[proofplan] The proof uses ellipticity to construct a pseudodifferential parametrix $Q \in \Psi^{-m}(M)$. The parametrix identities express $P$ as invertible up to smoothing remainders on the Sobolev spaces $H^{s+m}(M)$ and $H^s(M)$. On a compact manifold, smoothing remainders act compactly on Sobolev spaces. We then apply the abstract Hilbert-space criterion that a bounded operator with a two-sided inverse modulo compact operators is Fredholm. [/proofplan]

[step:Construct a Sobolev-bounded parametrix for $P$] Fix $s \in \mathbb{R}$. Define the Hilbert spaces \begin{align*} X := H^{s+m}(M), \qquad Y := H^s(M). \end{align*} By the Sobolev mapping theorem for pseudodifferential operators (citing a result not yet in the wiki: Sobolev mapping theorem for pseudodifferential operators), the [elliptic operator](/page/Elliptic%20Operator) $P \in \Psi^m(M)$ extends to a bounded [linear map](/page/Linear%20Map) \begin{align*} P:X \to Y. \end{align*} Since $P$ is elliptic, the parametrix theorem for elliptic pseudodifferential operators (citing a result not yet in the wiki: Parametrix theorem for elliptic pseudodifferential operators) gives an operator $Q \in \Psi^{-m}(M)$ and smoothing operators $R_1,R_2 \in \Psi^{-\infty}(M)$ such that, as operators on smooth functions, \begin{align*} QP = I - R_1 \end{align*} and \begin{align*} PQ = I - R_2. \end{align*} Here $I$ denotes the identity operator on the relevant function space. Again by the Sobolev mapping theorem, $Q$ extends to a bounded linear map \begin{align*} Q:Y \to X. \end{align*} Thus the two parametrix identities extend continuously to the Sobolev spaces as \begin{align*} QP = I_X - R_1:X \to X \end{align*} and \begin{align*} PQ = I_Y - R_2:Y \to Y, \end{align*} where $I_X:X \to X$ and $I_Y:Y \to Y$ are the identity maps. [/step]

[step:Identify the parametrix remainders as compact Sobolev operators]Because $R_1$ and $R_2$ are smoothing pseudodifferential operators, they improve Sobolev regularity by arbitrarily many derivatives. Since $M$ is compact, smoothing operators are compact on Sobolev spaces over compact manifolds (citing a result not yet in the wiki: Smoothing operators are compact on Sobolev spaces over compact manifolds). Hence \begin{align*} R_1:X \to X \end{align*} and \begin{align*} R_2:Y \to Y \end{align*} are compact linear operators.[/step]

[guided]The parametrix identities alone say that $Q$ is an inverse to $P$ up to the error terms $R_1$ and $R_2$. To turn this into a Fredholm statement, the crucial point is that these errors are compact on the precise Sobolev spaces in the theorem. The operators $R_1$ and $R_2$ belong to $\Psi^{-\infty}(M)$, which means that they are smoothing operators. Equivalently, for every real number $t \in \mathbb{R}$ and every real number $N \geq 0$, they act boundedly as maps \begin{align*} R_i:H^t(M) \to H^{t+N}(M) \end{align*} for $i \in \{1,2\}$. We use this with $t=s+m$ for $R_1$ and with $t=s$ for $R_2$. To see why compactness follows, choose any positive number $N>0$. The smoothing estimate gives bounded maps \begin{align*} R_1:H^{s+m}(M) \to H^{s+m+N}(M) \end{align*} and \begin{align*} R_2:H^s(M) \to H^{s+N}(M). \end{align*} On a compact manifold, the Sobolev embeddings \begin{align*} H^{s+m+N}(M) \hookrightarrow H^{s+m}(M) \end{align*} and \begin{align*} H^{s+N}(M) \hookrightarrow H^s(M) \end{align*} are compact for $N>0$. Therefore each $R_i$ factors as a bounded smoothing map followed by a compact Sobolev inclusion. A bounded map followed by a compact map is compact, so \begin{align*} R_1:H^{s+m}(M) \to H^{s+m}(M) \end{align*} and \begin{align*} R_2:H^s(M) \to H^s(M) \end{align*} are compact.[/guided]

[step:Use the compact-perturbation criterion to prove Fredholmness] We use the following Hilbert-space criterion. [claim:An operator with a two-sided inverse modulo compact operators is Fredholm] Let $X$ and $Y$ be Hilbert spaces, let $T:X \to Y$ and $S:Y \to X$ be bounded linear maps, and suppose there are compact operators $K_X:X \to X$ and $K_Y:Y \to Y$ such that \begin{align*} ST = I_X - K_X \end{align*} and \begin{align*} TS = I_Y - K_Y. \end{align*} Then $T:X \to Y$ is Fredholm. [/claim] [proof] First, for any [Hilbert space](/page/Hilbert%20Space) $H$ and [compact operator](/page/Compact%20Operator) $K:H \to H$, the operator $I_H-K$ has finite-dimensional kernel. Indeed, if $\ker(I_H-K)$ were infinite-dimensional, its closed unit ball would contain an infinite orthonormal sequence $(e_j)_{j=1}^{\infty}$. Since $e_j=Ke_j$ for every $j$, compactness of $K$ would force a subsequence of $(e_j)$ to converge in $H$, contradicting the fact that distinct orthonormal vectors have distance $\sqrt{2}$. The same compactness argument also gives that $\operatorname{Range}(I_H-K)$ is closed. Let $N:=\ker(I_H-K)$ and let $N^\perp$ denote its orthogonal complement in $H$. If the restriction \begin{align*} I_H-K:N^\perp \to H \end{align*} were not bounded below, there would be a sequence $(u_j)_{j=1}^{\infty}$ in $N^\perp$ with $\|u_j\|_H=1$ and $(I_H-K)u_j \to 0$ in $H$. Since $K$ is compact, after passing to a subsequence, $(Ku_j)$ converges in $H$. From $u_j=Ku_j+(I_H-K)u_j$, the same subsequence of $(u_j)$ converges to some $u \in H$. Passing to the limit gives $(I_H-K)u=0$, so $u \in N$. Since $N^\perp$ is closed, also $u \in N^\perp$. Hence $u=0$, contradicting $\|u\|_H=1$. Therefore there exists $c>0$ such that \begin{align*} \|(I_H-K)u\|_H \geq c\|u\|_H \end{align*} for every $u \in N^\perp$, and this lower bound implies that $\operatorname{Range}(I_H-K)$ is closed. Finally, the orthogonal complement of $\operatorname{Range}(I_H-K)$ is \begin{align*} \operatorname{Range}(I_H-K)^\perp=\ker(I_H-K)^*=\ker(I_H-K^*). \end{align*} The adjoint $K^*:H \to H$ is compact, so the first paragraph applied to $K^*$ shows that this orthogonal complement is finite-dimensional. Hence $I_H-K$ has closed range and finite-dimensional cokernel. Apply this conclusion first on $X$ with $K=K_X$. If $u \in \ker T$, then \begin{align*} 0=STu=(I_X-K_X)u. \end{align*} Thus $\ker T \subseteq \ker(I_X-K_X)$, and therefore $\ker T$ is finite-dimensional. Apply the conclusion next on $Y$ with $K=K_Y$. Since \begin{align*} I_Y-K_Y=TS, \end{align*} we have \begin{align*} \operatorname{Range}(I_Y-K_Y)\subseteq \operatorname{Range}T. \end{align*} The range of $I_Y-K_Y$ is closed and has finite codimension in $Y$. Therefore the larger subspace $\operatorname{Range}T$ also has finite codimension in $Y$, and a finite-codimensional subspace of a Hilbert space is closed. Hence $T$ has finite-dimensional kernel, closed range, and finite-dimensional cokernel. This is exactly the definition of a Fredholm operator. [/proof] We apply the claim with \begin{align*} T:=P:X \to Y, \qquad S:=Q:Y \to X, \qquad K_X:=R_1:X \to X, \qquad K_Y:=R_2:Y \to Y. \end{align*} The identities $QP=I_X-R_1$ and $PQ=I_Y-R_2$ provide the two-sided inverse modulo compact operators, and the previous step verified that $R_1$ and $R_2$ are compact. Therefore \begin{align*} P:H^{s+m}(M) \to H^s(M) \end{align*} is Fredholm. [/step]

[step:Conclude for every Sobolev index] The argument was carried out for an arbitrary fixed $s \in \mathbb{R}$. Hence, for every $s \in \mathbb{R}$, the elliptic pseudodifferential operator $P \in \Psi^m(M)$ defines a Fredholm operator \begin{align*} P:H^{s+m}(M) \to H^s(M). \end{align*} This proves the theorem. [/step]