# Lee's proof that the normal bundle is embedded

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

Suppose $M\subseteq \Bbb R^n$ is an embedded $m$-dimensional submanifold. For each $x\in M$, we define the normal space to $\pmb M$ at $\pmb x$ to be the $(n-m)$-dimensional subspace $N_xM\subseteq T_x\Bbb R^n$ 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 $T\Bbb R^n\approx \Bbb R^n\times\Bbb R^n$ consisting of vectors that are normal to $M$: $$NM = \big\{(x,v) \in \Bbb R^n\times\Bbb R^n : x\in M,\ v\in N_xM \big\}.$$

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 $M\subseteq \Bbb R^n$ is an embedded $m$-dimensional submanifold, then $NM$ is an embedded $n$-dimensional submanifold of $T\Bbb R^n\approx \Bbb R^n\times\Bbb R^n$.

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 $\big(u^1,\dots,u^n\big)$, so that $M\cap U$ is the set where $u^{m+1}=\dotsb=u^n=0$. At each point $x\in U$, the vectors $E_j|_x = (d\varphi_x)^{-1}\big(\partial/\partial u^j|_{\varphi(x)}\big)$ form a basis for $T_x\Bbb R^n$. We can expand each $E_j|_x$ in terms of the standard frame [emphasis added] as $$E_j\big|_x = E_j^i(x)\frac{\partial}{\partial x^i}\bigg|_x,$$ where each $E_j^i(x)$ is a partial derivative of $\varphi^{-1}$ evaluated at $\varphi(x)$, and thus is a smooth function of $x$.

• 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 $\partial/\partial x^j|_x = (d\varphi_x)^{-1}\big(\partial / \partial x^j|_{\varphi(x)}\big)$, 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.

• 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. Jul 27 '18 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 $u^j$. Aug 2 '18 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. Aug 2 '18 at 20:27

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 $$\big(u^i\big)$$, let's write $$\varphi = \big(\varphi^i\big)$$, and reserve the letters $$u^i$$ to denote the canonical coordinates on the codomain of $$\varphi$$. If $$p\in U$$, then $$E_j|_p = \frac{\partial}{\partial u^j}\bigg|_p \stackrel{\text{def}}{=} d\big(\varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg), \quad j=1,\dots,n,$$ and the $$E_j|_p$$ form a basis for $$T_p\mathbb{R}^n$$. Let us write the coordinate functions of the identity map of $$\mathbb{R}^n$$ as $$I = \big(\pi^i\big)$$, and use the letters $$x^i$$ for the canonical coordinates on the codomain of $$I$$, considered as a chart $$\big(\mathbb{R}^n,I\big)$$. Then we get another basis for $$T_p\mathbb{R}^n$$, the standard basis: $$\frac{\partial}{\partial x^i}\bigg|_p \stackrel{\text{def}}{=} d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg),\quad i = 1,\dots,n,$$ 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 &= d\big(\varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg) \\ &= d\big(I^{-1}\big)_{I(p)}\circ d\big(I\circ \varphi^{-1}\big)_{\varphi(p)}\bigg(\frac{\partial}{\partial u^j}\bigg|_{\varphi(p)}\bigg) \\ &= d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg) \\ &= \frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\,d\big(I^{-1}\big)_{I(p)}\bigg(\frac{\partial}{\partial x^i}\bigg|_{I(p)}\bigg) \\ &= \frac{\partial \big(\pi^i\circ \varphi^{-1}\big)}{\partial u^j}\big(\varphi(p)\big)\,\frac{\partial}{\partial x^i}\bigg|_{p}. \end{align*} Now the map $$\varphi\colon U\to \widehat U$$ is smooth (it is a diffeomorphism), and so is the partial derivative $$\frac{\partial (\pi^i\circ \varphi^{-1})}{\partial u^j}\colon \widehat U\to\mathbb{R}$$ of the transition map, so the composition $$E_j|_{\cdot}=\frac{\partial (\pi^i\circ \varphi^{-1})}{\partial u^j}\circ\varphi\colon U\to\mathbb{R}$$ is also smooth.

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 = (d\varphi_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).

• 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. Jul 26 '18 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. Jul 26 '18 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? Jul 26 '18 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). Jul 26 '18 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? Jul 26 '18 at 20:22

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 ${\partial\over\partial_i}\varphi^{-1}(x_1,..,x_m,0,..0)=E_i(x)$ is tangent to $M$. The standard frame is just the canonical basis. If $(x_1,...,x_n)$ are the coordinates $e_i={\partial\over\partial_i}(x_1,...,x_n)$.

• 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. Jul 26 '18 at 19:43