Question about proof of the Rank Theorem from Lee's Smooth Manifolds
$begingroup$
The Rank Theorem: Given a map, $F: M rightarrow N$ of constant rank, r,
there exist smooth charts $(U,phi)$ and $(V, psi)$ such that
$psi circ F circ phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$
In the book he defines, $phi: U_0 rightarrow tilde{U}_0$, where $tilde{U}_0$ is an open cube, with $phi(x,y) = (Q(x,y), y)$ where $x$ is short hand for
$x^1,...,x^r$ and $y$ is short hand for $x^{r+1},...,x^m$ which he relabels as $y^{r+1},...,y^m$, with $F(x,y) = (Q(x, y), R(x, y))$.
Then he arrives at
$$
D(F circ phi^{-1}) =
begin{pmatrix}
I &
0 \
frac{partial tilde{R^i}}{partial x^j} &
frac{partial tilde{R^i}}{partial y^j}
end{pmatrix}
$$
He mentions that it is necessary for $tilde{U}_0$ to be an open cube so that $tilde{R}$ is independent of $(y^1,...,y^{m-r})$, based on $D(F circ phi^{-1})$ above, but I thought the same held based on rank arguments alone.
He also uses the fact that $tilde{U}_0$ later on, where he says,
Because $tilde{U}_0$ is a cube and $F circ phi^{-1}$ has the form (4.6), it follows that
$F circ phi^{-1}(tilde{U}_0) subseteq V_0$
My question is, why is $tilde{U}_0$ being a cube needed for the above statements?
differential-geometry smooth-manifolds
asked Dec 7 '18 at 4:26
JeffJeff
11918
$endgroup$
add a comment |
$begingroup$
The Rank Theorem: Given a map, $F: M rightarrow N$ of constant rank, r,
there exist smooth charts $(U,phi)$ and $(V, psi)$ such that
$psi circ F circ phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$
In the book he defines, $phi: U_0 rightarrow tilde{U}_0$, where $tilde{U}_0$ is an open cube, with $phi(x,y) = (Q(x,y), y)$ where $x$ is short hand for
$x^1,...,x^r$ and $y$ is short hand for $x^{r+1},...,x^m$ which he relabels as $y^{r+1},...,y^m$, with $F(x,y) = (Q(x, y), R(x, y))$.
Then he arrives at
$$
D(F circ phi^{-1}) =
begin{pmatrix}
I &
0 \
frac{partial tilde{R^i}}{partial x^j} &
frac{partial tilde{R^i}}{partial y^j}
end{pmatrix}
$$
He mentions that it is necessary for $tilde{U}_0$ to be an open cube so that $tilde{R}$ is independent of $(y^1,...,y^{m-r})$, based on $D(F circ phi^{-1})$ above, but I thought the same held based on rank arguments alone.
He also uses the fact that $tilde{U}_0$ later on, where he says,
Because $tilde{U}_0$ is a cube and $F circ phi^{-1}$ has the form (4.6), it follows that
$F circ phi^{-1}(tilde{U}_0) subseteq V_0$
My question is, why is $tilde{U}_0$ being a cube needed for the above statements?
differential-geometry smooth-manifolds
asked Dec 7 '18 at 4:26
JeffJeff
11918
$endgroup$
add a comment |
$begingroup$
The Rank Theorem: Given a map, $F: M rightarrow N$ of constant rank, r,
there exist smooth charts $(U,phi)$ and $(V, psi)$ such that
$psi circ F circ phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$
In the book he defines, $phi: U_0 rightarrow tilde{U}_0$, where $tilde{U}_0$ is an open cube, with $phi(x,y) = (Q(x,y), y)$ where $x$ is short hand for
$x^1,...,x^r$ and $y$ is short hand for $x^{r+1},...,x^m$ which he relabels as $y^{r+1},...,y^m$, with $F(x,y) = (Q(x, y), R(x, y))$.
Then he arrives at
$$
D(F circ phi^{-1}) =
begin{pmatrix}
I &
0 \
frac{partial tilde{R^i}}{partial x^j} &
frac{partial tilde{R^i}}{partial y^j}
end{pmatrix}
$$
He mentions that it is necessary for $tilde{U}_0$ to be an open cube so that $tilde{R}$ is independent of $(y^1,...,y^{m-r})$, based on $D(F circ phi^{-1})$ above, but I thought the same held based on rank arguments alone.
He also uses the fact that $tilde{U}_0$ later on, where he says,
Because $tilde{U}_0$ is a cube and $F circ phi^{-1}$ has the form (4.6), it follows that
$F circ phi^{-1}(tilde{U}_0) subseteq V_0$
My question is, why is $tilde{U}_0$ being a cube needed for the above statements?
differential-geometry smooth-manifolds
asked Dec 7 '18 at 4:26
JeffJeff
11918
$endgroup$
The Rank Theorem: Given a map, $F: M rightarrow N$ of constant rank, r,
there exist smooth charts $(U,phi)$ and $(V, psi)$ such that
$psi circ F circ phi^{-1} (x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$
In the book he defines, $phi: U_0 rightarrow tilde{U}_0$, where $tilde{U}_0$ is an open cube, with $phi(x,y) = (Q(x,y), y)$ where $x$ is short hand for
$x^1,...,x^r$ and $y$ is short hand for $x^{r+1},...,x^m$ which he relabels as $y^{r+1},...,y^m$, with $F(x,y) = (Q(x, y), R(x, y))$.
Then he arrives at
$$
D(F circ phi^{-1}) =
begin{pmatrix}
I &
0 \
frac{partial tilde{R^i}}{partial x^j} &
frac{partial tilde{R^i}}{partial y^j}
end{pmatrix}
$$
He mentions that it is necessary for $tilde{U}_0$ to be an open cube so that $tilde{R}$ is independent of $(y^1,...,y^{m-r})$, based on $D(F circ phi^{-1})$ above, but I thought the same held based on rank arguments alone.
He also uses the fact that $tilde{U}_0$ later on, where he says,
Because $tilde{U}_0$ is a cube and $F circ phi^{-1}$ has the form (4.6), it follows that
$F circ phi^{-1}(tilde{U}_0) subseteq V_0$
My question is, why is $tilde{U}_0$ being a cube needed for the above statements?
differential-geometry smooth-manifolds
differential-geometry smooth-manifolds
asked Dec 7 '18 at 4:26
JeffJeff
11918
asked Dec 7 '18 at 4:26
JeffJeff
11918
edited Dec 7 '18 at 15:43
Jeff
asked Dec 7 '18 at 4:26
JeffJeff
11918
asked Dec 7 '18 at 4:26
JeffJeff
11918
asked Dec 7 '18 at 4:26
JeffJeff
11918
11918
add a comment |
add a comment |
1 Answer
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I'm struggling with the same result too.
I agree one does not need $tilde{U}_0$ to be a cube in order to the derivatives vanish.
For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,psi)$ despite the compositions.
Note that $psi(V_0)$ is a connected subset of $mathbb{R}^n$, which I think (haven't proof) is a cube.
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
$endgroup$
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
add a comment |
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$begingroup$
I'm struggling with the same result too.
I agree one does not need $tilde{U}_0$ to be a cube in order to the derivatives vanish.
For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,psi)$ despite the compositions.
Note that $psi(V_0)$ is a connected subset of $mathbb{R}^n$, which I think (haven't proof) is a cube.
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
$endgroup$
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
add a comment |
$begingroup$
I'm struggling with the same result too.
I agree one does not need $tilde{U}_0$ to be a cube in order to the derivatives vanish.
For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,psi)$ despite the compositions.
Note that $psi(V_0)$ is a connected subset of $mathbb{R}^n$, which I think (haven't proof) is a cube.
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
$endgroup$
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
add a comment |
$begingroup$
I'm struggling with the same result too.
I agree one does not need $tilde{U}_0$ to be a cube in order to the derivatives vanish.
For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,psi)$ despite the compositions.
Note that $psi(V_0)$ is a connected subset of $mathbb{R}^n$, which I think (haven't proof) is a cube.
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
$endgroup$
I'm struggling with the same result too.
I agree one does not need $tilde{U}_0$ to be a cube in order to the derivatives vanish.
For the second argument, I think the important fact about being a cube is to be connected, so you can be sure you stay in the domain of the chart $(V,psi)$ despite the compositions.
Note that $psi(V_0)$ is a connected subset of $mathbb{R}^n$, which I think (haven't proof) is a cube.
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
answered Dec 11 '18 at 14:37
Zuly SalinasZuly Salinas
62
62
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
add a comment |
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
I actually asked some people offline. I need to write up the answer. For the first part, the derivatives vanishing isn't sufficient to say that $tilde{R}(x,y) = tilde{R}(x,0)$.
$endgroup$
– Jeff
Dec 15 '18 at 2:46
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
For one, if you have something like a T-shaped region $tilde{R}$ might not even be defined at $tilde{R}(x,0)$. A second problem would be if $tilde{U_0}$ is disconnected. In this case, even though the derivative with respect to $y$ could be 0 in each region $tilde{R}(x,y)$ could be at different constant values with respect to $y$ in each region. The cube avoids these problems.
$endgroup$
– Jeff
Dec 15 '18 at 2:54
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
$begingroup$
You’re right, I found this interested fact to figure it out the importance of connected sets in the proof, and it’s related with the corollary following this theorem math.stackexchange.com/questions/252130/…
$endgroup$
– Zuly Salinas
Dec 27 '18 at 2:04
add a comment |
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