# ellipticity independent of metric?

I am new to the theory of pseudo-differential operators on compact manifolds, but I need to use a result related to this theory in a proof I'm working on. The problem is as follows: Let $(M,g)$ be a closed, compact Riemannian manifold. It is clear that $L_g:=(\Delta_g + 1)^k$, where $\Delta_g$ is the Laplace-Beltrami operator for the metric $g$, is an elliptic pseudo-differential operator for all positive real numbers $k$. Does it follow that $L_g$ is elliptic on $(M,h)$, where $h$ is a metric conformal to $g$?

This result is easy to prove (in the affirmative) for integral values of $k$, but I am uncomfortable with the sketch of a proof I have for non-integral $k$. Any thoughts on the matter would be appreciated. The main idea that I have to work with is my unproven notion that a symbol for a pseudo-differential operator is independent of the metric, but I having problems proving this idea in the manifold setting.

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## 1 Answer

Ellipticity of a given pseudo-differential operator is a metric independent condition. By a pseudodifferential operator I mean an operator $P: C^\infty(M)\to C^\infty(M)$ satisfying H\"{o}rmander's asymptotic conditions described in Definition 2.1 of

L. Hormander: Pseudodifferential operators, Comm. Pure Appl. Math., XVIII(1965), 501-517.

The attribute given signifies that we know $Pu$ for any $u\in C^\infty(M)$.

I think the confusion is about how you define the operator $L_s:=(\Delta_g+1)^s$. More importantly, the confusion can be traced to the fact that on a manifold the identification of a smooth function with a distribution is metric dependent.

More precisely, on a manifold there are two different types of objects: generalized functions (belong to the dual of the space of smooth densities) and distributions or generalized densities (belong to the dual of the space of smooth functions).

If we fix a volume form on the manifold then we can identify a distribution with a generalized function, but this identification depends on the choice of volume form. In the Euclidean case we do not pay much attention to this, because we have a God-given volume form. On manifolds one must be more careful. The kernel of an operator acting on functions is a distribution. (An elegant way out of this is to work with $1/2$-densities.)

Returning to the concrete case at hand, for $s<0$ the integral kernel of this $(1+\Delta_g)^s$ is the function

$$K_s(x,y)=\sum_k (\lambda_k+1)^s \Psi_k(x)\Psi_k(y),$$

where $(\Psi_k)$ is a unitary basis of $L^2(M,g)$ consisting of eigenfunction,

$$\Delta_g \Psi_k=\lambda_k\Psi_k.$$

If we want to reconstruct the action of $L_s$ on functions we need to think of $K_s$ as a distribution

$$L_su(x) = \int_M K_s(x,y) u(y) d V_g(y)$$

For $s>0$ you set $L_s=(L_{-s})^{-1}$. You see that the definition of $L_s$ depends on $g$. With this definition you do get a pseudo-differential operator which is elliptic.

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