Complex Topological K Theory and Bott Periodicity

By [F1], the first displayed equality holds exactly when some bundle $H$ satisfies the first stable-isomorphism display. This proves both directions of the first equivalence, including $H=0_X$ when no added summand is needed.

Apply [F2] to $H$. There are a bundle $H^{\mathrm{\prime }}$ and $N\ge 0$ with $H\oplus H^{\mathrm{\prime }}\cong {\epsilon }^N$. Adding $H^{\mathrm{\prime }}$ to both sides of the isomorphism in step 1.1 gives the trivial-stabilization display. Conversely, that display is the relation in step 1.1 with $H={\epsilon }^N$.

Set $F=F^{\mathrm{\prime }}=0_X$ in the proved equivalence. Then $[E]=[F]$ exactly when $E\oplus {\epsilon }^N\cong F\oplus {\epsilon }^N$ for some $N$, including $N=0$.

Pullback sends $[E]$ to $[f^{\ast }E]$ and preserves Whitney sums. By [F3] it extends uniquely to $f^{\ast }([E]-[F])=[f^{\ast }E]-[f^{\ast }F]$. Pullback also preserves tensor products and the trivial line, so this is a unital ring map. The canonical isomorphisms in [F1] give the identity and contravariant composition laws on bundle generators, hence on all virtual classes.

If $f_0\simeq f_1$, [F2] gives $f_0^{\ast }E\cong f_1^{\ast }E$ for every bundle $E$. The two induced maps therefore agree on all generators and, by the formula in step 1.1, on $K^0(Y)$. This is the sole use of AC.

If $f:(X,x_0)\to (Y,y_0)$ is based, then $f\circ i_{x_0}=i_{y_0}$. Step 1.1 gives $i_{x_0}^{\ast }{f}^{\ast }=i_{y_0}^{\ast }$, so $f^{\ast }$ carries the kernel in [F4] into the corresponding kernel.

Since $q\circ i$ is the constant map to the collapsed basepoint, [F1] gives $i^{\ast }{q}^{\ast }=0$ on reduced groups. Hence the image of $q^{\ast }$ lies in the kernel of $i^{\ast }$.

Let $u\in {\tilde{K}}^0(X)$ satisfy $i^{\ast }u=0$, and let ${\rho }_u:X\to \mathbb{Z}$ be its locally constant virtual-rank function. The nonzero-rank locus $C={\rho }_u^{-1}(\mathbb{Z}\setminus \{0\})$ is clopen and disjoint from $A$; put $D=X\setminus C$. Both are compact, and $q$ restricts to a homeomorphism $C\cong q(C)$, with $q(C)$ clopen in $X\mathrm{/}A$. Thus the summand $u{\mathrm{\mid }}_C$ already descends across $q$, and it remains to descend the zero-rank class $u{\mathrm{\mid }}_D$.

Write $u{\mathrm{\mid }}_D=[P]-[Q]$. Refine the finite clopen rank decompositions of $P$ and $Q$; zero virtual rank says their ranks agree on each piece. Apply [F3]'s complement theorem on every piece and enlarge the finitely many trivial ambient bundles to one common dimension $n$. The piecewise complements glue to a bundle $Q^{\mathrm{\prime }}$ with $Q\oplus Q^{\mathrm{\prime }}\cong {\epsilon }^n$, while $E=P\oplus Q^{\mathrm{\prime }}$ has the constant rank $n$. Hence $u{\mathrm{\mid }}_D=[E]-[{\epsilon }^n]$. Since $i^{\ast }u=0$, [F3]'s stable-isomorphism criterion on $A$ allows a further common trivial summand so that $E{\mathrm{\mid }}_A\cong A\times {\mathbb{C}}^n$ by an actual supplied trivialization $\tau$; rename the enlarged rank as $n$.

Extend the bundle map $\tau :A\times {\mathbb{C}}^n\to E{\mathrm{\mid }}_A$ to a neighborhood of $A$ in $D$. Concretely, choose finitely many bundle charts over $D$, express the finitely many frame vectors of $\tau$ by bounded real and imaginary coordinate functions on the closed chart pieces, extend those functions by [F4], and combine the local extensions by the finite partition in [F5]. The resulting $n$ sections agree with $\tau$ on $A$. Their exterior product is nonzero on $A$, so continuity gives an open neighborhood $U$ of $A$ in $D$ on which they are a frame. Thus $E{\mathrm{\mid }}_U$ has a trivialization extending $\tau$. Here [F6] discharges the DC hypotheses from the single assumption AC.

Form a bundle $\bar{E}$ on $q(D)=D\mathrm{/}A$: away from the collapsed point use the charts of $E$ on $D\setminus A$, and over $q(U)$ use the trivialization in step 3.1, identifying every fiber above $A$ with the same copy of ${\mathbb{C}}^n$. On overlaps the transition matrices are the old continuous ones, expressed in this frame, so the quotient charts glue to a rank-$n$ bundle. Pulling back gives $q^{\ast }\bar{E}\cong E$ over $D$, with the chosen trivialization over $A$. Therefore $v_D=[\bar{E}]-[{\epsilon }^n]$ is reduced at the quotient basepoint and pulls back to $u{\mathrm{\mid }}_D$. On the disjoint clopen part $q(C)$ put $v_C=((q{\mathrm{\mid }}_C{)}^{-1}{)}^{\ast }(u{\mathrm{\mid }}_C)$. The two virtual bundles assemble over the finite clopen decomposition $X\mathrm{/}A=q(C)⨿q(D)$ to a reduced class $v$ with $q^{\ast }v=u$. This proves $\ker i^{\ast }\subseteq \mathrm{im}{q}^{\ast }$ and, with step 1.1, exactness.

Replace $i$ by its mapping-cylinder inclusion, which is a closed cofibration with the same homotopy cofiber. The cone base inside each successive reduced mapping cone is again a closed cofibration of compact Hausdorff well-pointed CGWH spaces. Applying steps 1.1–4.1 at every stage gives exactness at every term. The quotient identifications in [F2] identify the successive quotients with reduced suspensions; using its reflection maps gives exactly the recorded signs $-\mathrm{\Sigma }i$, $-\mathrm{\Sigma }q$, and thereafter their suspended alternation. Homotopy invariance in [F1] transports exactness across these identifications.

A finite CW subcomplex inclusion satisfies the stated compactness, Hausdorff, CGWH, well-pointed, and closed-cofibration hypotheses, so the result specializes to finite CW pairs.

Continuous Gram–Schmidt on the ordered columns writes every $A\in {\mathrm{GL}}_n(\mathbb{C})$ uniquely as $A=QR$, where $Q\in U(n)$ and $R$ is upper triangular with positive real diagonal. No denominator vanishes because each initial set of columns is independent. The path $Q((1-t)R+tI)$ remains invertible and fixes $U(n)$ pointwise, so it is a deformation retraction of ${\mathrm{GL}}_n(\mathbb{C})$ onto $U(n)$.

The last-column map $SU(n)\to S^{2n-1}$ is locally trivial: near a chosen unit vector, continuous Gram–Schmidt completes that vector together with a fixed nearby frame, and multiplying the first completed vector by the inverse determinant puts the completion in $SU(n)$. Its fiber over the last basis vector is $SU(n-1)$. Thus $SU(n-1)\to SU(n)\to S^{2n-1}$ is a fibration.

Since $SU(1)$ is a point, induct simultaneously that $SU(n)$ is path-connected and simply connected. For $n\ge 2$, the sphere $S^{2n-1}$ is path-connected and has trivial fundamental group by [F3]. The pointed low-degree part of [F2], applied to step 1.2, first carries path-connectedness of the fiber and base to $SU(n)$ and then carries the inductive equality ${\pi }_1(SU(n-1))=0$ and ${\pi }_1(S^{2n-1})=0$ to ${\pi }_1(SU(n))=0$.

Determinant $U(n)\to U(1)$ is a fibration with fiber $SU(n)$ and section $s(z)=\mathrm{diag}(z,1,\dots ,1)$. By [F2] and step 2.1, ${\det }_{\ast }$ is injective on ${\pi }_1$, while the section makes it surjective. Step 1.1 transfers this isomorphism to ${\mathrm{GL}}_n(\mathbb{C})$ and ${\mathbb{C}}^{\times }$.

If a loop $g$ has determinant winding $k$, [F4] says $\det g$ is homotopic to $z\mapsto z^k$. Step 3.1 says that $g$ and $s(z^k)$ represent the same based homotopy class, which is precisely the displayed diagonal loop. Every construction was finite and explicit, so no choice principle was used.

Let $E$ have rank $n>0$. By [F1] it is clutched by a loop $g:S^1\to {\mathrm{GL}}_n(\mathbb{C})$. If $k$ is the winding number of $\det g$, [F2] deforms $g$ to $\mathrm{diag}(z^k,1,\dots ,1)$, so [F1] gives $E\cong {\gamma }^k\oplus {\epsilon }^{n-1}$. The rank-zero bundle is $0_{S^2}$. Consequently every virtual class is an integral combination of $1$ and powers of $[\gamma ]$.

The loops $\mathrm{diag}(z^2,1)$ and $\mathrm{diag}(z,z)$ have the same determinant. By [F2] they are homotopic, and [F1] gives ${\gamma }^2\oplus {\epsilon }^1\cong \gamma \oplus \gamma$. Hence $[\gamma {]}^2-2[\gamma ]+1=0$, or ${\beta }^2=0$. It follows algebraically that $[\gamma {]}^k=(1+\beta {)}^k=1+k\beta$ for every $k\in \mathbb{Z}$, since $(1+\beta {)}^{-1}=1-\beta$.

If $a+k\beta =0$, restriction to a point gives $a=0$. Then $k\beta =0$ implies $[{\gamma }^k]=1$ by step 1.2. By [F3], after adding the same trivial bundle, ${\gamma }^k$ and the trivial line are isomorphic. Their stabilized clutching determinants have winding numbers $k$ and $0$, so [F2] forces $k=0$. Thus $1$ and $\beta$ are additively independent. Together with steps 1.1–2.1 this proves the displayed ring presentation.

Basepoint restriction sends $1$ to $1\in \mathbb{Z}$ and $\beta =[\gamma ]-1$ to $0$, so its kernel is exactly $\mathbb{Z}\beta$. Reversing the hemisphere convention replaces $z$ by $z^{-1}$ and hence $\gamma$ by ${\gamma }^{\ast }$; step 1.2 gives $[{\gamma }^{\ast }]-1=(1-\beta )-1=-\beta$.

Let $D_{+}^2$ and $D_{-}^2$ be the closed hemispheres. Each inclusion $X\times \{0\}\hookrightarrow X\times D_{\pm }^2$ is a homotopy inverse to projection. By [F1], there are bundles $E_{\pm }$ on $X$ and isomorphisms $V{\mathrm{\mid }}_{X\times D_{\pm }^2}\cong {\mathrm{pr}}_X^{\ast }{E}_{\pm }$. In these trivializations, $V$ is obtained by an equatorial isomorphism $f(x,z):(E_{+}{)}_x\to (E_{-}{)}_x$.

At $z=1$, $f(x,1)$ is an isomorphism $E_{+}\to E_{-}$. Identify $E_{-}$ with $E=E_{+}$ by $f(x,1{)}^{-1}$. In the fixed coefficient convention the transition becomes $f(x,1{)}^{-1}f(x,z)$, which equals the identity at $z=1$. Thus $V=[E,f]$ with normalized $f$.

If $h_{\pm }$ and $h_{\pm }^{\mathrm{\prime }}$ are two normalized hemisphere trivializations of the same bundle, their ratios are maps $g_{\pm }:X\times D_{\pm }^2\to \mathrm{Aut}(E)$ with $g_{\pm }(x,1)=I$. The straight contraction of each disk to $1$ fixes $1$, so composing $g_{\pm }$ with it gives homotopies to the identity through maps still equal to $I$ at $1$. Applying these changing gauges to the equatorial transition gives a homotopy between the two normalized clutching maps.

For a virtual class, [F3] complements its negative bundle into a finite trivial bundle and then applies steps 1.1–3.1 to the resulting actual bundle; this is the optional stabilization in the statement.

A normalized homotopy $f_t$ glues, by [F2], a bundle on $X\times S^2\times I$. Its endpoint restrictions are isomorphic by [F1]. The normalization keeps the chosen common bundle and basepoint frame fixed, and adding trivial summands before the homotopy gives the same conclusion for stabilized data.

Fix a bundle chart over an open $U\subseteq X$ whose closure is compact and lies in a larger chart. For an integer $N\ge 1$, use the Fejér kernel $K_N(t)=N^{-1}\mathrm{\mid }1+e^{it}+\cdots +e^{i(N-1)t}{\mathrm{\mid }}^2$. It is nonnegative, has integral $2\pi$, and expands as ${\sum }_{\mathrm{\mid }j\mathrm{\mid }<N}(1-\mathrm{\mid }j\mathrm{\mid }\mathrm{/}N)e^{ijt}$. Entrywise integration in [F3] therefore defines on $U$ the Laurent polynomial $p_N(x,z)={\sum }_{\mathrm{\mid }j\mathrm{\mid }<N}(1-\mathrm{\mid }j\mathrm{\mid }\mathrm{/}N)c_j(x)z^j$, where $c_j(x)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(x,e^{it})e^{-ijt}dt$. Riemann-sum convergence uniform on compact chart closures makes every $c_j$ continuous in $x$.

Let $M$ bound the matrix entries of $f$ on the compact chart closure times $S^1$. Given $ϵ>0$, [F2] supplies $\delta >0$ such that $\parallel f(x,z{e}^{-it})-f(x,z)\parallel <ϵ\mathrm{/}2$ for $\mathrm{\mid }t\mathrm{\mid }<\delta$. On $\mathrm{\mid }t\mathrm{\mid }\ge \delta$, $K_N(t)\le (N{\sin }^2(\delta \mathrm{/}2){)}^{-1}$, so the integral of the tail times the bound $2M$ is below $ϵ\mathrm{/}2$ for all sufficiently large $N$. Since $p_N-f$ is the convolution of $f(x,z{e}^{-it})-f(x,z)$ with $K_N\mathrm{/}(2\pi )$, the short-arc and tail estimates prove $p_N\to f$ uniformly on that chart closure.

Choose finitely many such charts and a finite subordinate partition $\{{\varphi }_i\}$ by [F4]. In chart $i$, choose a Laurent approximant $p_i$ within a common tolerance. The section ${\varphi }_i{p}_i$ of $\mathrm{End}(E)$ has support inside its chart and extends by zero; hence $p={\sum }_i{\varphi }_i{p}_i$ is a global finite Laurent polynomial in $z$. Because ${\sum }_i{\varphi }_i=1$, the same tolerance bounds $p-f$ globally.

The automorphisms form an open subbundle of $\mathrm{End}(E)$: in a chart, invertibility is the open condition $\det \mathrm{\ne }0$. Compactness of $X\times S^1$ and the finite chart cover give a positive tolerance such that every section within that tolerance of $f$ is invertible, and every convex combination with $f$ remains within it. Choose $p$ accordingly. Since $f(x,1)=I$, $p(x,1)$ is invertible; put $q(x,z)=p(x,z)p(x,1{)}^{-1}$. Then $q$ is still Laurent polynomial, is normalized, and can be made arbitrarily close to $f$.

The straight-line family $h_s=(1-s)f+sq$ consists of automorphisms by step 4.1, depends continuously on $(x,z,s)$, and satisfies $h_s(x,1)=I$ for every $s$. It is the required normalized homotopy. For the rank-zero bundle the unique family is already polynomial, and for $X=\mathrm{\varnothing }$ every assertion is vacuous.

Let $f_t$ be a normalized automorphism homotopy. Apply steps 1.1–5.1 over the compact parameter space $X\times I$ to obtain a normalized Laurent family $p_t$ uniformly close to $f_t$. If prescribed normalized Laurent approximations $q_0,q_1$ were chosen sufficiently close at the endpoints, the straight segments from $q_0$ to $p_0$ and from $p_1$ to $q_1$ stay in the same open automorphism neighborhood and remain Laurent and normalized. Concatenating these with $p_t$ gives the required normalized Laurent-polynomial homotopy.

Multiplication gives $z^{r}f(z)={\sum }_{j=-r}^s{a}_j{z}^{j+r}$, whose exponents range from $0$ to $r+s$. On $\mathrm{\mid }z\mathrm{\mid }=1$ the scalar $z^r$ is nonzero, so $z^{r}f(z)$ remains an automorphism.

The line ${\gamma }^r$ has transition $z^r$. By [F1] and [F2], tensoring the bundle $[E,f]$ with ${\mathrm{pr}}_{S^2}^{\ast }{\gamma }^r$ multiplies its transition by $z^r$. Therefore $[E,z^{r}f]=[E,f][\gamma {]}^r$ in $K^0(X\times S^2)$.

By [F3], $[\gamma ]$ is a unit with $[\gamma {]}^{-r}=(1-\beta {)}^r=1-r\beta$. Multiplying the equality in step 2.1 by this unit gives $[E,f]=[E,z^{r}f][\gamma {]}^{-r}$, so clearing the negative powers loses no class information. This includes $r=0$, when no change occurs.

For $n=0$, take $L_{0}q=q=0\cdot z+a_0$ and add no summand. Suppose $n\ge 1$. On $(n+1)E$ define the block endomorphism. [F1, construct] $L_{n}q(z)=\left(\begin{array}{ccccc}I& -zI& 0& \cdots & 0\\ 0& I& -zI& \cdots & 0\\ \mathrm{⋮}& & \ddots & \ddots & \mathrm{⋮}\\ 0& \cdots & 0& I& -zI\\ a_n& a_{n-1}& \cdots & a_1& a_0\end{array}\right)\mathrm{.}$ Every entry is a finite polynomial in $z$ and the coefficient bundle maps, and only the superdiagonal entries depend on $z$. Thus $L_{n}q=a(x)z+b(x)$ varies continuously with $x$.

Starting with $L_{n}q$, add $z$ times column $1$ to column $2$, then $z$ times the new column $2$ to column $3$, and continue. The first $n$ rows become the first $n$ rows of the identity, while the final entry of the last row becomes $a_n{z}^n+\cdots +a_{1}z+a_0=q(z)$. Subtract suitable coefficient multiples of the first $n$ rows from the last row to clear its first $n$ entries. The resulting block matrix is $B(z)=\mathrm{diag}(I,\dots ,I,q(z))$.

Each column or row operation in step 2.1 is multiplication by an elementary triangular block matrix. Replacing its off-diagonal entry $c$ by $tc$, $0\le t\le 1$, is a path of invertible elementary matrices. Since $B(z)$ is invertible on $S^1$ by hypothesis, reversing the finite sequence gives a homotopy through invertible clutching maps from $B$ to $L_{n}q$. No fiber bases are selected globally: the block operations are bundle maps, and their invertibility can be checked in any local frame.

By [F1], $B$ clutches $[nE,I]\oplus [E,q]$. By [F2] and step 3.1, the corresponding stabilized bundles satisfy the following isomorphism. [F1, F2, step 3.1] $[E,q]\oplus [nE,I]\cong [(n+1)E,L_{n}q]\mathrm{.}$ This is the promised stable linearization. The matrix homotopy is a finite formula; AC is used only through [F2] to identify the endpoint bundles.

For $0\le t<1$, the fractional-linear map $z\mapsto (z+t)\mathrm{/}(1+tz)$ carries $S^1$ to itself, and $1+tz\mathrm{\ne }0$ there. Therefore $H_t(z)=(1+tz)(a(z+t)\mathrm{/}(1+tz)+b)=(a+tb)z+ta+b$ 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 $t_0<1$ for which $C=a+t_{0}b$ is invertible on every fiber.

Right multiplication of $H_{t_0}$ 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 $B=(t_{0}a+b)C^{-1}$. Put $A=-B$. Then $zI-A$ is invertible on $S^1$, so $A_x$ has no eigenvalue of modulus one.

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 $S^1$. Bézout polynomials for the relatively prime factors and Cayley–Hamilton give $V_{>}=\ker q_{>}(A)=\mathrm{im}{q}_{<}(A)$, $V_{<}=\ker q_{<}(A)=\mathrm{im}{q}_{>}(A)$, and $V=V_{>}\oplus V_{<}$. 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.

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 $q\mathrm{/}\mathrm{\mid }q\mathrm{\mid }$. 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.

On $E_{>}$, the family $tzI-A{\mathrm{\mid }}_{E_{>}}$ is invertible for $0\le t\le 1$, since every eigenvalue of $A{\mathrm{\mid }}_{E_{>}}$ has modulus greater than one; it deforms $zI-A$ to the constant $-A$, which [F1] identifies with $I$. On $E_{<}$, $zI-tA{\mathrm{\mid }}_{E_{<}}$ stays invertible because all eigenvalues have modulus less than one; it deforms $zI-A$ to $zI$. Hence [F1] gives $[E,zI-A]\cong [E_{>},I]\oplus [E_{<},z]$.

Applying [F3] and the convention [F4] to step 5.1 gives the stated $K$-class, hence a combination of $1$ and $\beta$. 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.