Connection/Curvature as a matrix of Real valued forms
$begingroup$
Let $P(M,G)$ be a principal $G$ bundle. Let $omega$ be a connection $1$ form on $P(M,G)$. This is a $mathfrak{g}$ valued $1$ form on $P$ i.e., for each $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$.
This Wikipedia page says we can think of $omega$ as a matrix of $1$-forms (I believe they mean the usual real valued $1$ forms). I do not understand what they mean here.
I think they are assuming $mathfrak{g}$ is sub algebra of $M(n,mathbb{R})$ for some $n$.
Then, given $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$. So, for $vin V$, $omega(p)(v)$ is a real valued $ntimes n $ matrix.
Suppose $omega(p)(v)=[a_{ij}]_{1leq i,jleq n}$. If we vary $v$ along $T_pP$, what we get are functions $a_{ij}:T_pPrightarrow mathbb{R}$.
With this, we have $omega(p)(v)=[a_{ij}(v)]_{1leq i,jleq n}$, here $a_{ij}$ are maps $T_pPrightarrow mathbb{R}$.
Ignoring $v$, we see that
$omega(p)=[a_{ij}]_{1leq i,jleq n}$ where $a_{ij}:T_pPrightarrow mathbb{R}$.
So, once we fix $pin P$, we have $a_{ij}:T_pPrightarrow mathbb{R}$. To show dependence on $p$, we write $a_{ij}(p):T_pPrightarrow mathbb{R}$.
So, for each $pin P$, we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. This is what a $mathbb{R}$ valued $1$ form on $P$ comes with. For each $pin P$, it comes with a map $T_pPrightarrow mathbb{R}$.
See $a_{ij}$ as a $1$ form on $mathbb{R}$ as for each $pin P$ we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. So, we have $n^2$ real valued $1$ forms on $P$. We can denote this by $omega=[a_{ij}]$ or in a more better way $omega=[omega_{ij}]$ (replaced the notation $a_{ij}$ with $omega_{ij}$).
Is this the matrix of $1$-forms they are talking about?
If this is true, same is the case with curvature form. Given $pin P$ we have $Omega(p):T_pPtimes T_pPrightarrow mathfrak{g}$. Again, for each pair $(v_1,v_2)$ we have a matrix $Omega(p)(v_1,v_2)=[a_{ij}]$.
Same thing as above, we can write $Omega=[Omega_{ij}]$ where $Omega_{ij}:Prightarrow Lambda^2 TP$ given by $Omega_{ij}(p):T_pPtimes T_pPrightarrow mathbb{R}$, $Omega(ij)(p)(v_1,v_2)$ is a $i,j$ element in in the matrix $Omega(p)(v_1,v_2)$ (being an element of $mathfrak{g}$).
Is this what it mean to say a connection form is given by matrix of $1$-forms?
differential-geometry curvature connections principal-bundles
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
add a comment |
$begingroup$
Let $P(M,G)$ be a principal $G$ bundle. Let $omega$ be a connection $1$ form on $P(M,G)$. This is a $mathfrak{g}$ valued $1$ form on $P$ i.e., for each $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$.
This Wikipedia page says we can think of $omega$ as a matrix of $1$-forms (I believe they mean the usual real valued $1$ forms). I do not understand what they mean here.
I think they are assuming $mathfrak{g}$ is sub algebra of $M(n,mathbb{R})$ for some $n$.
Then, given $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$. So, for $vin V$, $omega(p)(v)$ is a real valued $ntimes n $ matrix.
Suppose $omega(p)(v)=[a_{ij}]_{1leq i,jleq n}$. If we vary $v$ along $T_pP$, what we get are functions $a_{ij}:T_pPrightarrow mathbb{R}$.
With this, we have $omega(p)(v)=[a_{ij}(v)]_{1leq i,jleq n}$, here $a_{ij}$ are maps $T_pPrightarrow mathbb{R}$.
Ignoring $v$, we see that
$omega(p)=[a_{ij}]_{1leq i,jleq n}$ where $a_{ij}:T_pPrightarrow mathbb{R}$.
So, once we fix $pin P$, we have $a_{ij}:T_pPrightarrow mathbb{R}$. To show dependence on $p$, we write $a_{ij}(p):T_pPrightarrow mathbb{R}$.
So, for each $pin P$, we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. This is what a $mathbb{R}$ valued $1$ form on $P$ comes with. For each $pin P$, it comes with a map $T_pPrightarrow mathbb{R}$.
See $a_{ij}$ as a $1$ form on $mathbb{R}$ as for each $pin P$ we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. So, we have $n^2$ real valued $1$ forms on $P$. We can denote this by $omega=[a_{ij}]$ or in a more better way $omega=[omega_{ij}]$ (replaced the notation $a_{ij}$ with $omega_{ij}$).
Is this the matrix of $1$-forms they are talking about?
If this is true, same is the case with curvature form. Given $pin P$ we have $Omega(p):T_pPtimes T_pPrightarrow mathfrak{g}$. Again, for each pair $(v_1,v_2)$ we have a matrix $Omega(p)(v_1,v_2)=[a_{ij}]$.
Same thing as above, we can write $Omega=[Omega_{ij}]$ where $Omega_{ij}:Prightarrow Lambda^2 TP$ given by $Omega_{ij}(p):T_pPtimes T_pPrightarrow mathbb{R}$, $Omega(ij)(p)(v_1,v_2)$ is a $i,j$ element in in the matrix $Omega(p)(v_1,v_2)$ (being an element of $mathfrak{g}$).
Is this what it mean to say a connection form is given by matrix of $1$-forms?
differential-geometry curvature connections principal-bundles
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22
add a comment |
$begingroup$
Let $P(M,G)$ be a principal $G$ bundle. Let $omega$ be a connection $1$ form on $P(M,G)$. This is a $mathfrak{g}$ valued $1$ form on $P$ i.e., for each $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$.
This Wikipedia page says we can think of $omega$ as a matrix of $1$-forms (I believe they mean the usual real valued $1$ forms). I do not understand what they mean here.
I think they are assuming $mathfrak{g}$ is sub algebra of $M(n,mathbb{R})$ for some $n$.
Then, given $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$. So, for $vin V$, $omega(p)(v)$ is a real valued $ntimes n $ matrix.
Suppose $omega(p)(v)=[a_{ij}]_{1leq i,jleq n}$. If we vary $v$ along $T_pP$, what we get are functions $a_{ij}:T_pPrightarrow mathbb{R}$.
With this, we have $omega(p)(v)=[a_{ij}(v)]_{1leq i,jleq n}$, here $a_{ij}$ are maps $T_pPrightarrow mathbb{R}$.
Ignoring $v$, we see that
$omega(p)=[a_{ij}]_{1leq i,jleq n}$ where $a_{ij}:T_pPrightarrow mathbb{R}$.
So, once we fix $pin P$, we have $a_{ij}:T_pPrightarrow mathbb{R}$. To show dependence on $p$, we write $a_{ij}(p):T_pPrightarrow mathbb{R}$.
So, for each $pin P$, we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. This is what a $mathbb{R}$ valued $1$ form on $P$ comes with. For each $pin P$, it comes with a map $T_pPrightarrow mathbb{R}$.
See $a_{ij}$ as a $1$ form on $mathbb{R}$ as for each $pin P$ we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. So, we have $n^2$ real valued $1$ forms on $P$. We can denote this by $omega=[a_{ij}]$ or in a more better way $omega=[omega_{ij}]$ (replaced the notation $a_{ij}$ with $omega_{ij}$).
Is this the matrix of $1$-forms they are talking about?
If this is true, same is the case with curvature form. Given $pin P$ we have $Omega(p):T_pPtimes T_pPrightarrow mathfrak{g}$. Again, for each pair $(v_1,v_2)$ we have a matrix $Omega(p)(v_1,v_2)=[a_{ij}]$.
Same thing as above, we can write $Omega=[Omega_{ij}]$ where $Omega_{ij}:Prightarrow Lambda^2 TP$ given by $Omega_{ij}(p):T_pPtimes T_pPrightarrow mathbb{R}$, $Omega(ij)(p)(v_1,v_2)$ is a $i,j$ element in in the matrix $Omega(p)(v_1,v_2)$ (being an element of $mathfrak{g}$).
Is this what it mean to say a connection form is given by matrix of $1$-forms?
differential-geometry curvature connections principal-bundles
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
Let $P(M,G)$ be a principal $G$ bundle. Let $omega$ be a connection $1$ form on $P(M,G)$. This is a $mathfrak{g}$ valued $1$ form on $P$ i.e., for each $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$.
This Wikipedia page says we can think of $omega$ as a matrix of $1$-forms (I believe they mean the usual real valued $1$ forms). I do not understand what they mean here.
I think they are assuming $mathfrak{g}$ is sub algebra of $M(n,mathbb{R})$ for some $n$.
Then, given $pin P$, we have $omega(p):T_pPrightarrow mathfrak{g}$. So, for $vin V$, $omega(p)(v)$ is a real valued $ntimes n $ matrix.
Suppose $omega(p)(v)=[a_{ij}]_{1leq i,jleq n}$. If we vary $v$ along $T_pP$, what we get are functions $a_{ij}:T_pPrightarrow mathbb{R}$.
With this, we have $omega(p)(v)=[a_{ij}(v)]_{1leq i,jleq n}$, here $a_{ij}$ are maps $T_pPrightarrow mathbb{R}$.
Ignoring $v$, we see that
$omega(p)=[a_{ij}]_{1leq i,jleq n}$ where $a_{ij}:T_pPrightarrow mathbb{R}$.
So, once we fix $pin P$, we have $a_{ij}:T_pPrightarrow mathbb{R}$. To show dependence on $p$, we write $a_{ij}(p):T_pPrightarrow mathbb{R}$.
So, for each $pin P$, we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. This is what a $mathbb{R}$ valued $1$ form on $P$ comes with. For each $pin P$, it comes with a map $T_pPrightarrow mathbb{R}$.
See $a_{ij}$ as a $1$ form on $mathbb{R}$ as for each $pin P$ we have $a_{ij}(p):T_pPrightarrow mathbb{R}$. So, we have $n^2$ real valued $1$ forms on $P$. We can denote this by $omega=[a_{ij}]$ or in a more better way $omega=[omega_{ij}]$ (replaced the notation $a_{ij}$ with $omega_{ij}$).
Is this the matrix of $1$-forms they are talking about?
If this is true, same is the case with curvature form. Given $pin P$ we have $Omega(p):T_pPtimes T_pPrightarrow mathfrak{g}$. Again, for each pair $(v_1,v_2)$ we have a matrix $Omega(p)(v_1,v_2)=[a_{ij}]$.
Same thing as above, we can write $Omega=[Omega_{ij}]$ where $Omega_{ij}:Prightarrow Lambda^2 TP$ given by $Omega_{ij}(p):T_pPtimes T_pPrightarrow mathbb{R}$, $Omega(ij)(p)(v_1,v_2)$ is a $i,j$ element in in the matrix $Omega(p)(v_1,v_2)$ (being an element of $mathfrak{g}$).
Is this what it mean to say a connection form is given by matrix of $1$-forms?
differential-geometry curvature connections principal-bundles
differential-geometry curvature connections principal-bundles
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
203119
203119
$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22
add a comment |
$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22
$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22
add a comment |
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$begingroup$
This is correct when $G$ is a matrix group, in particular, a subgroup of $GL(n,Bbb R)$ in the cases you've been writing.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 6:14
$begingroup$
@TedShifrin Sir, Thank you.. Ok.. Can you give a reference where this is done like this or is it something that is known to everyone but no one writes it..
$endgroup$
– Praphulla Koushik
Dec 23 '18 at 6:22