Spin and Spin-c obstruction and torsor classification

By [F2], replace the supplied bundle by a smooth oriented isometric model; this preserves its characteristic classes, frame lifts, and twisting actions. Use the cover isomorphism in [F4] to transport the positive-square convention in [F3] to the negative-square convention in [F1], and apply [F3] to that smooth model. A Spin lift exists exactly when $w_2(V)=0$, and its classes then form the stated $H^1(M;{\mathbb{F}}_2)$ torsor, with action by principal sign bundles. Transporting through the fixed frame isometry proves precisely the same assertions for the original bundle. No new smoothness hypothesis on that original bundle is used.

For completeness identify this integral-lift condition with the integral Bockstein, using [F9]. Represent $w_2(V)$ by a mod-two cocycle $z$ and choose an integer lift $\tilde{z}$ of its values on singular simplices under [A1]. Since $\delta z=0$, write $\delta \tilde{z}=2b$ uniquely with $b$ an integer degree-three cochain. Integer cochain groups have no two-torsion, so $\delta b=0$; by definition $[b]={\beta }_2[z]$. If $[b]=0$, write $b=\delta a$; then $\tilde{z}-2a$ is an integer degree-two cocycle reducing to $z$. Conversely suppose an integer cocycle $u$ has reduction cohomologous to $z$, say ${\rho }_{2}u-z=\delta v$. Lift the mod-two degree-one cochain $v$ to an integer cochain $\tilde{v}$ under [A1]. The cocycle $u-\delta \tilde{v}$ reduces exactly to $z$ and is an integer cochain lift with zero coboundary, so the Bockstein vanishes. Different choices of integer cochain lift change $b$ by an integer coboundary; changing the mod-two representative is handled by the same correction. Hence ${\beta }_2{w}_2(V)=0$ exactly when $w_2(V)$ has an integral cohomology lift.

Given a Spin-c lift $P_c$ let $L=\det (P_c)$, the associated line for $[s,z]\mapsto z^2$. By [F4], it determines a Spin lift of $V\oplus L_{\mathbb{R}}$. Smooth both summands by [F2] if necessary, and use step 1.1 and [F7] to obtain $0=w_2(V\oplus L_{\mathbb{R}})=w_2(V)+w_2(L_{\mathbb{R}})=w_2(V)+{\rho }_2{c}_1(L)\mathrm{.}$ Both first Stiefel–Whitney classes vanish because the two summands are oriented, so the Whitney formula has no cross term. Thus every determinant satisfies ${\rho }_2{c}_1(L)=w_2(V)$.

Conversely, suppose $c\in H^2(M;\mathbb{Z})$ reduces to $w_2(V)$. By [F5]–[F6], choose a complex line $L$ with $c_1(L)=c$ componentwise, give it a Hermitian metric using a subordinate partition, and use its smooth model from [F2]. Then [F7] gives $w_2(V\oplus L_{\mathbb{R}})=w_2(V)+{\rho }_{2}c=0$. By step 1.1 this direct sum has a Spin lift. Its restriction to the block-adapted frames, as proved in [F4], gives a Spin-c lift of $V$ with determinant $L$. Thus existence of an integral lift of $w_2(V)$ is sufficient, including arbitrary integral torsion classes.

By [F5] every principal lift under consideration is numerable: its trivializing cover admits a subordinate smooth partition of unity on the smooth base. Once a Spin-c lift exists, [F8] therefore applies on every path-connected component. It gives a free transitive action by complex line bundles, identified through [F6] with $H^2(-;\mathbb{Z})$. This concerns isomorphisms over the fixed frame bundle, as in [F1]; it does not assert that the determinant class distinguishes all Spin-c lifts. In fact central twisting by a line $K$ multiplies determinant transitions by the squares of the transitions of $K$, directly from the character $z\mapsto z^2$, so the determinant changes by $K^{\otimes 2}$.

Components of a manifold are open and path connected. Singular chains on their disjoint union form the direct sum of their chain complexes, so singular cochains, cocycles and cohomology in each fixed degree are their products. Bundle lifts and line twists also glue uniquely across the open disjoint components, and full AC chooses one initial lift in each nonempty lift set when necessary. Thus the componentwise existence and torsor conclusions give exactly the claimed groups and actions on all of $M$. For the empty base both lift sets consist of their empty object and each cohomology group is zero. Combining steps 1.1, 1.2, 2.1, 2.2 and 3.1 proves the complete Statement with no chosen preferred lift.