Chart Transformation in $T^*_x M$
I am confused about a statement in my differential geometry script. It states:
$(U,varphi = (x_1,...,x_n))$ and $(U,psi = (y_1,...,y_n))$ are charts of a smooth manifold, then $$(dy_i)_x=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(x_1,...,x_n)(dx_alpha)_x$$ holds for the base in case of a change of charts.
I tried to recreate the formula with our definitions but im confused about the (x_1,...,x_n) in the formula, how do these get there and what does it state?
My try was:
begin{align*} d(y_i)_x & =d(x_icirc psi^{-1} circ psi)_x\
&= d(x_icirc psi^{-1})_{psi (x)}circ(sum^n_{alpha=1}d(y_alpha)_xcdot e_alpha)\
& =sum^n_{alpha=1} frac{partial(x_icirc psi^{-1})}{partial y_alpha} (psi(x))dy_alpha\
& =sum^n_{alpha=1} frac{partial(varphicirc psi^{-1})_i}{partial y_alpha} (psi(x))dy_alpha &(?)\
end{align*}
But there is no $(x_1,...,x_n)$ and i have no idea how i could get it.
differential-geometry co-tangent-space
asked Dec 26 '18 at 21:16
Timmathstf
307
add a comment |
I am confused about a statement in my differential geometry script. It states:
$(U,varphi = (x_1,...,x_n))$ and $(U,psi = (y_1,...,y_n))$ are charts of a smooth manifold, then $$(dy_i)_x=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(x_1,...,x_n)(dx_alpha)_x$$ holds for the base in case of a change of charts.
I tried to recreate the formula with our definitions but im confused about the (x_1,...,x_n) in the formula, how do these get there and what does it state?
My try was:
begin{align*} d(y_i)_x & =d(x_icirc psi^{-1} circ psi)_x\
&= d(x_icirc psi^{-1})_{psi (x)}circ(sum^n_{alpha=1}d(y_alpha)_xcdot e_alpha)\
& =sum^n_{alpha=1} frac{partial(x_icirc psi^{-1})}{partial y_alpha} (psi(x))dy_alpha\
& =sum^n_{alpha=1} frac{partial(varphicirc psi^{-1})_i}{partial y_alpha} (psi(x))dy_alpha &(?)\
end{align*}
But there is no $(x_1,...,x_n)$ and i have no idea how i could get it.
differential-geometry co-tangent-space
asked Dec 26 '18 at 21:16
Timmathstf
307
add a comment |
I am confused about a statement in my differential geometry script. It states:
$(U,varphi = (x_1,...,x_n))$ and $(U,psi = (y_1,...,y_n))$ are charts of a smooth manifold, then $$(dy_i)_x=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(x_1,...,x_n)(dx_alpha)_x$$ holds for the base in case of a change of charts.
I tried to recreate the formula with our definitions but im confused about the (x_1,...,x_n) in the formula, how do these get there and what does it state?
My try was:
begin{align*} d(y_i)_x & =d(x_icirc psi^{-1} circ psi)_x\
&= d(x_icirc psi^{-1})_{psi (x)}circ(sum^n_{alpha=1}d(y_alpha)_xcdot e_alpha)\
& =sum^n_{alpha=1} frac{partial(x_icirc psi^{-1})}{partial y_alpha} (psi(x))dy_alpha\
& =sum^n_{alpha=1} frac{partial(varphicirc psi^{-1})_i}{partial y_alpha} (psi(x))dy_alpha &(?)\
end{align*}
But there is no $(x_1,...,x_n)$ and i have no idea how i could get it.
differential-geometry co-tangent-space
asked Dec 26 '18 at 21:16
Timmathstf
307
I am confused about a statement in my differential geometry script. It states:
$(U,varphi = (x_1,...,x_n))$ and $(U,psi = (y_1,...,y_n))$ are charts of a smooth manifold, then $$(dy_i)_x=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(x_1,...,x_n)(dx_alpha)_x$$ holds for the base in case of a change of charts.
I tried to recreate the formula with our definitions but im confused about the (x_1,...,x_n) in the formula, how do these get there and what does it state?
My try was:
begin{align*} d(y_i)_x & =d(x_icirc psi^{-1} circ psi)_x\
&= d(x_icirc psi^{-1})_{psi (x)}circ(sum^n_{alpha=1}d(y_alpha)_xcdot e_alpha)\
& =sum^n_{alpha=1} frac{partial(x_icirc psi^{-1})}{partial y_alpha} (psi(x))dy_alpha\
& =sum^n_{alpha=1} frac{partial(varphicirc psi^{-1})_i}{partial y_alpha} (psi(x))dy_alpha &(?)\
end{align*}
But there is no $(x_1,...,x_n)$ and i have no idea how i could get it.
differential-geometry co-tangent-space
differential-geometry co-tangent-space
asked Dec 26 '18 at 21:16
Timmathstf
307
asked Dec 26 '18 at 21:16
Timmathstf
307
edited Dec 26 '18 at 21:53
asked Dec 26 '18 at 21:16
Timmathstf
307
asked Dec 26 '18 at 21:16
Timmathstf
307
asked Dec 26 '18 at 21:16
Timmathstf
307
307
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It is just the chain rule! Call $f = psi circ varphi^{-1}$. If $varphi: U to tilde{U}$ and $psi : U to tilde{V}$ then $f : tilde{U} to tilde{V}$ and
$$
{rm d}y_i = sum_j frac{partial f(x_1,cdots, x_n)}{partial x_j}{rm d}x_j
$$
where $x in tilde{U}$, $y in tilde{V}$ and $y = f(x)$
You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold
answered Dec 26 '18 at 22:00
caverac
13.6k21030
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
add a comment |
The book is saying that
$$(dy_i)_p=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p))(dx_alpha)_p.$$
[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $varphi(p) in mathbb R^n$ is the representation of $p$ in ${ x_alpha}$ coordinates, while $psi(p) in mathbb R^n$ is the representation of $p$ in ${ y_i }$ coordinates, and $psi circ varphi^{-1} : mathbb R^n to mathbb R^n$ is the transition map from ${ x_alpha }$ coordinates to ${ y_i }$ coordinates.]
Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors ${ frac{partial}{partial x_alpha}vert_p} $ at $p$. Specifically, we want to show that
$$ (dy_i)_p left( frac{partial }{partial x_alpha}vert_pright) = frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p)) (star)$$
This follows immediately from the definitions: for any function $f : M to mathbb R$, the definition of $df$ says
begin{align} (df)_p left( frac{partial }{partial x_alpha}vert_pright)
&= frac{partial left( f circ varphi^{-1}right) }{partial x_alpha} (varphi(p)).end{align}
Applying this with $y_i$, which is the $i$th component of $psi$, we get $(star)$, as required.
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is just the chain rule! Call $f = psi circ varphi^{-1}$. If $varphi: U to tilde{U}$ and $psi : U to tilde{V}$ then $f : tilde{U} to tilde{V}$ and
$$
{rm d}y_i = sum_j frac{partial f(x_1,cdots, x_n)}{partial x_j}{rm d}x_j
$$
where $x in tilde{U}$, $y in tilde{V}$ and $y = f(x)$
You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold
answered Dec 26 '18 at 22:00
caverac
13.6k21030
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
add a comment |
It is just the chain rule! Call $f = psi circ varphi^{-1}$. If $varphi: U to tilde{U}$ and $psi : U to tilde{V}$ then $f : tilde{U} to tilde{V}$ and
$$
{rm d}y_i = sum_j frac{partial f(x_1,cdots, x_n)}{partial x_j}{rm d}x_j
$$
where $x in tilde{U}$, $y in tilde{V}$ and $y = f(x)$
You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold
answered Dec 26 '18 at 22:00
caverac
13.6k21030
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
add a comment |
It is just the chain rule! Call $f = psi circ varphi^{-1}$. If $varphi: U to tilde{U}$ and $psi : U to tilde{V}$ then $f : tilde{U} to tilde{V}$ and
$$
{rm d}y_i = sum_j frac{partial f(x_1,cdots, x_n)}{partial x_j}{rm d}x_j
$$
where $x in tilde{U}$, $y in tilde{V}$ and $y = f(x)$
You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold
answered Dec 26 '18 at 22:00
caverac
13.6k21030
It is just the chain rule! Call $f = psi circ varphi^{-1}$. If $varphi: U to tilde{U}$ and $psi : U to tilde{V}$ then $f : tilde{U} to tilde{V}$ and
$$
{rm d}y_i = sum_j frac{partial f(x_1,cdots, x_n)}{partial x_j}{rm d}x_j
$$
where $x in tilde{U}$, $y in tilde{V}$ and $y = f(x)$
You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold
answered Dec 26 '18 at 22:00
caverac
13.6k21030
edited Dec 27 '18 at 10:44
answered Dec 26 '18 at 22:00
caverac
13.6k21030
answered Dec 26 '18 at 22:00
caverac
13.6k21030
answered Dec 26 '18 at 22:00
caverac
13.6k21030
13.6k21030
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
add a comment |
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
Yes, but i'm confused about the $(x_1,...,x_n)$ vector in the mix. It doesn't makes sense to me, i think there should be a point as input shouldn't it?
– Timmathstf
Dec 26 '18 at 22:05
1
1
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
@Timmathstf You can go directly from $tilde{U}$ to $tilde{V}$ without passing through the manifold, so need to use the concept of point at all
– caverac
Dec 26 '18 at 22:08
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
Thank you, that makes sense! Mabe i was a bit too pedantic when looking at that strang formula...
– Timmathstf
Dec 26 '18 at 22:11
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
@Timmathstf I put together a small sketch that may help you see it more clearly
– caverac
Dec 27 '18 at 10:45
add a comment |
The book is saying that
$$(dy_i)_p=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p))(dx_alpha)_p.$$
[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $varphi(p) in mathbb R^n$ is the representation of $p$ in ${ x_alpha}$ coordinates, while $psi(p) in mathbb R^n$ is the representation of $p$ in ${ y_i }$ coordinates, and $psi circ varphi^{-1} : mathbb R^n to mathbb R^n$ is the transition map from ${ x_alpha }$ coordinates to ${ y_i }$ coordinates.]
Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors ${ frac{partial}{partial x_alpha}vert_p} $ at $p$. Specifically, we want to show that
$$ (dy_i)_p left( frac{partial }{partial x_alpha}vert_pright) = frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p)) (star)$$
This follows immediately from the definitions: for any function $f : M to mathbb R$, the definition of $df$ says
begin{align} (df)_p left( frac{partial }{partial x_alpha}vert_pright)
&= frac{partial left( f circ varphi^{-1}right) }{partial x_alpha} (varphi(p)).end{align}
Applying this with $y_i$, which is the $i$th component of $psi$, we get $(star)$, as required.
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
add a comment |
The book is saying that
$$(dy_i)_p=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p))(dx_alpha)_p.$$
[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $varphi(p) in mathbb R^n$ is the representation of $p$ in ${ x_alpha}$ coordinates, while $psi(p) in mathbb R^n$ is the representation of $p$ in ${ y_i }$ coordinates, and $psi circ varphi^{-1} : mathbb R^n to mathbb R^n$ is the transition map from ${ x_alpha }$ coordinates to ${ y_i }$ coordinates.]
Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors ${ frac{partial}{partial x_alpha}vert_p} $ at $p$. Specifically, we want to show that
$$ (dy_i)_p left( frac{partial }{partial x_alpha}vert_pright) = frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p)) (star)$$
This follows immediately from the definitions: for any function $f : M to mathbb R$, the definition of $df$ says
begin{align} (df)_p left( frac{partial }{partial x_alpha}vert_pright)
&= frac{partial left( f circ varphi^{-1}right) }{partial x_alpha} (varphi(p)).end{align}
Applying this with $y_i$, which is the $i$th component of $psi$, we get $(star)$, as required.
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
add a comment |
The book is saying that
$$(dy_i)_p=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p))(dx_alpha)_p.$$
[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $varphi(p) in mathbb R^n$ is the representation of $p$ in ${ x_alpha}$ coordinates, while $psi(p) in mathbb R^n$ is the representation of $p$ in ${ y_i }$ coordinates, and $psi circ varphi^{-1} : mathbb R^n to mathbb R^n$ is the transition map from ${ x_alpha }$ coordinates to ${ y_i }$ coordinates.]
Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors ${ frac{partial}{partial x_alpha}vert_p} $ at $p$. Specifically, we want to show that
$$ (dy_i)_p left( frac{partial }{partial x_alpha}vert_pright) = frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p)) (star)$$
This follows immediately from the definitions: for any function $f : M to mathbb R$, the definition of $df$ says
begin{align} (df)_p left( frac{partial }{partial x_alpha}vert_pright)
&= frac{partial left( f circ varphi^{-1}right) }{partial x_alpha} (varphi(p)).end{align}
Applying this with $y_i$, which is the $i$th component of $psi$, we get $(star)$, as required.
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
The book is saying that
$$(dy_i)_p=sum^n_{alpha=1}frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p))(dx_alpha)_p.$$
[To avoid confusion, I'm using the letter $p$ to denote the point on the manifold $M$. So $varphi(p) in mathbb R^n$ is the representation of $p$ in ${ x_alpha}$ coordinates, while $psi(p) in mathbb R^n$ is the representation of $p$ in ${ y_i }$ coordinates, and $psi circ varphi^{-1} : mathbb R^n to mathbb R^n$ is the transition map from ${ x_alpha }$ coordinates to ${ y_i }$ coordinates.]
Perhaps the best way to prove this equation is to let $(dy_i)_p$ act on the basis of a tangent vectors ${ frac{partial}{partial x_alpha}vert_p} $ at $p$. Specifically, we want to show that
$$ (dy_i)_p left( frac{partial }{partial x_alpha}vert_pright) = frac{partial(psi circ varphi^{-1})_i}{partial x_alpha}(varphi(p)) (star)$$
This follows immediately from the definitions: for any function $f : M to mathbb R$, the definition of $df$ says
begin{align} (df)_p left( frac{partial }{partial x_alpha}vert_pright)
&= frac{partial left( f circ varphi^{-1}right) }{partial x_alpha} (varphi(p)).end{align}
Applying this with $y_i$, which is the $i$th component of $psi$, we get $(star)$, as required.
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
answered Dec 26 '18 at 22:05
Kenny Wong
17.7k21338
17.7k21338
add a comment |
add a comment |
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