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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.

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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 –  Piero D'Ancona Jun 18 '12 at 7:33
    
Thanks Piero, in fact I had overlooked this method because I hadn't tried it with non-self adjoint operators! –  Eric Jun 18 '12 at 15:20
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3 Answers

up vote 3 down vote accepted

I think you may be looking for this paper:

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

Abstract

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.

http://www.pnas.org/content/79/2/710.short

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).

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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.

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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, $a_2\ge 0,\quad a_2+\Re a_1\ge 0$.

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The constant C in the Fefferman-Phong inequality is positive .. –  user36539 Sep 9 '13 at 19:27
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