How does push-forward of inverse map look like?
up vote
0
down vote
favorite
In the post Differential of the multiplication and inverse maps on a Lie group, second answer, appears the following solution for the push-forward of the inverse map:
$iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (ell_{a^{-1}})_* Y_a tag1$
In the second answer it is explained that applying the chain rule to
$iota(x) = r_{a^{-1}} circ iota circ l_{a^{-1}}(x) tag2$
you obtain Eq. (1). I guess that the chain rule has to be applied by derivation respect to some parameter $t$. Doing that I get:
$$frac{diota(x)}{dt} = frac{dr_{a^{-1}}(iota(l_a^{-1}(x)))}{d(iota(l_a^{-1}(x)))}cdot frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}cdot frac{dl_{a^{-1}}(x)}{dx}cdot frac{dx}{dt} = r_{a^{-1}}frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}l_{a^{-1}}frac{dx}{dt} tag3$$
And I don't know how to continue, so, I don't see how Eq. (1) can be obtained by this method.
differential-geometry lie-groups lie-algebras
asked Nov 15 at 16:35
Vicky
1387
add a comment |
up vote
0
down vote
favorite
In the post Differential of the multiplication and inverse maps on a Lie group, second answer, appears the following solution for the push-forward of the inverse map:
$iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (ell_{a^{-1}})_* Y_a tag1$
In the second answer it is explained that applying the chain rule to
$iota(x) = r_{a^{-1}} circ iota circ l_{a^{-1}}(x) tag2$
you obtain Eq. (1). I guess that the chain rule has to be applied by derivation respect to some parameter $t$. Doing that I get:
$$frac{diota(x)}{dt} = frac{dr_{a^{-1}}(iota(l_a^{-1}(x)))}{d(iota(l_a^{-1}(x)))}cdot frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}cdot frac{dl_{a^{-1}}(x)}{dx}cdot frac{dx}{dt} = r_{a^{-1}}frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}l_{a^{-1}}frac{dx}{dt} tag3$$
And I don't know how to continue, so, I don't see how Eq. (1) can be obtained by this method.
differential-geometry lie-groups lie-algebras
asked Nov 15 at 16:35
Vicky
1387
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the post Differential of the multiplication and inverse maps on a Lie group, second answer, appears the following solution for the push-forward of the inverse map:
$iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (ell_{a^{-1}})_* Y_a tag1$
In the second answer it is explained that applying the chain rule to
$iota(x) = r_{a^{-1}} circ iota circ l_{a^{-1}}(x) tag2$
you obtain Eq. (1). I guess that the chain rule has to be applied by derivation respect to some parameter $t$. Doing that I get:
$$frac{diota(x)}{dt} = frac{dr_{a^{-1}}(iota(l_a^{-1}(x)))}{d(iota(l_a^{-1}(x)))}cdot frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}cdot frac{dl_{a^{-1}}(x)}{dx}cdot frac{dx}{dt} = r_{a^{-1}}frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}l_{a^{-1}}frac{dx}{dt} tag3$$
And I don't know how to continue, so, I don't see how Eq. (1) can be obtained by this method.
differential-geometry lie-groups lie-algebras
asked Nov 15 at 16:35
Vicky
1387
In the post Differential of the multiplication and inverse maps on a Lie group, second answer, appears the following solution for the push-forward of the inverse map:
$iota_{*,a}(Y_a) = -(r_{a^{-1}})_* (ell_{a^{-1}})_* Y_a tag1$
In the second answer it is explained that applying the chain rule to
$iota(x) = r_{a^{-1}} circ iota circ l_{a^{-1}}(x) tag2$
you obtain Eq. (1). I guess that the chain rule has to be applied by derivation respect to some parameter $t$. Doing that I get:
$$frac{diota(x)}{dt} = frac{dr_{a^{-1}}(iota(l_a^{-1}(x)))}{d(iota(l_a^{-1}(x)))}cdot frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}cdot frac{dl_{a^{-1}}(x)}{dx}cdot frac{dx}{dt} = r_{a^{-1}}frac{d(iota(l_a^{-1}(x)))}{d(l_{a^{-1}}(x))}l_{a^{-1}}frac{dx}{dt} tag3$$
And I don't know how to continue, so, I don't see how Eq. (1) can be obtained by this method.
differential-geometry lie-groups lie-algebras
differential-geometry lie-groups lie-algebras
asked Nov 15 at 16:35
Vicky
1387
asked Nov 15 at 16:35
Vicky
1387
edited Nov 15 at 17:45
asked Nov 15 at 16:35
Vicky
1387
asked Nov 15 at 16:35
Vicky
1387
asked Nov 15 at 16:35
Vicky
1387
1387
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17
add a comment |
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999933%2fhow-does-push-forward-of-inverse-map-look-like%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
The chain rule for pushforwards is just that $(F circ G)_{ast} = F_{ast} circ G_{ast}$.
– Travis
Nov 15 at 16:47
How do you prove it? Sorry, I'm a roockie in this branch of maths
– Vicky
Nov 15 at 16:49
In coordinates this is just the multivariable chain rule.
– Travis
Nov 15 at 17:17