Complex Spin-c Thom orientation from the equivariant spinor symbol

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.