# 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. – Nick L Oct 19 '20 at 15:39

## 2 Answers

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.

• thank you for your answer! One question though: what do you mean when you say that the action is compatible with the Clifford-multiplication? Does it commute? Is it rather to be understood as a sort of homomorphism? – Kafka91 Oct 28 '20 at 8:10
• 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. – Sebastian Oct 28 '20 at 8:29

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 a Dirac operator and all the structure, possibly after resolving the space. For instance Hartmann, Lesch, and Vertman - On the domain of Dirac and Laplace type operators on stratified spaces and Albin and Gell-Redman - The index formula for families of Dirac type operators on pseudomanifolds.

• Thanks, this helped! – Kafka91 Oct 28 '20 at 8:11