# Range of Lifted Differential Operators

Let $D:C^\infty M \rightarrow C^\infty M$ a differential operator on a Riemannian manifold $M$. Define its maximal extension as closed operator $$D_\max:L^2M\supset \mathrm{dom}(D_\max):=\{u\in L^2M\vert Du\in L^2M\} \rightarrow L^2M, u \mapsto Du$$

Further suppose that $p\colon\hat M\rightarrow M$ is a finite-sheeted Riemannian covering and that $\hat D$ is the lift of $D$, i.e. $\hat Dp^*u=p^*Du$ for $u\in C^\infty M$. Similar to above define $\hat D_\max$ as closed operator in $L^2 \hat M$.

Question: If $D_\max$ has closed range, does also $\hat D_\max$ have closed range?

Below are some thoughts and proof attempts:

Examples and alternative Characterization:

If $M$ is compact, then closedness of range is often trivial. Let's look at some non-compact examples. We use the following characterization:

"A closed operator $A$ has closed range iff there is an "Poincaré-inequality" of the form $\Vert A u \Vert_{L^2} \ge C \Vert u - Qu\Vert_{L^2}$, where $Q\in \mathcal{B}(L^2M)$ is the orthogonal projection onto $\ker A$. "

E.g. if $M$ is a bounded open subset of euclidean space (with $\partial \bar M$ not too terrible) and $D= \nabla$, then the Poincaré-Wirtinger inequality tells us that $D_\max$ has closed range. However, if $M=\mathbb{R}^n$ then $\nabla$ does not have closed range. These two examples show that the closed range assumption is really a global condition, which makes it harder to deal with covering spaces. Also it suggests that $p$ better be finite-sheeted, otherwise we might for example run into trouble when lifting an operator from the torus to the euclidean plane.

Lifting the property for invariant functions:

Let $H_0\subset L^2\hat M$ be the subspace of functions which are invariant under the action of $\mathrm{Aut}(p)$, equivalently $H_0=\{p^*u \vert u\in L^2 M\}$. It's easy to see that the the Poincaré-inequality from $M$ lifts to functions in $H_0 \cap \mathrm{dom }(\hat D_\max)$. But does this imply that the Poincaré-inquality holds for all functions?

In an attempt to understand this, let's consider a $2$-sheeted cover, for simplicity $p\colon S^2\rightarrow \mathbb{R}P^2$. Then $\mathrm{Aut}(p)$ is generated by the antipodal map $a$ and we have $$L^2\hat M = \ker (a^* -1 ) \oplus \ker (a^* + 1 )=: H_0 \oplus H_1.$$ It'd be great if he had an isomorphism $T\colon H_0 \xrightarrow{\sim}H_1$ that commutes with $D$ (on say smooth functions). But this seems to be impossible. Finding $T$ which is an iso is not hard: Let $\chi$ the characteristic function of the upper half plane, then $Tu=\chi u - a^*(\chi u)$ defines an iso, but does not even preserve the smooth subspace. If we want smooth functions to be preserved, we immediately think of a multiplication operator $Tu=\varphi u$, for an antipodal map $\varphi: S^2\rightarrow S^1$ (it has to be nowhere-vanishing, so we might as well map into the circle). But this is forbidden by Borsuk-Ulam.

Using partitions of $1$?

Denote $\hat u=\vert\mathrm{Aut}(p) \vert ^{-1}\cdot \sum_{a\in \mathrm{Aut}(p)}a^*u\in H_0$ and suppose $f_1 + \dots + f_N =1$ is a partition of unity such that $a^* \mathrm{supp} f_i \cap \mathrm{supp} f_i = \emptyset$ if $a\neq \mathrm{id}$. Then all information about $u \in L^2\hat M$ is contained in the collection $(f_1 u)\hat{~},\dots , (f_1 u)\hat{~}$. For each of those we have a Poincarè-inequality and I hoped to generalize it to $u$, but for that one needs control on $\Vert df_i \Vert_{L^\infty}$ (which in general isn't even finite).