Mixing
Why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $?
up vote
0
down vote
favorite
In Euclidean space $ mathbb{R}^n $, let $ X_j=frac{partial}{partial x_j} $, and for every $ X, Yinmathfrak{X}(mathbb{R}^n) $, if
$$ X=sum_{i}f_iX_i=(f_1, f_2, cdots, f_n),quad Y=sum_{j}g_jX_j=(g_1, g_2, cdots, g_n), $$
then
begin{align*}
nabla_XY&=sum_jX(g_j)X_j+sum_{i, j}f_inabla_{X_i}X_j=sum_jX(g_j)X_j\
&=(X(g_1), X(g_2),cdots, X(g_n))\
&=operatorname{d} Y(X).
end{align*}
This is from my textbook. I don't understand why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $? ($ operatorname{d} $ is the tangent map).
differential-geometry riemannian-geometry
asked yesterday
Philip
1,046315
add a comment |
up vote
0
down vote
favorite
In Euclidean space $ mathbb{R}^n $, let $ X_j=frac{partial}{partial x_j} $, and for every $ X, Yinmathfrak{X}(mathbb{R}^n) $, if
$$ X=sum_{i}f_iX_i=(f_1, f_2, cdots, f_n),quad Y=sum_{j}g_jX_j=(g_1, g_2, cdots, g_n), $$
then
begin{align*}
nabla_XY&=sum_jX(g_j)X_j+sum_{i, j}f_inabla_{X_i}X_j=sum_jX(g_j)X_j\
&=(X(g_1), X(g_2),cdots, X(g_n))\
&=operatorname{d} Y(X).
end{align*}
This is from my textbook. I don't understand why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $? ($ operatorname{d} $ is the tangent map).
differential-geometry riemannian-geometry
asked yesterday
Philip
1,046315
1
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
1
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Euclidean space $ mathbb{R}^n $, let $ X_j=frac{partial}{partial x_j} $, and for every $ X, Yinmathfrak{X}(mathbb{R}^n) $, if
$$ X=sum_{i}f_iX_i=(f_1, f_2, cdots, f_n),quad Y=sum_{j}g_jX_j=(g_1, g_2, cdots, g_n), $$
then
begin{align*}
nabla_XY&=sum_jX(g_j)X_j+sum_{i, j}f_inabla_{X_i}X_j=sum_jX(g_j)X_j\
&=(X(g_1), X(g_2),cdots, X(g_n))\
&=operatorname{d} Y(X).
end{align*}
This is from my textbook. I don't understand why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $? ($ operatorname{d} $ is the tangent map).
differential-geometry riemannian-geometry
asked yesterday
Philip
1,046315
In Euclidean space $ mathbb{R}^n $, let $ X_j=frac{partial}{partial x_j} $, and for every $ X, Yinmathfrak{X}(mathbb{R}^n) $, if
$$ X=sum_{i}f_iX_i=(f_1, f_2, cdots, f_n),quad Y=sum_{j}g_jX_j=(g_1, g_2, cdots, g_n), $$
then
begin{align*}
nabla_XY&=sum_jX(g_j)X_j+sum_{i, j}f_inabla_{X_i}X_j=sum_jX(g_j)X_j\
&=(X(g_1), X(g_2),cdots, X(g_n))\
&=operatorname{d} Y(X).
end{align*}
This is from my textbook. I don't understand why $ (X(g_1), X(g_2),cdots, X(g_n))=operatorname{d}Y(X) $? ($ operatorname{d} $ is the tangent map).
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked yesterday
Philip
1,046315
asked yesterday
Philip
1,046315
edited 23 hours ago
asked yesterday
Philip
1,046315
asked yesterday
Philip
1,046315
asked yesterday
Philip
1,046315
1,046315
1
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
1
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago
add a comment |
1
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
1
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago
1
1
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
1
1
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.
Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$
Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.
Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$
Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.
Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$
Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.
Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$
Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In $M=mathbb{R}^n$ we have the global chart given by the identity map $phi:xin M mapsto x in mathbb{R}^n$. For every point $p$ in $M$ we have the basis of the tangent vectors given by this chart $frac{partial}{partial varphi_i}|_p$ with $1leq i leq n$. They are the usuar directional derivatives, $frac{partial}{partial varphi_i}|_p=frac{partial}{partial x_i}|_p$. So for every smooth function $f:Mto mathbb{R}$ and for every $i$, we have $frac{partial}{partial varphi_i}|_p(f)=frac{partial f}{partial x_i}(p)$. In your notation, $X_i=frac{partial}{partial varphi_i}|_p$.
Every smooth vector field $X$ in $mathfrak{X}(M)$ can be writen univocally as $X=sum_i^ng_ifrac{partial}{partial varphi_i}$ where $g_i:Mto mathbb{R}$ is smooth and is given by $g_i(p)=X_p(varphi_i)$ for every $i$. So we have an isomorphism
$$
X in mathfrak{X}(M)mapsto (X(varphi_1),ldots,X(varphi_n)) in mathcal{C}^infty(M,mathbb{R}^n)
$$
with inverse
$$
(g_1,dots,g_n) in mathcal{C}^infty(M,mathbb{R}^n) mapsto sum_i^ng_ifrac{partial}{partial varphi_i} in mathfrak{X}(M)
$$
Under the identifications above, consider a smooth vector field $Y=(g_1,ldots,g_n)$. It is just a smooth function $mathbb{R}^ntomathbb{R}^n$ so you can compute its differential map (or tangent map). For every $p$ in $M$, $d_pY: T_pM to T_{Y(p)}mathbb{R}^n$ is given by $d_pY(X_p)(f)=X_p(fcirc Y)$ for every $f$ in $mathcal{C}^infty(M)$. We can again use the identifications $T_pMcong mathbb{R}^n$ and $T_{Y(p)}congmathbb{R}^n$. Under this identifications $d_pY$ is just the linear endomorphism of $mathbb{R}^n$ with the differential matrix of $Y$ at $p$ as asociated matrix. That explain $(X(g_1),ldots,X(g_n))=(dg_1(X),ldots,dg_n(X))=dY(X)$.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 hours ago
Dante Grevino
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 hours ago
Dante Grevino
1463
answered 18 hours ago
Dante Grevino
1463
1463
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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%2f3004891%2fwhy-xg-1-xg-2-cdots-xg-n-operatornamedyx%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
1
Your notation is nonstandard. Try clarifying what your notation in the post means (and I suspect, you'll answer your own question in the process).
– Matt
yesterday
1
That's OK if you identify smooth vector fields on $mathbb{R}^n$ with smooth functions $mathbb{R}^ntomathbb{R}^n$. What's your definition for tangent map?
– Dante Grevino
yesterday
@DanteGrevino I am afraid I don't know and that's where my confusion is from.
– Philip
23 hours ago