Mixing
{Arnold Dynamical System I, 1992} f diffeomorphism => derivative of f isomorphism
$begingroup$
Let $f$ be a diffeomorphism of $M$ onto $N$ (both $M$,$N$ are smooth manifold). Prove that the mapping $f_{*x}$ is a vector-space isomorphism
between $T_xM$ and $T_xN$.
Here is a proof:
Let $x=f^{-1}(y) in M,, yin N$ and $hin M, kin N$ and I denote $U=f_{*x}$ and $V=f^{-1}_{x*}$. We know that $U,V$ are linear maps.
We have:
$$
f(x+h)=f(x)+Uh+o(|h|)
$$
$$
f^{-1}(y+k)=f^{-1}(y)+Vk+o(|k|)
$$
Then
$$
y+k = f(f^{-1}(y)+Vk+o(|k|))=y+U(Vk+o(|k|)) +o(|Vk+o(|k|)|)
$$
which yields
$$
k=UVk
$$
for every $k$ so $U$ is invertible and $U^{-1}=V$.
ordinary-differential-equations derivatives proof-verification differential-geometry vector-spaces
asked Jan 14 at 8:55
SmiliaSmilia
667617
$endgroup$
add a comment |
$begingroup$
Let $f$ be a diffeomorphism of $M$ onto $N$ (both $M$,$N$ are smooth manifold). Prove that the mapping $f_{*x}$ is a vector-space isomorphism
between $T_xM$ and $T_xN$.
Here is a proof:
Let $x=f^{-1}(y) in M,, yin N$ and $hin M, kin N$ and I denote $U=f_{*x}$ and $V=f^{-1}_{x*}$. We know that $U,V$ are linear maps.
We have:
$$
f(x+h)=f(x)+Uh+o(|h|)
$$
$$
f^{-1}(y+k)=f^{-1}(y)+Vk+o(|k|)
$$
Then
$$
y+k = f(f^{-1}(y)+Vk+o(|k|))=y+U(Vk+o(|k|)) +o(|Vk+o(|k|)|)
$$
which yields
$$
k=UVk
$$
for every $k$ so $U$ is invertible and $U^{-1}=V$.
ordinary-differential-equations derivatives proof-verification differential-geometry vector-spaces
asked Jan 14 at 8:55
SmiliaSmilia
667617
$endgroup$
$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49
add a comment |
$begingroup$
Let $f$ be a diffeomorphism of $M$ onto $N$ (both $M$,$N$ are smooth manifold). Prove that the mapping $f_{*x}$ is a vector-space isomorphism
between $T_xM$ and $T_xN$.
Here is a proof:
Let $x=f^{-1}(y) in M,, yin N$ and $hin M, kin N$ and I denote $U=f_{*x}$ and $V=f^{-1}_{x*}$. We know that $U,V$ are linear maps.
We have:
$$
f(x+h)=f(x)+Uh+o(|h|)
$$
$$
f^{-1}(y+k)=f^{-1}(y)+Vk+o(|k|)
$$
Then
$$
y+k = f(f^{-1}(y)+Vk+o(|k|))=y+U(Vk+o(|k|)) +o(|Vk+o(|k|)|)
$$
which yields
$$
k=UVk
$$
for every $k$ so $U$ is invertible and $U^{-1}=V$.
ordinary-differential-equations derivatives proof-verification differential-geometry vector-spaces
asked Jan 14 at 8:55
SmiliaSmilia
667617
$endgroup$
Let $f$ be a diffeomorphism of $M$ onto $N$ (both $M$,$N$ are smooth manifold). Prove that the mapping $f_{*x}$ is a vector-space isomorphism
between $T_xM$ and $T_xN$.
Here is a proof:
Let $x=f^{-1}(y) in M,, yin N$ and $hin M, kin N$ and I denote $U=f_{*x}$ and $V=f^{-1}_{x*}$. We know that $U,V$ are linear maps.
We have:
$$
f(x+h)=f(x)+Uh+o(|h|)
$$
$$
f^{-1}(y+k)=f^{-1}(y)+Vk+o(|k|)
$$
Then
$$
y+k = f(f^{-1}(y)+Vk+o(|k|))=y+U(Vk+o(|k|)) +o(|Vk+o(|k|)|)
$$
which yields
$$
k=UVk
$$
for every $k$ so $U$ is invertible and $U^{-1}=V$.
ordinary-differential-equations derivatives proof-verification differential-geometry vector-spaces
ordinary-differential-equations derivatives proof-verification differential-geometry vector-spaces
asked Jan 14 at 8:55
SmiliaSmilia
667617
asked Jan 14 at 8:55
SmiliaSmilia
667617
asked Jan 14 at 8:55
SmiliaSmilia
667617
asked Jan 14 at 8:55
SmiliaSmilia
667617
asked Jan 14 at 8:55
SmiliaSmilia
667617
667617
$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49
add a comment |
$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49
$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Denote by $g=f^{-1}$ we have using the chain rule,
$$
d(gcirc f)[x](h) = dg[f(x)](df[x](h))
$$
which gives
$$
h = V(Uh)
$$
hence $U^{-1}=V$.
answered Jan 14 at 9:38
SmiliaSmilia
667617
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Denote by $g=f^{-1}$ we have using the chain rule,
$$
d(gcirc f)[x](h) = dg[f(x)](df[x](h))
$$
which gives
$$
h = V(Uh)
$$
hence $U^{-1}=V$.
answered Jan 14 at 9:38
SmiliaSmilia
667617
$endgroup$
add a comment |
$begingroup$
Denote by $g=f^{-1}$ we have using the chain rule,
$$
d(gcirc f)[x](h) = dg[f(x)](df[x](h))
$$
which gives
$$
h = V(Uh)
$$
hence $U^{-1}=V$.
answered Jan 14 at 9:38
SmiliaSmilia
667617
$endgroup$
add a comment |
$begingroup$
Denote by $g=f^{-1}$ we have using the chain rule,
$$
d(gcirc f)[x](h) = dg[f(x)](df[x](h))
$$
which gives
$$
h = V(Uh)
$$
hence $U^{-1}=V$.
answered Jan 14 at 9:38
SmiliaSmilia
667617
$endgroup$
Denote by $g=f^{-1}$ we have using the chain rule,
$$
d(gcirc f)[x](h) = dg[f(x)](df[x](h))
$$
which gives
$$
h = V(Uh)
$$
hence $U^{-1}=V$.
answered Jan 14 at 9:38
SmiliaSmilia
667617
answered Jan 14 at 9:38
SmiliaSmilia
667617
answered Jan 14 at 9:38
SmiliaSmilia
667617
answered Jan 14 at 9:38
SmiliaSmilia
667617
667617
add a comment |
add a comment |
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$begingroup$
You could use the chain rule and prove the theorem in one line.
$endgroup$
– Amitai Yuval
Jan 14 at 9:14
$begingroup$
Oh yeah indeed !
$endgroup$
– Smilia
Jan 14 at 9:36
$begingroup$
Your operations only make sense if the manifolds are affine spaces. That could be accomplished locally by considering charts of the manifolds. On curved space in general you can not simply add a tangent vector to a point.
$endgroup$
– LutzL
Jan 14 at 9:49