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Feb 26, 2015 at 7:20 comment added Bombyx mori It is not immediately clear to me how you may work with eta invariant if there is no spin structure or other geometric structure available. You can work with eta invariants for a generalized Dirac operator, not necessarily a classical Dirac operator. But how this is related to Pin structure is quite involved and unclear to me.
Apr 15, 2014 at 23:26 comment added Zitao Wang the eta invariant may depend on the (s)pin structure. So it is still possible to write it as a local expression in terms of the characteristic numbers and the (s)pin structure.
Apr 15, 2014 at 21:03 comment added Zitao Wang Thanks for the clarification. By Corollary 5.2 of Stolz's, in the case when M is orientable, the signature of the manifold is related to the eta invariant of the Dirac operator in a simple way. Namely, $\eta(M,g,\phi)=1/16\text{sign}(M) \text{mod}2\mathbb{Z}$. So the fact that the signature is computable in terms of the Pontryagin numbers imply that the eta invariant of the Dirac operator be computable in terms of Pontryagin numbers. So it would be nice to see analogous things to hold for $pin^{+}$ manifolds.
Apr 15, 2014 at 19:50 history answered Paul Siegel CC BY-SA 3.0