research37 min studyLesson 969 in Mathematics

Families index and parameter dependence

Replace jumping kernels by a stabilized finite-rank index bundle and retain topology invisible to the numerical index.

On this page

What you’ll understand

  • Construct an index bundle by finite-dimensional augmentation over a compact parameter space.
  • Explain why varying kernel dimensions do not define an ordinary vector bundle.
  • Compute a rank-zero nontrivial family class and a product-family twist.

For one operator the index is an integer. For a continuous family, those integers are only the ranks of a richer K-theory class over the parameter space. Kernels may appear or disappear, so writing a difference of kernel bundles can be meaningless without a constant-rank hypothesis. Stabilization repairs the construction by adding a finite-dimensional space that makes the entire family surjective, yielding a genuine finite-rank bundle whose virtual difference persists through the jumps.

Stabilize before taking a bundle difference

Let B be a compact parameter space and T_b a norm-continuous family of Fredholm maps between Hilbert bundles, locally trivialized continuously. Around each parameter a finite-dimensional target correction makes T_b surjective. Compactness permits finitely many such corrections to be combined into a continuous map J from a fixed trivial bundle B×C^N into the targets, so the augmented family (T,J) is surjective everywhere. Its kernels then form a genuine finite-rank vector bundle K.

The index bundle is [K]−[B×C^N] in complex K⁰(B). Changing the correction or enlarging it changes both terms by compatible stabilizations and leaves the virtual class unchanged. When the original kernels and cokernels have locally constant dimensions, they do form bundles and give the familiar difference. Without that hypothesis, the notation [ker T]−[coker T] can only be shorthand for the stabilized class, not a literal subtraction of two ordinary bundles.

T~b(u,v)=Tbu+Jbv surjective,K=ker⁡T~,Ind⁡(T)=[K]−[C‾ N]∈K0(B)\widetilde T_b(u,v)=T_bu+J_bv\text{ surjective},\qquad K=\ker\widetilde T,\qquad \operatorname{Ind}(T)=[K]-[\underline{\mathbb C}^{\,N}]\in K^0(B)
The correction has finite dimension and the augmented kernel has locally constant finite rank. Compactness of B gives a common finite correction.↔ Scroll wide equations horizontally.

A geometric family has a vertical index theorem

For a smooth proper submersion π:Z→B with closed even-dimensional fibers, choose a smoothly varying vertical Riemannian spin structure and compatible bundle connections. The vertical twisted chiral Dirac operators form an elliptic family. Their analytic K-class equals the topological family pushforward. After applying the rational Chern character, its class is integration along the fibers of vertical  times ch(E). The vertical tangent bundle and fiber orientation are essential: they are not the full tangent bundle of the total space.

The cohomological identity determines all rational even-degree components on B, not only degree zero. Positive-degree components can be nonzero even when every numerical index vanishes. Conversely, rational differential forms may miss torsion in the integral K-class. Superconnections provide differential-form representatives and control variation, but the K-theoretic index is defined without requiring constant-dimensional eigenspaces. Smoothness and the closed-fiber assumption prevent boundary domains or noncompact ends from being silently introduced into this theorem.

ch⁡(Ind⁡DE+)=π∗ ⁣(A^(T(Z/B))ch⁡(E))∈Heven(B;Q)\operatorname{ch}(\operatorname{Ind}D_E^+)=\pi_*\!\left(\widehat A(T(Z/B))\operatorname{ch}(E)\right)\in H^{\mathrm{even}}(B;\mathbb Q)
This is the cohomological families spin Dirac formula. Fiber integration uses the chosen vertical orientation; the full K-class contains more integral information.↔ Scroll wide equations horizontally.

Finite families already reveal the issue

For a map between finite-rank bundles E and F, each fiber map is Fredholm and its index bundle is [E]−[F], whether or not its rank is constant. Thus even finite-dimensional examples can have jumping kernels and a nontrivial family index. Adding an identity on a trivial infinite-dimensional Hilbert bundle embeds such an example into the usual Hilbert-space Fredholm setting without changing the finite defects or virtual class.

The ranks of the virtual class are the ordinary indices at the parameters. They are locally constant, but equal rank does not mean equal K-class. A line bundle minus a trivial line has rank zero and may have a nonzero first Chern class. This is the family counterpart of the earlier relative-symbol example: topology lives in how vector spaces fit together, rather than just their dimensions. It is also why inspecting zero eigenvalues one parameter at a time cannot recover the complete index bundle.

A jumping kernel with a nonzero rank-zero class

Let L be the tautological line over CP¹, with c₁(L)=−u and ∫u=1. For a line ℓ⊂C², define T_ℓ:ℓ→C by projection to the first coordinate. Compute pointwise defects and the index bundle.

  1. Except at the line span(0,1), projection is a nonzero map between one-dimensional spaces and hence an isomorphism. At that exceptional line it is zero, so both kernel and cokernel have dimension one.

  2. The numerical index is zero everywhere, but the kernels and cokernels do not form vector bundles of constant rank over CP¹. For this finite bundle map the stabilized index is [L]−[1].

  3. Since u²=0 on CP¹, ch(L)=exp(−u)=1−u. Therefore ch(Ind T)=−u, which is nonzero. The index bundle has rank zero but is not the zero K-class; adding a trivial Hilbert identity preserves the example.

The result

A constant numerical index does not determine a family class. The nontrivial line-bundle topology survives even though each fiber has index zero.

Explore the model

Interactive lab

Interactive lab · matrix

A restricted family with no kernel jumps

Pause and Reset, then set a = 0.25 and b = 0: the selected matrix has singular values one and 0.25. Reset and set both a = 1 and b = 0: both singular values are one. Every supported matrix is invertible and has zero finite-dimensional index. The family cannot reach a=0 or represent the tautological line over CP¹, so it illustrates only why uniform invertibility gives trivial defects, not the jumping or twisting example itself.

A restricted family with no kernel jumpsThe 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. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

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 forA family of vector spaces needs global gluing data; a few scalar ranks cannot identify its K-class.

Check your understanding

What is valid when kernel dimensions jump in a continuous Fredholm family over compact B?

Try it yourself

Take the product family of de Rham parity operators on a genus-three surface over B=CP¹. Twist all vertical forms by a base line bundle A with c₁(A)=2u. Compute the index bundle, its rank and its degree-two Chern-character component.

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 surface has two even and six odd harmonic dimensions. Tensor these constant harmonic spaces with A; this is a de Rham family, so use its Euler index rather than the spin  density.

Compare with the solution

The harmonic bundles are A⊕A in even parity and six copies of A in odd parity. Thus Ind D⁺=2[A]−6[A]=−4[A]. Since u²=0, ch(A)=exp(2u)=1+2u, giving ch(Ind D⁺)=−4−8u. Its rank is minus four at every parameter, while its degree-two component is minus eight times u. The same result follows by multiplying the fiber Euler characteristic χ(Σ₃)=−4 by the base twist. The numerical index alone sees only minus four and misses the base-dependent Chern information.

Take it with you

  • Jumping kernels require stabilization before defining an ordinary index bundle.
  • A families index lies in integral complex K⁰(B), with numerical indices as its ranks.
  • Fiberwise characteristic integration computes rational family information beyond degree zero.

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