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
16419
$endgroup$
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
16419
$endgroup$
$begingroup$
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
16419
$endgroup$
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
16419
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
16419
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
16419
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
16419
asked Dec 23 '18 at 2:02
Praphulla KoushikPraphulla Koushik
16419
16419
$begingroup$
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050007%2fconnection-curvature-as-a-matrix-of-real-valued-forms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050007%2fconnection-curvature-as-a-matrix-of-real-valued-forms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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