Mixing
Holonomy algebra acts on curvature tensors
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I am trying to read a work that involves affine connections, and every time I am in doubt about something I get in trouble, because all the references I find consider connections on principal bundles, and the time I would spend studying them - I don't even know the definitions - to be able to understand the explanations is so big that I cannot do it. I wuold like to know, then, if someone could help me with an explanation that doesn't involve the notion of a connection on a principal bundle.
The current issue is the following:
Given a manifold with an affine conncetion $nabla$ whose curvature tensor is $R$, why does $nabla R = 0$ imply $(B cdot R)(X, Y) := -R(X, BY)-R(BX, Y)+BR(X, Y)-R(X, Y)B = 0 $ for every $X, Y in T_pM$ and $B$ in the Lie algebra of the holonomy group $Hol_p(nabla)$?
In the work I am reading, the author says that $(Bcdot R)$ is the "natural action of the holonomy Lie algebra on curvature tensors".
Thank you very much!
connections holonomy
asked 2 days ago
Valentino
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I am trying to read a work that involves affine connections, and every time I am in doubt about something I get in trouble, because all the references I find consider connections on principal bundles, and the time I would spend studying them - I don't even know the definitions - to be able to understand the explanations is so big that I cannot do it. I wuold like to know, then, if someone could help me with an explanation that doesn't involve the notion of a connection on a principal bundle.
The current issue is the following:
Given a manifold with an affine conncetion $nabla$ whose curvature tensor is $R$, why does $nabla R = 0$ imply $(B cdot R)(X, Y) := -R(X, BY)-R(BX, Y)+BR(X, Y)-R(X, Y)B = 0 $ for every $X, Y in T_pM$ and $B$ in the Lie algebra of the holonomy group $Hol_p(nabla)$?
In the work I am reading, the author says that $(Bcdot R)$ is the "natural action of the holonomy Lie algebra on curvature tensors".
Thank you very much!
connections holonomy
asked 2 days ago
Valentino
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday
add a comment |
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0
down vote
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up vote
0
down vote
favorite
I am trying to read a work that involves affine connections, and every time I am in doubt about something I get in trouble, because all the references I find consider connections on principal bundles, and the time I would spend studying them - I don't even know the definitions - to be able to understand the explanations is so big that I cannot do it. I wuold like to know, then, if someone could help me with an explanation that doesn't involve the notion of a connection on a principal bundle.
The current issue is the following:
Given a manifold with an affine conncetion $nabla$ whose curvature tensor is $R$, why does $nabla R = 0$ imply $(B cdot R)(X, Y) := -R(X, BY)-R(BX, Y)+BR(X, Y)-R(X, Y)B = 0 $ for every $X, Y in T_pM$ and $B$ in the Lie algebra of the holonomy group $Hol_p(nabla)$?
In the work I am reading, the author says that $(Bcdot R)$ is the "natural action of the holonomy Lie algebra on curvature tensors".
Thank you very much!
connections holonomy
asked 2 days ago
Valentino
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to read a work that involves affine connections, and every time I am in doubt about something I get in trouble, because all the references I find consider connections on principal bundles, and the time I would spend studying them - I don't even know the definitions - to be able to understand the explanations is so big that I cannot do it. I wuold like to know, then, if someone could help me with an explanation that doesn't involve the notion of a connection on a principal bundle.
The current issue is the following:
Given a manifold with an affine conncetion $nabla$ whose curvature tensor is $R$, why does $nabla R = 0$ imply $(B cdot R)(X, Y) := -R(X, BY)-R(BX, Y)+BR(X, Y)-R(X, Y)B = 0 $ for every $X, Y in T_pM$ and $B$ in the Lie algebra of the holonomy group $Hol_p(nabla)$?
In the work I am reading, the author says that $(Bcdot R)$ is the "natural action of the holonomy Lie algebra on curvature tensors".
Thank you very much!
connections holonomy
connections holonomy
asked 2 days ago
Valentino
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Valentino
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Valentino
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Valentino
11
asked 2 days ago
Valentino
11
11
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Valentino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday
add a comment |
I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday
I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday
I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday
add a comment |
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I forgot to say: I now that $nabla$ is symplectic for some symplectic form $omega$ on $M$, so that $R$ satisfies, for instance, the first and the second Bianchi's identities.
– Valentino
yesterday