Mixing
Coordinate representation of component functions of a diffeomorphism
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $p in M$. Choose a smooth chart $(U, phi)$ around $p$ and then we have $phi : U to widehat{U} = phi[U]$ to be a diffeomorphism with
$phi(a) = (x^1(a), dots, x^n(a))$ with $x^i : U to mathbb{R}$ being the component functions.
Now we have the coordinate repsentation of $x^j$ to be $widehat{x^j} = x^j circ phi^{-1} : widehat{U} to mathbb{R}$, but does $widehat{x^j}(a^1, dots, a^n) = a^j$ for all $(a^1, dots, a^n) in widehat{U}$?
I was told that this is the case, but I'm having some trouble seeing how to show this since $widehat{x^j}(a^1, dots, a^n) = x^j(phi^{-1}(a^1, dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.
differential-geometry manifolds
edited Jan 5 at 5:18
stressed out
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $p in M$. Choose a smooth chart $(U, phi)$ around $p$ and then we have $phi : U to widehat{U} = phi[U]$ to be a diffeomorphism with
$phi(a) = (x^1(a), dots, x^n(a))$ with $x^i : U to mathbb{R}$ being the component functions.
Now we have the coordinate repsentation of $x^j$ to be $widehat{x^j} = x^j circ phi^{-1} : widehat{U} to mathbb{R}$, but does $widehat{x^j}(a^1, dots, a^n) = a^j$ for all $(a^1, dots, a^n) in widehat{U}$?
I was told that this is the case, but I'm having some trouble seeing how to show this since $widehat{x^j}(a^1, dots, a^n) = x^j(phi^{-1}(a^1, dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.
differential-geometry manifolds
edited Jan 5 at 5:18
stressed out
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
$endgroup$
$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
1
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50
add a comment |
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $p in M$. Choose a smooth chart $(U, phi)$ around $p$ and then we have $phi : U to widehat{U} = phi[U]$ to be a diffeomorphism with
$phi(a) = (x^1(a), dots, x^n(a))$ with $x^i : U to mathbb{R}$ being the component functions.
Now we have the coordinate repsentation of $x^j$ to be $widehat{x^j} = x^j circ phi^{-1} : widehat{U} to mathbb{R}$, but does $widehat{x^j}(a^1, dots, a^n) = a^j$ for all $(a^1, dots, a^n) in widehat{U}$?
I was told that this is the case, but I'm having some trouble seeing how to show this since $widehat{x^j}(a^1, dots, a^n) = x^j(phi^{-1}(a^1, dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.
differential-geometry manifolds
edited Jan 5 at 5:18
stressed out
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
$endgroup$
Let $M$ be a smooth manifold of dimension $n$ and let $p in M$. Choose a smooth chart $(U, phi)$ around $p$ and then we have $phi : U to widehat{U} = phi[U]$ to be a diffeomorphism with
$phi(a) = (x^1(a), dots, x^n(a))$ with $x^i : U to mathbb{R}$ being the component functions.
Now we have the coordinate repsentation of $x^j$ to be $widehat{x^j} = x^j circ phi^{-1} : widehat{U} to mathbb{R}$, but does $widehat{x^j}(a^1, dots, a^n) = a^j$ for all $(a^1, dots, a^n) in widehat{U}$?
I was told that this is the case, but I'm having some trouble seeing how to show this since $widehat{x^j}(a^1, dots, a^n) = x^j(phi^{-1}(a^1, dots, a^n))$ and since $x^j$ could be any (smooth) function I don't see how I could show the above.
differential-geometry manifolds
differential-geometry manifolds
edited Jan 5 at 5:18
stressed out
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
edited Jan 5 at 5:18
stressed out
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
edited Jan 5 at 5:18
stressed out
4,2691533
edited Jan 5 at 5:18
stressed out
4,2691533
edited Jan 5 at 5:18
stressed out
4,2691533
4,2691533
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
asked Jan 5 at 4:06
PerturbativePerturbative
4,20611551
4,20611551
$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
1
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50
add a comment |
$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
1
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50
$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
1
1
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $Usubseteq mathbb{R}^n$ and a homeomorphism $phi: U to widehat{U}=phi[U]$ where $pinwidehat{U} subseteq M$ is an open set in $M$. Now, we have some standard functions on $mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows:
$$x^i:Uto mathbb{R}$$
$$x^i(u^1,cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $phi$ and I'm going to call them $widehat{x^i}$ by defining $widehat{x^i}=x^i circ phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p in M$, then $M$ has some coordinates like $(a^1,cdots,a^n)$ which is available after introducing $phi$. More precisely, $phi^{-1}(p)=(a^1,cdots,a^n)$
In our notation, $widehat{x^j}(p)=x^jcircphi^{-1}(p)=x^j(a^1,cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $phi$) to refer to the points of $M$ like that. I hope everything is clear now.
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
$endgroup$
add a comment |
$begingroup$
So if I have a function from some set $X$ into $mathbb{R}^k$, $f : X to mathbb{R}^k$ then the component functions $f^i : X to mathbb{R}$ satisfy $f(p) = (f^1(p), dots, f^n(p))$ for all $p in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(pi_i circ f)(p) = f^i(p)$ for all $p in X$ and for all $ 1leq i leq k$. Note that $pi_i$ here means the $i^{text{th}}$ projection function from $mathbb{R}^k$ to $mathbb{R}$.
Now translating the above to the question at hand, I have $x^i : U to mathbb{R}$ to be the component functions of the homeomorphism $phi : U to phi[U] subseteq mathbb{R}^n$. Hence $(pi_j circ phi)(p) = x^j(p)$ for all $p in U$.
Now choose $(a^1, dots, a^n) in widehat{U} = phi[U]$. I will show that $widehat{x^j}(a^1, dots, a^n) = a^j$. Observe then that
begin{align*}
widehat{x^j}(a^1, dots, a^n) &= x^j(phi^{-1}(a^1, dots, a^n)) \
&= (pi_j circ phi)(phi^{-1}(a^1, dots, a^n)) \
&= pi_j(phi(phi^{-1}(a^1, dots, a^n))) \
&= pi_j(a^1, dots, a^n) \
&= a^j
end{align*}
as desired.
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $Usubseteq mathbb{R}^n$ and a homeomorphism $phi: U to widehat{U}=phi[U]$ where $pinwidehat{U} subseteq M$ is an open set in $M$. Now, we have some standard functions on $mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows:
$$x^i:Uto mathbb{R}$$
$$x^i(u^1,cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $phi$ and I'm going to call them $widehat{x^i}$ by defining $widehat{x^i}=x^i circ phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p in M$, then $M$ has some coordinates like $(a^1,cdots,a^n)$ which is available after introducing $phi$. More precisely, $phi^{-1}(p)=(a^1,cdots,a^n)$
In our notation, $widehat{x^j}(p)=x^jcircphi^{-1}(p)=x^j(a^1,cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $phi$) to refer to the points of $M$ like that. I hope everything is clear now.
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
$endgroup$
add a comment |
$begingroup$
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $Usubseteq mathbb{R}^n$ and a homeomorphism $phi: U to widehat{U}=phi[U]$ where $pinwidehat{U} subseteq M$ is an open set in $M$. Now, we have some standard functions on $mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows:
$$x^i:Uto mathbb{R}$$
$$x^i(u^1,cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $phi$ and I'm going to call them $widehat{x^i}$ by defining $widehat{x^i}=x^i circ phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p in M$, then $M$ has some coordinates like $(a^1,cdots,a^n)$ which is available after introducing $phi$. More precisely, $phi^{-1}(p)=(a^1,cdots,a^n)$
In our notation, $widehat{x^j}(p)=x^jcircphi^{-1}(p)=x^j(a^1,cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $phi$) to refer to the points of $M$ like that. I hope everything is clear now.
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
$endgroup$
add a comment |
$begingroup$
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $Usubseteq mathbb{R}^n$ and a homeomorphism $phi: U to widehat{U}=phi[U]$ where $pinwidehat{U} subseteq M$ is an open set in $M$. Now, we have some standard functions on $mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows:
$$x^i:Uto mathbb{R}$$
$$x^i(u^1,cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $phi$ and I'm going to call them $widehat{x^i}$ by defining $widehat{x^i}=x^i circ phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p in M$, then $M$ has some coordinates like $(a^1,cdots,a^n)$ which is available after introducing $phi$. More precisely, $phi^{-1}(p)=(a^1,cdots,a^n)$
In our notation, $widehat{x^j}(p)=x^jcircphi^{-1}(p)=x^j(a^1,cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $phi$) to refer to the points of $M$ like that. I hope everything is clear now.
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
$endgroup$
I believe that definitions might differ in different textbooks, but here's my understanding of the situation. I will try to be consistent with your notations as much as I can.
You have an open set $Usubseteq mathbb{R}^n$ and a homeomorphism $phi: U to widehat{U}=phi[U]$ where $pinwidehat{U} subseteq M$ is an open set in $M$. Now, we have some standard functions on $mathbb{R}^n$ that I'm going to call them "coordinate functions" that are defined as follows:
$$x^i:Uto mathbb{R}$$
$$x^i(u^1,cdots,u^n)=u^i$$
I can define a similar set of coordinate functions on my manifold using $phi$ and I'm going to call them $widehat{x^i}$ by defining $widehat{x^i}=x^i circ phi^{-1}$. That's all that there's to it. Nothing more.
Now if $p in M$, then $M$ has some coordinates like $(a^1,cdots,a^n)$ which is available after introducing $phi$. More precisely, $phi^{-1}(p)=(a^1,cdots,a^n)$
In our notation, $widehat{x^j}(p)=x^jcircphi^{-1}(p)=x^j(a^1,cdots,a^n)=a^j$ which is what you wanted to show, I believe.
Your confusion probably stems from the fact that you're assuming that the points of the manifold $M$ are $n$-tuples, ignoring the fact that you need to first introduce a parametrization (i.e. a homeomorphism $phi$) to refer to the points of $M$ like that. I hope everything is clear now.
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
edited Jan 5 at 5:01
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
answered Jan 5 at 4:50
stressed outstressed out
4,2691533
4,2691533
add a comment |
add a comment |
$begingroup$
So if I have a function from some set $X$ into $mathbb{R}^k$, $f : X to mathbb{R}^k$ then the component functions $f^i : X to mathbb{R}$ satisfy $f(p) = (f^1(p), dots, f^n(p))$ for all $p in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(pi_i circ f)(p) = f^i(p)$ for all $p in X$ and for all $ 1leq i leq k$. Note that $pi_i$ here means the $i^{text{th}}$ projection function from $mathbb{R}^k$ to $mathbb{R}$.
Now translating the above to the question at hand, I have $x^i : U to mathbb{R}$ to be the component functions of the homeomorphism $phi : U to phi[U] subseteq mathbb{R}^n$. Hence $(pi_j circ phi)(p) = x^j(p)$ for all $p in U$.
Now choose $(a^1, dots, a^n) in widehat{U} = phi[U]$. I will show that $widehat{x^j}(a^1, dots, a^n) = a^j$. Observe then that
begin{align*}
widehat{x^j}(a^1, dots, a^n) &= x^j(phi^{-1}(a^1, dots, a^n)) \
&= (pi_j circ phi)(phi^{-1}(a^1, dots, a^n)) \
&= pi_j(phi(phi^{-1}(a^1, dots, a^n))) \
&= pi_j(a^1, dots, a^n) \
&= a^j
end{align*}
as desired.
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
$endgroup$
add a comment |
$begingroup$
So if I have a function from some set $X$ into $mathbb{R}^k$, $f : X to mathbb{R}^k$ then the component functions $f^i : X to mathbb{R}$ satisfy $f(p) = (f^1(p), dots, f^n(p))$ for all $p in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(pi_i circ f)(p) = f^i(p)$ for all $p in X$ and for all $ 1leq i leq k$. Note that $pi_i$ here means the $i^{text{th}}$ projection function from $mathbb{R}^k$ to $mathbb{R}$.
Now translating the above to the question at hand, I have $x^i : U to mathbb{R}$ to be the component functions of the homeomorphism $phi : U to phi[U] subseteq mathbb{R}^n$. Hence $(pi_j circ phi)(p) = x^j(p)$ for all $p in U$.
Now choose $(a^1, dots, a^n) in widehat{U} = phi[U]$. I will show that $widehat{x^j}(a^1, dots, a^n) = a^j$. Observe then that
begin{align*}
widehat{x^j}(a^1, dots, a^n) &= x^j(phi^{-1}(a^1, dots, a^n)) \
&= (pi_j circ phi)(phi^{-1}(a^1, dots, a^n)) \
&= pi_j(phi(phi^{-1}(a^1, dots, a^n))) \
&= pi_j(a^1, dots, a^n) \
&= a^j
end{align*}
as desired.
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
$endgroup$
add a comment |
$begingroup$
So if I have a function from some set $X$ into $mathbb{R}^k$, $f : X to mathbb{R}^k$ then the component functions $f^i : X to mathbb{R}$ satisfy $f(p) = (f^1(p), dots, f^n(p))$ for all $p in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(pi_i circ f)(p) = f^i(p)$ for all $p in X$ and for all $ 1leq i leq k$. Note that $pi_i$ here means the $i^{text{th}}$ projection function from $mathbb{R}^k$ to $mathbb{R}$.
Now translating the above to the question at hand, I have $x^i : U to mathbb{R}$ to be the component functions of the homeomorphism $phi : U to phi[U] subseteq mathbb{R}^n$. Hence $(pi_j circ phi)(p) = x^j(p)$ for all $p in U$.
Now choose $(a^1, dots, a^n) in widehat{U} = phi[U]$. I will show that $widehat{x^j}(a^1, dots, a^n) = a^j$. Observe then that
begin{align*}
widehat{x^j}(a^1, dots, a^n) &= x^j(phi^{-1}(a^1, dots, a^n)) \
&= (pi_j circ phi)(phi^{-1}(a^1, dots, a^n)) \
&= pi_j(phi(phi^{-1}(a^1, dots, a^n))) \
&= pi_j(a^1, dots, a^n) \
&= a^j
end{align*}
as desired.
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
$endgroup$
So if I have a function from some set $X$ into $mathbb{R}^k$, $f : X to mathbb{R}^k$ then the component functions $f^i : X to mathbb{R}$ satisfy $f(p) = (f^1(p), dots, f^n(p))$ for all $p in X$. I didn't realize at the time of posting the question but what this means is that the component functions $f^i$ are characterized by the fact that $(pi_i circ f)(p) = f^i(p)$ for all $p in X$ and for all $ 1leq i leq k$. Note that $pi_i$ here means the $i^{text{th}}$ projection function from $mathbb{R}^k$ to $mathbb{R}$.
Now translating the above to the question at hand, I have $x^i : U to mathbb{R}$ to be the component functions of the homeomorphism $phi : U to phi[U] subseteq mathbb{R}^n$. Hence $(pi_j circ phi)(p) = x^j(p)$ for all $p in U$.
Now choose $(a^1, dots, a^n) in widehat{U} = phi[U]$. I will show that $widehat{x^j}(a^1, dots, a^n) = a^j$. Observe then that
begin{align*}
widehat{x^j}(a^1, dots, a^n) &= x^j(phi^{-1}(a^1, dots, a^n)) \
&= (pi_j circ phi)(phi^{-1}(a^1, dots, a^n)) \
&= pi_j(phi(phi^{-1}(a^1, dots, a^n))) \
&= pi_j(a^1, dots, a^n) \
&= a^j
end{align*}
as desired.
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
answered Jan 11 at 23:29
PerturbativePerturbative
4,20611551
4,20611551
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$begingroup$
I think your notation is confusing. I might be wrong, but intuitively from your definitions it seems that $hat{x^j}$ is just supposed to give you the j-th coordinate of the vector $x=(x^1,cdots,x^n)$ in your co-domain $hat{U}$. And $x^j$ isn't just any smooth function. It's the $j$-th coordinate function for $U$. Just like $hat{x^j}$ but instead of acting on $hat{U}$, it acts on vectors in your domain $U$. You're overcomplicating it, I think.
$endgroup$
– stressed out
Jan 5 at 4:29
$begingroup$
Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
$endgroup$
– Ted Shifrin
Jan 5 at 6:19
1
$begingroup$
Also, it's meaningless to say that $phi$ is a diffeomorphism. You have no differential structure on $M$ by default. $phi$ must be a homeomorphism.
$endgroup$
– stressed out
Jan 6 at 10:50