Mixing
Coordinate independence of connections
$begingroup$
So I am trying to prove the following:
Let $V rightarrow M$ be a vector bundle $nabla$ a connection on $V$. Then there is a unique sequence of linear maps
$$ Omega^0(M;V) xrightarrow{nabla} Omega^1(M;V) xrightarrow{nabla} cdots $$
such that $nabla$ coincdies with the connection for $p=0$ and such that $$nabla (w wedge s ) = dw wedge s + (-1)^{|w|} w wedge nabla s (*) $$
Where $Omega^k(M;V) := Omega^k(M) otimes_{Omega^0(M)} Omega^0(V)$, where $Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $Omega^0(V)$ are the smooth sections of $V rightarrow M$.
A connection $nabla:Omega^0(M;V) rightarrow Omega^1(M;V)$ is defined to be a map that satisfies
$$ nabla (fs) = df otimes s + f nabla s $$
So I wanted to define $nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = sum w_i otimes e_i$ and we must have
$$ nabla s = sum_i dw_i otimes e_i + w _i wedge nabla e_i $$
The problem is I could not show this is coordinate independent.
My failed attempt:
given another local frame $f_1, ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i = sum A_{ij} f_j $ where $A_{ij} in C^infty(U)$.
Then
begin{align*}
sum_i Big(dw_i otimes sum A_{ij} f_j +w_i wedge nabla (sum A_{ij} f_j ) Big) &= sum_i Big(sum_j A_{ij} dw_i otimes f_j+ w_i wedge (sum_j dA_{ij} otimes f_j + A_{ij} nabla f_j ) Big)
end{align*}
differential-geometry algebraic-topology connections
asked Jan 17 at 8:18
CL.CL.
2,2352925
$endgroup$
add a comment |
$begingroup$
So I am trying to prove the following:
Let $V rightarrow M$ be a vector bundle $nabla$ a connection on $V$. Then there is a unique sequence of linear maps
$$ Omega^0(M;V) xrightarrow{nabla} Omega^1(M;V) xrightarrow{nabla} cdots $$
such that $nabla$ coincdies with the connection for $p=0$ and such that $$nabla (w wedge s ) = dw wedge s + (-1)^{|w|} w wedge nabla s (*) $$
Where $Omega^k(M;V) := Omega^k(M) otimes_{Omega^0(M)} Omega^0(V)$, where $Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $Omega^0(V)$ are the smooth sections of $V rightarrow M$.
A connection $nabla:Omega^0(M;V) rightarrow Omega^1(M;V)$ is defined to be a map that satisfies
$$ nabla (fs) = df otimes s + f nabla s $$
So I wanted to define $nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = sum w_i otimes e_i$ and we must have
$$ nabla s = sum_i dw_i otimes e_i + w _i wedge nabla e_i $$
The problem is I could not show this is coordinate independent.
My failed attempt:
given another local frame $f_1, ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i = sum A_{ij} f_j $ where $A_{ij} in C^infty(U)$.
Then
begin{align*}
sum_i Big(dw_i otimes sum A_{ij} f_j +w_i wedge nabla (sum A_{ij} f_j ) Big) &= sum_i Big(sum_j A_{ij} dw_i otimes f_j+ w_i wedge (sum_j dA_{ij} otimes f_j + A_{ij} nabla f_j ) Big)
end{align*}
differential-geometry algebraic-topology connections
asked Jan 17 at 8:18
CL.CL.
2,2352925
$endgroup$
add a comment |
$begingroup$
So I am trying to prove the following:
Let $V rightarrow M$ be a vector bundle $nabla$ a connection on $V$. Then there is a unique sequence of linear maps
$$ Omega^0(M;V) xrightarrow{nabla} Omega^1(M;V) xrightarrow{nabla} cdots $$
such that $nabla$ coincdies with the connection for $p=0$ and such that $$nabla (w wedge s ) = dw wedge s + (-1)^{|w|} w wedge nabla s (*) $$
Where $Omega^k(M;V) := Omega^k(M) otimes_{Omega^0(M)} Omega^0(V)$, where $Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $Omega^0(V)$ are the smooth sections of $V rightarrow M$.
A connection $nabla:Omega^0(M;V) rightarrow Omega^1(M;V)$ is defined to be a map that satisfies
$$ nabla (fs) = df otimes s + f nabla s $$
So I wanted to define $nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = sum w_i otimes e_i$ and we must have
$$ nabla s = sum_i dw_i otimes e_i + w _i wedge nabla e_i $$
The problem is I could not show this is coordinate independent.
My failed attempt:
given another local frame $f_1, ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i = sum A_{ij} f_j $ where $A_{ij} in C^infty(U)$.
Then
begin{align*}
sum_i Big(dw_i otimes sum A_{ij} f_j +w_i wedge nabla (sum A_{ij} f_j ) Big) &= sum_i Big(sum_j A_{ij} dw_i otimes f_j+ w_i wedge (sum_j dA_{ij} otimes f_j + A_{ij} nabla f_j ) Big)
end{align*}
differential-geometry algebraic-topology connections
asked Jan 17 at 8:18
CL.CL.
2,2352925
$endgroup$
So I am trying to prove the following:
Let $V rightarrow M$ be a vector bundle $nabla$ a connection on $V$. Then there is a unique sequence of linear maps
$$ Omega^0(M;V) xrightarrow{nabla} Omega^1(M;V) xrightarrow{nabla} cdots $$
such that $nabla$ coincdies with the connection for $p=0$ and such that $$nabla (w wedge s ) = dw wedge s + (-1)^{|w|} w wedge nabla s (*) $$
Where $Omega^k(M;V) := Omega^k(M) otimes_{Omega^0(M)} Omega^0(V)$, where $Omega^k(M)$ are the smooth $k$-forms ($k=0$ give smooth functions), and $Omega^0(V)$ are the smooth sections of $V rightarrow M$.
A connection $nabla:Omega^0(M;V) rightarrow Omega^1(M;V)$ is defined to be a map that satisfies
$$ nabla (fs) = df otimes s + f nabla s $$
So I wanted to define $nabla$ locally. Since given a local frame $e_i$ for $V$, we may write $s = sum w_i otimes e_i$ and we must have
$$ nabla s = sum_i dw_i otimes e_i + w _i wedge nabla e_i $$
The problem is I could not show this is coordinate independent.
My failed attempt:
given another local frame $f_1, ldots, f_r$ of $V$. Suppose $e = Af$, so $e_i = sum A_{ij} f_j $ where $A_{ij} in C^infty(U)$.
Then
begin{align*}
sum_i Big(dw_i otimes sum A_{ij} f_j +w_i wedge nabla (sum A_{ij} f_j ) Big) &= sum_i Big(sum_j A_{ij} dw_i otimes f_j+ w_i wedge (sum_j dA_{ij} otimes f_j + A_{ij} nabla f_j ) Big)
end{align*}
differential-geometry algebraic-topology connections
differential-geometry algebraic-topology connections
asked Jan 17 at 8:18
CL.CL.
2,2352925
asked Jan 17 at 8:18
CL.CL.
2,2352925
asked Jan 17 at 8:18
CL.CL.
2,2352925
asked Jan 17 at 8:18
CL.CL.
2,2352925
asked Jan 17 at 8:18
CL.CL.
2,2352925
2,2352925
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ sum_{i}sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right),$$
which is precisely the expression you want.
Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ nabla s = sum_i dw_i otimes e_i + (-1)^k w_i wedge nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i wedge dA_{ij} = d(w_iA_{ij})$.
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
$endgroup$
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
add a comment |
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1 Answer
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$begingroup$
Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ sum_{i}sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right),$$
which is precisely the expression you want.
Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ nabla s = sum_i dw_i otimes e_i + (-1)^k w_i wedge nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i wedge dA_{ij} = d(w_iA_{ij})$.
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
$endgroup$
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
add a comment |
$begingroup$
Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ sum_{i}sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right),$$
which is precisely the expression you want.
Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ nabla s = sum_i dw_i otimes e_i + (-1)^k w_i wedge nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i wedge dA_{ij} = d(w_iA_{ij})$.
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
$endgroup$
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
add a comment |
$begingroup$
Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ sum_{i}sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right),$$
which is precisely the expression you want.
Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ nabla s = sum_i dw_i otimes e_i + (-1)^k w_i wedge nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i wedge dA_{ij} = d(w_iA_{ij})$.
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
$endgroup$
Since $ A_{ij} dw_i + w_i dA_{ij} = d(w_i A_{ij})$ for all $i$ and $j$, the expression on the right-hand side reduces to
$$ sum_{i}sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right),$$
which is precisely the expression you want.
Edit: If the $w_i$'s are $k$-forms (rather than just smooth functions), then you would need to introduce an extra sign in your proposed definition:
$$ nabla s = sum_i dw_i otimes e_i + (-1)^k w_i wedge nabla e_i.$$
You'll end up with a corresponding factor of $(-1)^k$ on the right-hand side of your equation. Fortunately, everything works out because $A_{ij} dw_i + (-1)^k w_i wedge dA_{ij} = d(w_iA_{ij})$.
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
edited Jan 17 at 9:01
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
answered Jan 17 at 8:35
Kenny WongKenny Wong
19.1k21440
19.1k21440
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
add a comment |
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
Sorry, I am still unclear, I thought my desired expression is $$ sum dw_j otimes f_j + w_j nabla f_j$$ I thought $A_{ij}$ should only come in play to the sections $Omega^0(V)$ and I still use the same local coordinates for $Omega^1(M)$.
$endgroup$
– CL.
Jan 17 at 8:42
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
@CL. No, $s = sum_{j} w'_j otimes f_j$, where $w'_j := sum_j w_i A_{ij} $. So you should be aiming for $s = sum_j left( dw'_j otimes f_j + w'_j nabla f_j right) = sum_j left( d(w_i A_{ij}) otimes f_j + (w_i A_{ij}) nabla f_j right)$
$endgroup$
– Kenny Wong
Jan 17 at 8:46
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
Yes, thanks a lot.
$endgroup$
– CL.
Jan 17 at 8:54
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
$begingroup$
thanks for the edit too .
$endgroup$
– CL.
Jan 22 at 8:32
add a comment |
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