Covariant derivative induced by Levi-Civita connection and compatibility with Lie brackets The...
How can a day be of 24 hours?
What steps are necessary to read a Modern SSD in Medieval Europe?
Could a dragon use its wings to swim?
How can the PCs determine if an item is a phylactery?
Oldie but Goldie
Gauss' Posthumous Publications?
Calculating discount not working
Is a linearly independent set whose span is dense a Schauder basis?
Does int main() need a declaration on C++?
Why do we say “un seul M” and not “une seule M” even though M is a “consonne”?
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
What happens if you break a law in another country outside of that country?
Another proof that dividing by 0 does not exist -- is it right?
Gödel's incompleteness theorems - what are the religious implications?
Is it reasonable to ask other researchers to send me their previous grant applications?
Arrows in tikz Markov chain diagram overlap
Is there a rule of thumb for determining the amount one should accept for a settlement offer?
Early programmable calculators with RS-232
Which acid/base does a strong base/acid react when added to a buffer solution?
Is it correct to say moon starry nights?
Is it possible to create a QR code using text?
Why did early computer designers eschew integers?
Car headlights in a world without electricity
Shortening a title without changing its meaning
Covariant derivative induced by Levi-Civita connection and compatibility with Lie brackets
The Next CEO of Stack OverflowLie bracket of canonical vectors on tangent space to a point on a manifold is zero.Expression for Levi-Civita ConnectionShape Operators and Symmetric Linear TransformationsParallel translation via $e$-connectionDerivations on a ManifoldLee's Riemannian Manifold, Zero curvature implies flatnesslocal expression for affine connectionsAbout a Morse function on the euclidean $n$-sphere.Compute $nabla_{alpha '} V$ in local coordinateVolume form is parallel with respect to Levi-Civita connectionAre Christoffel symbols structure coefficients?
$begingroup$
Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
$$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.
$ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $
$D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$
$D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$
$D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.
where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $
I want to prove
$$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$
Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
$$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
$$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
$$= X(V^j) partial_j - Y(U^i) partial_i$$
The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
$$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
$$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
$$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
$$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
Where is my mistake?
differential-geometry riemannian-geometry
$endgroup$
add a comment |
$begingroup$
Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
$$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.
$ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $
$D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$
$D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$
$D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.
where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $
I want to prove
$$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$
Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
$$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
$$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
$$= X(V^j) partial_j - Y(U^i) partial_i$$
The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
$$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
$$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
$$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
$$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
Where is my mistake?
differential-geometry riemannian-geometry
$endgroup$
add a comment |
$begingroup$
Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
$$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.
$ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $
$D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$
$D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$
$D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.
where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $
I want to prove
$$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$
Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
$$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
$$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
$$= X(V^j) partial_j - Y(U^i) partial_i$$
The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
$$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
$$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
$$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
$$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
Where is my mistake?
differential-geometry riemannian-geometry
$endgroup$
Let $M$ be a Riemannian manifold with Levi-Civita connection $ nabla$. Let $S$ be a differentiable manifold and $ varphi : S to M $ be a $C^{infty}$ immersion. Let
$$ D : TS times { text{vector fields along } varphi } to TM $$ s.t.
$ (v,X) mapsto D_v(X) in T_{varphi(pi(v))} M $
$D_{alpha v_1 + beta v_2}(X) = alpha D_{v_1}(X) + beta D_{v_2}(X)$
$D_v( X+Y) = D_v(X)+D_v(Y)$ and $ D_v(fX) = f(pi(v))D_v(X) + v(f)X_{pi(v)}$
$D_v (varphi^* (X) ) = nabla_{dvarphi_{pi(v)}(v)} (X) $ for any $X$ vector field in $M$.
where $ pi : TM to M$ is the natural projection $ v in T_pM mapsto p $ and $varphi^*(X) = X circ varphi $
I want to prove
$$ D_X( dvarphi(Y)) - D_Y( dvarphi(X)) = dvarphi ( [X,Y] )$$
Being $dvarphi(X)$ and $d varphi(Y)$ vector fields along $varphi$ I can write
$$ dvarphi(X) = U^i partial_i quad quad dvarphi(Y) = V^j partial_j $$
where $partial_i bigr |_p = frac{partial}{partial x^i} bigr |_{varphi(p)}$ for each $p in S$ and $ { x^1, dots x^n } $ are coordinates near $varphi(p)$ in $M$. By linearity and using Leibniz rule the LHS becomes
$$ V^jD_X(partial_j) + X(V^j) partial_j - U^iD_Y(partial_i) - Y(U^i) partial_i =$$
$$ = V^j U^i nabla_{frac{partial}{partial x^i}} (frac{partial}{partial x^j}) + X(V^j) partial_j - V^j U^i nabla_{frac{partial}{partial x^j}} (frac{partial}{partial x^i})- Y(U^i) partial_i$$
$$= X(V^j) partial_j - Y(U^i) partial_i$$
The RHS is, for any $p in S$ and $f in C^{infty}_{varphi(p)}M$
$$(dvarphi ( [X,Y] ))_p (f) = dvarphi_p ( [X,Y]_p) (f) = [X,Y]_p (f circ varphi) $$
$$= X_p(Y(f circ varphi)) - Y_p(X(f circ varphi)) = X_p(dvarphi(Y)(f)) - Y_p(dvarphi(X)(f))$$
$$= X_p( V^j frac{partial f}{partial x^j} bigr |_{varphi( ) } ) - Y_p( U^i frac{partial f}{partial x^i} bigr |_{varphi( ) } ) $$ $$= X_p(V^j)frac{partial f}{partial x^j} bigr |_{varphi(p) } - Y_p(U^i)frac{partial f}{partial x^i} bigr |_{varphi(p) } + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
$$= LHS + V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) }) - U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) $$
Where is my mistake?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Feb 4 at 15:12
J. W. Tanner
4,2361320
4,2361320
asked Feb 4 at 14:49
Bremen000Bremen000
517310
517310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is no mistake. Actually the remaining term is zero:
We have
begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
=V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
=V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}
Similar
$$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3099933%2fcovariant-derivative-induced-by-levi-civita-connection-and-compatibility-with-li%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no mistake. Actually the remaining term is zero:
We have
begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
=V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
=V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}
Similar
$$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.
$endgroup$
add a comment |
$begingroup$
There is no mistake. Actually the remaining term is zero:
We have
begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
=V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
=V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}
Similar
$$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.
$endgroup$
add a comment |
$begingroup$
There is no mistake. Actually the remaining term is zero:
We have
begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
=V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
=V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}
Similar
$$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.
$endgroup$
There is no mistake. Actually the remaining term is zero:
We have
begin{align}V_p^jX_p ( frac{partial f}{partial x^j} bigr |_{varphi( ) })=V_p^jX_p ( frac{partial f }{partial x^j} circvarphi)
=V_p^j(dvarphi (X))_p ( frac{partial f}{partial x^j} )
=V_p^j U^i_pfrac{partial }{partial x^i}bigr |_{varphi(p) } ( frac{partial f}{partial x^j} )end{align}
Similar
$$U_p^iY_p ( frac{partial f}{partial x^i} bigr |_{varphi( ) }) =U_p^i V^j_pfrac{partial }{partial x^j}bigr |_{varphi(p) } ( frac{partial f }{partial x^i} )$$
so the remaining term is given by
$$V^j_pU^i_pleft(left[ frac{partial }{partial x^i}, frac{partial }{partial x^j}right]_{varphi(p)}(f)right)=0$$
since on canonical tangent vectors given by a coordinate chart the Lie bracket vanishes.
answered Mar 18 at 2:04
triitrii
81817
81817
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3099933%2fcovariant-derivative-induced-by-levi-civita-connection-and-compatibility-with-li%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