Exterior derivative is linear PDO of order $1$
$begingroup$
I am using this definition of PDO on page 19.
I want to verify my understanding by the example of exetior derivative.
The exterior derivative $d:Omega^p(M) rightarrow Omega^{p+1}(M)$ can locally be given by, for a $p$-form,
$$ w = sum_{|I|=p} a_I dx_I mapsto sum_{|I|=p} sum_{j=1}^n frac{partial{a_I}}{partial x_j} dx_j wedge dx_I$$
where $I$ subset of ${0,ldots, n }$, $n = dim M$.
Claim: $d$ is a linear PDO of order $1$.
The definition of PDO requiries that locally $P$ looks like
$$f mapsto sum _{|alpha| le k } A^alpha (y) frac{partial^{alpha}}{partial x_alpha} f(y).$$
So I think it is not obvious what $A^alpha$s are where $alpha = (0,ldots, 1, ldots, 0)$, $1$ in the $j$th position.
In these coordinates. $frac{partial w}{partial x_j}$
is $sum_{|I|=p} frac{partial a_I}{partial x_j} dx_I$Depending on $n$, $A^alpha$ is a matrix with values in $pm1$, $0$.
Is my line of thought correct?
differential-geometry differential-topology
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
$endgroup$
add a comment |
$begingroup$
I am using this definition of PDO on page 19.
I want to verify my understanding by the example of exetior derivative.
The exterior derivative $d:Omega^p(M) rightarrow Omega^{p+1}(M)$ can locally be given by, for a $p$-form,
$$ w = sum_{|I|=p} a_I dx_I mapsto sum_{|I|=p} sum_{j=1}^n frac{partial{a_I}}{partial x_j} dx_j wedge dx_I$$
where $I$ subset of ${0,ldots, n }$, $n = dim M$.
Claim: $d$ is a linear PDO of order $1$.
The definition of PDO requiries that locally $P$ looks like
$$f mapsto sum _{|alpha| le k } A^alpha (y) frac{partial^{alpha}}{partial x_alpha} f(y).$$
So I think it is not obvious what $A^alpha$s are where $alpha = (0,ldots, 1, ldots, 0)$, $1$ in the $j$th position.
In these coordinates. $frac{partial w}{partial x_j}$
is $sum_{|I|=p} frac{partial a_I}{partial x_j} dx_I$Depending on $n$, $A^alpha$ is a matrix with values in $pm1$, $0$.
Is my line of thought correct?
differential-geometry differential-topology
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
$endgroup$
$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21
add a comment |
$begingroup$
I am using this definition of PDO on page 19.
I want to verify my understanding by the example of exetior derivative.
The exterior derivative $d:Omega^p(M) rightarrow Omega^{p+1}(M)$ can locally be given by, for a $p$-form,
$$ w = sum_{|I|=p} a_I dx_I mapsto sum_{|I|=p} sum_{j=1}^n frac{partial{a_I}}{partial x_j} dx_j wedge dx_I$$
where $I$ subset of ${0,ldots, n }$, $n = dim M$.
Claim: $d$ is a linear PDO of order $1$.
The definition of PDO requiries that locally $P$ looks like
$$f mapsto sum _{|alpha| le k } A^alpha (y) frac{partial^{alpha}}{partial x_alpha} f(y).$$
So I think it is not obvious what $A^alpha$s are where $alpha = (0,ldots, 1, ldots, 0)$, $1$ in the $j$th position.
In these coordinates. $frac{partial w}{partial x_j}$
is $sum_{|I|=p} frac{partial a_I}{partial x_j} dx_I$Depending on $n$, $A^alpha$ is a matrix with values in $pm1$, $0$.
Is my line of thought correct?
differential-geometry differential-topology
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
$endgroup$
I am using this definition of PDO on page 19.
I want to verify my understanding by the example of exetior derivative.
The exterior derivative $d:Omega^p(M) rightarrow Omega^{p+1}(M)$ can locally be given by, for a $p$-form,
$$ w = sum_{|I|=p} a_I dx_I mapsto sum_{|I|=p} sum_{j=1}^n frac{partial{a_I}}{partial x_j} dx_j wedge dx_I$$
where $I$ subset of ${0,ldots, n }$, $n = dim M$.
Claim: $d$ is a linear PDO of order $1$.
The definition of PDO requiries that locally $P$ looks like
$$f mapsto sum _{|alpha| le k } A^alpha (y) frac{partial^{alpha}}{partial x_alpha} f(y).$$
So I think it is not obvious what $A^alpha$s are where $alpha = (0,ldots, 1, ldots, 0)$, $1$ in the $j$th position.
In these coordinates. $frac{partial w}{partial x_j}$
is $sum_{|I|=p} frac{partial a_I}{partial x_j} dx_I$Depending on $n$, $A^alpha$ is a matrix with values in $pm1$, $0$.
Is my line of thought correct?
differential-geometry differential-topology
differential-geometry differential-topology
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
edited Dec 7 '18 at 22:21
CL.
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
asked Dec 7 '18 at 15:47
CL.CL.
2,2382825
2,2382825
$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21
add a comment |
$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21
$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21
$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21
add a comment |
1 Answer
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$begingroup$
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,ldots,x_n$, we can write
$$dleft(sum_{|I|=p}a_Idx_Iright)=sum_{j=1}^ndx_jwedgeleft[frac{partial}{partial x_j}sum_{|I|=p}a_Idx_Iright]=sum_{j=1}^nA^jfrac{partial}{partial x_j}left(sum_{|I|=p}a_Idx_Iright) .$$
We have $n$ different $binom{n}{p+1}timesbinom{n}{p}$ matrices $A^1,ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of ${1,ldots,n}$. For $I,Jsubset{1,ldots,n}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $pm 1$ if $J=Icup{j}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
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add a comment |
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$begingroup$
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,ldots,x_n$, we can write
$$dleft(sum_{|I|=p}a_Idx_Iright)=sum_{j=1}^ndx_jwedgeleft[frac{partial}{partial x_j}sum_{|I|=p}a_Idx_Iright]=sum_{j=1}^nA^jfrac{partial}{partial x_j}left(sum_{|I|=p}a_Idx_Iright) .$$
We have $n$ different $binom{n}{p+1}timesbinom{n}{p}$ matrices $A^1,ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of ${1,ldots,n}$. For $I,Jsubset{1,ldots,n}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $pm 1$ if $J=Icup{j}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,ldots,x_n$, we can write
$$dleft(sum_{|I|=p}a_Idx_Iright)=sum_{j=1}^ndx_jwedgeleft[frac{partial}{partial x_j}sum_{|I|=p}a_Idx_Iright]=sum_{j=1}^nA^jfrac{partial}{partial x_j}left(sum_{|I|=p}a_Idx_Iright) .$$
We have $n$ different $binom{n}{p+1}timesbinom{n}{p}$ matrices $A^1,ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of ${1,ldots,n}$. For $I,Jsubset{1,ldots,n}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $pm 1$ if $J=Icup{j}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
$endgroup$
add a comment |
$begingroup$
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,ldots,x_n$, we can write
$$dleft(sum_{|I|=p}a_Idx_Iright)=sum_{j=1}^ndx_jwedgeleft[frac{partial}{partial x_j}sum_{|I|=p}a_Idx_Iright]=sum_{j=1}^nA^jfrac{partial}{partial x_j}left(sum_{|I|=p}a_Idx_Iright) .$$
We have $n$ different $binom{n}{p+1}timesbinom{n}{p}$ matrices $A^1,ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of ${1,ldots,n}$. For $I,Jsubset{1,ldots,n}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $pm 1$ if $J=Icup{j}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
$endgroup$
Your thoughts look correct. I'll just (attempt) to set the record straight.
In local coordinates $x_1,ldots,x_n$, we can write
$$dleft(sum_{|I|=p}a_Idx_Iright)=sum_{j=1}^ndx_jwedgeleft[frac{partial}{partial x_j}sum_{|I|=p}a_Idx_Iright]=sum_{j=1}^nA^jfrac{partial}{partial x_j}left(sum_{|I|=p}a_Idx_Iright) .$$
We have $n$ different $binom{n}{p+1}timesbinom{n}{p}$ matrices $A^1,ldots A^n$, which we can conveniently index by $p-$ and $(p+1)-$element subsets of ${1,ldots,n}$. For $I,Jsubset{1,ldots,n}$ with $|I|=p$, $|J|=p+1$, the $(J,I)$-entry of $A^j$ will be $pm 1$ if $J=Icup{j}$ (with the sign of the entry depending on $I$ in the obvious way), and $0$ otherwise.
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
answered Dec 8 '18 at 21:29
AweyganAweygan
14.2k21441
14.2k21441
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$begingroup$
Sorry, it is the one in page 19. I edited.
$endgroup$
– CL.
Dec 7 '18 at 22:21