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Let $\Phi: M\to S^n$ be a map from an even-dimensional, $\dim M=n$, spin manifold $M$ with the boundary $\partial M$ to a unit sphere. And $\Phi$ is locally constant near $\partial M$. If we take a point $p_0\in \Phi(\partial M)\subseteq S^n$ and let $N=S^n\setminus B_{r}(p_0)$, i.e., $N$ is obtained from $S^n$ by removing an open ball $B_\delta(p_0)$ centered at $p_0$ with sufficiently small radius $r>0$. Then $N$ is topologically a ball of dimension $n-1$ with boundary $S^{n-1}$. We now pass to doubles and extend $\Phi$ to a spin map $F:{\rm d}M\to {\rm d}N$. Consider the following two twisted Dirac operators acting on twisted spinor bundles $$\mathcal{D}_{M}^{\Phi^*\mathcal{S}(TN)}:\Gamma\big(\mathcal{S}(TM)\widehat{\otimes} \Phi^*\mathcal{S}(TN)\big)\to \Gamma\big(\mathcal{S}(TM)\widehat{\otimes} \Phi^*\mathcal{S}(TN)\big) $$ and $$\mathcal{D}_{M}^{\Phi^*\mathcal{S}(TS^n)}:\Gamma\big(\mathcal{S}(TM)\widehat{\otimes}\Phi^*\mathcal{S}(TS^n)\big)\to \Gamma\big(\mathcal{S}(TM)\widehat{\otimes} \Phi^*\mathcal{S}(TN)\big) $$

here $\widehat{\otimes}$ is the graded tensor product.

My questions are the following:

(1) Do we have $$ \Phi^*\mathcal{S}(TN)=\Phi^*\mathcal{S}(TS^n)? $$ If not, what is the difference term between them?

(2) Do we have $$ {\rm index}\big(\mathcal{D}_{M}^{+,\Phi^*\mathcal{S}(TN)}\big)={\rm index}\big(\mathcal{D}_{M}^{+,\Phi^*\mathcal{S}(TS^n)}\big)? $$ If not, what is the difference term between them?

(3) What local boundary conditions are added so that we have $$ {\rm index}\big(\mathcal{D}_{{\rm d}M}^{+,F^*\mathcal{S}(T{\rm d}N)}\big)={\rm index}\big(\mathcal{D}_{M}^{+,\Phi^*\mathcal{S}(TN)}\big)? $$

Could you give me some help with some ideas and details? Thanks in advance.

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  • $\begingroup$ I don't fully understand questions (1) and (2). $S(TN)$ is not defined on the full image of $\Phi$, so how do you take the pullback? On the other hand, the manifold $dM$ seems to have an orientation reversing symmetry (swap the halves of the double) that lifts to the spinor bundles. I would therefore expect the index on the left to vanish. $\endgroup$ Commented Oct 20, 2022 at 14:32

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