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For an oriented Riemannian manifold $(M^n,g)$ with spin structure, one can define the spinor bundle $\pi_g:\mathbf{S}_g\to M$.

The space of metrics is convex. So if $g_t=(1-t)g_0+tg_1$ is a family of metrics, I think all spinor bundles $\pi_{g_t}:\mathbf{S}_{g_t}\to M$ are isomorphic as smooth vector bundles over $M$. So what exactly is the dependence of the spinor bundle on the choice of metric?

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    $\begingroup$ As you say, the space of metrics is convex, hence contractible, so from a homotopy-theoretic point of view the metric is basically unique. You really only need the spin structure to define the spinor bundle. $\endgroup$ – Qiaochu Yuan Jun 30 '18 at 16:05

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