Index of twisted Dirac operator I do not understand a step in the proof of the Lemma 11.4.1 in the book "Analytic K-Homology" by Higson, Roe.
Let $S$ be a Dirac bundle over a closed manifold $M$ and $D$ the corresponding Dirac operator. Let $E$ be a Hermitian vector bundle over $M$. Then $S \otimes E$ is again a Dirac bundle, so let $D_E$ be the corresponding twisted Dirac operator.
Let $P \colon M \to M_n(\mathbb{C})$ be the projection-valued function determining $E$. Let $D_n$ be the operator $1 \otimes D$ on $H_n := \mathbb{C}^n \otimes L^2(S)$. Let $\chi$ be a normalizing function.

Then it is claimed that $\operatorname{Index}(\chi(D_E))$ equals the index of the operator $\chi(PD_nP)$ on the space $PH_n$.

What bothers me is that the operator  $PD_nP$ does not depend on the chosen metric / connection on $E$, whereas the operator $D_E$ does depend on it. And it is never mentioned in the book that the index of the twisted operator $D_E$ is independent of this choices.
 A: Actually, the datum $P : M \to M_n(\mathbb{C})$ defines both a metric $(\cdot,\cdot)^P$ and a connection $\nabla^P$ on $E$:


*

*If $P\xi$, $P\eta \in P(C^\infty(M) \otimes \mathbb{C}^n) = \Gamma(E)$, then $$(\xi,\eta)^P_x := \langle P(x)\xi(x), P(x)\eta(x)\rangle$$ for $\langle \cdot,\cdot \rangle$ the inner product on $\mathbb{C}^n$.

*If $P\xi \in P(C^\infty(M) \otimes \mathbb{C}^n) = \Gamma(E)$, then $$\nabla^P P\xi := (1_{\Omega^1(M)} \otimes P) d P\xi,$$ where $d$ is the exterior derivative on $C^\infty(M)$ acting on $C^\infty(M) \otimes \mathbb{C}^n$ as $d \otimes 1_n$.
One then has that $PD_n P$ is, in fact, the twisted Dirac operator of the twisted spinor bundle $S \otimes E$, for $E$ endowed instead with the metric $(\cdot,\cdot)^P$ and connection $\nabla^P$. In particular, the fact that $\operatorname{Ind}(D_E) = \operatorname{Ind}(PD_n P)$ shows that $\operatorname{Ind}(D_E)$ is independent of the choice of metric and connection on $E$.
If you want another perspective on this issue, take a look at Chapter 11 of Roe's Elliptic operators, topology and asymptotic methods, which, in general, offers a very nice, concretely global-analytic complement to Higson--Roe. 
So, one can define a continuous family of Dirac operators on a vector bundle $F \to M$ (e.g., your $F = S \otimes E$) as a one-parameter family $(g_t;c_t,(\cdot,\cdot)_t,\nabla^t)$, where for each $t \in [0,1]$,


*

*$g_t$ is a Riemannian metric on $M$;

*$c_t$ is a Clifford action, $(\cdot,\cdot)_t$ a Hermitian metric, and $\nabla^t$ a Clifford connection, making $F$ into a Dirac bundle over $(M,g_t)$;


for each $t$, then, let $D_t$ be the Dirac operator of the Dirac bundle $(F,(\cdot,\cdot)_t,c_t,\nabla^t) \to (M,g_t)$. One can then show that the $D_t$ satisfy elliptic estimates uniformly in $t$, and hence prove by functional-analytic arguments [Roe, Prop. 11.13] that $t \mapsto \operatorname{Ind}(D_t)$ must therefore be constant. In your particular context, $g$ and $c$ are constant, particularly since the spinor bundle $S$ is fixed, but given your vector bundle $E$, you'd need to construct $(\cdot,\cdot)_t$ and $\nabla^t$ to appy Roe's proposition. Then main ideas are these:


*

*If $(\cdot,\cdot)_0$ and $(\cdot,\cdot)_1$ are two smooth metrics on $E$, then there exists a fibre-wise positive-definite (with respect to $(\cdot,\cdot)_0$) $T \in \operatorname{End}(E)$ such that $(\xi,\eta)_1 = (\xi,T\eta)_0$ for all $\xi,\eta \in \Gamma(E)$. Since $T$ is fibrewise positive-definite with respect to $(\cdot,\cdot)_0$, you can define $\log T \in \operatorname{End}(E)$ with $T_x = \exp((\log T)_x)$ for each $x$; then $(\cdot,\cdot)_t$ can be defined by $$(\xi(x),\eta(x))_{t,x} := (\xi(x),\exp(t(\log T)_x)\eta(x))_{0,x}.$$

*If the metric $(\cdot,\cdot)$ is fixed, and $\nabla$ and $\nabla^\prime$ are two metric connections on $E$, then $M := \nabla^\prime - \nabla$ is a skew-adjoint $\operatorname{End}(E)$-valued $1$-form, and hence you can take $\nabla^t := \nabla + t M$.


These ingredients then allow you to define the corresponding continuous family of Dirac operators on $S \otimes E$.
