Connection $1$-form acting on vector fields
$begingroup$
I'm reading this paper about the c-map between special Kähler manifolds and Hyperkähler manifolds and in the introduction the authors talk about the cotangent bundle as a certain associated bundle of a principal GL(M)-bundle. Hence, the consider a connection $1$-form $omega$ which, by definition, satisfies some properties.
The problem is that the authors use a technical language I'm not used to. So I've tried to write $omega$ acting over vector fields. My procedure is as follows:
When I studied linear connections, I undesrtood the connection $1$-form as $n^2$ $1$-forms instead of an $operatorname{End}(TM)$-valued $1$-form. In components, those $1$-forms read as
$$ omega_i^j = Gamma_{ik}^j theta^k . $$
So, what I want to do is to define $omega$ as $operatorname{End}(TM)$-valued $1$-form using the above $omega_i^j$.
Hence, for each vector field $X$, I define $omega(X):TMrightarrow TM$ as
$$ omega(X)(Y)=nabla_XY - X(Y^j)e_j .$$
This expression is $C^infty(M)$-linear in $X$ trivially but also in $Y$, so it might be valid. And that is what I ask to you. IS the above expression a correct definition for a connection $1$-form? And, It would be possible to get a similar expression but without coordinates (I would like to remove the term $X(Y^j)e_j$?
Thanks
EDIT:
Let me clarify my notation. $nabla$ refers to a linear connection. I think it should be a connection on the principal bundle but since the paper is focus on special Kähler manifolds I think it will reduce to a linear connection at the end. $Gamma_{in}^k$ denotes the symbols of the connection, the Christoffel symbols. Finally, ${e_i}$ is a frame and ${theta^j}$ is the dual frame.
differential-geometry connections principal-bundles kahler-manifolds
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
$endgroup$
|
show 1 more comment
$begingroup$
I'm reading this paper about the c-map between special Kähler manifolds and Hyperkähler manifolds and in the introduction the authors talk about the cotangent bundle as a certain associated bundle of a principal GL(M)-bundle. Hence, the consider a connection $1$-form $omega$ which, by definition, satisfies some properties.
The problem is that the authors use a technical language I'm not used to. So I've tried to write $omega$ acting over vector fields. My procedure is as follows:
When I studied linear connections, I undesrtood the connection $1$-form as $n^2$ $1$-forms instead of an $operatorname{End}(TM)$-valued $1$-form. In components, those $1$-forms read as
$$ omega_i^j = Gamma_{ik}^j theta^k . $$
So, what I want to do is to define $omega$ as $operatorname{End}(TM)$-valued $1$-form using the above $omega_i^j$.
Hence, for each vector field $X$, I define $omega(X):TMrightarrow TM$ as
$$ omega(X)(Y)=nabla_XY - X(Y^j)e_j .$$
This expression is $C^infty(M)$-linear in $X$ trivially but also in $Y$, so it might be valid. And that is what I ask to you. IS the above expression a correct definition for a connection $1$-form? And, It would be possible to get a similar expression but without coordinates (I would like to remove the term $X(Y^j)e_j$?
Thanks
EDIT:
Let me clarify my notation. $nabla$ refers to a linear connection. I think it should be a connection on the principal bundle but since the paper is focus on special Kähler manifolds I think it will reduce to a linear connection at the end. $Gamma_{in}^k$ denotes the symbols of the connection, the Christoffel symbols. Finally, ${e_i}$ is a frame and ${theta^j}$ is the dual frame.
differential-geometry connections principal-bundles kahler-manifolds
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
$endgroup$
$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
1
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55
|
show 1 more comment
$begingroup$
I'm reading this paper about the c-map between special Kähler manifolds and Hyperkähler manifolds and in the introduction the authors talk about the cotangent bundle as a certain associated bundle of a principal GL(M)-bundle. Hence, the consider a connection $1$-form $omega$ which, by definition, satisfies some properties.
The problem is that the authors use a technical language I'm not used to. So I've tried to write $omega$ acting over vector fields. My procedure is as follows:
When I studied linear connections, I undesrtood the connection $1$-form as $n^2$ $1$-forms instead of an $operatorname{End}(TM)$-valued $1$-form. In components, those $1$-forms read as
$$ omega_i^j = Gamma_{ik}^j theta^k . $$
So, what I want to do is to define $omega$ as $operatorname{End}(TM)$-valued $1$-form using the above $omega_i^j$.
Hence, for each vector field $X$, I define $omega(X):TMrightarrow TM$ as
$$ omega(X)(Y)=nabla_XY - X(Y^j)e_j .$$
This expression is $C^infty(M)$-linear in $X$ trivially but also in $Y$, so it might be valid. And that is what I ask to you. IS the above expression a correct definition for a connection $1$-form? And, It would be possible to get a similar expression but without coordinates (I would like to remove the term $X(Y^j)e_j$?
Thanks
EDIT:
Let me clarify my notation. $nabla$ refers to a linear connection. I think it should be a connection on the principal bundle but since the paper is focus on special Kähler manifolds I think it will reduce to a linear connection at the end. $Gamma_{in}^k$ denotes the symbols of the connection, the Christoffel symbols. Finally, ${e_i}$ is a frame and ${theta^j}$ is the dual frame.
differential-geometry connections principal-bundles kahler-manifolds
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
$endgroup$
I'm reading this paper about the c-map between special Kähler manifolds and Hyperkähler manifolds and in the introduction the authors talk about the cotangent bundle as a certain associated bundle of a principal GL(M)-bundle. Hence, the consider a connection $1$-form $omega$ which, by definition, satisfies some properties.
The problem is that the authors use a technical language I'm not used to. So I've tried to write $omega$ acting over vector fields. My procedure is as follows:
When I studied linear connections, I undesrtood the connection $1$-form as $n^2$ $1$-forms instead of an $operatorname{End}(TM)$-valued $1$-form. In components, those $1$-forms read as
$$ omega_i^j = Gamma_{ik}^j theta^k . $$
So, what I want to do is to define $omega$ as $operatorname{End}(TM)$-valued $1$-form using the above $omega_i^j$.
Hence, for each vector field $X$, I define $omega(X):TMrightarrow TM$ as
$$ omega(X)(Y)=nabla_XY - X(Y^j)e_j .$$
This expression is $C^infty(M)$-linear in $X$ trivially but also in $Y$, so it might be valid. And that is what I ask to you. IS the above expression a correct definition for a connection $1$-form? And, It would be possible to get a similar expression but without coordinates (I would like to remove the term $X(Y^j)e_j$?
Thanks
EDIT:
Let me clarify my notation. $nabla$ refers to a linear connection. I think it should be a connection on the principal bundle but since the paper is focus on special Kähler manifolds I think it will reduce to a linear connection at the end. $Gamma_{in}^k$ denotes the symbols of the connection, the Christoffel symbols. Finally, ${e_i}$ is a frame and ${theta^j}$ is the dual frame.
differential-geometry connections principal-bundles kahler-manifolds
differential-geometry connections principal-bundles kahler-manifolds
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
edited Dec 23 '18 at 22:50
Dog_69
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
asked Dec 23 '18 at 19:52
Dog_69Dog_69
6001523
6001523
$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
1
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55
|
show 1 more comment
$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
1
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55
$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
1
1
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55
|
show 1 more comment
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$begingroup$
What is $nabla$ here? It might help to go back and say what the $theta^k$ are, too.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:15
1
$begingroup$
The expression you wrote is correct. The connection $1$-form is not intrinsically well defined on $M$, which means that it has no coordinate-free expression. It is, however, well-defined on the associated frame bundle.
$endgroup$
– Amitai Yuval
Dec 23 '18 at 22:16
$begingroup$
@Ted. Right. I'll add an edit with all the information. Sorry.
$endgroup$
– Dog_69
Dec 23 '18 at 22:41
$begingroup$
@Amitai Thanks for the comment, at least I was right. Could you write the expression you are referring to, please? Is something like $R_g^*omega=g^{-1}omega g$, for $gin GL(M)$? Or are you referring to other? Thanks again.
$endgroup$
– Dog_69
Dec 23 '18 at 22:46
$begingroup$
So the connection $nabla$ is given by $nabla e_i = sum omega_i^jotimes e_j$. This specifies $omega(e_i)$ in the notation you were using; we input a section of $TM$ and get out a $TM$-valued $1$-form.
$endgroup$
– Ted Shifrin
Dec 23 '18 at 22:55