Definition of Differential operator
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Definition 2.2, page 19 Let $M$ be a smooth manifold and $E_i rightarrow M$ be two smooth vector bundles. A PDO $P:Gamma (M,E_0) rightarrow Gamma(M,E_1)$ of order $k$ is a a linear map which satisfies the following properties:
$P$ is local in the sense that if $s in Gamma(M,E_0)$ vanishes on an open subset $U subseteq M$, then so does $Ps$.
If $x:U rightarrow Bbb R^n$ is a chart, $phi_i: E_i Big|_U rightarrow U times Bbb K ^{p_i}$ a trivialization, then the localizaed operator $phi_1 circ P circ phi_0^{-1}$ can be written as
$$ (phi_1 circ P circ phi_0^{-1}) (f)(y) = sum_{|alpha| le k } A^{(alpha)}(y) frac{partial^alpha}{partial x_alpha} f(y) $$
for each $f in C^infty(U, Bbb K^{p_0})$ where $A^alpha:U rightarrow M_{p_1,p_0}(Bbb K)$. is a smooth function.
I returned to this definition after working on it for a while. I am confused now - why doesn't 2 => 1?
differential-geometry differential-topology differential-operators
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
$endgroup$
add a comment |
$begingroup$
Definition 2.2, page 19 Let $M$ be a smooth manifold and $E_i rightarrow M$ be two smooth vector bundles. A PDO $P:Gamma (M,E_0) rightarrow Gamma(M,E_1)$ of order $k$ is a a linear map which satisfies the following properties:
$P$ is local in the sense that if $s in Gamma(M,E_0)$ vanishes on an open subset $U subseteq M$, then so does $Ps$.
If $x:U rightarrow Bbb R^n$ is a chart, $phi_i: E_i Big|_U rightarrow U times Bbb K ^{p_i}$ a trivialization, then the localizaed operator $phi_1 circ P circ phi_0^{-1}$ can be written as
$$ (phi_1 circ P circ phi_0^{-1}) (f)(y) = sum_{|alpha| le k } A^{(alpha)}(y) frac{partial^alpha}{partial x_alpha} f(y) $$
for each $f in C^infty(U, Bbb K^{p_0})$ where $A^alpha:U rightarrow M_{p_1,p_0}(Bbb K)$. is a smooth function.
I returned to this definition after working on it for a while. I am confused now - why doesn't 2 => 1?
differential-geometry differential-topology differential-operators
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
$endgroup$
add a comment |
$begingroup$
Definition 2.2, page 19 Let $M$ be a smooth manifold and $E_i rightarrow M$ be two smooth vector bundles. A PDO $P:Gamma (M,E_0) rightarrow Gamma(M,E_1)$ of order $k$ is a a linear map which satisfies the following properties:
$P$ is local in the sense that if $s in Gamma(M,E_0)$ vanishes on an open subset $U subseteq M$, then so does $Ps$.
If $x:U rightarrow Bbb R^n$ is a chart, $phi_i: E_i Big|_U rightarrow U times Bbb K ^{p_i}$ a trivialization, then the localizaed operator $phi_1 circ P circ phi_0^{-1}$ can be written as
$$ (phi_1 circ P circ phi_0^{-1}) (f)(y) = sum_{|alpha| le k } A^{(alpha)}(y) frac{partial^alpha}{partial x_alpha} f(y) $$
for each $f in C^infty(U, Bbb K^{p_0})$ where $A^alpha:U rightarrow M_{p_1,p_0}(Bbb K)$. is a smooth function.
I returned to this definition after working on it for a while. I am confused now - why doesn't 2 => 1?
differential-geometry differential-topology differential-operators
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
$endgroup$
Definition 2.2, page 19 Let $M$ be a smooth manifold and $E_i rightarrow M$ be two smooth vector bundles. A PDO $P:Gamma (M,E_0) rightarrow Gamma(M,E_1)$ of order $k$ is a a linear map which satisfies the following properties:
$P$ is local in the sense that if $s in Gamma(M,E_0)$ vanishes on an open subset $U subseteq M$, then so does $Ps$.
If $x:U rightarrow Bbb R^n$ is a chart, $phi_i: E_i Big|_U rightarrow U times Bbb K ^{p_i}$ a trivialization, then the localizaed operator $phi_1 circ P circ phi_0^{-1}$ can be written as
$$ (phi_1 circ P circ phi_0^{-1}) (f)(y) = sum_{|alpha| le k } A^{(alpha)}(y) frac{partial^alpha}{partial x_alpha} f(y) $$
for each $f in C^infty(U, Bbb K^{p_0})$ where $A^alpha:U rightarrow M_{p_1,p_0}(Bbb K)$. is a smooth function.
I returned to this definition after working on it for a while. I am confused now - why doesn't 2 => 1?
differential-geometry differential-topology differential-operators
differential-geometry differential-topology differential-operators
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
asked Dec 10 '18 at 9:07
CL.CL.
2,2532825
2,2532825
add a comment |
add a comment |
1 Answer
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The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $P$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $Usubset M$ and sections $s_1,s_2$ of $E_0$, the fact that $s_1|_U=s_2|_U$ implies $P(s_1)|_U=P(s_2)|_U$. Thus, there is a well defined operator $P|_U:Gamma(U,E_0)toGamma(U,E_1)$ defined by choosing extensions of locally defined sections, applying $P$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $P$ to $U$ and (1) is needed for this restriction to be sufficiently closely related to $P$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.
In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
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$begingroup$
The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $P$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $Usubset M$ and sections $s_1,s_2$ of $E_0$, the fact that $s_1|_U=s_2|_U$ implies $P(s_1)|_U=P(s_2)|_U$. Thus, there is a well defined operator $P|_U:Gamma(U,E_0)toGamma(U,E_1)$ defined by choosing extensions of locally defined sections, applying $P$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $P$ to $U$ and (1) is needed for this restriction to be sufficiently closely related to $P$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.
In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $P$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $Usubset M$ and sections $s_1,s_2$ of $E_0$, the fact that $s_1|_U=s_2|_U$ implies $P(s_1)|_U=P(s_2)|_U$. Thus, there is a well defined operator $P|_U:Gamma(U,E_0)toGamma(U,E_1)$ defined by choosing extensions of locally defined sections, applying $P$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $P$ to $U$ and (1) is needed for this restriction to be sufficiently closely related to $P$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.
In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $P$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $Usubset M$ and sections $s_1,s_2$ of $E_0$, the fact that $s_1|_U=s_2|_U$ implies $P(s_1)|_U=P(s_2)|_U$. Thus, there is a well defined operator $P|_U:Gamma(U,E_0)toGamma(U,E_1)$ defined by choosing extensions of locally defined sections, applying $P$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $P$ to $U$ and (1) is needed for this restriction to be sufficiently closely related to $P$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.
In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
$endgroup$
The problem is that without imposing (1) it does not make much sense to talk about (2). Statement (1) essentialy means that $P$ can be localized in a rather naive sense without problems. The point is that (1) implies that for any open subset $Usubset M$ and sections $s_1,s_2$ of $E_0$, the fact that $s_1|_U=s_2|_U$ implies $P(s_1)|_U=P(s_2)|_U$. Thus, there is a well defined operator $P|_U:Gamma(U,E_0)toGamma(U,E_1)$ defined by choosing extensions of locally defined sections, applying $P$ and restricting the result. The condition in (2) actually implicitly uses the restriction of $P$ to $U$ and (1) is needed for this restriction to be sufficiently closely related to $P$ itself to be useful. In the context you are studying, you can think about pseudodifferential operators as examples of operators for which localization is a much more subtle issue.
In fact, it is evan true that for linear operators (1) implies (2), by the so-called Peetre-theorem.
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
answered Dec 10 '18 at 11:17
Andreas CapAndreas Cap
11.2k923
11.2k923
add a comment |
add a comment |
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