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2

The action $T\colon S\to S$ is compatible with the Clifford-multiplication, hence the decomposition into $\pm1$ eigenspaces is preserved. Moreover, the spin connection is also preserved, and by the definition of the Dirac operator this should solve question 1.

2

Posted from the comments (1 2 3), by request. Of course, the spinor formula itself arguably is an explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries, as you suggest—in which case I imagine one can prove rigorously that the answer is ‘no’. If some special formulæ are instead of interest, Jessica Fintzen, ...

1

For question 2, the original manifold M will be a double cover of the quotient space since it's an involution, and if you have an operator that commutes with the involution on the covering space, it should descend to the quotient space. If there are "nice" fixed points one can get an orbifold or a stratified space, where you can still make sense of ...

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