Boundary operators and the Atiyah–Patodi–Singer index
Fix a product collar and spectral domain, then account for boundary spectral asymmetry and zero modes using reduced eta.
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The question for today
What you’ll understand
- Specify the product collar, tangential operator and APS domain with a declared sign convention.
- Define eta by analytic continuation and retain its zero-mode correction.
- Calculate a shifted-circle eta invariant and prove the index change from allowing boundary zero modes.
A boundary creates infinitely many potential boundary data, so interior ellipticity alone is not a Fredholm boundary problem. Atiyah–Patodi–Singer conditions use the spectrum of a tangential operator to select half of those data. The resulting index equals a bulk characteristic integral corrected by boundary spectral asymmetry. Zero eigenvalues require their own term: replacing reduced eta by half of eta alone changes the answer exactly where the spectral domain is most sensitive.
The collar fixes the spectral sign
Let M be a compact oriented even-dimensional Riemannian spin manifold with boundary Y, with metric and twisting connection of product type on a collar. Orient Y by the outward-normal-first rule and choose an inward distance coordinate u≥0. Write D⁺=J(∂_u+B), where J is inward Clifford multiplication from S⁺ to S⁻ and B is the self-adjoint tangential operator on S⁺ restricted to Y. This factorization fixes B's sign; using an outward coordinate without changing the formulas would reverse it.
Use the domain of H¹ positive spinors whose H^(1/2) trace is annihilated by the spectral projection Π_(≥0)(B). Thus negative boundary modes are allowed, while zero and positive modes are forbidden. With this domain D⁺ to L² negative spinors is Fredholm, and its adjoint has the corresponding complementary boundary condition, not an independently imposed copy of the same one. The spectral projection is global along Y, which makes this a nonlocal elliptic boundary condition.
Eta is a regularized spectral imbalance
For the closed odd-dimensional boundary Dirac operator, define eta initially by summing sign(λ)|λ|^(−s) over nonzero eigenvalues for sufficiently large real part of s. The sum has meromorphic continuation and is regular at zero in this Dirac setting. Its value there is η(B), while h(B) is the ordinary finite dimension of the zero eigenspace. The reduced invariant is the real number (η+h)/2 for this index formula; taking it only modulo integers would lose the exact integer index.
With the declared APS convention, the twisted spin formula is the bulk integral of  times ch(E) minus reduced eta. Product geometry removes additional boundary transgression terms; a nonproduct metric or connection requires appropriate corrections or a deformation argument that accounts for them. Eta is not the difference of two separately finite counts of positive and negative modes. Most boundary spectra have infinitely many of each sign, and analytic continuation is essential even in a simple circle example.
Crossings and changed cuts must be tracked
If the boundary operator stays invertible along a smooth variation, its spectral projection varies continuously in the appropriate calculus. At a zero crossing, the strict negative-mode domain changes by a finite-dimensional space. The bulk and eta terms encode this change through spectral flow. Allowing zero modes by replacing Π_(≥0) with Π_(>0) enlarges the domain by h(B) dimensions and increases the index by that amount. The replacement formula therefore has η−h rather than η+h in the boundary numerator.
Reversing B alone sends η to its negative while keeping h unchanged, so reduced eta does not simply change sign when a kernel is present. Reversing the geometric orientation also changes chirality, bulk orientation and boundary conventions, which must be handled together. These observations matter when gluing manifolds or comparing formulas from different sources. They also explain why a formally attractive expression containing only the bulk integral is generally not an integer on a manifold with boundary.
Work it through
A shifted circle and its zero-mode limit
For B_a=−i∂θ+a on periodic functions on a circle of length 2π, compute eta and reduced eta for 0<a<1, evaluate a=1/4, and compare with a=0.
The eigenvalues are n+a. For 0<a<1, positive terms are k+a and the absolute negative terms are k+1−a, with k≥0. Thus η_a(s)=ζ(s,a)−ζ(s,1−a) in the convergent half-plane, with ζ the Hurwitz zeta function.
Using its continued value ζ(0,q)=1/2−q gives η_a(0)=1−2a. There is no zero eigenvalue, so reduced eta is 1/2−a. At a=1/4 these are 1/2 and 1/4 respectively; the allowed negative modes have n≤−1.
At a=0 the nonzero spectrum pairs symmetrically, giving eta zero, but the constant mode gives h=1 and reduced eta 1/2. The limit from positive a agrees with 1/2, whereas the limit from negative a is −1/2. The one-unit jump records the crossing.
The eta correction detects a spectral asymmetry invisible to the principal symbol, and the kernel term is necessary at the integer shift.
The living notebook
Explore the model
Interactive lab · spectral
A finite cancellation is not eta regularization
Pause and Reset, then set mode count N = 1 and decay p = 0; at x=π/2 and t=0 the sum is one. Reset and set both N = 3 and p = 0: the sum is zero because 1+0−1 cancels. This instrument is a finite sine sum, with no shifted eigenvalue or zeta-continuation control. Use it to distinguish ordinary finite cancellation from the analytic continuation required to define eta, rather than treating its zero sum as an APS invariant.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. 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 forFinite mode arithmetic cannot replace a specified boundary spectral domain and a regularized eta invariant.
Pause & reason
Check your understanding
Your turn
Try it yourself
For a product-type APS problem with h(B)=3 and index −2, replace the boundary condition Π_(≥0)trace=0 by Π_(>0)trace=0. Prove the new index and explain the corresponding eta formula.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The new domain contains the old one with a three-dimensional quotient, supplied by boundary zero eigensections. Use the exact sequence for the kernel and cokernel of a finite-dimensional domain extension.
Compare with the solution
The Sobolev trace extension theorem realizes every boundary zero eigensection, so the enlarged domain modulo the old domain has dimension three. For the two Fredholm realizations, the exact sequence 0→ker D_old→ker D_new→C³→coker D_old→coker D_new→0 shows, by its alternating dimension sum, that ind D_new−ind D_old=3. The new index is therefore one. Algebraically the new boundary term is (η−h)/2 rather than (η+h)/2, increasing the bulk-minus-boundary expression by h=3. This argument does not require every added boundary mode to extend to a homogeneous interior solution: the cokernel part of the exact sequence accounts for that possibility.
Keep the idea
Take it with you
- APS Fredholmness includes a product collar, a tangential sign and a specified spectral domain.
- Reduced eta combines regularized spectral asymmetry with the boundary zero-mode dimension.
- Changing the boundary cut is a finite-dimensional domain change with a computable index shift.
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