A question about Dirac operators Let $D$ be a Dirac operator on spinor bundle $S$ over even-dimensional non-compact spin manifold $X$,
$$
\left<s_1,s_2\right>_{L_2}
 = \int_X \left<s_1,s_2\right> \quad \forall s_1,s_2\in\Gamma_{cpt}(S)
$$
My question is:

if there exists a constant $C>0$ such that
$$||Ds||\geq C ||s||$$
holds outside some compact subset $K\subset X$, why are the Dirac operator $D$,$D^+$, and $D^-$ Fredholm operators?

Could you please give me some help with the details? Thanks!
 A: The right hypotheses read as follows. Let $M$ be a Riemannian manifold and let $D: \Gamma (M;E) \to \Gamma (M;E)$ be a Dirac operator; that is, $D$ is formally selfadjoint and the symbol satisfies $| smb_D (\xi)^2 | = |\xi|^2$ for each $\xi \in T^{\ast} M$.
We make the following two assumptions:
(a) $M$ is complete.
(b): there is a compact $K \subset M$ and $c>0$ be such that for all $u \in \Gamma_c (M;E)$ with $supp(u) \cap K = \emptyset$, we have $\|Du\| \geq c\|u\|$.
Hypothesis (a) implies that the closure of the unbounded operator $D: \Gamma_c (M;E) \to L^2 (M;E)$, also denoted $D: dom (D) \to L^2 (M;E)$, is selfadjoint. This is the (classical) Chernoff-Wolf Theorem; I refer to Higson-Roe, ''Analytic K-homology'', Proposition 10.2.10 for the proof (which does not use ellipticity).
Theorem: if (a) and (b) are satisfied, the closure of $D$ is Fredholm.
You get the statement you want if you allow $E$ to have a grading $\iota$, i.e. a selfadjoint isometry with $D \iota + \iota D=0$. Then you split $E=E^+ \oplus E^-$; $D$ falls apart into two operators $D^+:\Gamma (M;E^+)$ to $\Gamma (M;E^-)$ and $D^-:\Gamma (M;E^-)$ to $\Gamma (M;E^+)$, and it follows that the closures of $D^+$ and $D^-$ are both Fredholm. You get $ind (D^-)=-ind(D^+)$.
Proof: an unbounded selfadjoint operator $D: dom (D) \to L^2 (M;E)$ is Fredholm if and only if there is $b>0$ such that the spectral projection $\chi_{[-b,b]}(D)$ has finite rank (equivalently, is compact). Let $V_b:= im (\chi_{[-b,b]}(D)) \subset dom (D)$.
Lemma: for each $f \in C^\infty_c (M)$, the composition $V_b \subset L^2 (M;E) \stackrel{f}{\to} L^2 (M;E)$ is compact (using shorthand notation for the multiplication operator by $f$).
Proof: let $\varphi \in C_0 (\mathbb{R})$ with $\varphi|_{[b-b,]}=1$ and use Proposition 10.5.2 of the book quoted above. This does not use hypothesis (b). qed
Now let $\beta>0$ be such that $b+\beta < c$. Due to the completeness of $M$, there is $\mu \in C^\infty_c (M)$ with $0 \leq \mu \leq 1$, $\mu|_K=1$ and $\|[D,\mu]\|\leq \beta$.
For $u \in V_b$, we get the estimate
$$
\|u\| \leq \| \mu u\| + \| (1-\mu)u\|\leq \| \mu u\| +\frac{1}{c} \| D(1-\mu)u\| \leq 
\| \mu u\| +\frac{1}{c} \| [D,1-\mu]u\| + \frac{1}{c} \| (1-\mu) Du\|\leq 
$$
$$
\leq \| \mu u\| +\frac{\beta}{c} \| u\| + \frac{1}{c} \|  Du\|\leq 
 \| \mu u\| +\frac{b+\beta}{c} \| u\| .
$$
Since $\frac{b+\beta}{c}<1$, we may subtract $\frac{b+\beta}{c} \| u\|$ and divide by $(1-\frac{b+\beta}{c})$. The result is an estimate
$$
\| u \| \leq C \| \mu u\|
$$
for all $u \in V_b$.
To conclude, let $u_n \in V_b$ be a bounded sequence. After passage to a subsequence, $\mu u_n$ converges, and above estimate then proves that $u_n$ is a Cauchy sequence. Hence $dim (V_b)<\infty$. qed
