Timeline for Essential self-adjointness of differential operators on compact manifolds

9 events

when toggle format	what		by	license	comment
Dec 4, 2016 at 21:17	answer	added	Raziel		timeline score: 2
Dec 17, 2011 at 14:16	history	edited	Daniel Tausk	CC BY-SA 3.0	added 705 characters in body
Dec 17, 2011 at 14:05	vote	accept	Daniel Tausk
Dec 15, 2011 at 6:00	answer	added	Terry Tao		timeline score: 10
Nov 24, 2010 at 4:15	comment	added	Daniel Tausk		Correcting small imprecision of my previous comment: if $L$ is first-order (symmetric) then $L=-i\big(X+\frac12\mathrm{div}(X)\big)+V$, with $V$ the multiplication operator by a smooth real-valued function. Since $V$ is bounded self-adjoint, the conclusion is the same...
Nov 23, 2010 at 19:33	comment	added	Daniel Tausk		($F_t$ denotes the flow of $X$).
Nov 23, 2010 at 19:33	comment	added	Daniel Tausk		By the way, the answer is also affirmative if $L$ is first-order. In that case, $L=-i\big(X+\frac12\mathrm{div}(X)\big)$, with $X$ a smooth vector field in $M$. Since $M$ is compact, $X$ is complete and we obtain $L$ as the generator of a one-parameter unitary group $U_t(f)=(f\circ F_t)\sqrt{\det\mathrm{d}F_t}$, which leaves $C^\infty(M)$ invariant. Thus $L$ is essentially self-adjoint on $C^\infty(M)$.
Nov 23, 2010 at 18:12	history	edited	Daniel Tausk	CC BY-SA 2.5	edited body
Nov 23, 2010 at 17:45	history	asked	Daniel Tausk	CC BY-SA 2.5