# Elliptic operator becomes Fredholm

Let $X$ be a Riemannian $n$-manifold with tubular end $\mathbb R^+\times Y$, where $Y$ is a closed $n-1$-manifold. Suppose $L:L^{p,w}_2(X)\to L^{p,w}(X)$ is the Laplacian operator which is translation invariant on the cylindrical end; here $L^{p,w}_i$ is a weighted $L^{p}_i$ space, i.e. $w$ is a function on $X$ such that on the cylindrical end part $w(t,y)=\sigma t$ and $L^{p,w}_i=e^{-w}L^P_i$.

Why is it the case that, if $L$ acting on $L^{p,w}_2(\mathbb R\times Y)\to L^{p,w}(\mathbb R\times Y)$ is Fredholm, then $L\colon L^{p,w}_2(X)\to L^{p,w}(X)$ is Fredholm?

I saw such a statement (Lemma 2.4) in Frøyshov's paper: K A Frøyshov, Monopoles over 4–manifolds containing long necks. I, Geom. Topol. 9 (2005) 1–93 MR2115668.

Assume $L$ acting on functional space over the cylinderal end is Fredholm, then restricted to this space it has finite dimensional kernel and cokernel. The kernel of functions $L$ acting on the functional space $X$ has to be a subspace of the above space after restriction, since every function on $X$ can be restricted on the cylinderal end. Therefore the issue is really when $Lf$ vanishes for $f$ defined on the cylinderal end, whether $f$ can be "extended" to a function $g$ on $X$, such that the dimension of $g|_{\partial X}=f$ is finite inside of certain functional space. The same reasoning goes for the cokernel if you pick up the complex squre root of the elliptic operator so that $L=PP^{*}$ and consider $\ker P^{*}$.

We thus reduce the problem to the classical boundary problem $Lg=0$ and $g=f$ given in the cylinderal end. I think this is where classical theory like trace operators, Fredholm alternative, etc enters in (see Evans). I do not really know what $L^{p,w}_{2}$ is so I cannot give an answer, but I think in general the answer is yes if the boundary is nice enough and one works with Sobolev spaces. In our case $Y$ should be a smooth submanifold. If you are into index theory, the "buzzword" to look for is the "Calderon projector" and "analytic Fredholm theory".

Also, the paper linked in the paper you cited seems to be quite self-contained and readable.

• $L^{p,w}_2$ is the $W^p_{i,w}$ is the paper you cited.
– DLIN
Commented Oct 18, 2017 at 1:35
• @DLIN: Maybe I am not getting your idea correctly. Do you mean your $L^{p,w}_{2}$ is just the space in page 410 of the paper? Can you elaborate your difficulty in following the author's proof in the paper you linked? Commented Oct 18, 2017 at 1:50
• Just this party, I do not understand why it is enough to say a operator $L$ on $X$ is Fredholm, if we can say that $L$ on the "twice cylindrical end" is Fredholm.
– DLIN
Commented Oct 18, 2017 at 5:16
• I do not know what you mean by "twice cylinderal end". Actually I found the proof in the paper to be quite readable, so I am a bit confused. Commented Oct 18, 2017 at 13:20
• not being an expert in the field so I am not sure, but do you mean "Calderon projector" instead of what you wrote? Commented Oct 18, 2017 at 15:00

To my mind, this can be done by a simple parametrix patching argument (I'm mimicking here the proof of 14.2.1 in Kronheimer-Mrowka's Monopoles and Three-Manifolds, but I am sure it was known much earlier).

Embed $X \setminus (3,\infty) \times Y$ into some closed $4$-manifold $Z$, isometrically. The Laplacian $L_Z:L_2^{p}(Z) \to L^{p}(Z)$ is Fredholm by elliptic theory. Take a parametrix $P_Z:L^p(Z) \to L_2^{p}(Z)$, i.e. an operator such that $P_Z L_Z - 1$ and $L_ZP_Z-1$ are compact. Similarly, for the Laplacian $L_{t}:L_2^{p,w}(\mathbb{R}\times Y) \to L^{p,w}(\mathbb{R} \times Y)$ one can onlo find $P_{t}$ such that $P_{t} L_{t}-1$ and $L_{t}P_{t}-1$ are compact.

Take $\beta_t, \beta_Z$ to be a smooth partition of unity subordinate to the covering of $X$ by $(1,\infty) \times Y$ and $X \setminus [2,\infty) \times Y$. Let $\eta_Z$ be a smooth function on $Z$ which is $1$ on the support of $\beta_Z$ and has support in $X \setminus [3,\infty) \times Y$; let $\eta_t$ be a smooth function on $\mathbb{R} \times Y$ which is $1$ on the support of $\beta_t$ and is supported in $(0,\infty) \times Y$. This way, any function $f$ on $X$ can be expressed as the sum of $\beta_Z f$ and $\beta_t f$, which extend by $0$ to functions on $Z$ and $\mathbb{R} \times Y$ respectively. Vice versa, functions $g$, $h$ on $Z$, $\mathbb{R} \times Y$ respectively may be "glued together" to form a function on $X$ given by $\eta_Z g + \eta_t h$.

Now, we construct $P:L^{p,w}(X) \to L_2^{p,w}(X)$ via $$P(u) = \eta_Z P_Z (\beta_Z u) + \eta_t P_t(\beta_t u).$$

It is straightforward to check that $LP-1$ and $PL-1$ are compact. For instance, in $$LP(u) = L \eta_Z P_Z (\beta_Z u) + L \eta_t P_t (\beta_t u),$$ the first term can be expressed as $$\eta_Z L P_Z (\beta_Z u) + K(u)$$ for some compact operator $K$ since derivatives of $\eta_Z$ have compact support and are smooth, and the inclusions $L_i^{p} \to L^{p}$ are compact for $i=1,2$ when restricted to compact subsets (to be precise, submanifolds of codimension $0$). However, $\eta_Z L P_Z (\beta_Z u) = \eta_Z L_Z P_Z(\beta_Z u) = \eta_Z \beta_Z u + K'(u) = \beta_Z u + K'(u)$ where $K'$ is some compact operator, and one sees that $LP(u) = u + \tilde K(u)$ for some compact $\tilde K$.

The compactness of $PL-1$ is proven almost identically.