Dimension of kernel of a operator
$begingroup$
This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.
The context:
I want to show
If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} rightarrow W^l$ has kernel whose dimension is independent of $l$.
Black box terminology explanation:
Definition: Let $E_i rightarrow M$ be two vector bundles, $P:Gamma(M,E_0) rightarrow Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as
$$ Pf = sum_{|alpha| le k } A^alpha(y) frac{partial^alpha}{partial x_alpha} f(y).$$
Definition 2: Let $E rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $Gamma(M,E)$ with respect to this norm.
Let us suppose $P:W^{k+l} rightarrow W^k$ is well defined and:
Theorem: Let $Pu=f$, $f in W^l$, $u in W^r$ for some integer $r$, then $u in W^{l+k}$.
It is claimed that then we have the kernel is independent of $l$.
How does this follow?
Reference: Pg 48-49.
functional-analysis differential-geometry sobolev-spaces vector-bundles
asked Jan 4 at 7:25
CL.CL.
2,2712925
$endgroup$
add a comment |
$begingroup$
This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.
The context:
I want to show
If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} rightarrow W^l$ has kernel whose dimension is independent of $l$.
Black box terminology explanation:
Definition: Let $E_i rightarrow M$ be two vector bundles, $P:Gamma(M,E_0) rightarrow Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as
$$ Pf = sum_{|alpha| le k } A^alpha(y) frac{partial^alpha}{partial x_alpha} f(y).$$
Definition 2: Let $E rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $Gamma(M,E)$ with respect to this norm.
Let us suppose $P:W^{k+l} rightarrow W^k$ is well defined and:
Theorem: Let $Pu=f$, $f in W^l$, $u in W^r$ for some integer $r$, then $u in W^{l+k}$.
It is claimed that then we have the kernel is independent of $l$.
How does this follow?
Reference: Pg 48-49.
functional-analysis differential-geometry sobolev-spaces vector-bundles
asked Jan 4 at 7:25
CL.CL.
2,2712925
$endgroup$
$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12
add a comment |
$begingroup$
This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.
The context:
I want to show
If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} rightarrow W^l$ has kernel whose dimension is independent of $l$.
Black box terminology explanation:
Definition: Let $E_i rightarrow M$ be two vector bundles, $P:Gamma(M,E_0) rightarrow Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as
$$ Pf = sum_{|alpha| le k } A^alpha(y) frac{partial^alpha}{partial x_alpha} f(y).$$
Definition 2: Let $E rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $Gamma(M,E)$ with respect to this norm.
Let us suppose $P:W^{k+l} rightarrow W^k$ is well defined and:
Theorem: Let $Pu=f$, $f in W^l$, $u in W^r$ for some integer $r$, then $u in W^{l+k}$.
It is claimed that then we have the kernel is independent of $l$.
How does this follow?
Reference: Pg 48-49.
functional-analysis differential-geometry sobolev-spaces vector-bundles
asked Jan 4 at 7:25
CL.CL.
2,2712925
$endgroup$
This question is simply applying a theorem. But I do not understand how . One can treat most of the content as black box. I will provide the definitions.
The context:
I want to show
If $M$ is a compact $n$-dimensional manifold, $P$ an ellipitic differential operator of order $k$, then $P:W^{k+l} rightarrow W^l$ has kernel whose dimension is independent of $l$.
Black box terminology explanation:
Definition: Let $E_i rightarrow M$ be two vector bundles, $P:Gamma(M,E_0) rightarrow Gamma(M,E_1)$ is an elliptic differential operator of order $k$, if locally $P$ can be written as
$$ Pf = sum_{|alpha| le k } A^alpha(y) frac{partial^alpha}{partial x_alpha} f(y).$$
Definition 2: Let $E rightarrow M$ be a complex vector bundle over a compact $n$-dimensional manifold. Then we can give the space of sections $Gamma(M,E)$ a Sobolev norm. We denote $W^k$ be the completion of $Gamma(M,E)$ with respect to this norm.
Let us suppose $P:W^{k+l} rightarrow W^k$ is well defined and:
Theorem: Let $Pu=f$, $f in W^l$, $u in W^r$ for some integer $r$, then $u in W^{l+k}$.
It is claimed that then we have the kernel is independent of $l$.
How does this follow?
Reference: Pg 48-49.
functional-analysis differential-geometry sobolev-spaces vector-bundles
functional-analysis differential-geometry sobolev-spaces vector-bundles
asked Jan 4 at 7:25
CL.CL.
2,2712925
asked Jan 4 at 7:25
CL.CL.
2,2712925
edited Jan 4 at 7:39
CL.
asked Jan 4 at 7:25
CL.CL.
2,2712925
asked Jan 4 at 7:25
CL.CL.
2,2712925
asked Jan 4 at 7:25
CL.CL.
2,2712925
2,2712925
$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12
add a comment |
$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12
$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12
add a comment |
1 Answer
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Suppose $u in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 in W^{ell}$ for each $ell,$ so the elliptic regularity theorem gives $u in W^{k+ell}$ for all $ell.$ So if $N_r subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} rightarrow W^r,$ we get $N_r subset N_{ell}$ for each $ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.
answered Jan 4 at 8:44
ktoiktoi
2,4161618
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Suppose $u in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 in W^{ell}$ for each $ell,$ so the elliptic regularity theorem gives $u in W^{k+ell}$ for all $ell.$ So if $N_r subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} rightarrow W^r,$ we get $N_r subset N_{ell}$ for each $ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.
answered Jan 4 at 8:44
ktoiktoi
2,4161618
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add a comment |
$begingroup$
Suppose $u in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 in W^{ell}$ for each $ell,$ so the elliptic regularity theorem gives $u in W^{k+ell}$ for all $ell.$ So if $N_r subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} rightarrow W^r,$ we get $N_r subset N_{ell}$ for each $ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.
answered Jan 4 at 8:44
ktoiktoi
2,4161618
$endgroup$
add a comment |
$begingroup$
Suppose $u in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 in W^{ell}$ for each $ell,$ so the elliptic regularity theorem gives $u in W^{k+ell}$ for all $ell.$ So if $N_r subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} rightarrow W^r,$ we get $N_r subset N_{ell}$ for each $ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.
answered Jan 4 at 8:44
ktoiktoi
2,4161618
$endgroup$
Suppose $u in W^{k+r}$ lies in the kernel of $P,$ so $Pu = 0.$ Then as $0 in W^{ell}$ for each $ell,$ so the elliptic regularity theorem gives $u in W^{k+ell}$ for all $ell.$ So if $N_r subset W^{k+r}$ is the null space of $P$ viewed as an operator $W^{k+r} rightarrow W^r,$ we get $N_r subset N_{ell}$ for each $ell.$ As $r$ was also arbitrary, we see that $N_r$ is independent of $r$ and hence its dimension also is.
answered Jan 4 at 8:44
ktoiktoi
2,4161618
answered Jan 4 at 8:44
ktoiktoi
2,4161618
answered Jan 4 at 8:44
ktoiktoi
2,4161618
answered Jan 4 at 8:44
ktoiktoi
2,4161618
2,4161618
add a comment |
add a comment |
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$begingroup$
It may be worth pointing out that your definition for an elliptic operator is wrong (what you wrote is just the definition of a linear differential operator).
$endgroup$
– Amitai Yuval
Jan 4 at 20:36
$begingroup$
Of course, a linear differential operator is said to be elliptic if its principal symbol satisfies a non-degeneracy condition.
$endgroup$
– Amitai Yuval
Jan 4 at 20:37
$begingroup$
Oh yes, thanks a lot. I will add this.
$endgroup$
– CL.
Jan 4 at 23:12