Lee's proof that the normal bundle is embedded











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First I'll fix some of the definitions and context for the forthcoming question.



Suppose $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $xin M$, we define the normal space to $pmb M$ at $pmb x$ to be the $(n-m)$-dimensional subspace $N_xMsubseteq T_xBbb R^n$ consisting of all vectors that are orthogonal to $T_xM$ with respect to the Euclidean dot product. The normal bundle of $pmb{M}$, denoted by $NM$, is the subset of $TBbb R^napprox Bbb R^ntimesBbb R^n$ consisting of vectors that are normal to $M$:
$$
NM = big{(x,v) in Bbb R^ntimesBbb R^n : xin M, vin N_xM big}.
$$



The statement and part of the proof of Theorem 6.23 in Lee's Introduction to Smooth Manifolds is reproduced below:




Theorem 6.23. If $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $TBbb R^napprox Bbb R^ntimesBbb R^n$.



Proof. Let $x_0$ be any point of $M$, and let $(U,varphi)$ be a slice chart for $M$ in $Bbb R^n$ centered at $x_0$. Write $widehat U = varphi(U)subseteq Bbb R^n$, and write the coordinate functions of $varphi$ as $big(u^1,dots,u^nbig)$, so that $Mcap U$ is the set where $u^{m+1}=dotsb=u^n=0$. At each point $xin U$, the vectors $E_j|_x = (dvarphi_x)^{-1}big(partial/partial u^j|_{varphi(x)}big)$ form a basis for $T_xBbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as
$$
E_jbig|_x = E_j^i(x)frac{partial}{partial x^i}bigg|_x,
$$
where each $E_j^i(x)$ is a partial derivative of $varphi^{-1}$ evaluated at $varphi(x)$, and thus is a smooth function of $x$.




The proof goes on, but my question is about this part of the proof.




  • Is $partial/partial u^j|_{varphi(x)}$ literally the partial derivative operator in the $e_j = (0,dots,0,underbrace{1}_{text{$j$th component}},0,dots,0)$ direction? If not, what is it precisely, using Lee's notation?

  • I suspect that if $partial/partial u^j|_{varphi(x)}$ is not literally the partial derivative operator in the $e_j$ direction, then
    $partial/partial x^j|_x = (dvarphi_x)^{-1}big(partial / partial x^j|_{varphi(x)}big)$, where $partial / partial x^j|_{varphi(x)}$ is
    literally the partial derivative operator in the $e_j$ direction, and that $partial/partial u^j|_{varphi(x)}$ must be something else.

  • What is $E^i_j(x)$ explicitly?


Note that Lee hasn't yet defined standard frame at this point in the text, so I suppose a complete answer to this question would also address what the standard frame actually is.










share|cite|improve this question




















  • 3




    Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
    – Jack Lee
    Jul 27 at 21:07










  • @JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
    – AOrtiz
    Aug 2 at 17:38












  • Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
    – Jack Lee
    Aug 2 at 20:27

















up vote
1
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First I'll fix some of the definitions and context for the forthcoming question.



Suppose $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $xin M$, we define the normal space to $pmb M$ at $pmb x$ to be the $(n-m)$-dimensional subspace $N_xMsubseteq T_xBbb R^n$ consisting of all vectors that are orthogonal to $T_xM$ with respect to the Euclidean dot product. The normal bundle of $pmb{M}$, denoted by $NM$, is the subset of $TBbb R^napprox Bbb R^ntimesBbb R^n$ consisting of vectors that are normal to $M$:
$$
NM = big{(x,v) in Bbb R^ntimesBbb R^n : xin M, vin N_xM big}.
$$



The statement and part of the proof of Theorem 6.23 in Lee's Introduction to Smooth Manifolds is reproduced below:




Theorem 6.23. If $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $TBbb R^napprox Bbb R^ntimesBbb R^n$.



Proof. Let $x_0$ be any point of $M$, and let $(U,varphi)$ be a slice chart for $M$ in $Bbb R^n$ centered at $x_0$. Write $widehat U = varphi(U)subseteq Bbb R^n$, and write the coordinate functions of $varphi$ as $big(u^1,dots,u^nbig)$, so that $Mcap U$ is the set where $u^{m+1}=dotsb=u^n=0$. At each point $xin U$, the vectors $E_j|_x = (dvarphi_x)^{-1}big(partial/partial u^j|_{varphi(x)}big)$ form a basis for $T_xBbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as
$$
E_jbig|_x = E_j^i(x)frac{partial}{partial x^i}bigg|_x,
$$
where each $E_j^i(x)$ is a partial derivative of $varphi^{-1}$ evaluated at $varphi(x)$, and thus is a smooth function of $x$.




The proof goes on, but my question is about this part of the proof.




  • Is $partial/partial u^j|_{varphi(x)}$ literally the partial derivative operator in the $e_j = (0,dots,0,underbrace{1}_{text{$j$th component}},0,dots,0)$ direction? If not, what is it precisely, using Lee's notation?

  • I suspect that if $partial/partial u^j|_{varphi(x)}$ is not literally the partial derivative operator in the $e_j$ direction, then
    $partial/partial x^j|_x = (dvarphi_x)^{-1}big(partial / partial x^j|_{varphi(x)}big)$, where $partial / partial x^j|_{varphi(x)}$ is
    literally the partial derivative operator in the $e_j$ direction, and that $partial/partial u^j|_{varphi(x)}$ must be something else.

  • What is $E^i_j(x)$ explicitly?


Note that Lee hasn't yet defined standard frame at this point in the text, so I suppose a complete answer to this question would also address what the standard frame actually is.










share|cite|improve this question




















  • 3




    Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
    – Jack Lee
    Jul 27 at 21:07










  • @JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
    – AOrtiz
    Aug 2 at 17:38












  • Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
    – Jack Lee
    Aug 2 at 20:27















up vote
1
down vote

favorite









up vote
1
down vote

favorite











First I'll fix some of the definitions and context for the forthcoming question.



Suppose $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $xin M$, we define the normal space to $pmb M$ at $pmb x$ to be the $(n-m)$-dimensional subspace $N_xMsubseteq T_xBbb R^n$ consisting of all vectors that are orthogonal to $T_xM$ with respect to the Euclidean dot product. The normal bundle of $pmb{M}$, denoted by $NM$, is the subset of $TBbb R^napprox Bbb R^ntimesBbb R^n$ consisting of vectors that are normal to $M$:
$$
NM = big{(x,v) in Bbb R^ntimesBbb R^n : xin M, vin N_xM big}.
$$



The statement and part of the proof of Theorem 6.23 in Lee's Introduction to Smooth Manifolds is reproduced below:




Theorem 6.23. If $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $TBbb R^napprox Bbb R^ntimesBbb R^n$.



Proof. Let $x_0$ be any point of $M$, and let $(U,varphi)$ be a slice chart for $M$ in $Bbb R^n$ centered at $x_0$. Write $widehat U = varphi(U)subseteq Bbb R^n$, and write the coordinate functions of $varphi$ as $big(u^1,dots,u^nbig)$, so that $Mcap U$ is the set where $u^{m+1}=dotsb=u^n=0$. At each point $xin U$, the vectors $E_j|_x = (dvarphi_x)^{-1}big(partial/partial u^j|_{varphi(x)}big)$ form a basis for $T_xBbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as
$$
E_jbig|_x = E_j^i(x)frac{partial}{partial x^i}bigg|_x,
$$
where each $E_j^i(x)$ is a partial derivative of $varphi^{-1}$ evaluated at $varphi(x)$, and thus is a smooth function of $x$.




The proof goes on, but my question is about this part of the proof.




  • Is $partial/partial u^j|_{varphi(x)}$ literally the partial derivative operator in the $e_j = (0,dots,0,underbrace{1}_{text{$j$th component}},0,dots,0)$ direction? If not, what is it precisely, using Lee's notation?

  • I suspect that if $partial/partial u^j|_{varphi(x)}$ is not literally the partial derivative operator in the $e_j$ direction, then
    $partial/partial x^j|_x = (dvarphi_x)^{-1}big(partial / partial x^j|_{varphi(x)}big)$, where $partial / partial x^j|_{varphi(x)}$ is
    literally the partial derivative operator in the $e_j$ direction, and that $partial/partial u^j|_{varphi(x)}$ must be something else.

  • What is $E^i_j(x)$ explicitly?


Note that Lee hasn't yet defined standard frame at this point in the text, so I suppose a complete answer to this question would also address what the standard frame actually is.










share|cite|improve this question















First I'll fix some of the definitions and context for the forthcoming question.



Suppose $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $xin M$, we define the normal space to $pmb M$ at $pmb x$ to be the $(n-m)$-dimensional subspace $N_xMsubseteq T_xBbb R^n$ consisting of all vectors that are orthogonal to $T_xM$ with respect to the Euclidean dot product. The normal bundle of $pmb{M}$, denoted by $NM$, is the subset of $TBbb R^napprox Bbb R^ntimesBbb R^n$ consisting of vectors that are normal to $M$:
$$
NM = big{(x,v) in Bbb R^ntimesBbb R^n : xin M, vin N_xM big}.
$$



The statement and part of the proof of Theorem 6.23 in Lee's Introduction to Smooth Manifolds is reproduced below:




Theorem 6.23. If $Msubseteq Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $TBbb R^napprox Bbb R^ntimesBbb R^n$.



Proof. Let $x_0$ be any point of $M$, and let $(U,varphi)$ be a slice chart for $M$ in $Bbb R^n$ centered at $x_0$. Write $widehat U = varphi(U)subseteq Bbb R^n$, and write the coordinate functions of $varphi$ as $big(u^1,dots,u^nbig)$, so that $Mcap U$ is the set where $u^{m+1}=dotsb=u^n=0$. At each point $xin U$, the vectors $E_j|_x = (dvarphi_x)^{-1}big(partial/partial u^j|_{varphi(x)}big)$ form a basis for $T_xBbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as
$$
E_jbig|_x = E_j^i(x)frac{partial}{partial x^i}bigg|_x,
$$
where each $E_j^i(x)$ is a partial derivative of $varphi^{-1}$ evaluated at $varphi(x)$, and thus is a smooth function of $x$.




The proof goes on, but my question is about this part of the proof.




  • Is $partial/partial u^j|_{varphi(x)}$ literally the partial derivative operator in the $e_j = (0,dots,0,underbrace{1}_{text{$j$th component}},0,dots,0)$ direction? If not, what is it precisely, using Lee's notation?

  • I suspect that if $partial/partial u^j|_{varphi(x)}$ is not literally the partial derivative operator in the $e_j$ direction, then
    $partial/partial x^j|_x = (dvarphi_x)^{-1}big(partial / partial x^j|_{varphi(x)}big)$, where $partial / partial x^j|_{varphi(x)}$ is
    literally the partial derivative operator in the $e_j$ direction, and that $partial/partial u^j|_{varphi(x)}$ must be something else.

  • What is $E^i_j(x)$ explicitly?


Note that Lee hasn't yet defined standard frame at this point in the text, so I suppose a complete answer to this question would also address what the standard frame actually is.







differential-geometry proof-explanation smooth-manifolds






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edited Jul 26 at 20:24

























asked Jul 26 at 17:59









AOrtiz

10.3k21240




10.3k21240








  • 3




    Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
    – Jack Lee
    Jul 27 at 21:07










  • @JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
    – AOrtiz
    Aug 2 at 17:38












  • Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
    – Jack Lee
    Aug 2 at 20:27
















  • 3




    Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
    – Jack Lee
    Jul 27 at 21:07










  • @JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
    – AOrtiz
    Aug 2 at 17:38












  • Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
    – Jack Lee
    Aug 2 at 20:27










3




3




Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
– Jack Lee
Jul 27 at 21:07




Thanks for pointing out that I used the term "frame" before defining it. That was a typo -- it should have said "standard coordinate basis," not "standard coordinate frame." I've added a correction to my online list.
– Jack Lee
Jul 27 at 21:07












@JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
– AOrtiz
Aug 2 at 17:38






@JackLee Would you mind explaining what the $partial/partial u^j|_{varphi(x)}$ are? I thought I understood what this meant, but I realize I don't know how to actually define this notation since I'm used to seeing $partial/partial x^j|_{varphi(x)}$ where $x^j$ is literally the $j$th standard coordinate. I am just not sure how to interpret this with $u^j$. Intuitively I believe it's supposed to be partial differentiation in the "direction" of $u^j$.
– AOrtiz
Aug 2 at 17:38














Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
– Jack Lee
Aug 2 at 20:27






Yes, $partial/partial u^j|_{varphi(x)}$ is literally the partial derivative in the $u^j$ direction, evaluated at the point $varphi(x)$. See Corollary 3.3 on page 54.
– Jack Lee
Aug 2 at 20:27












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For your first question, just look at $Bbb{S}^2 subset Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(theta,phi,rho)$ form a slice chart for $Bbb{S}^2$, but we knew that not one of the basis vector ${partial_{theta}, partial_{phi},partial_rho}$ is equal to any of ${partial_x,partial_y,partial_z}$. (This is exercise 5.10 actually).



This is obvious since both are different charts $(U,x^i)$ and $(V,widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $partial_{x^i}|_p = partial widetilde{x}^j/partial x^i (hat{p}) , partial_{widetilde{x}^j}|_p$.



To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U to Bbb{R}$ is the $k$-th coordinate function; so
begin{align}
E^k_j(x) &= E_j|_x x^k = (dvarphi_x)^{-1} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)}Bigg) x^k \ &= d(varphi^{-1})_{varphi(x)} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)} Bigg) x^k = frac{partial}{partial u^j}Big|_{varphi(x)} (x^k circ varphi^{-1}) \&= frac{partial big(varphi^{-1}big)^k}{partial u^j} (varphi(x)).
end{align}



Therefore $$E_j|_x = frac{partial big(varphi^{-1}big)^i}{partial u^j} (varphi(x)) frac{partial}{partial x^i}Big|_{x}$$



Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $Bbb{R}^n$ (identity map).






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  • Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
    – AOrtiz
    Jul 26 at 19:52










  • It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
    – Kelvin Lois
    Jul 26 at 20:02












  • But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
    – AOrtiz
    Jul 26 at 20:05












  • Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
    – Kelvin Lois
    Jul 26 at 20:07












  • Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
    – AOrtiz
    Jul 26 at 20:22




















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Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.



Instead of writing the coordinate functions of $varphi$ as $big(u^ibig)$, let's write $varphi = big(varphi^ibig)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $varphi$. If $pin U$, then
$$
E_j|_p = frac{partial}{partial u^j}bigg|_p stackrel{text{def}}{=} dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg), quad j=1,dots,n,
$$

and the $E_j|_p$ form a basis for $T_pmathbb{R}^n$. Let us write the coordinate functions of the identity map of $mathbb{R}^n$ as $I = big(pi^ibig)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $big(mathbb{R}^n,Ibig)$. Then we get another basis for $T_pmathbb{R}^n$, the standard basis:
$$
frac{partial}{partial x^i}bigg|_p stackrel{text{def}}{=} dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg),quad i = 1,dots,n,
$$

so we can express $E_j|_p$ in terms of the standard basis:
$$
E_j|_p = E_j^i(p)frac{partial}{partial x^i}bigg|_p.
$$

Now we can follow Lee on page 64 to get
begin{align*}
frac{partial}{partial u^j}bigg|_p &= dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
&= dbig(I^{-1}big)_{I(p)}circ dbig(Icirc varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
&= dbig(I^{-1}big)_{I(p)}bigg(frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big)frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
&= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
&= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),frac{partial}{partial x^i}bigg|_{p}.
end{align*}

Now the map $varphicolon Uto widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $frac{partial (pi^icirc varphi^{-1})}{partial u^j}colon widehat Utomathbb{R}$ of the transition map, so the composition $E_j|_{cdot}=frac{partial (pi^icirc varphi^{-1})}{partial u^j}circvarphicolon Utomathbb{R}$ is also smooth.






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    The map $varphi$ is a local diffeomorphism. Suppose $varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $mathbb{R}^n$, $varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${partialoverpartial_i}varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={partialoverpartial_i}(x_1,...,x_n)$.






    share|cite|improve this answer





















    • Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
      – AOrtiz
      Jul 26 at 19:43













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    For your first question, just look at $Bbb{S}^2 subset Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(theta,phi,rho)$ form a slice chart for $Bbb{S}^2$, but we knew that not one of the basis vector ${partial_{theta}, partial_{phi},partial_rho}$ is equal to any of ${partial_x,partial_y,partial_z}$. (This is exercise 5.10 actually).



    This is obvious since both are different charts $(U,x^i)$ and $(V,widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $partial_{x^i}|_p = partial widetilde{x}^j/partial x^i (hat{p}) , partial_{widetilde{x}^j}|_p$.



    To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U to Bbb{R}$ is the $k$-th coordinate function; so
    begin{align}
    E^k_j(x) &= E_j|_x x^k = (dvarphi_x)^{-1} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)}Bigg) x^k \ &= d(varphi^{-1})_{varphi(x)} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)} Bigg) x^k = frac{partial}{partial u^j}Big|_{varphi(x)} (x^k circ varphi^{-1}) \&= frac{partial big(varphi^{-1}big)^k}{partial u^j} (varphi(x)).
    end{align}



    Therefore $$E_j|_x = frac{partial big(varphi^{-1}big)^i}{partial u^j} (varphi(x)) frac{partial}{partial x^i}Big|_{x}$$



    Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $Bbb{R}^n$ (identity map).






    share|cite|improve this answer























    • Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
      – AOrtiz
      Jul 26 at 19:52










    • It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
      – Kelvin Lois
      Jul 26 at 20:02












    • But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
      – AOrtiz
      Jul 26 at 20:05












    • Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
      – Kelvin Lois
      Jul 26 at 20:07












    • Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
      – AOrtiz
      Jul 26 at 20:22

















    up vote
    2
    down vote













    For your first question, just look at $Bbb{S}^2 subset Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(theta,phi,rho)$ form a slice chart for $Bbb{S}^2$, but we knew that not one of the basis vector ${partial_{theta}, partial_{phi},partial_rho}$ is equal to any of ${partial_x,partial_y,partial_z}$. (This is exercise 5.10 actually).



    This is obvious since both are different charts $(U,x^i)$ and $(V,widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $partial_{x^i}|_p = partial widetilde{x}^j/partial x^i (hat{p}) , partial_{widetilde{x}^j}|_p$.



    To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U to Bbb{R}$ is the $k$-th coordinate function; so
    begin{align}
    E^k_j(x) &= E_j|_x x^k = (dvarphi_x)^{-1} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)}Bigg) x^k \ &= d(varphi^{-1})_{varphi(x)} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)} Bigg) x^k = frac{partial}{partial u^j}Big|_{varphi(x)} (x^k circ varphi^{-1}) \&= frac{partial big(varphi^{-1}big)^k}{partial u^j} (varphi(x)).
    end{align}



    Therefore $$E_j|_x = frac{partial big(varphi^{-1}big)^i}{partial u^j} (varphi(x)) frac{partial}{partial x^i}Big|_{x}$$



    Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $Bbb{R}^n$ (identity map).






    share|cite|improve this answer























    • Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
      – AOrtiz
      Jul 26 at 19:52










    • It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
      – Kelvin Lois
      Jul 26 at 20:02












    • But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
      – AOrtiz
      Jul 26 at 20:05












    • Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
      – Kelvin Lois
      Jul 26 at 20:07












    • Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
      – AOrtiz
      Jul 26 at 20:22















    up vote
    2
    down vote










    up vote
    2
    down vote









    For your first question, just look at $Bbb{S}^2 subset Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(theta,phi,rho)$ form a slice chart for $Bbb{S}^2$, but we knew that not one of the basis vector ${partial_{theta}, partial_{phi},partial_rho}$ is equal to any of ${partial_x,partial_y,partial_z}$. (This is exercise 5.10 actually).



    This is obvious since both are different charts $(U,x^i)$ and $(V,widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $partial_{x^i}|_p = partial widetilde{x}^j/partial x^i (hat{p}) , partial_{widetilde{x}^j}|_p$.



    To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U to Bbb{R}$ is the $k$-th coordinate function; so
    begin{align}
    E^k_j(x) &= E_j|_x x^k = (dvarphi_x)^{-1} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)}Bigg) x^k \ &= d(varphi^{-1})_{varphi(x)} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)} Bigg) x^k = frac{partial}{partial u^j}Big|_{varphi(x)} (x^k circ varphi^{-1}) \&= frac{partial big(varphi^{-1}big)^k}{partial u^j} (varphi(x)).
    end{align}



    Therefore $$E_j|_x = frac{partial big(varphi^{-1}big)^i}{partial u^j} (varphi(x)) frac{partial}{partial x^i}Big|_{x}$$



    Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $Bbb{R}^n$ (identity map).






    share|cite|improve this answer














    For your first question, just look at $Bbb{S}^2 subset Bbb{R}^3$. The spherical coordinates $(u^1,u^2,u^3)=(theta,phi,rho)$ form a slice chart for $Bbb{S}^2$, but we knew that not one of the basis vector ${partial_{theta}, partial_{phi},partial_rho}$ is equal to any of ${partial_x,partial_y,partial_z}$. (This is exercise 5.10 actually).



    This is obvious since both are different charts $(U,x^i)$ and $(V,widetilde{x}^i)$, and we don't expect their basis to be equal, but related by a coordinate transformation rule $partial_{x^i}|_p = partial widetilde{x}^j/partial x^i (hat{p}) , partial_{widetilde{x}^j}|_p$.



    To obtain explicit form of $E^i_j(x)$, just carry out the computation $E_j|_x (x^k)$, where $x^k : U to Bbb{R}$ is the $k$-th coordinate function; so
    begin{align}
    E^k_j(x) &= E_j|_x x^k = (dvarphi_x)^{-1} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)}Bigg) x^k \ &= d(varphi^{-1})_{varphi(x)} Bigg(frac{partial}{partial u^j}Big|_{varphi(x)} Bigg) x^k = frac{partial}{partial u^j}Big|_{varphi(x)} (x^k circ varphi^{-1}) \&= frac{partial big(varphi^{-1}big)^k}{partial u^j} (varphi(x)).
    end{align}



    Therefore $$E_j|_x = frac{partial big(varphi^{-1}big)^i}{partial u^j} (varphi(x)) frac{partial}{partial x^i}Big|_{x}$$



    Standard frame is not yet defined until Ch.8 (about vector field). Here standard frame is a coordinate frame of the standard chart for $Bbb{R}^n$ (identity map).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 14 at 6:49









    AOrtiz

    10.3k21240




    10.3k21240










    answered Jul 26 at 19:44









    Kelvin Lois

    3,0722822




    3,0722822












    • Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
      – AOrtiz
      Jul 26 at 19:52










    • It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
      – Kelvin Lois
      Jul 26 at 20:02












    • But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
      – AOrtiz
      Jul 26 at 20:05












    • Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
      – Kelvin Lois
      Jul 26 at 20:07












    • Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
      – AOrtiz
      Jul 26 at 20:22




















    • Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
      – AOrtiz
      Jul 26 at 19:52










    • It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
      – Kelvin Lois
      Jul 26 at 20:02












    • But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
      – AOrtiz
      Jul 26 at 20:05












    • Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
      – Kelvin Lois
      Jul 26 at 20:07












    • Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
      – AOrtiz
      Jul 26 at 20:22


















    Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
    – AOrtiz
    Jul 26 at 19:52




    Thanks for the answer. Can you speak more to what $partial/partial u^j|_{varphi(x)}$ actually is? It seems like you might be familiar with Lee's notations.
    – AOrtiz
    Jul 26 at 19:52












    It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
    – Kelvin Lois
    Jul 26 at 20:02






    It is just a basis of $T_{varphi(x)}Bbb{R}^n$ (vector space of derivations). Note that these basis are generally different for different charts.
    – Kelvin Lois
    Jul 26 at 20:02














    But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
    – AOrtiz
    Jul 26 at 20:05






    But there is more information in it than that, right? For instance, Lee mentions that the $u^j$ are coordinate functions of $varphi$, so how are $partial/partial u^j$ defined in terms of $varphi$ and $partial/partial x^j$, where $partial/partial x^j$ is literally the partial derivative operator with respect to the $j$th standard basis direction?
    – AOrtiz
    Jul 26 at 20:05














    Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
    – Kelvin Lois
    Jul 26 at 20:07






    Yes. The $partial/partial u^j$ is the directional derivatives operator $textbf{in the coordinates}$ $(u^1,dots,u^n)$ of $Bbb{R}^n$. $partial_{x^i}$ different from $partial_{u^i}$ because they are two different coorrdinates (charts).
    – Kelvin Lois
    Jul 26 at 20:07














    Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
    – AOrtiz
    Jul 26 at 20:22






    Thanks for sticking with me on this. Your most recent comment doesn't address how to define $partial/partial u^j$ in terms of $varphi$ and $x^j$. Can you speak to that?
    – AOrtiz
    Jul 26 at 20:22












    up vote
    2
    down vote



    accepted










    Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.



    Instead of writing the coordinate functions of $varphi$ as $big(u^ibig)$, let's write $varphi = big(varphi^ibig)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $varphi$. If $pin U$, then
    $$
    E_j|_p = frac{partial}{partial u^j}bigg|_p stackrel{text{def}}{=} dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg), quad j=1,dots,n,
    $$

    and the $E_j|_p$ form a basis for $T_pmathbb{R}^n$. Let us write the coordinate functions of the identity map of $mathbb{R}^n$ as $I = big(pi^ibig)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $big(mathbb{R}^n,Ibig)$. Then we get another basis for $T_pmathbb{R}^n$, the standard basis:
    $$
    frac{partial}{partial x^i}bigg|_p stackrel{text{def}}{=} dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg),quad i = 1,dots,n,
    $$

    so we can express $E_j|_p$ in terms of the standard basis:
    $$
    E_j|_p = E_j^i(p)frac{partial}{partial x^i}bigg|_p.
    $$

    Now we can follow Lee on page 64 to get
    begin{align*}
    frac{partial}{partial u^j}bigg|_p &= dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
    &= dbig(I^{-1}big)_{I(p)}circ dbig(Icirc varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
    &= dbig(I^{-1}big)_{I(p)}bigg(frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big)frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
    &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
    &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),frac{partial}{partial x^i}bigg|_{p}.
    end{align*}

    Now the map $varphicolon Uto widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $frac{partial (pi^icirc varphi^{-1})}{partial u^j}colon widehat Utomathbb{R}$ of the transition map, so the composition $E_j|_{cdot}=frac{partial (pi^icirc varphi^{-1})}{partial u^j}circvarphicolon Utomathbb{R}$ is also smooth.






    share|cite|improve this answer



























      up vote
      2
      down vote



      accepted










      Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.



      Instead of writing the coordinate functions of $varphi$ as $big(u^ibig)$, let's write $varphi = big(varphi^ibig)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $varphi$. If $pin U$, then
      $$
      E_j|_p = frac{partial}{partial u^j}bigg|_p stackrel{text{def}}{=} dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg), quad j=1,dots,n,
      $$

      and the $E_j|_p$ form a basis for $T_pmathbb{R}^n$. Let us write the coordinate functions of the identity map of $mathbb{R}^n$ as $I = big(pi^ibig)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $big(mathbb{R}^n,Ibig)$. Then we get another basis for $T_pmathbb{R}^n$, the standard basis:
      $$
      frac{partial}{partial x^i}bigg|_p stackrel{text{def}}{=} dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg),quad i = 1,dots,n,
      $$

      so we can express $E_j|_p$ in terms of the standard basis:
      $$
      E_j|_p = E_j^i(p)frac{partial}{partial x^i}bigg|_p.
      $$

      Now we can follow Lee on page 64 to get
      begin{align*}
      frac{partial}{partial u^j}bigg|_p &= dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
      &= dbig(I^{-1}big)_{I(p)}circ dbig(Icirc varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
      &= dbig(I^{-1}big)_{I(p)}bigg(frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big)frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
      &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
      &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),frac{partial}{partial x^i}bigg|_{p}.
      end{align*}

      Now the map $varphicolon Uto widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $frac{partial (pi^icirc varphi^{-1})}{partial u^j}colon widehat Utomathbb{R}$ of the transition map, so the composition $E_j|_{cdot}=frac{partial (pi^icirc varphi^{-1})}{partial u^j}circvarphicolon Utomathbb{R}$ is also smooth.






      share|cite|improve this answer

























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.



        Instead of writing the coordinate functions of $varphi$ as $big(u^ibig)$, let's write $varphi = big(varphi^ibig)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $varphi$. If $pin U$, then
        $$
        E_j|_p = frac{partial}{partial u^j}bigg|_p stackrel{text{def}}{=} dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg), quad j=1,dots,n,
        $$

        and the $E_j|_p$ form a basis for $T_pmathbb{R}^n$. Let us write the coordinate functions of the identity map of $mathbb{R}^n$ as $I = big(pi^ibig)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $big(mathbb{R}^n,Ibig)$. Then we get another basis for $T_pmathbb{R}^n$, the standard basis:
        $$
        frac{partial}{partial x^i}bigg|_p stackrel{text{def}}{=} dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg),quad i = 1,dots,n,
        $$

        so we can express $E_j|_p$ in terms of the standard basis:
        $$
        E_j|_p = E_j^i(p)frac{partial}{partial x^i}bigg|_p.
        $$

        Now we can follow Lee on page 64 to get
        begin{align*}
        frac{partial}{partial u^j}bigg|_p &= dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
        &= dbig(I^{-1}big)_{I(p)}circ dbig(Icirc varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
        &= dbig(I^{-1}big)_{I(p)}bigg(frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big)frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
        &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
        &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),frac{partial}{partial x^i}bigg|_{p}.
        end{align*}

        Now the map $varphicolon Uto widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $frac{partial (pi^icirc varphi^{-1})}{partial u^j}colon widehat Utomathbb{R}$ of the transition map, so the composition $E_j|_{cdot}=frac{partial (pi^icirc varphi^{-1})}{partial u^j}circvarphicolon Utomathbb{R}$ is also smooth.






        share|cite|improve this answer














        Kelvin Lois' answer essentially got me "unstuck," but I want to write up my own answer to this question, since I find that by writing things out in detail, not making any of the "usual" identifications makes it easier for me to understand, and perhaps it may help others too.



        Instead of writing the coordinate functions of $varphi$ as $big(u^ibig)$, let's write $varphi = big(varphi^ibig)$, and reserve the letters $u^i$ to denote the canonical coordinates on the codomain of $varphi$. If $pin U$, then
        $$
        E_j|_p = frac{partial}{partial u^j}bigg|_p stackrel{text{def}}{=} dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg), quad j=1,dots,n,
        $$

        and the $E_j|_p$ form a basis for $T_pmathbb{R}^n$. Let us write the coordinate functions of the identity map of $mathbb{R}^n$ as $I = big(pi^ibig)$, and use the letters $x^i$ for the canonical coordinates on the codomain of $I$, considered as a chart $big(mathbb{R}^n,Ibig)$. Then we get another basis for $T_pmathbb{R}^n$, the standard basis:
        $$
        frac{partial}{partial x^i}bigg|_p stackrel{text{def}}{=} dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg),quad i = 1,dots,n,
        $$

        so we can express $E_j|_p$ in terms of the standard basis:
        $$
        E_j|_p = E_j^i(p)frac{partial}{partial x^i}bigg|_p.
        $$

        Now we can follow Lee on page 64 to get
        begin{align*}
        frac{partial}{partial u^j}bigg|_p &= dbig(varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
        &= dbig(I^{-1}big)_{I(p)}circ dbig(Icirc varphi^{-1}big)_{varphi(p)}bigg(frac{partial}{partial u^j}bigg|_{varphi(p)}bigg) \
        &= dbig(I^{-1}big)_{I(p)}bigg(frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big)frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
        &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),dbig(I^{-1}big)_{I(p)}bigg(frac{partial}{partial x^i}bigg|_{I(p)}bigg) \
        &= frac{partial big(pi^icirc varphi^{-1}big)}{partial u^j}big(varphi(p)big),frac{partial}{partial x^i}bigg|_{p}.
        end{align*}

        Now the map $varphicolon Uto widehat U$ is smooth (it is a diffeomorphism), and so is the partial derivative $frac{partial (pi^icirc varphi^{-1})}{partial u^j}colon widehat Utomathbb{R}$ of the transition map, so the composition $E_j|_{cdot}=frac{partial (pi^icirc varphi^{-1})}{partial u^j}circvarphicolon Utomathbb{R}$ is also smooth.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 14 at 6:52

























        answered Nov 14 at 6:38









        AOrtiz

        10.3k21240




        10.3k21240






















            up vote
            0
            down vote













            The map $varphi$ is a local diffeomorphism. Suppose $varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $mathbb{R}^n$, $varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${partialoverpartial_i}varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={partialoverpartial_i}(x_1,...,x_n)$.






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            • Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
              – AOrtiz
              Jul 26 at 19:43

















            up vote
            0
            down vote













            The map $varphi$ is a local diffeomorphism. Suppose $varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $mathbb{R}^n$, $varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${partialoverpartial_i}varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={partialoverpartial_i}(x_1,...,x_n)$.






            share|cite|improve this answer





















            • Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
              – AOrtiz
              Jul 26 at 19:43















            up vote
            0
            down vote










            up vote
            0
            down vote









            The map $varphi$ is a local diffeomorphism. Suppose $varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $mathbb{R}^n$, $varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${partialoverpartial_i}varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={partialoverpartial_i}(x_1,...,x_n)$.






            share|cite|improve this answer












            The map $varphi$ is a local diffeomorphism. Suppose $varphi(x) =0$,as you guessed, let $e_1,...,e_n$ be a basis of $mathbb{R}^n$, $varphi^{-1}(x_1,..,x_m,0,..,0)$ is in $M$, this implies that ${partialoverpartial_i}varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={partialoverpartial_i}(x_1,...,x_n)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 26 at 18:46









            Tsemo Aristide

            54.3k11344




            54.3k11344












            • Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
              – AOrtiz
              Jul 26 at 19:43




















            • Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
              – AOrtiz
              Jul 26 at 19:43


















            Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
            – AOrtiz
            Jul 26 at 19:43






            Thanks for the answer. I am not familiar with the notation $partial/partial_i$. In Lee's notation, the letter that follows the $partial$ in the "denominator" carries nonzero information that this answer is currently obscuring. The standard frame is just the canonical basis of what? There are a couple of different tangent spaces in the discussion, and some of them are "identified" with one another, but I would like to keep them all distinct to understand what's really being said.
            – AOrtiz
            Jul 26 at 19:43




















             

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