Chart Transformation in $T^*_x M$












2














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.










share|cite|improve this question





























    2














    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.










    share|cite|improve this question



























      2












      2








      2







      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.










      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 21:53

























      asked Dec 26 '18 at 21:16









      Timmathstf

      307




      307






















          2 Answers
          2






          active

          oldest

          votes


















          1














          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 $y = f(x)$



          enter image description here



          You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold






          share|cite|improve this answer























          • 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



















          1














          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.






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053329%2fchart-transformation-in-t-x-m%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            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 $y = f(x)$



            enter image description here



            You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold






            share|cite|improve this answer























            • 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
















            1














            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 $y = f(x)$



            enter image description here



            You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold






            share|cite|improve this answer























            • 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














            1












            1








            1






            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 $y = f(x)$



            enter image description here



            You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold






            share|cite|improve this answer














            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 $y = f(x)$



            enter image description here



            You see from this diagram that $f = psi circ varphi^{-1}$ connects directly $tilde{U}$ and $tilde{V}$ without going through the manifold







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 27 '18 at 10:44

























            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


















            • 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











            1














            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.






            share|cite|improve this answer


























              1














              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.






              share|cite|improve this answer
























                1












                1








                1






                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.






                share|cite|improve this answer












                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 26 '18 at 22:05









                Kenny Wong

                17.7k21338




                17.7k21338






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3053329%2fchart-transformation-in-t-x-m%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Human spaceflight

                    Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                    張江高科駅