Why is a connection on the bundle $SO(M)$ metric compatible?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
If we have an orientable manifold $M$ with a metric $g$ and signature $(r, s)$, we can define the principal-$SO(r, s)$ bundle $SO(M)$, the bundle of orthonormal frames of $TM$. This is a subset of the principal-$GL(n, mathbbR)$ bundle $Fr(M)$, the frame bundle of $M$.
As a Koszul connection $nabla$, any such connection is metric compatible. That is, for all $X, Y, Z in mathrmVect(M)$, we have
$$X g(Y, Z) = g(nabla_X Y, Z) + g(Y, nabla_Z X).$$
My question is this: how can we prove that all connections on the bundle $SO(M)$ are metric compatible?
differential-geometry riemannian-geometry connections principal-bundles tangent-bundle
add a comment |Â
up vote
1
down vote
favorite
If we have an orientable manifold $M$ with a metric $g$ and signature $(r, s)$, we can define the principal-$SO(r, s)$ bundle $SO(M)$, the bundle of orthonormal frames of $TM$. This is a subset of the principal-$GL(n, mathbbR)$ bundle $Fr(M)$, the frame bundle of $M$.
As a Koszul connection $nabla$, any such connection is metric compatible. That is, for all $X, Y, Z in mathrmVect(M)$, we have
$$X g(Y, Z) = g(nabla_X Y, Z) + g(Y, nabla_Z X).$$
My question is this: how can we prove that all connections on the bundle $SO(M)$ are metric compatible?
differential-geometry riemannian-geometry connections principal-bundles tangent-bundle
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we have an orientable manifold $M$ with a metric $g$ and signature $(r, s)$, we can define the principal-$SO(r, s)$ bundle $SO(M)$, the bundle of orthonormal frames of $TM$. This is a subset of the principal-$GL(n, mathbbR)$ bundle $Fr(M)$, the frame bundle of $M$.
As a Koszul connection $nabla$, any such connection is metric compatible. That is, for all $X, Y, Z in mathrmVect(M)$, we have
$$X g(Y, Z) = g(nabla_X Y, Z) + g(Y, nabla_Z X).$$
My question is this: how can we prove that all connections on the bundle $SO(M)$ are metric compatible?
differential-geometry riemannian-geometry connections principal-bundles tangent-bundle
If we have an orientable manifold $M$ with a metric $g$ and signature $(r, s)$, we can define the principal-$SO(r, s)$ bundle $SO(M)$, the bundle of orthonormal frames of $TM$. This is a subset of the principal-$GL(n, mathbbR)$ bundle $Fr(M)$, the frame bundle of $M$.
As a Koszul connection $nabla$, any such connection is metric compatible. That is, for all $X, Y, Z in mathrmVect(M)$, we have
$$X g(Y, Z) = g(nabla_X Y, Z) + g(Y, nabla_Z X).$$
My question is this: how can we prove that all connections on the bundle $SO(M)$ are metric compatible?
differential-geometry riemannian-geometry connections principal-bundles tangent-bundle
asked Aug 1 at 6:30
user1379857
1765
1765
add a comment |Â
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868778%2fwhy-is-a-connection-on-the-bundle-som-metric-compatible%23new-answer', 'question_page');
);
Post as a guest
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
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
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