research31 min studyLesson 962 in Mathematics

Index stability and homotopy invariance

Track a Fredholm index under norm-continuous deformation while allowing kernel and cokernel dimensions to jump together.

On this page

What you’ll understand

  • State norm-continuous Fredholm homotopy invariance precisely.
  • Distinguish compact perturbation invariance from arbitrary large perturbations.
  • Compute shift defects and an explicit paired jump of kernel and cokernel.

An index survives many changes that make the individual solution spaces difficult to follow. This stability is not a claim that every eigenvalue or kernel stays fixed. It says that the difference between two finite defects is locally constant on the space of Fredholm operators. The topology of that operator space and the requirement that the entire path remain Fredholm are the mechanisms behind a useful invariant rather than incidental technical qualifications.

Use a topology that controls the whole operator

For bounded operators between fixed Hilbert spaces, the Fredholm set is open in operator norm and the integer-valued index is locally constant there. Therefore a norm-continuous path T_t of Fredholm operators, with t in a connected interval, has constant index. A familiar finite-dimensional matrix path is a special case, but infinite-dimensional spaces allow components labeled by different indices. Pointwise or strong convergence is weaker than norm convergence and does not by itself give this conclusion.

Differential operators of a fixed order can be compared as bounded maps between fixed Sobolev spaces. Smoothly varying coefficients and a continuously elliptic principal symbol on a compact closed manifold give an appropriate continuous Fredholm family. For unbounded L² realizations one instead needs a controlled common domain or a suitable graph topology; merely writing a parameter in a differential expression does not specify such a family. Boundary conditions that vary must be included in the continuity analysis.

t⟼Tt∈Fred⁡(H0,H1) norm-continuous⟹ind⁡Tt=ind⁡T0t\longmapsto T_t\in\operatorname{Fred}(H_0,H_1)\text{ norm-continuous}\quad\Longrightarrow\quad\operatorname{ind}T_t=\operatorname{ind}T_0
The statement assumes that every parameter lies in the Fredholm set and that the Hilbert-space identifications are fixed or continuously trivialized.↔ Scroll wide equations horizontally.

A compact perturbation cannot change the difference

If K is compact and T is Fredholm, then every T+tK is Fredholm and has the same index as T. One proof passes to operators modulo compact operators: T has an inverse in that quotient, and adding K does not change its quotient class. Finite-rank corrections are especially concrete examples. On a closed manifold, changing only lower-order terms of an order-m elliptic operator gives a compact perturbation from H^s to H^(s−m), because the extra regularity feeds through a compact Sobolev inclusion.

An arbitrary bounded perturbation is different. Small perturbations preserve Fredholmness by openness, while a large one can leave the Fredholm set. A nonzero index obstructs invertibility throughout its component because an invertible operator has both defects zero. The converse fails: index zero does not produce an inverse automatically. It removes this particular numerical obstruction, leaving a finite-dimensional obstruction that a suitable finite-rank correction can remove for an individual Fredholm map.

K compact, T Fredholm⟹ind⁡(T+K)=ind⁡T,ind⁡(ST)=ind⁡S+ind⁡TK\text{ compact},\ T\text{ Fredholm}\quad\Longrightarrow\quad\operatorname{ind}(T+K)=\operatorname{ind}T,\qquad\operatorname{ind}(ST)=\operatorname{ind}S+\operatorname{ind}T
Composition requires compatible Hilbert spaces and both maps Fredholm. Compactness is the decisive hypothesis for an unrestricted perturbation size.↔ Scroll wide equations horizontally.

Kernel jumps are compatible with a stable index

A kernel dimension is upper semicontinuous in a suitable Fredholm family: additional zero modes can appear at exceptional parameters. The cokernel adjusts so that their difference remains fixed. In finite dimensions rank-nullity makes this cancellation transparent. In infinite dimensions the same phenomenon occurs inside the finite-dimensional blocks left after one splits off uniformly invertible complements. It is precisely these finite blocks that underlie both local constancy and the later construction of a families index.

The unilateral shift gives an infinite-dimensional example with a genuinely nonzero index. Sending the first basis vector to the second and continuing upward is an isometry with no kernel, but its image misses the first coordinate. Its cokernel has dimension one. Repeated shifts miss more coordinates, and compact modifications can move the missing-coordinate conditions without changing their number minus any new kernel. This makes an abstract quotient argument visible as ordinary linear constraints on sequences.

A rank-one change to a double shift

On ℓ²(N₀), let S e_j=e_(j+1) and P₀u=u₀e₀. Compute the index of T_t=S²+tP₀ for real t, including t=0.

  1. For u=(u₀,u₁,…), the image has coordinates (tu₀,0,u₀,u₁,…). If this sequence is zero, its third and subsequent coordinates force every u_j to vanish. Thus the kernel is zero for every t.

  2. A target sequence y belongs to the image exactly when y₁=0 and y₀=t y₂. These are two independent continuous linear conditions, so the image is closed and has codimension two.

  3. Consequently ind T_t=0−2=−2 for all t. This agrees with ind S²=2 ind S=−2 and compact perturbation invariance, since tP₀ has rank at most one. No member of this path can be invertible.

The result

The finite-rank perturbation moves the range constraints but preserves the index minus two. The calculation verifies closed range as well as both defect dimensions.

Explore the model

Interactive lab

Interactive lab · matrix

Stay inside an invertible component

Pause and Reset, then set a = 1 and b = −1; the selected matrix has determinant one. Reset and set both a = 1 and b = 1; determinant remains one while the shear reverses. Predict the same singular values (1+√5)/2 and (√5−1)/2 in both runs. Playback interpolates geometry, while readouts describe the selected matrix. This finite invertible family illustrates a deformation; it cannot display a nonzero infinite-dimensional Fredholm index.

Stay inside an invertible componentThe 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 path must remain in the allowed operator class before an invariant can be transported along it.

Check your understanding

What can happen along a norm-continuous path of Fredholm operators?

Try it yourself

Let P₀ be the rank-one orthogonal projection above and R_t=I−2tP₀ for 0≤t≤1. Find its kernel, cokernel and index for every t, and explain the exceptional parameter.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Split ℓ² into the first coordinate and its orthogonal complement. R_t is scalar on the first summand and identity on the second.

Compare with the solution

For t≠1/2, the scalar 1−2t is nonzero, so R_t is invertible with inverse acting by (1−2t)⁻¹ on e₀ and identity on its complement. Its kernel and cokernel vanish. At t=1/2, R_t kills e₀ and maps onto e₀-perpendicular, a closed codimension-one subspace. Both defects have dimension one, and its index is still zero. The inverse norm diverges as t approaches 1/2 even though Fredholmness and index persist there. Thus a stable index gives no uniform inverse estimate and no constant-rank kernel bundle.

Take it with you

  • Norm-continuous paths preserve index while they remain Fredholm.
  • Compact perturbations may alter solution spaces but preserve their dimension difference.
  • A nonzero index obstructs invertibility; a zero index is not an inverse estimate.

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