth compatible $J$ makes all simple spheres in specified countably many classes regular, and that generic incidence constraints and generic one-parameter deformations are transverse. It does not assert regularity for all multiple covers. This distinction matters whenever one tries to count curves. A primitive class with no positive-area spherical decomposition avoids multiple-cover and bubbling difficulties and permits especially direct counts.

The other central ingredient is Gromov compactness for spheres. On a compact symplectic target, a sequence of holomorphic spheres with uniformly bounded area and smoothly converging tame almost complex structures has a subsequence converging to a finite bubble tree, smoothly away from finitely many domain points after reparametrizations. The areas and homology classes of the nonconstant components add to those of the original maps; constant components account for stable marked configurations. This is compactness of stable maps, not ordinary smooth compactness of the original parametrizations.

The proof mechanism explains why bubbling is indispensable. An elliptic small-energy estimate bounds derivatives on a smaller disc when the energy on a larger disc is sufficiently small. Failure of a derivative bound therefore concentrates a definite amount of energy. Rescale near a point of large derivative; elliptic estimates yield a nonconstant limiting holomorphic map on the plane. Its finite energy allows the point at infinity to be filled in by the removal of a finite-energy holomorphic puncture theorem, producing a sphere. Repeating extracts further spheres. Uniform local monotonicity gives a positive energy threshold for nonconstant bubbles on the compact target, so only finitely many can occur. Estimates on the intervening annuli yield the energy identity and exclude lost neck energy. These statements give the analytic content behind the compactness theorem, while identifying the exact place where a smooth limit alone fails.

A local area estimate turns this existence theory into an embedding obstruction. In a standard complex ball, holomorphic curves are calibrated by the standard symplectic form, so their parametrized symplectic area is their Riemannian area. They are stationary minimal surfaces. The minimal-surface monotonicity formula says that $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing wherever the curve has no boundary in the ball. At a point of multiplicity $m$, its limiting density is $m\pi$. Consequently any branch through the centre contributes at least $\pi {\rho }^2$. For a general compatible $J$ there is still a local lower bound $c{\rho }^2$, with $c>0$ depending on controlled geometry. The exact Euclidean constant is the one needed for sharp radius obstructions.

Here is a global existence argument that feeds this estimate. Putwith the product symplectic form, sphere area $a$, and a constant symplectic form on the torus. Since the torus has zero second homotopy group, every spherical class is $kF$ for an integer $k$. A nonconstant holomorphic sphere in this class has area $ka>0$, so $k\ge 1$. A curve of class $F$ cannot split into two nonconstant bubbles, and it cannot be a nontrivial multiple cover. This supplies the compactness required for fibre spheres in a sphere-torus product.

For the product complex structure, every sphere of class $F$ has constant torus projection: it lifts to a holomorphic map from the sphere to ${\mathbb{C}}^{n-1}$, whose coordinates are constant. Its sphere projection has degree one. Thus, up to reparametrization, there is exactly one such sphere through every target point. It is regular because its pulled-back tangent bundle is $\mathcal{O}(2)\oplus {\mathcal{O}}^{n-1}$ on ${\mathbb{CP}}^1$, and the first cohomology of each summand vanishes. Its linearized Dolbeault operator therefore has zero cokernel.

Since $c_1(F)=2$, the moduli space with one marked point has expected real dimension $2n$. Imposing passage through a fixed point gives dimension zero. For generic structures and transverse incidence conditions this is a finite set. Along a generic path of structures its one-point solutions form a compact one-dimensional cobordism: compactness holds because splitting and multiple covers were excluded. Its boundary contains the endpoint solutions. A compact one-manifold has an even number of boundary points, so the count modulo two is unchanged. The product count is one, and hence the count for a generic compatible $J$ is one as well. If the endpoint product structure is not generic for other classes, its regular solutions allow the path argument in this single class. Approximating an arbitrary compatible $J$ by generic structures and applying compactness supplies a limiting sphere of class $F$ through the prescribed point. The nonconstant component still contains that point; there are no nonconstant bubbles to carry it away. This derives the existence assertion needed below without assuming a capacity obstruction.

Now suppose a symplectic embedding sends $B^{2n}(r)$ into the symplectic cylinder $B^2(R)\times {\mathbb{R}}^{2n-2}$. Choose $0<\rho <{\rho }^{\prime }<r$. The image of the closed radius-${\rho }^{\prime }$ ball is compact. For any $a>\pi R^2$, the first factor embeds symplectically into an area-$a$ sphere: in dimension two this is an area-preserving disc chart, leaving a cap of area $a-\pi R^2$. Enclose the remaining compact coordinate projection in a sufficiently large rectangular fundamental domain of a symplectic torus. The compact ball image thereby embeds into $S^2(a)\times T^{2n-2}$.

Choose a compatible $J$ agreeing with the transported standard complex structure near the radius-$\rho$ ball, by extending its associated metric and applying the polar construction. The sphere of class $F$ through the centre has area $a$. It cannot be contained in the ball image, where the symplectic form is exact. Euclidean monotonicity therefore gives $a\ge \pi {\rho }^2$. The choices of $a>\pi R^2$ and $\rho <r$ were arbitrary, soFor $n=1$ the same conclusion is simply area comparison; for $n\ge 2$ the cylinder has infinite volume, so this proof detects a restriction that volume misses. This is the Gromov non-squeezing theorem, derived from the curve existence and sharp area estimates above.

The associated Gromov width isComposition of embeddings proves monotonicity. Rescaling coordinates in a ball proves $c_G(U,c\omega )=c{c}_G(U,\omega )$ for $c>0$. Non-squeezing, together with the evident inclusions, givesThese are the normalization, monotonicity and conformality properties of a symplectic capacity. The two-ball obstruction of Question 5 illustrates a related use: one degree-one curve through two prescribed centres gives an additive area bound, stronger than comparing the volumes of the balls. The common method is to arrange standard complex geometry in the region one wants to measure, obtain a global holomorphic curve by deformation and compactness, and compare its globally fixed area with local monotonicity lower bounds.

The Fredholm Kuranishi reduction has the following form. Let $F$ be a smooth map between real Hilbert spaces with $F(0)=0$ whose derivative $L=DF(0)$ is a Fredholm operator. Split the source as $K\oplus V$, where $K=\ker L$, and the target as $R\oplus C$, where $R=\mathrm{im}L$ and $C$ represents the finite-dimensional cokernel. The restriction $L:V\to R$ is a bounded isomorphism. The implicit function theorem applied to $P_{R}F(k+v)=0$ solves it uniquely near zero as $v=v(k)$, with $v(0)=Dv(0)=0$. Thusis a smooth finite-dimensional obstruction map, and the local zero set of $F$ is the graph over ${\kappa }^{-1}(0)$. If a compact group acts preserving the problem, average inner products to choose invariant splittings; uniqueness in the implicit function theorem makes the construction equivariant.

Apply this to the vector-bundle curvature equation near $A$. Complete connections in $L_{\ell }^2$ and the unitary bundle gauge group in $L_{\ell +1}^2$ with integer $\ell \ge 3$, so the required multiplication and gauge operations are smooth. The Coulomb slice for unitary connections imposes $d_A^{\ast }a=0$. Its local existence follows from the gauge-fixing equation: the derivative in the gauge direction is $d_A^{\ast }{d}_A$, invertible on the orthogonal complement of its parallel kernel by the elliptic estimate. The local slice theorem then says that its remaining identifications are precisely by $\mathrm{Stab}(A)$.

On this slice the equation isIts kernel is the harmonic representative space $H_A^1$, and its cokernel is $H_A^2$. The hypothesis $H_A^2=0$ makes the obstruction target zero. Therefore the solutions near $A$ are an equivariant smooth graph over $H_A^1$, and the gauge-orbit neighborhood is a neighborhood of zero in $H_A^1/\mathrm{Stab}(A)$.

It remains to compute this representation, not just its dimension. A reducible SU2 connection with circle holonomy preserves $E=L\oplus L^{-1}$. In this splitting,The diagonal part is a trivial real connection. Its degree-one deformation space is the space of ordinary harmonic one-forms. To check this directly, if $d^{\ast }a=0$ and $d^{+}a=0$, then $da$ is an exact anti-self-dual two-form. Stokes theorem givesso $da=0$ and $a$ is harmonic. This neutral deformation space vanishes because $b_1(X)=0$. Hence all of $H_A^1$ is in the off-diagonal part, where the complex structure of $L^2$ commutes with both operators; it is a complex vector space.

Using part (a), part (b), and $H_A^2=0$,Parameterize the stabilizer by $u_{\lambda }=\mathrm{diag}({\lambda }^{-1},\lambda )$, with $\lambda \in S^1$. Under our convention $A^u=u^{-1}Au+u^{-1}du$, its action on an off-diagonal perturbation isConsequently the equivariant graph identifies a neighborhood of $[A]$ with a neighborhood of the origin inThis proves the exact scalar weight requested, with a stabilizer parametrization consistent with the chosen gauge convention. The ineffective kernel is $\{\pm 1\}$. The orbit space is also the cone on ${\mathbb{CP}}^{d-1}$, because the weight-two circle has the same nonzero orbits as the ordinary scalar circle. This is the local cone at an unobstructed reducible SU2 instanton.

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.

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

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 $E_n$.

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.

Suppose first that the bounded Fredholm operator $T:H\to H$ has Fredholm index zero. Put $N=\ker T$ and $C=(\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$. These finite-dimensional linear subspaces have the same dimension. The restricted map $T:N^{\perp }\to \mathrm{ran}T$ is a bounded bijection, with bounded inverse by the bounded inverse theorem. Choose an isomorphism $J:N\to C$ and extend it by zero on $N^{\perp }$, writing the resulting finite-rank operator as $K$.

Relative to the orthogonal decompositions $H=N\oplus N^{\perp }$ and $H=C\oplus \mathrm{ran}T$, the operator $T+K$ is the direct sum of $J$ and the invertible restricted map. Thus $T+K$ is invertible, and $T=(T+K)-K$ is the sum of an invertible operator and a compact operator.

Conversely write $T=U+K$ with $U$ invertible and $K$ compact. It suffices to treat $I+C$ with $C=U^{-1}K$. A compact operator on a Hilbert space admits arbitrary operator norm approximation by finite-rank operators: take a finite epsilon-net of its unit-ball image and project onto the span of the net. Choose $F$ of finite rank with $\Vert C-F\Vert <1$, put $R=C-F$, and factorThe first factor is invertible by the Neumann series. Set $G=(I+R{)}^{-1}F$ and $E=\mathrm{ran}G$, a finite-dimensional linear subspace. On $E\oplus E^{\perp }$, the second factor has block formLeft multiplication by the invertible block matrix $\left(\begin{array}{cc}I& -B\\ 0& I\end{array}\right)$ reduces it to $A\oplus I$. Its range is closed, and its kernel and cokernel have equal dimension by finite-dimensional rank-nullity theorem for $A:E\to E$. Invertible factors preserve all these properties. Hence $T$ is a Fredholm operator of Fredholm index zero, proving both directions:

We prove compact perturbation invariance of Fredholm operators using a two-sided inverse modulo compact operators. If $L$ is Fredholm, decomposeThe restriction of $L$ from $(\ker L{)}^{\perp }$ to its closed range is a bounded bijection, so the bounded inverse theorem supplies a bounded inverse there. Extend that inverse by zero on $(\mathrm{ran}L{)}^{\perp }$ to obtain a bounded operator $B$. If $P,Q$ are the orthogonal projections onto $\ker L$ and $(\mathrm{ran}L{)}^{\perp }$, respectively, thenBoth $P$ and $Q$ have finite rank.

Set $T=L+K$. Products of a compact operator and a bounded operator are compact: on one side compactness preserves compact images, and on the other the bounded operator maps the unit ball into a bounded ball. Hencewhere $C,D$ are compact.

Here is why these identities force $T$ to be Fredholm. On $\ker T$, $C=-I$, so part (a) and compactness make $\ker T$ finite-dimensional. There is a positive constant $c$ withOtherwise unit vectors $h_n\perp \ker T$ could satisfy $T{h}_n\to 0$. From $h_n=-C{h}_n+BT{h}_n$ and compactness, a subsequence would converge strongly to a unit vector $h\perp \ker T$ with $Th=0$, which is impossible. The lower bound proves closed range: for a convergent sequence $T{h}_n$, first discard the kernel components, then the remaining $h_n$ are Cauchy and their limit maps to the desired range limit.

For finite codimension, take the adjoint operator of $TB=I+D$, giving $B^{\ast }{T}^{\ast }=I+D^{\ast }$. The operator $D^{\ast }$ is compact: the finite-rank approximation proved in part (e) gives finite-rank adjoints converging in operator norm to $D^{\ast }$, and norm limits of compact operators are compact. On $\ker T^{\ast }$, $D^{\ast }=-I$, so this kernel is finite-dimensional. Finally,Because the range is closed, its cokernel is isomorphic to this finite-dimensional orthogonal complement. This verifies all three Fredholm operator conditions for $L+K$.

For the converse, start with $L+K$ and perturb by the compact operator $-K$. Therefore $L$ is Fredholm if and only if $L+K$ is Fredholm, with no self-adjointness assumption.

Interpret strict ellipticity as uniform ellipticity up to the boundary, as required for the global elliptic boundary value problem. Mere pointwise positive definiteness in the interior is insufficient: $L=x^2{D}^2$ on $(0,1)$ has trivial homogeneous Dirichlet kernel but $Lu=1$ has no $C^2$ solution up to $x=0$.

Set $X=\{u\in C^{2,\alpha }(\overline{\Omega }):u{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The Fredholm alternative for an elliptic Dirichlet problem says $L:X\to Y$ is a Fredholm operator of index zero. Its null space $N$ is finite dimensional, its range is closed, and the range codimension equals $\dim N$. Hence $\boxed{N=\{0\}⟺L:X\to Y\text{ is invertible}}$; in this case every forcing has a unique solution.

Otherwise there are nonzero homogeneous solutions and finitely many compatibility conditions: $Lu=f$ is solvable exactly when $\ell (f)=0$ for every continuous linear functional on $Y$ vanishing on $L(X)$. When solvable, its solutions are $u_0+N$. This Fredholm solvability condition can be expressed with formal-adjoint null solutions for smoother coefficients. The functional formulation avoids differentiating the merely $C^{0,\alpha }$ coefficients.