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Linear clutching splits into spectral subbundles

Statement

Assume AC. Suppose a(x)z+b(x) is a linear clutching automorphism of E for every x∈X and ∣z∣=1. A disk-automorphism homotopy and a constant change of hemisphere frame reduce it to zI−Ax. The generalized eigenspaces of Ax with eigenvalues outside and inside the unit circle form complementary subbundles E> and E<. In the fixed convention,

[E,zI−A]=[E>,I]⊕[E<,z],

so its K-class is pr⁡X∗[E>]+pr⁡X∗[E<] pr⁡S2∗[γ]. The construction preserves direct sums.

Facts & Assumptions

Given: AC, compact Hausdorff X, and the displayed linear clutching family obtained after Polynomial clutching families stabilize to linear clutching.

[F1]

A constant bundle automorphism extends over a hemisphere and hence can be absorbed by changing a clutching trivialization (Clutching construction for bundles over a suspension).

[F2]

Winding number is invariant under homotopy through nonzero loops and is additive under products (Winding number identifies the fundamental group of C times with the integers).

[F3]

External product identifies [E,I] with the pullback from X and [E,z] with the pullback of E tensored by the Hopf line (External product in complex K-theory).

[F4]

The Hopf convention fixes z as γ and β=[γ]−1 (Hopf-line calculation of K⁰(S²)).

[A1]

AC is inherited from [F3] and [F4]; the finite-dimensional spectral construction itself makes no selections.

Proof

technique · direct
1.1constructalgebra

For 0≤t<1, the fractional-linear map z↦(z+t)/(1+tz) carries S1 to itself, and 1+tz≠0 there. Therefore Ht(z)=(1+tz)(a(z+t)/(1+tz)+b)=(a+tb)z+ta+b is a homotopy through linear clutching automorphisms from az+b. At t=1, the coefficient a+b is the original automorphism evaluated at z=1. Openness of bundle automorphisms and compactness of X give t0<1 for which C=a+t0b is invertible on every fiber.

2.1F1step 1.1algebra

Right multiplication of Ht0 by the constant automorphism C−1 does not change the glued bundle by [F1]. Since scalar z commutes with C, it gives zI+B, where B=(t0a+b)C−1. Put A=−B. Then zI−A is invertible on S1, so Ax has no eigenvalue of modulus one.

3.1step 2.1algebra

For one fiber V, factor the characteristic polynomial of A as q=q>q<, with the roots of the monic factors respectively outside and inside S1. Bézout polynomials for the relatively prime factors and Cayley–Hamilton give V>=ker⁡q>(A)=im⁡q<(A), V<=ker⁡q<(A)=im⁡q>(A), and V=V>⊕V<. Both spaces are A-invariant and have precisely the indicated generalized eigenvalues. This also proves uniqueness: any invariant splitting with the same spectral locations is annihilated by the corresponding factor and therefore equals these kernels.

4.1F2step 3.1algebra

These fiber splittings vary continuously. Around each root cluster choose a small circle disjoint from all roots. For a sufficiently small change of the polynomial, the straight-line change stays nonzero on each circle, so [F2] preserves the winding number of q/∣q∣. Factoring q into linear factors shows this winding is exactly the number of enclosed roots counted with multiplicity. Thus the inside and outside monic factors vary continuously in their coefficients. In a local frame choose vectors whose images under q<(A) and q>(A) form the bases in step 3.1; the same determinant remains nonzero nearby. Their images therefore give local frames for E> and E<, proving that the fiberwise spaces are complementary subbundles.

5.1F1step 4.1algebra

On E>, the family tzI−A∣E> is invertible for 0≤t≤1, since every eigenvalue of A∣E> has modulus greater than one; it deforms zI−A to the constant −A, which [F1] identifies with I. On E<, zI−tA∣E< stays invertible because all eigenvalues have modulus less than one; it deforms zI−A to zI. Hence [F1] gives [E,zI−A]≅[E>,I]⊕[E<,z].

6.1F3F4A1step 3.1step 5.1∎

Applying [F3] and the convention [F4] to step 5.1 gives the stated K-class, hence a combination of 1 and β. For a block direct sum, the characteristic polynomial factors and the unique inside/outside invariant splitting in step 3.1 is the direct sum of the individual splittings, so the construction is additive. Rank zero gives two zero subbundles and the same formula.

Depends on

Dependency tree · 2 levels

Sources