Let $(M,g)$ be a Riemannian Manifold and $L^2$ the Hilbert space given by the volume form associated to the metric. Let $L_0^2$ be the subspace which is orthogonal to the constant functions. When is a pseudodifferential operator on $M$ a positive operator on $L^2_0$?

For second order operators the Laplacian $\Delta$ is the main example.

For order zero, the obvious examples are multiplication by $f$ where $f \in C^\infty(M)$ is a smooth function and $f > 0$. Conversely if $f < 0$ anywhere then it is clear that the multiplication operator is not positive.

If $A$ is positive on $L^2_0$ then

$(\Delta^{p/2} A \Delta^{p/2} v, v) = (A \Delta^{p/2} v, \Delta^{p/2} v) > 0$

for $v \in L^2_0$ non-zero. So we can use the Laplacian as a sort of natural way to change the order of a given positive operator. Note that the principle symbol of such an operator is $||\xi||^{p}\sigma(A)(x,\xi)$.

How else can I construct more positive pseudodifferential operators? So far I can only come up with operators whose symbols in a fiber look like $||\xi||^{p}f(x)$. I am looking for "more interesting" symbols, such as those whose restriction to the co-sphere at a point is non-constant.

Ideally of course I would just like a global criterion for a symbol to quantize to a positive operator, but something tells me that this is a hard problem. If it is any easier, I would also be interested in specific examples, like the sphere with the round metric.

  • 1
    $\begingroup$ Of course you already considered the squares of pseudodifferential operators $A^*A$ and for some reason they do not offer enough examples for your goals $\endgroup$ Jun 18, 2012 at 7:33
  • $\begingroup$ Thanks Piero, in fact I had overlooked this method because I hadn't tried it with non-self adjoint operators! $\endgroup$
    – Eric
    Jun 18, 2012 at 15:20

3 Answers 3


I think you may be looking for this paper:

Symplectic geometry and positivity of pseudo-differential operators C. Fefferman† and D. H. Phong


In this paper we establish positivity for pseudo-differential operators under a condition that is essentially also necessary. The proof is based on a microlocalization procedure and a geometric lemma.


Basically, you must require that the principal symbol be positive except in a set of small symplectic capacity (it cannot contain a symplectically embedded unit cube).


If $A$ is a symmetric partial differential operator of order $2k$ on a compact manifold whose principal symbol is positive definite, then for $\lambda\gg 0$ the operator $A+\lambda$ is positive definite. This follows by using the theory of pseudo-differential operators with parameters discussed for example in Shubin's book.


Let $A$ be a selfadjoint (pseudo)differential operator of order 2 on $(M,g)$ with a nonnegative symbol. It is a consequence of the Fefferman-Phong inequality that $A$ is semi-bounded from below, i.e. $A+C\ge 0$, where $C$ is a constant.

Now you could object that the total symbol is not invariantly defined: true but considering that $A$ acts on half-densities (identified with functions on a Riemannian manifold), you get that $$ a_2+ \Re a_1 $$ is indeed invariantly defined. Here the symbol of $A$ is $a_2+a_1+r_0$, where $a_j$ is of order $j$, $r_0$ of order 0, $a_2\ge 0,\quad a_2+\Re a_1\ge 0$.

  • $\begingroup$ The constant C in the Fefferman-Phong inequality is positive .. $\endgroup$
    – user36539
    Sep 9, 2013 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.