Mixing
Transformation of local connection form in a representation
$begingroup$
In this question I am trying to build up the theory of connections but without actually using fiber bundles, working solely with local objects on the base, however I am stuck on something.
For the sake of simplicity I will restrict my attention to an open set $Usubseteq M$ which is a trivialization domain for the underlying principal bundle. All objects are assumed to be defined there.
Let $G$ be a matrix Lie group with Lie algebra $mathfrak g$, and let $V$ be a finite dimensional vector space, and $rho:Grightarrowtext{GL}(V)$ be a representation. The induced Lie algebra representation will be denoted as $rho_ast:mathfrak grightarrowmathfrak{gl}(V)=text{End}(V)$.
A differential $k$-form $lambda$ with values in $V$ is said to be a tensorial differential form of type $rho$, if for a "gauge transformation" $Lambda:Urightarrow G$ it transforms as $$ lambda^prime=rho(Lambda^{-1})lambda . $$ A tensorial $0$-form of type $rho$ will be called a field of type $rho$.
A (local) connection is a $mathfrak g$-valued $1$-form on $U$, $omega$, which transforms as $$ omega^prime=text{Ad}(Lambda^{-1})omega+Lambda^{-1}dLambda=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda. $$
If $psi$ is a field of type $rho$, it's covariant derivative is defined as $$ Dpsi=dpsi+rho_ast(omega)psi. $$
I want to show that this is a tensorial 1-form of type $rho$.
If the connection form in the representation $rho$ transforms as $$ rho_ast(omega)^prime=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)+rho(Lambda)^{-1}drho(Lambda), $$ then we have $$ (Dpsi)^prime=d(rho(Lambda)^{-1}psi)+rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)rho(Lambda)^{-1}psi+rho(Lambda)^{-1}drho(Lambda)rho(Lambda)^{-1}psi \ =rho(Lambda^{-1})(dpsi+rho_ast(omega)psi)=rho(Lambda^{-1})Dpsi, $$
so this is precisely the needed transformation law.
So I'd like to show that the transformation law $omega^prime=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda$ induces the needed transformation law in the "represented" connection form $rho_ast(omega)$.
Because of the linearity of the Lie algebra representation, the two terms in the transformation of the connection form can be treated separately.
For the adjoint transformation term we have $$ rho_ast(Lambda^{-1}omegaLambda)=frac{d}{dt}|_{t=0}rho(Lambda^{-1}e^{omega t}Lambda)=frac{d}{dt}|_{t=0}rho(Lambda)^{-1}rho(e^{omega t})rho(Lambda)=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda), $$ where the homomorphism property of $rho$ had been used, and the exponential of $omega$ is formal in the sense that it is just a composition with $omega$, and to properly understand it, one may plug in an arbitrary vector field to $omega$.
I have issues with the Maurer-Cartan term $Lambda^{-1}dLambda$, however, as here the "formal exponential" interpretation seems iffy, and I don't know how to handle the presence of $d$.
The only thing I can do is to assume there is a Lie algebra valued function $A:Urightarrowmathfrak g$ such that $Lambda=e^A$. Because then the Maurer-Cartan term is $$ Lambda^{-1}dLambda=e^{-A}de^A=Lambda^{-1}dALambda=text{Ad}(Lambda^{-1})dA. $$
Then we can use the previous result to show that $$ rho_ast(Lambda^{-1}dLambda)=rho_ast(text{Ad}(Lambda^{-1})dA)=rho(Lambda)^{-1}rho_ast(dA)rho(Lambda)=rho(Lambda)^{-1}drho_ast(A)rho(Lambda) \ =rho(Lambda)^{-1}drho_ast(A)rho(e^A)=rho(Lambda)^{-1}drho_ast(A)e^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)} \ =rho(Lambda)^{-1}drho(e^A)=rho(Lambda)^{-1}drho(Lambda), $$ which is the desired expression.
Question 1: Is my assumption that there is an $A:Urightarrowmathfrak g$ such that $Lambda=e^A$ correct? Seems like a restrictive assumption to me, even if physicists do things like that all the time.
Question 2: Is there an elementary derivation of this (elementary = uses matrix groups without left-invariant vector fields or anything like that) that doesn't rely on this assumption, or any other derivation that is more appealing than what I have done here?
differential-geometry lie-groups lie-algebras connections
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
$endgroup$
add a comment |
$begingroup$
In this question I am trying to build up the theory of connections but without actually using fiber bundles, working solely with local objects on the base, however I am stuck on something.
For the sake of simplicity I will restrict my attention to an open set $Usubseteq M$ which is a trivialization domain for the underlying principal bundle. All objects are assumed to be defined there.
Let $G$ be a matrix Lie group with Lie algebra $mathfrak g$, and let $V$ be a finite dimensional vector space, and $rho:Grightarrowtext{GL}(V)$ be a representation. The induced Lie algebra representation will be denoted as $rho_ast:mathfrak grightarrowmathfrak{gl}(V)=text{End}(V)$.
A differential $k$-form $lambda$ with values in $V$ is said to be a tensorial differential form of type $rho$, if for a "gauge transformation" $Lambda:Urightarrow G$ it transforms as $$ lambda^prime=rho(Lambda^{-1})lambda . $$ A tensorial $0$-form of type $rho$ will be called a field of type $rho$.
A (local) connection is a $mathfrak g$-valued $1$-form on $U$, $omega$, which transforms as $$ omega^prime=text{Ad}(Lambda^{-1})omega+Lambda^{-1}dLambda=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda. $$
If $psi$ is a field of type $rho$, it's covariant derivative is defined as $$ Dpsi=dpsi+rho_ast(omega)psi. $$
I want to show that this is a tensorial 1-form of type $rho$.
If the connection form in the representation $rho$ transforms as $$ rho_ast(omega)^prime=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)+rho(Lambda)^{-1}drho(Lambda), $$ then we have $$ (Dpsi)^prime=d(rho(Lambda)^{-1}psi)+rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)rho(Lambda)^{-1}psi+rho(Lambda)^{-1}drho(Lambda)rho(Lambda)^{-1}psi \ =rho(Lambda^{-1})(dpsi+rho_ast(omega)psi)=rho(Lambda^{-1})Dpsi, $$
so this is precisely the needed transformation law.
So I'd like to show that the transformation law $omega^prime=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda$ induces the needed transformation law in the "represented" connection form $rho_ast(omega)$.
Because of the linearity of the Lie algebra representation, the two terms in the transformation of the connection form can be treated separately.
For the adjoint transformation term we have $$ rho_ast(Lambda^{-1}omegaLambda)=frac{d}{dt}|_{t=0}rho(Lambda^{-1}e^{omega t}Lambda)=frac{d}{dt}|_{t=0}rho(Lambda)^{-1}rho(e^{omega t})rho(Lambda)=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda), $$ where the homomorphism property of $rho$ had been used, and the exponential of $omega$ is formal in the sense that it is just a composition with $omega$, and to properly understand it, one may plug in an arbitrary vector field to $omega$.
I have issues with the Maurer-Cartan term $Lambda^{-1}dLambda$, however, as here the "formal exponential" interpretation seems iffy, and I don't know how to handle the presence of $d$.
The only thing I can do is to assume there is a Lie algebra valued function $A:Urightarrowmathfrak g$ such that $Lambda=e^A$. Because then the Maurer-Cartan term is $$ Lambda^{-1}dLambda=e^{-A}de^A=Lambda^{-1}dALambda=text{Ad}(Lambda^{-1})dA. $$
Then we can use the previous result to show that $$ rho_ast(Lambda^{-1}dLambda)=rho_ast(text{Ad}(Lambda^{-1})dA)=rho(Lambda)^{-1}rho_ast(dA)rho(Lambda)=rho(Lambda)^{-1}drho_ast(A)rho(Lambda) \ =rho(Lambda)^{-1}drho_ast(A)rho(e^A)=rho(Lambda)^{-1}drho_ast(A)e^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)} \ =rho(Lambda)^{-1}drho(e^A)=rho(Lambda)^{-1}drho(Lambda), $$ which is the desired expression.
Question 1: Is my assumption that there is an $A:Urightarrowmathfrak g$ such that $Lambda=e^A$ correct? Seems like a restrictive assumption to me, even if physicists do things like that all the time.
Question 2: Is there an elementary derivation of this (elementary = uses matrix groups without left-invariant vector fields or anything like that) that doesn't rely on this assumption, or any other derivation that is more appealing than what I have done here?
differential-geometry lie-groups lie-algebras connections
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
$endgroup$
$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44
add a comment |
$begingroup$
In this question I am trying to build up the theory of connections but without actually using fiber bundles, working solely with local objects on the base, however I am stuck on something.
For the sake of simplicity I will restrict my attention to an open set $Usubseteq M$ which is a trivialization domain for the underlying principal bundle. All objects are assumed to be defined there.
Let $G$ be a matrix Lie group with Lie algebra $mathfrak g$, and let $V$ be a finite dimensional vector space, and $rho:Grightarrowtext{GL}(V)$ be a representation. The induced Lie algebra representation will be denoted as $rho_ast:mathfrak grightarrowmathfrak{gl}(V)=text{End}(V)$.
A differential $k$-form $lambda$ with values in $V$ is said to be a tensorial differential form of type $rho$, if for a "gauge transformation" $Lambda:Urightarrow G$ it transforms as $$ lambda^prime=rho(Lambda^{-1})lambda . $$ A tensorial $0$-form of type $rho$ will be called a field of type $rho$.
A (local) connection is a $mathfrak g$-valued $1$-form on $U$, $omega$, which transforms as $$ omega^prime=text{Ad}(Lambda^{-1})omega+Lambda^{-1}dLambda=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda. $$
If $psi$ is a field of type $rho$, it's covariant derivative is defined as $$ Dpsi=dpsi+rho_ast(omega)psi. $$
I want to show that this is a tensorial 1-form of type $rho$.
If the connection form in the representation $rho$ transforms as $$ rho_ast(omega)^prime=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)+rho(Lambda)^{-1}drho(Lambda), $$ then we have $$ (Dpsi)^prime=d(rho(Lambda)^{-1}psi)+rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)rho(Lambda)^{-1}psi+rho(Lambda)^{-1}drho(Lambda)rho(Lambda)^{-1}psi \ =rho(Lambda^{-1})(dpsi+rho_ast(omega)psi)=rho(Lambda^{-1})Dpsi, $$
so this is precisely the needed transformation law.
So I'd like to show that the transformation law $omega^prime=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda$ induces the needed transformation law in the "represented" connection form $rho_ast(omega)$.
Because of the linearity of the Lie algebra representation, the two terms in the transformation of the connection form can be treated separately.
For the adjoint transformation term we have $$ rho_ast(Lambda^{-1}omegaLambda)=frac{d}{dt}|_{t=0}rho(Lambda^{-1}e^{omega t}Lambda)=frac{d}{dt}|_{t=0}rho(Lambda)^{-1}rho(e^{omega t})rho(Lambda)=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda), $$ where the homomorphism property of $rho$ had been used, and the exponential of $omega$ is formal in the sense that it is just a composition with $omega$, and to properly understand it, one may plug in an arbitrary vector field to $omega$.
I have issues with the Maurer-Cartan term $Lambda^{-1}dLambda$, however, as here the "formal exponential" interpretation seems iffy, and I don't know how to handle the presence of $d$.
The only thing I can do is to assume there is a Lie algebra valued function $A:Urightarrowmathfrak g$ such that $Lambda=e^A$. Because then the Maurer-Cartan term is $$ Lambda^{-1}dLambda=e^{-A}de^A=Lambda^{-1}dALambda=text{Ad}(Lambda^{-1})dA. $$
Then we can use the previous result to show that $$ rho_ast(Lambda^{-1}dLambda)=rho_ast(text{Ad}(Lambda^{-1})dA)=rho(Lambda)^{-1}rho_ast(dA)rho(Lambda)=rho(Lambda)^{-1}drho_ast(A)rho(Lambda) \ =rho(Lambda)^{-1}drho_ast(A)rho(e^A)=rho(Lambda)^{-1}drho_ast(A)e^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)} \ =rho(Lambda)^{-1}drho(e^A)=rho(Lambda)^{-1}drho(Lambda), $$ which is the desired expression.
Question 1: Is my assumption that there is an $A:Urightarrowmathfrak g$ such that $Lambda=e^A$ correct? Seems like a restrictive assumption to me, even if physicists do things like that all the time.
Question 2: Is there an elementary derivation of this (elementary = uses matrix groups without left-invariant vector fields or anything like that) that doesn't rely on this assumption, or any other derivation that is more appealing than what I have done here?
differential-geometry lie-groups lie-algebras connections
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
$endgroup$
In this question I am trying to build up the theory of connections but without actually using fiber bundles, working solely with local objects on the base, however I am stuck on something.
For the sake of simplicity I will restrict my attention to an open set $Usubseteq M$ which is a trivialization domain for the underlying principal bundle. All objects are assumed to be defined there.
Let $G$ be a matrix Lie group with Lie algebra $mathfrak g$, and let $V$ be a finite dimensional vector space, and $rho:Grightarrowtext{GL}(V)$ be a representation. The induced Lie algebra representation will be denoted as $rho_ast:mathfrak grightarrowmathfrak{gl}(V)=text{End}(V)$.
A differential $k$-form $lambda$ with values in $V$ is said to be a tensorial differential form of type $rho$, if for a "gauge transformation" $Lambda:Urightarrow G$ it transforms as $$ lambda^prime=rho(Lambda^{-1})lambda . $$ A tensorial $0$-form of type $rho$ will be called a field of type $rho$.
A (local) connection is a $mathfrak g$-valued $1$-form on $U$, $omega$, which transforms as $$ omega^prime=text{Ad}(Lambda^{-1})omega+Lambda^{-1}dLambda=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda. $$
If $psi$ is a field of type $rho$, it's covariant derivative is defined as $$ Dpsi=dpsi+rho_ast(omega)psi. $$
I want to show that this is a tensorial 1-form of type $rho$.
If the connection form in the representation $rho$ transforms as $$ rho_ast(omega)^prime=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)+rho(Lambda)^{-1}drho(Lambda), $$ then we have $$ (Dpsi)^prime=d(rho(Lambda)^{-1}psi)+rho(Lambda)^{-1}rho_ast(omega)rho(Lambda)rho(Lambda)^{-1}psi+rho(Lambda)^{-1}drho(Lambda)rho(Lambda)^{-1}psi \ =rho(Lambda^{-1})(dpsi+rho_ast(omega)psi)=rho(Lambda^{-1})Dpsi, $$
so this is precisely the needed transformation law.
So I'd like to show that the transformation law $omega^prime=Lambda^{-1}omegaLambda+Lambda^{-1}dLambda$ induces the needed transformation law in the "represented" connection form $rho_ast(omega)$.
Because of the linearity of the Lie algebra representation, the two terms in the transformation of the connection form can be treated separately.
For the adjoint transformation term we have $$ rho_ast(Lambda^{-1}omegaLambda)=frac{d}{dt}|_{t=0}rho(Lambda^{-1}e^{omega t}Lambda)=frac{d}{dt}|_{t=0}rho(Lambda)^{-1}rho(e^{omega t})rho(Lambda)=rho(Lambda)^{-1}rho_ast(omega)rho(Lambda), $$ where the homomorphism property of $rho$ had been used, and the exponential of $omega$ is formal in the sense that it is just a composition with $omega$, and to properly understand it, one may plug in an arbitrary vector field to $omega$.
I have issues with the Maurer-Cartan term $Lambda^{-1}dLambda$, however, as here the "formal exponential" interpretation seems iffy, and I don't know how to handle the presence of $d$.
The only thing I can do is to assume there is a Lie algebra valued function $A:Urightarrowmathfrak g$ such that $Lambda=e^A$. Because then the Maurer-Cartan term is $$ Lambda^{-1}dLambda=e^{-A}de^A=Lambda^{-1}dALambda=text{Ad}(Lambda^{-1})dA. $$
Then we can use the previous result to show that $$ rho_ast(Lambda^{-1}dLambda)=rho_ast(text{Ad}(Lambda^{-1})dA)=rho(Lambda)^{-1}rho_ast(dA)rho(Lambda)=rho(Lambda)^{-1}drho_ast(A)rho(Lambda) \ =rho(Lambda)^{-1}drho_ast(A)rho(e^A)=rho(Lambda)^{-1}drho_ast(A)e^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)}=rho(Lambda)^{-1}de^{rho_ast(A)} \ =rho(Lambda)^{-1}drho(e^A)=rho(Lambda)^{-1}drho(Lambda), $$ which is the desired expression.
Question 1: Is my assumption that there is an $A:Urightarrowmathfrak g$ such that $Lambda=e^A$ correct? Seems like a restrictive assumption to me, even if physicists do things like that all the time.
Question 2: Is there an elementary derivation of this (elementary = uses matrix groups without left-invariant vector fields or anything like that) that doesn't rely on this assumption, or any other derivation that is more appealing than what I have done here?
differential-geometry lie-groups lie-algebras connections
differential-geometry lie-groups lie-algebras connections
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
asked Jan 21 at 11:43
Bence RacskóBence Racskó
3,403823
3,403823
$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44
add a comment |
$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44
$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44
$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44
add a comment |
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$begingroup$
A matrix exponential cannot be differentiated as $de^A=dAe^A$, so this derivation is wrong.
$endgroup$
– Bence Racskó
Jan 21 at 18:44