Real Clifford Modules, KO Theory, and ABS Orientations

If $X=\mathrm{\varnothing }$, the conclusion is vacuous; assume $X$ is nonempty. For each $i$ define the $i$th component of $\mathrm{\Phi }:E\to X\times {\mathbb{R}}^{rs}$ over $x$ by $\sqrt{{\rho }_i(x)}{\tau }_i(v)$ when $x\in U_i$, and by zero when $x\mathrm{\notin }{U}_i$. The real square-root function is continuous on $[0,\mathrm{\infty })$ (away from zero, $\mathrm{\mid }\sqrt{a}-\sqrt{b}\mathrm{\mid }=\mathrm{\mid }a-b\mathrm{\mid }\mathrm{/}(\sqrt{a}+\sqrt{b})$; at zero, $0\le \sqrt{a}<\epsilon$ whenever $0\le a<{\epsilon }^2$). Because $\mathrm{supp}({\rho }_i)\subseteq U_i$, every point outside $U_i$ has a neighborhood on which ${\rho }_i=0$; hence extending the chart component by zero is continuous there, and within $U_i$ it is a product of continuous coordinate functions. Hence each component, and therefore $\mathrm{\Phi }$, is continuous. It is fiberwise linear.

If $v\in E_x$ is nonzero, the partition identity gives some $i$ with ${\rho }_i(x)>0$. Then $x\in U_i$ and [F1] makes ${\tau }_i(v)\mathrm{\ne }0$, so the $i$th component of ${\mathrm{\Phi }}_x(v)$ is nonzero. Thus each ${\mathrm{\Phi }}_x$ is injective and has rank $r$.

In local bundle coordinates, the matrix of $\mathrm{\Phi }$ has rank $r$ at every point. At any chosen $x$, some $r\times r$ minor is nonzero; continuity keeps that minor nonzero on a neighborhood. After reordering the target coordinates, the image there is the graph of a continuous matrix over ${\mathbb{R}}^r$. Thus $E^{\mathrm{\prime }}=\mathrm{\Phi }(E)$ is a rank-$r$ subbundle of the trivial bundle.

On a neighborhood with a continuous frame matrix $Q$ for $E^{\mathrm{\prime }}$, the orthogonal projection onto $E^{\mathrm{\prime }}$ is $P=Q(Q^{\mathsf{T}}Q{)}^{-1}{Q}^{\mathsf{T}}$. The Gram matrix $Q^{\mathsf{T}}Q$ is positive definite and invertible; its inverse varies continuously because each inverse entry is a cofactor divided by the nonzero determinant, so $P$ is continuous. Therefore $I-P$ has locally constant rank $rs-r$ and its image is a subbundle $F=E^{\mathrm{\prime }\perp }$. Fiberwise orthogonal decomposition gives $E^{\mathrm{\prime }}\oplus F=X\times {\mathbb{R}}^{rs}$, and $\mathrm{\Phi }$ identifies $E$ with $E^{\mathrm{\prime }}$. If $r=0$, take $E^{\mathrm{\prime }}=0$ and $F=X\times {\mathbb{R}}^0$.

Let $C$ be one critical component at level $a$. In a tubular chart write $f-a=q+R$, where $q$ is the fiber Hessian quadratic form and $R$ has order at least three in the normal variable $z$. Set $f_s=q+sR$. Its vertical differential has the form $A_s(x,z)z$, where $A_s$ is the smooth symmetric operator obtained by integrating the vertical Hessian along $tz$, $0\le t\le 1$. Since the normal Hessian is invertible, $A_s$ is uniformly invertible on a small tube over compact $C$. Write $R={\sum }_i{z}_i{R}_i$ with $R_i(x,z)={\int }_0^1{\mathrm{\partial }}_{i}R(x,tz)dt$, and set $W_s=-A_s^{-1}(R_i{)}_i$. Then $d{f}_s(W_s)=-R$. Radial integration and metric duality make this construction intrinsic on the normal bundle.

We establish the compact smoothing control needed below. By [F6] choose a proper embedding $j:M\to {\mathbb{R}}^n$. Its image $S$ is closed: intersecting a closed subset of $M$ with the inverse image of a compact ball gives a compact set, whose image is closed in that ball. Apply [F1] to $S\subseteq {\mathbb{R}}^n$; composing the inverse tube map with its bundle projection gives a smooth retraction $R:U\to S$ on an open neighborhood $U$ of $S$. Let $g:{\mathbb{R}}^q\to M$ be continuous, let $Q$ be compact, and fix $\eta >0$. Compactness of $j(g(Q))$ and continuity of $R$, which fixes $S$, give $\delta >0$ such that $p+w\in U$ and $\mathrm{\parallel }R(p+w)-p\mathrm{\parallel }<\eta$ whenever $p\in j(g(Q))$ and $\mathrm{\parallel }w\mathrm{\parallel }<\delta$. By [F5] choose smooth $h:{\mathbb{R}}^q\to {\mathbb{R}}^n$ with $\mathrm{\parallel }h-jg\mathrm{\parallel }<\delta$. Then $j^{-1}R(jg+t(h-jg))$ is a continuous homotopy on $Q$, stays within ambient distance $\eta$ of $jg$, and has smooth endpoint wherever it is defined on an open neighborhood. If a smooth cutoff $\chi$ has compact support in $\mathrm{int}Q$, replace $t$ by $t\chi (x)$ and set the homotopy equal to $g$ off $Q$. It glues continuously, is fixed off $\mathrm{supp}\chi$, and its endpoint is smooth near $\{\chi =1{\}}^{\circ }$. Such cutoffs are supplied by [F7]. Thus any positive compact margins from chart boundaries and closed obstructions can be preserved throughout this localized homotopy; no global pointwise-control assertion for manifold-valued approximation is needed.

The flow of ${\mathrm{\partial }}_s+W_s$ fixes the zero section and exists for $0\le s\le 1$ on a uniformly smaller tube, because $W_s$ is of order at least two and $C$ is compact. Along it $d(f_s)\mathrm{/}ds=0$, giving a bundle normal form. Split the normal Hessian into negative and positive spectral subbundles and rescale by its positive square roots, obtaining $f=a-\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }v{\mathrm{\mid }}^2$ on ${\nu }^{-}\oplus {\nu }^{+}$. The spectral projections are smooth locally: choose disjoint contours separating the positive and negative spectra and integrate the matrix resolvent. The positive square root is smooth by the inverse theorem applied to $T\mapsto T^2$, whose derivative $H\mapsto TH+HT$ is invertible for positive definite $T$. These local descriptions agree by uniqueness.

Let $g:D^q\to M$ be a parameter cell with $q<k=\mathrm{rank}{\nu }^{-}$, whose boundary maps into the lower sublevel. In a closed Morse block let $\mathrm{\Sigma }$ be the co-core $u=0$, including its closed positive disk; it is compact and disjoint from the lower sublevel. Thus $g^{-1}(\mathrm{\Sigma })$ is compact in the cell interior. Choose nested compact neighborhoods of this preimage, away from the fixed boundary collar, mapping into a slightly larger Morse tube. The transition annulus misses $\mathrm{\Sigma }$ by a positive distance. Extend $g$ to ${\mathbb{R}}^q$ by radial retraction to the cell. The compact controlled homotopy of step 1.2 gives a smooth map arbitrarily close to $g$ there. Use a cutoff in the homotopy parameter, equal to one near $g^{-1}(\mathrm{\Sigma })$ and zero off the larger neighborhood. Sufficiently small error preserves the Morse tube and avoids $\mathrm{\Sigma }$ on the transition annulus; the resulting map is smooth near every possible intersection and unchanged on the collar. On a neighborhood of that compact intersection write it as $(b,u,v)$. Choose finitely many local frames for $b^{\ast }{\nu }^{-}$ with bump functions supported away from the collar, whose products $s_1,\dots ,s_N$ span its fibers near the intersection. The family $u_a=u+\sum a_i{s}_i$ is transverse to zero in the parameter directions there. Its incidence manifold has dimension $q+N-k<N$, so [F3] gives arbitrarily small parameters $a$ for which $u_a$ has no zero there. On the remaining compact parts of the supports over the positive disk the unperturbed $u$ is bounded away from zero; small parameters create no new intersection. The homotopy $u_{ta}$ is fixed on the collar and stays in the larger tube. Hence the cell misses the entire closed co-core. Choose the block for this avoidance slightly larger than the one subsequently used for the stopped flow, so that any new points near its entrance face also miss the co-core.

Fix a critical value $c$ and choose $\epsilon >0$ so that $c\pm \epsilon$ are regular and the compact band between them contains exactly the critical manifolds at $c$. Choose mutually disjoint Morse tubes as in step 2.1, and shrink $\epsilon$ so that each contains the closed block $H=\{\mathrm{\mid }u{\mathrm{\mid }}^2\le 2\epsilon ,\mathrm{\mid }v{\mathrm{\mid }}^2\le \epsilon \}$. This block lies in $M^{c+\epsilon }$; its $u$-face lies in $M^{c-\epsilon }$. Use $X=(2u,-2v)$ near each block and patch to $-\mathrm{grad}f$ elsewhere in the band. Convex combinations still satisfy $df(X)<0$ at every regular point. Put $B=M^{c-\epsilon }\cup \bigcup H$. Outside the interior of $B$ in the band there are no critical points, and this closed set is compact, so $-df(X)$ has a positive lower bound there. Every trajectory starting outside $B$ therefore reaches $B$ in bounded time. It enters a block only through its $v$-face, where $d\mathrm{\mid }v{\mathrm{\mid }}^2(X)=-4\mathrm{\mid }v{\mathrm{\mid }}^2<0$, or reaches the regular lower level, where $df(X)<0$. These transverse entrance equations and uniqueness of the flow make the first hitting time continuous, including where the two faces meet; it tends to zero at an entrance face. Stopping the flow on $B$ consequently gives a deformation of $M^{c+\epsilon }$ onto $B$, fixed on $B$. Only the compact band moves. Critical-free gaps between the chosen levels are treated by the same normalized descending flow.

The block is $D({\nu }^{-}){\times }_{C}D({\nu }^{+})$ after rescaling its radii. Its intersection $A_H$ with the lower sublevel is $\{\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }v{\mathrm{\mid }}^2\ge \epsilon \}\cap H$. If the negative rank is positive, $u\mathrm{\ne }0$ on $A_H$, and increasing $\mathrm{\mid }u\mathrm{\mid }$ linearly to $\sqrt{2\epsilon }$ at fixed $v$ retracts $A_H$ onto the $u$-face. More explicitly, in each negative radial fiber $A_H$ is the collar $\sqrt{\epsilon +\mathrm{\mid }v{\mathrm{\mid }}^2}\le \mathrm{\mid }u\mathrm{\mid }\le \sqrt{2\epsilon }$; this description is invariant under orthogonal changes of frames. Thus the union $B$ is the lower sublevel with these thickened negative disks attached through collars of their sphere faces. Removing the collars and contracting the positive disk factors gives, up to homotopy relative to the lower sublevel, attachment of $D({\nu }^{-})$ along $S({\nu }^{-})$. The collar formulas also cover zero positive rank; for zero negative rank $A_H$ is empty and the block is a new minimum neighborhood, retracted by $v\mapsto (1-t)v$. To pass a parameter cell through a positive-negative-rank block, first deform into $B$ by step 3.2 and apply step 3.1 to avoid the closed co-core $u=0$ in a slightly larger block, with the field still $(2u,-2v)$ there. Make the perturbations sufficiently small to stay below a slightly larger regular upper level with no additional critical values. Along the resulting compact family in a block, $\mathrm{\mid }u\mathrm{\mid }$ is bounded below away from zero until it exits, and the explicit flow $(e^{2t}u,e^{-2t}v)$ reaches the $u$-face in uniformly bounded time. The flow then reaches the lower sublevel. It is fixed on parameter points already in that sublevel. The slightly enlarged upper sublevel retracts onto the original upper sublevel through its critical-free intervening band, fixed on the latter. Composing the perturbation homotopy with this retraction keeps its target in the original upper sublevel. This proves the relative passage for a cell of dimension less than the negative rank.

Perform the relative passage on characteristic disks in cell order, extending boundary homotopies by [F4]. The finite number of cells and compact parameter images make all the required crossing times finite. Iterating across regular critical-free gaps and the finitely many isolated critical values proves passage to the lower sublevel whenever all crossed negative ranks are greater than $d$; components sharing a critical value are handled together in disjoint tubes. If the remaining lower sublevel is a sufficiently small neighborhood of the minimum locus, its normal contraction from step 4.1 completes deformation into that locus. No infinite-time convergence is asserted.

Write the upper component as $g(x,v)$ over $D_{+}V$, with $g_0(x)=g(x,0)$. Stabilize $\varphi$ by the pulled-back isomorphism $g_0^{-1}$ on ${\pi }^{\ast }F$. Its upper gauge is $U=\mathrm{diag}(g,g_0^{-1})$. Radial contraction of the disk deforms this to $\mathrm{diag}(g_0,g_0^{-1})$. For any orthogonal endomorphism $a$ of $F$, set $R_t=\left(\begin{array}{cc}\cos tI& -\sin tI\\ \sin tI& \cos tI\end{array}\right)\mathrm{.}$ The path $\mathrm{diag}(a,I)R_t\mathrm{diag}(I,a^{-1})R_t^{-1}$, $0\le t\le \pi \mathrm{/}2$, starts at $\mathrm{diag}(a,a^{-1})$ and ends at $I$. Hence there is an explicit orthogonal homotopy $U_s$ from $U$ to $I$ over the entire upper disk.

Let $\beta$ be the stabilized target clutching map. Use $U^{-1}$ as an automorphism of its upper disk bundle. The required lower automorphism on the equator is $\beta U^{-1}{\beta }^{-1}$. Extend it over the lower disk by $h_{-}(x,rv)=\beta (x,v)U_{1-r}(x,v{)}^{-1}\beta (x,v{)}^{-1}(r>0),h_{-}(x,0)=I\mathrm{.}$ Because $U_1=I$, orthogonality gives uniform continuity at the center, even though $\beta$ need not extend there. The two automorphisms obey the equatorial clutching relation and therefore define a genuine automorphism of the fixed target bundle. Composing it with the stabilized $\varphi$ gives upper gauge $U^{-1}U=I$, as required.

If $X$ is empty, the claim is vacuous; assume it is nonempty. Each closed cell of $X$ is the continuous image of a closed disk and hence compact. Since there are finitely many cells, $X$ is a finite union of compact closed-cell images, so it is compact; [F1] gives Hausdorffness.

By [F2] and step 1.1, $X$ is paracompact and Hausdorff. The bundle's local trivializations form an open cover of $X$. Compactness gives a finite subcover. Apply [F3] to this cover: its AC and DC hypotheses hold because AC is given and [F4] derives DC. The resulting locally finite partition has only finitely many nonzero members: choose for each point a neighborhood meeting finitely many supports, then take a finite subcover of those neighborhoods by compactness; the union of the finite index sets is finite. Thus the chart subcover and partition give a finite numeration.

Apply [F5] to this finite numeration. It supplies a fiberwise-linear embedding $E\hookrightarrow X\times {\mathbb{R}}^N$ and the subbundle $F$ orthogonal to its image in the standard Euclidean metric. The fiberwise orthogonal direct sum is the whole trivial bundle, so $E\oplus F\cong X\times {\mathbb{R}}^N$. AC is used to obtain the partition in [F3] and, through [F4], to discharge its DC hypothesis in step 2.1.

Trace cyclicity gives invariance under both left and right translations. For left-invariant fields the Koszul identity reduces to ${\mathrm{\nabla }}_{X}Y=[X,Y]\mathrm{/}2$: substituting this expression gives zero torsion and metric compatibility, since $⟨[X,Y],Z⟩=-⟨Y,[X,Z]⟩$. It follows by differentiating that geodesics are $\exp (tA)g$, and that parallel translation in a left-trivialized frame is $\exp (-t{\mathrm{ad}}_A\mathrm{/}2)$. Its curvature is $R(X,Y)Z=-{\mathrm{ad}}_{[X,Y]}Z\mathrm{/}4$, so $R(Y,X)X=-{\mathrm{ad}}_X^{2}Y\mathrm{/}4$. The factor $1\mathrm{/}4$ is retained throughout; rescaling the trace inner product does not change this connection.

The maps $r(g)=-g^{-1}$ and $r_i(g)=J_{i}g{J}_i$ are commuting involutive isometries. Their fixed equations are $g^2=-I$ and $g{J}_i=-J_{i}g$. Their common fixed set is smooth near each fixed point: in a normal-coordinate ball, an isometry fixing the center acts by its orthogonal derivative because it preserves geodesics and their initial velocities; hence the fixed set is the exponential of the common fixed linear subspace. The same description proves total geodesy. Its components are closed in the compact group. The ambient geodesic symmetry at a common fixed point commutes with each defining involution, so restricts to its component, proving that the component is symmetric.

Differentiating $(\exp (tA)J_j{)}^2=-I$ at $t=0$ gives $A{J}_j=-J_{j}A$; differentiating anticommutation with $J_i$ gives $A{J}_i=J_{i}A$ for $i<j$. Conversely these relations imply the two defining equations for every $t$, and the continuous curve stays in the component through $J_j$. The fixed-space description in step 1.2 shows that these are exactly its tangent directions.

In the real orthogonal two-plane decomposition of a skew generator, $\exp (A)=-I$ forces all rotation angles to be odd multiples of $\pi$. Consequently every geodesic from $I$ to $-I$ has length at least $\pi$ for the normalized metric. The group $\mathrm{S}\mathrm{O}(p)$ is a nonempty connected compact boundaryless manifold; [F2] supplies a minimizing geodesic, so its distance between these endpoints is at least $\pi$. Every orthogonal complex structure $J$ gives $\cos (\pi t)I+\sin (\pi t)J$, of length $\pi$ and midpoint $J$. This proves the first midpoint assertion. Inductively let $K$ anticommute with $J_1,\dots ,J_j$ and set $J^{\mathrm{\prime }}=K{J}_j^{-1}$. Then $(J^{\mathrm{\prime }}{)}^2=-I$, $J^{\mathrm{\prime }}$ commutes with $J_i$ for $i<j$, and it anticommutes with $J_j$. Step 2.1 gives a path $\exp (\pi t{J}^{\mathrm{\prime }})J_j$ in $P_j$ with midpoint $K$ and endpoints $\pm J_j$; it has length $\pi$.

The component $P_j$ is nonempty, connected, compact and boundaryless by step 1.2; its metric is therefore complete. Since AC implies the countable-choice premise in [F2], Hopf–Rinow supplies a minimizing path in $P_j$ between $J_j$ and $-J_j$ whenever those endpoints lie in that component. Total geodesy makes it $\exp (tA)J_j$, with $\exp (A)=-I$ and the tangent relations of step 2.1. Set $J^{\mathrm{\prime }}=\exp (A\mathrm{/}2)$. The odd rotation angles give $(J^{\mathrm{\prime }}{)}^2=-I$; the tangent relations imply that $J^{\mathrm{\prime }}$ anticommutes with $J_j$ and commutes with its predecessors. Its midpoint $K=J^{\mathrm{\prime }}{J}_j$ therefore anticommutes with all $J_1,\dots ,J_j$. The path constructed from this $K$ in step 3.1 also lies in $P_j$ and has length $\pi$, so the minimum length is $\pi$ and the midpoint correspondence is exact. Taking components gives the assertion. The symmetry at $J_j$ is $g\mapsto J_j{g}^{-1}{J}_j$, whose formula is unchanged on replacing $J_j$ by $-J_j$; hence the indicated antipodes are poles.

In an oriented orthonormal frame the generators anticommute and each has square $-1$. Reversing the product order takes $n(n-1)\mathrm{/}2=2m(4m-1)$ swaps, an even number; multiplying the $n$ generator squares contributes $(-1{)}^{4m}=1$. Thus ${\omega }^2=1$. For any orthonormal frame, pairwise anticommutation makes the product equal to the alternating product $\mathrm{Alt}(v_1,\dots ,v_n)=\frac{1}{n!}{\sum }_{\sigma }\mathrm{sgn}(\sigma )c(v_{\sigma (1)})\cdots c(v_{\sigma (n)})$. This is an alternating multilinear function of the frame vectors. An oriented orthogonal change has determinant $+1$, so it leaves this top alternating product unchanged. Hence $\omega$ is frame-independent and, in local frames, varies continuously.

Moving any unit vector $v$ past the other $n-1$ orthogonal generators gives $\omega c(v)=(-1{)}^{n-1}c(v)\omega =-c(v)\omega$. If the module also has a parity grading, $\omega$ has even parity because $n$ is even, so it preserves ${\mathrm{\Lambda }}^0$ and ${\mathrm{\Lambda }}^1$. The continuous maps $p_{\pm }=(I\pm \omega )\mathrm{/}2$ are idempotents with $p_{+}{p}_{-}=0$ and $p_{+}+p_{-}=I$. In any local frame their matrix entries vary continuously, and the trace of an idempotent equals its rank, so each image rank is locally constant. At a point choose a basis of the image; applying $p_{\pm }$ to the same frame vectors remains independent on a neighborhood and spans the image there. Thus their images are complementary subbundles (graded when a parity grading is supplied). The anticommutation identity makes $c(v)$ interchange their images, and $c(v{)}^{-1}=-c(v)$ for $\parallel v\parallel =1$, so the interchange is an isomorphism.

Pull the two eigenbundles back to the two closed disk halves of $S(\mathbb{R}\oplus V)$. Thicken the halves slightly so their overlap is an equatorial collar; radial projection in each fiber identifies that collar continuously with $S(V)$, so extend the transition $c(v)$ constantly along the radial coordinate. The opposite transition is $-c(v)$, its inverse, and the two-chart cocycle is therefore satisfied. The quotient gluing theorem [F4] gives a vector bundle, and its relative disk/sphere triple is exactly $({\mathrm{\Lambda }}^{+},c(v),{\mathrm{\Lambda }}^{-})$. The volume eigenspaces are defined by $\omega$, whereas an additional parity grading is the grading operator in [F1]; these are distinct constructions, and no identification between them is assumed.

Start in dimension two with $S=\mathbb{C}\oplus \mathbb{C}$ of even/odd parity and $c(e_1)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, $c(e_2)=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$. Their squares are $-1$ and they anticommute. For dimension $2l$ take the graded tensor of $l$ such modules, and put the $a$th plane's generators after the grading operators on preceding factors. Thus $c(v{)}^2=-\parallel v{\parallel }^2$ and the complex dimension is $2^l$. The two-dimensional matrices and their products span $M_2(\mathbb{C})$; their graded tensors span $M_{2^l}(\mathbb{C})$, so this is the complex spinor module, with even and odd parts of equal dimension for $l>0$. For $l=0$ take a single even complex line.

A spin element $s$ acts by its even Clifford representation $\rho (s)$. On the product group in [F1] define ${\rho }^c(s,z)=z\rho (s)$, with central weight one. The kernel element $(-1,-1)$ acts as $(-1)(-I)=I$, so this representation descends. It preserves grading because spin elements are even. For $g=[s,z]$ and any vector $v$, ${\rho }^c(g)c(v){\rho }^c(g{)}^{-1}=\rho (s)c(v)\rho (s{)}^{-1}=c(\lambda (s)v)$: the central scalar cancels, and Clifford conjugation gives precisely the oriented vector action. This verifies equivariance including the $U(1)$ factor; using weight zero would not descend.

Form the associated bundles $S^{\pm }=P{\times }_{{\rho }^c}{S}^{\pm }$. Step 2.1 makes $c(v):{\pi }^{\ast }{S}^{+}\to {\pi }^{\ast }{S}^{-}$ globally well-defined. It is invertible off the zero section with inverse $-c(v)\mathrm{/}\parallel v{\parallel }^2$, and hence defines the relative triple and class ${\tau }_V^c$. This uses neither a global complex structure on $V$ nor a choice of local half-angle functions on overlaps. Pullback of the given principal lift pulls back the whole symbol, proving naturality of the class.

On the fiber of a two-plane, the even-to-odd symbol is $x+iy$. On the equatorial unit circle this is exactly $z$, so clutching gives $[\gamma ]-1={\beta }_c$ with [F2]'s sign convention. On ${\mathbb{R}}^{2l}$ the tensor construction in step 1.1 gives the total symbol $c(v)\otimes 1+\mathrm{\Gamma }\otimes c(w)$ successively. Its square is the negative sum of squared lengths. The two-term tensor calculation identifies its relative class with the external product of the plane symbols: extend each two-term differential by zero, take $d=d_1\otimes 1+\mathrm{\Gamma }\otimes d_2$, and pass from the total complex to its even/odd symbol by $d-d^{\ast }$. This is exactly the displayed total Clifford symbol. In the even order $(S^{+}{T}^{+},S^{-}{T}^{-})$ and odd order $(S^{-}{T}^{+},S^{+}{T}^{-})$, its block is $\left(\begin{array}{cc}a& b^{\ast }\\ b& -a^{\ast }\end{array}\right)$. The self-adjoint relative tensor block is $\left(\begin{array}{cc}a& -b^{\ast }\\ b& a^{\ast }\end{array}\right)$; right multiplication by the global even-bundle automorphism $\mathrm{diag}(I,-I)$ identifies them. Thus using the skew-adjoint symbol changes no relative product class. Where either factor is invertible the tensor complex is contractible, and the total symbol has contracting inverse $-D\mathrm{/}(\parallel v{\parallel }^2+\parallel w{\parallel }^2)$ on the boundary. Thus the fiber class is ${\beta }_c^{⊠l}$ and product with it is an isomorphism in every degree by [F2].

Filter the finite CW pair by its relative skeleta $A=X_{-1}\subset X_0\subset \cdots \subset X_N=X$. The Thom quotients inherit the corresponding cofibrations. Pullback of the underlying real vector bundle to a cell's characteristic disk is trivial by finite-bundle homotopy invariance [F4], with an orientation-preserving trivialization. In this trivialization the global class lies on $D_{+}^k\wedge S^{2l}$. Restriction to the center is an isomorphism by the disk contraction, so the class is the pullback of its fiber class from step 4.1. This argument requires no principal-bundle trivialization. Excision and collapse of the attaching sphere identify the cell source with $K^{j-k}(\mathrm{p}\mathrm{t})$ and the Thom-cell target with $K^{j-k-2l}(\mathrm{p}\mathrm{t})$. The map is external product by the fiber class of step 4.1, hence is an isomorphism. The finitely many cells give a direct sum of these isomorphisms. The disk frame need not agree with any frame on the preceding skeleton: its boundary is collapsed, and the global class already supplies the gluing compatibility.

Cup product on the right by the fixed global class gives a morphism of the long exact sequences of $(X_k,X_{k-1},A)$ and their Thom quotients. It commutes with the boundary because in the cylinder construction the symbol is pulled back unchanged along the cylinder; tensoring then restricting to its boundary is the same as restricting then tensoring. The needed complex exact sequences are [F3] in every degree, not an assertion that unsuspended degree-zero exactness already desuspends. The five consecutive terms with central term $K^j(X_k,A)$ have outer isomorphisms by step 5.1 and the induction hypothesis for $X_{k-1}$ in degrees $j-1$ and $j$. Exactness proves the central map injective and surjective by lifting and correcting through the adjacent terms. Starting with $(A,A)$ proves the Thom isomorphism for all $j$.

For two even-rank bundles the direct-sum lift is induced by $([s,z],[t,w])\mapsto [st,zw]$. Changing either representative by $(-1,-1)$ does not change the image class, so this is well-defined. On spinors the product action is $(z\rho (s))\widehat{\otimes }(w\rho (t))=zw(\rho (s)\widehat{\otimes }\rho (t))$, exactly the weight-one representation of the direct-sum Spin-c group. Thus the tensor of the associated spinor bundles is the associated spinor bundle of the sum, and its total symbol is the one in step 4.1. The tensor-symbol product calculation proves ${\tau }_{V\oplus V^{\mathrm{\prime }}}^c={\tau }_V^c{\tau }_{V^{\mathrm{\prime }}}^c$ under the usual Thom identification. Together with step 3.1 this proves both naturality and direct-sum compatibility of the isomorphisms.

If the Hilbert space is zero-dimensional, the unique path is the unique endomorphism path and the assertion is vacuous. Otherwise, by [F1] and [F2], choose an orthonormal eigenbasis of $H$. Every eigenvalue $\lambda$ is real because self-adjointness gives $⟨Hv,v⟩=⟨v,Hv⟩$, while $Hv=\lambda v$ makes the two sides conjugates; since $H$ is gapped, $\lambda \mathrm{\ne }0$. On its eigenspace $H_t$ acts by $(1-t)\lambda +t\mathrm{sgn}(\lambda )$, which is positive for $\lambda >0$ and negative for $\lambda <0$ for every $t\in [0,1]$. Thus $H_t$ has no zero eigenvalue.

Each eigenvalue expression is continuous in $t$, so the diagonal operator path $H_t$ is continuous. At $t=1$, its eigenvalues are $\pm 1$, giving $H_1^{\ast }=H_1$ and $H_1^2=I$.

The symmetry relation implies $S\mathrm{sign}(H)S^{-1}=\mathrm{sign}(\sigma H)=\sigma \mathrm{sign}(H)$: $S$ sends a $\lambda$-eigenvector to a $\sigma \lambda$-eigenvector, and the scalar values involved are real, so the statement holds for antiunitary as well as unitary $S$. Therefore $S{H}_t{S}^{-1}=(1-t)\sigma H+t\sigma \mathrm{sign}(H)=\sigma H_t$ for every $t$.