Laplacian on non-compact domains I consider non-compact domains $\Omega$ with cylindrical ends. For example, $\Omega$ has a cover $\Omega_0 \cup (0,\infty) \times M$, where $\Omega_0$ has finite measure and $M$ is a compact manifold.
By Lockhart-McOwen's results, we know that to study the Laplacian on such domains one needs to use weighted Sobolev spaces. What about $1+\Delta$? For example, on Euclidean space $\mathbf{R}^n$, we know $1+\Delta:W^{2,p}\to L^p$ is an isomorphism. Is there a similar result for manifolds with cylindrical ends?
 A: There is indeed a pseudodifferential calculus that can handle this situation. It is called SG calculus, see e.g. here, or scattering calculus, see e.g. here. Symbol estimates (on $\mathbb R^n$) are
$$
  |\partial_x^\alpha \partial_\xi^\beta a(x,\xi)| \leq C_{\alpha\beta}
  (1+|x|)^{p-|\alpha|} (1+|\xi|)^{m-|\beta|}.
$$ 
This calculus can be introduced on manifolds $\boldsymbol X$ with cylindrical ends (also called SG or scattering manifolds). Pseudodifferential operators act between weighted function spaces. The Fredholm property of an operator $A$ is equivalent to its ellipticity, where the latter means (for a classical operator $A$) pointwise invertibility of (each component of) the triplet
$$
  (\sigma_\psi^m(A),\sigma_e^p(A),\sigma_{e,\psi}^{p,m}(A))
$$ 
(i.e., it is in fact ellipticity in a calculus with symbolic structure). In case $X=\mathbb R^n$ and $A=a(x,D)$, 


*

*$\sigma_\psi^m(A)(x,\xi) = \lim_{\lambda\to\infty} \lambda^{-m} a(x,\lambda \xi)\,$ for $(x,\xi)\in \mathbb R^n\times(\mathbb R^n\setminus0)$ is the usual principal symbol, 

*$\sigma_e^p(A)(x,\xi) = \lim_{\mu\to\infty}\mu^{-p} a(\mu x,\xi)\,$ for $(x,\xi)\in (\mathbb R^n\setminus0)\times\mathbb R^n$ is the principal "exit" symbol,

*$\sigma_{e,\psi}^{p,m}(A)$ arises as a compatibility condition between the former two:
$$
  \sigma_{e,\psi}^{p,m}(A)(x,\xi) =  \lim_{\mu\to\infty}
  \mu^{-p} \sigma_\psi^m(A)(\mu x,\xi) = \lim_{\lambda\to\infty}
  \lambda^{-m} \sigma_e^p(A)(x,\lambda \xi), 
$$
$(x,\xi)\in (\mathbb R^n\setminus0)\times(\mathbb R^n\setminus0)$.
In your example, the operator $1-\Delta$ is SG elliptic, while the operator $-\Delta$ is not.
EDIT 


*

*If one is particularly interested in the operator $1-\Delta$, with $\Delta=\Delta_g$ and $g$ being an SG metric on $X$, then one immediately gets that $1-\Delta\colon H^2(X) \longrightarrow L^2(X)$ is an isomorphism (because $1-\Delta$ is a positive selfadjoint operator in $L^2(X)$ with domain $H^2(X)$), and then - employing the SG calculus - one concludes that $1-\Delta\colon W^{2,p}(X) \longrightarrow L^p(X)$ is an isomorphism for all $1<p<\infty$ (here, the spaces $W^{\sigma,p}(X)$ for $\sigma\in\mathbb R$, $1<p<\infty$ are defined with the help of the metric $g$). 

*Lockhard and McOwen actually use a different compactification of $\mathbb R^n$. They consider infinity as a conic point. In this case, the function spaces are different. When these are defined with the help of a cone metric $g$, then there is a discrete set of $p$ for which $1-\Delta\colon W^{2,p}(X) \longrightarrow L^p(X)$ is not even a Fredholm operator. Whenever one passes over such an exceptional $p$, the index for the remaining $p$ is going to change. A good reference can be found here. 
