# Induced action by an involution on spinor bundle and Dirac operator

Let $$M$$ be a $$4n$$-dimensional spin manifold with a fixed Riemannian metric $$g$$. Let $$S$$ be a spinor bundle over $$M$$ and fix the Riemannian connection on it. There is a decomposition $$S=S^+\oplus S^-$$, where $$S^\pm$$ is the $$\pm$$1-eigenbundle with respect to the Clifford multiplication by $$\omega=e_1\cdot ...\cdot e_{4n}$$, where $$\{e_1,...,e_{4n}\}$$ is a positively oriented orthonormal basis of $$T_xM$$ at every $$x\in M$$.

Now let $$t:M\rightarrow M$$ be an orientation and spin structure preserving involution (fixed-point free) which is an isometry w.r.t. the metric $$g$$. This involution lifts to an action $$T:S\rightarrow S$$ on the spinor bundle.

Question 1: How does one show that $$T$$ preserves the above $$\mathbb{Z}_2$$-grading and that it commutes with the Dirac operator $$D:\Gamma(S)\rightarrow \Gamma(S)$$ defined by $$D\sigma=\sum e_j\cdot \nabla_{e_j}\sigma$$, where $$\nabla$$ is the covariant derivative associated to the Riemannian connection?

Question 2: How does the above Dirac operator induce a Dirac operator on the quotient manifold $$M/t$$?

• For the quotient to be a manifold you should require that $t$ doesn't have fixed points. Oct 19, 2020 at 15:39

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.
• Your involution lifts to an action $Dt\colon TM\to TM$. Then, you get a commuting diagram of the Clifford multiplication before and after the action of $t$ on both bundles. Oct 28, 2020 at 8:29