Elliptic symbols and Fredholm operators
Choose Sobolev realizations before relating an invertible principal symbol to a finite analytic index.
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The question for today
What you’ll understand
- Specify the domain and target of an elliptic Fredholm realization.
- Explain how a parametrix gives closed range and finite defects.
- Compute circle kernels and show how boundary domains change an index.
The index measures a difference between two failures of invertibility: solutions of the homogeneous equation and obstructions to solving the inhomogeneous equation. Its definition is analytic, but ellipticity makes that difference accessible to topology. Before using either word, specify the spaces on which the operator acts. The same differential expression with different endpoint conditions can have different indices, even though its interior principal symbol never changes.
The principal symbol controls high frequencies
Let M be a smooth compact manifold without boundary and E,F smooth finite-rank complex Hermitian bundles. A differential operator P of positive order m extends continuously from H^s(M;E) to H^(s−m)(M;F). Our symbol convention replaces each derivative ∂_j by iξ_j in its highest-order terms. Ellipticity means this bundle map is invertible at every nonzero real covector. It concerns high-frequency directions and does not say that P itself has no kernel.
For example, the symbol of −i∂θ on a circle is ξ, which is nonzero away from the zero covector, but constant functions remain in the kernel. Lower-order terms can move these exceptional modes without changing the principal symbol. Conversely, checking the determinant of the coefficient at one point is insufficient on a manifold: invertibility must hold in every base point and every nonzero cotangent direction, with the bundles and coordinate transformations accounted for.
A parametrix leaves only compact errors
Elliptic pseudodifferential calculus constructs an operator Q of order −m such that QP and PQ equal the appropriate identities modulo smoothing operators. On a compact manifold those remainders act compactly on the relevant Sobolev spaces. Atkinson's theorem then makes P Fredholm: its kernel is finite dimensional, its image is closed, and its cokernel is finite dimensional. Closed range is part of the conclusion and cannot be omitted from a general definition of Fredholmness.
Elliptic regularity makes homogeneous solutions smooth, so the kernel and index do not depend on the chosen Sobolev exponent s. The cokernel is naturally identified with the smooth kernel of the formal adjoint using the Hermitian metrics, density and distributional pairing. At the convenient realization H^m to L² this is the L² adjoint equation, with the correct adjoint domain. One must not simply subtract dimensions of unrelated null spaces obtained from independently chosen boundary conditions.
Compactness and a domain are essential hypotheses
On a noncompact manifold an invertible principal symbol alone gives no global Fredholm theorem. Spectrum can accumulate near zero through behavior at infinity, and compact Sobolev embeddings can fail. On a compact manifold with boundary, the interior symbol likewise does not specify a Fredholm problem. The trace space and an elliptic boundary condition supply additional data. Later lessons use spectral boundary conditions, whose global nature is precisely what permits certain Dirac problems to be Fredholm.
A formally self-adjoint elliptic operator on a closed manifold has equal kernel and cokernel dimensions and therefore index zero. Interesting nonzero indices usually arise by separating a graded operator into a map from its positive bundle to its negative bundle. Even when the numerical index vanishes, neither invertibility nor the absence of solutions follows. The circle example provides a calibration: a zero index can accompany either a one-dimensional kernel or no kernel at all.
Work it through
Move a circle eigenvalue through zero
On the circle of length 2π with periodic sections, compute the index of P_a=−i∂θ+a from H¹ to L² for a=0 and a=1/2.
The Fourier modes exp(inθ) have eigenvalues n+a for every integer n. The principal symbol is ξ for both parameters, so ellipticity is unchanged.
At a=0 the kernel consists of constants. The range is the closed subspace of functions with zero mean, hence the cokernel also has dimension one. The index is 1−1=0.
At a=1/2 no eigenvalue vanishes. Dividing Fourier coefficients by n+1/2 gives an inverse from L² to H¹ because (1+n²)/(n+1/2)² is uniformly bounded. Both defects vanish and the index is again zero.
The index is zero in both cases, although the actual solution spaces change. The Sobolev inverse estimate is part of the invertibility claim.
The living notebook
Explore the model
Interactive lab · matrix
Invertibility versus conditioning
Pause and Reset, then set diagonal scale a = 1 and shear b = 0: both singular values are one. Reset and set both a = 0.25 and b = 0: predict singular values one and 0.25 and condition number four. Every supported matrix remains invertible. This finite matrix illustrates a lower bound on singular values; it does not establish a global elliptic estimate or represent the circle operator's exceptional constant mode.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forAn invertible high-frequency symbol and the full operator's kernel are different pieces of information.
Pause & reason
Check your understanding
Your turn
Try it yourself
For D=d/dx on [0,1], compute its index first with domain H¹ and then with domain H¹₀, both targeting L². Explain why the unchanged interior symbol does not force the indices to agree.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Integrate a proposed right-hand side. In one dimension H¹₀ imposes both endpoint traces, and the integral between them is constrained.
Compare with the solution
On H¹, every f in L² has a primitive u(x)=∫₀ˣf, so D is surjective; its kernel is the one-dimensional constants. Its index is one. On H¹₀ the only constant satisfying both endpoint conditions is zero. The range is exactly {f:∫₀¹f=0}: necessity follows by the fundamental theorem for Sobolev functions, and sufficiency follows from the same primitive, whose two traces now vanish. This range is closed with codimension one, so the index is minus one. Both are bounded Fredholm maps from their stated domains, but they encode different boundary problems.
Keep the idea
Take it with you
- Specify Sobolev spaces, compactness and boundary data before computing an index.
- A compact parametrix error yields closed range and finite defects.
- Index zero allows nonzero kernels and does not imply invertibility.
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