Coordinate representation of component functions of a diffeomorphism












-1














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







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  • 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.
    – stressed out
    yesterday












  • Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
    – Ted Shifrin
    yesterday










  • 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.
    – stressed out
    14 hours ago
















-1














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







share|cite|improve this question
























  • 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.
    – stressed out
    yesterday












  • Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
    – Ted Shifrin
    yesterday










  • 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.
    – stressed out
    14 hours ago














-1












-1








-1







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







share|cite|improve this question















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









stressed out

3,9941533




3,9941533










asked yesterday









Perturbative

4,13511450




4,13511450












  • 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.
    – stressed out
    yesterday












  • Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
    – Ted Shifrin
    yesterday










  • 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.
    – stressed out
    14 hours ago


















  • 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.
    – stressed out
    yesterday












  • Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
    – Ted Shifrin
    yesterday










  • 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.
    – stressed out
    14 hours ago
















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.
– stressed out
yesterday






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.
– stressed out
yesterday














Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
– Ted Shifrin
yesterday




Note that $(x^1,dots,x^n) = phi$, so you're just doing the $j$th component of $phicircphi^{-1}$.
– Ted Shifrin
yesterday












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.
– stressed out
14 hours ago




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.
– stressed out
14 hours ago










1 Answer
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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.






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    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.






    share|cite|improve this answer




























      2














      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.






      share|cite|improve this answer


























        2












        2








        2






        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.






        share|cite|improve this answer














        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        stressed out

        3,9941533




        3,9941533






























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