Criteria for Positivity of Pseudoddifferential Operators on Manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:23:38Z http://mathoverflow.net/feeds/question/99867 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99867/criteria-for-positivity-of-pseudoddifferential-operators-on-manifolds Criteria for Positivity of Pseudoddifferential Operators on Manifolds Eric 2012-06-18T01:52:18Z 2012-06-18T20:30:45Z <p>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 $L^2_0$?</p> <p>For second order operators the Laplacian $\Delta$ is the main example.</p> <p>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 $f &lt; 0$ anywhere then it is clear that the multiplication operator is not positive.</p> <p>If $A$ is positive on $L^2_0$ then</p> <p>$(\Delta^{p/2} A \Delta^{p/2} v, v) = (A \Delta^{p/2} v, \Delta^{p/2} v) > 0$</p> <p>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 $||\xi||^{p}\sigma(A)(x,\xi)$.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/99867/criteria-for-positivity-of-pseudoddifferential-operators-on-manifolds/99881#99881 Answer by Liviu Nicolaescu for Criteria for Positivity of Pseudoddifferential Operators on Manifolds Liviu Nicolaescu 2012-06-18T10:09:10Z 2012-06-18T10:09:10Z <p>If $A$ is a symmetric partial <em>differential</em> 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.</p> http://mathoverflow.net/questions/99867/criteria-for-positivity-of-pseudoddifferential-operators-on-manifolds/99909#99909 Answer by alvarezpaiva for Criteria for Positivity of Pseudoddifferential Operators on Manifolds alvarezpaiva 2012-06-18T15:27:16Z 2012-06-18T15:27:16Z <p>I think you may be looking for this paper:</p> <p>Symplectic geometry and positivity of pseudo-differential operators C. Feffermanâ€  and D. H. Phong</p> <p>Abstract</p> <p>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.</p> <p><a href="http://www.pnas.org/content/79/2/710.short" rel="nofollow">http://www.pnas.org/content/79/2/710.short</a></p> <p>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). </p> http://mathoverflow.net/questions/99867/criteria-for-positivity-of-pseudoddifferential-operators-on-manifolds/99943#99943 Answer by Bazin for Criteria for Positivity of Pseudoddifferential Operators on Manifolds Bazin 2012-06-18T20:30:45Z 2012-06-18T20:30:45Z <p>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.</p> <p>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, $a_2\ge 0,\quad a_2+\Re a_1\ge 0$.</p>