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I'd like to show that the normal bundle of $S^1$ is trivial. That is, I want to find a homeomorphism $\varphi : NS^1 \to S^1 \times \mathbb R$ such that $\varphi \mid_{N_s S^1}$ is linear for every $s \in S^1$ where $S^1$ comes with the embedding $\theta \mapsto (\sin \theta, \cos \theta)$.

Points in $NS^1$ are of the form $(s,v)$ where $s \in S^1$ and $v \in T\mathbb R^n_{(\sin s, \cos s)}$

I'm a bit confused about how to write down the map. It seems to be $(s,y) \mapsto (s,y)$ but is this really correct?

In "pictures": I want $\varphi$ to take what looks like a "flower (in $\mathbb R^2$) with petals that are infinite lines" and map it homeomorphically into the cylinder $S^1 \times \mathbb R$.

Thanks for your help.

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You should specify that you want the normal bundle of the standard embedding $S^1 \subset \mathbb{R}^2$. Then, work through the definitions I gave in my answer on your previous question to see exactly how you should be expressing the points of $NS^1$. –  Aaron Mazel-Gee Sep 11 '12 at 21:25
    
Once you know that the normal bundle is a line bundle, that is, its fiber is $1$-dimensional, then it suffices to show that there exists a nonzero section of $s : S^1 \rightarrow NS^1$. –  student Sep 11 '12 at 22:19
    
@AaronMazel-Gee I'm trying but I changed my definitions. I am restricting my computations to the case where $M= \mathbb R^n$. Now the points $v$ (or are they vectors?) are in a tangent plane, tangent to $\mathbb R^n$. Does this make any sense at all? –  Matt N. Sep 12 '12 at 14:47
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It's easier to write down a section... But first you should write down the normal bundle explicitly (what you write is a bit incomplete). An explicit description is in this comment of our last long exchange and in the notation given there $\nu\colon S^1 \to NS^1 \subset \mathbb{R}^2 \times \mathbb{R}^2, \, x \mapsto (x,x)$ will do (and towards the end of that exchange I tell you how to translate that into a trivialization of $NS^1$). –  t.b. Sep 12 '12 at 14:53
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9999 :-) ${}{}{}$ –  Asaf Karagila Sep 24 '12 at 21:30
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1 Answer

up vote 2 down vote accepted

Thank you, t.b., with your help I managed the following:

Claim: The normal bundle of $S^1 \hookrightarrow \mathbb R^2$ is trivial.

Proof: $ N_s S^1 = \{ \lambda s \mid \lambda \in \mathbb R \}$ and $$NS^1 = \bigsqcup_{s \in S^1} N_s S^1 = \{ (s, \lambda s) \mid s \in S^1 , \lambda \in \mathbb R\}$$ Now we want to give a bundle isomorphism $\varphi : S^1 \times \mathbb R \to NS^1$. Define $\varphi : (s, \lambda ) \mapsto (s, \lambda s)$. It is clear that this map is continuous, bijective and that its inverse is also continuous. Hence $\varphi$ is a homeomorphism. It is also clear that for fixed $s$, $\varphi$ is linear. Hence $\varphi$ is fibrewise linear and hence a bundle isomorphism. Hence $NS^1$ is trivial.


Alternatively, one can use the following definitions:

$(\tt{Def})$ A section of a normal bundle $\pi : NS \to S$, where $S$ is a submanifold of some manifold $M$ is a map $s: S^1 \to NS^1$ such that $\pi \circ s = \mathrm{id}_S$.

$(\tt{Def})$ The normal bundle of a dimension $k$ submanifold $S$ of $\mathbb R^n$ is trivial if and only if there are $k$ sections $s_1 , \dots , s_{k}: S \to NS$ such that $\mathrm{span}\{ s_1 (s), \dots, s_{k}(s)\} = N_sS$ for all $s \in S$.

In the case of $S^1 \hookrightarrow \mathbb R^2$ with the bundle projection $\pi : NS^1 \to S^1, (s,\lambda s) \mapsto s$, $n=2$, $k = n-k = 1$ hence it is enough to give one section $s_1 : S^1 \to NS^1$ such that $\mathrm{span}(s_1(s)) = N_s S^1$ for all $s \in S^1$. Define $s_1: s \mapsto (s,s)$.


The corresponding for the tangent bundle:

$(\tt{Def})$ Let $M$ be a smooth manifold. We define $C^\infty (M) = \{ f: M \to \mathbb R \mid f \circ \varphi^{-1} \text{ is } \infty \text{ many times diff'able for every chart } \varphi : U \to \mathbb R^n \} $. Let $x \in M$. Then a derivation at $x$ is a linear map $D: C^\infty (M) \to \mathbb R$ such that for all $f,g \in C^\infty (M)$ we have $D(fg) = D(f) \cdot g + f \cdot D(g)$. Then the tangent space of $M$ at $x$, $T_x M$, is the set of all derivations at $x$.

$(\tt{Def})$ The tangent bundle $TM \xrightarrow{\pi} M$ of a smooth manifold $M$ is the disjoint union of its tangent spaces, that is, $$ TM = \bigsqcup_{x \in M} T_x M = \{ (x, v) \mid x \in M, v \text{ a vector in the tangent space } T_x M \}$$ where $T_x M $ is the tangent space to $M$ at $x$.

$(\tt{Def})$ Let $M$ be an $n$-manifold. Then we call $TM \xrightarrow{\pi} M$ trivial if there exist $n$ sections $s_1, \dots , s_n : TM \to M$ such that $\mathrm{span} \{s_1(x) , \dots , s_n(x) \} = TM$.

Can we alternatively define it to be trivial if there exists a (fibrewise) linear homeomorphism $\varphi : TM \to M \times \mathbb R^{N-n}$?

Claim: The tangent bundle of $S^1 \xrightarrow{\mathrm{id}} \mathbb R^2$ is trivial.

Proof:

Note: $T_s S^1$ is a line through the origin.

Let $s$ be a point on $S^1$. Then the tangent at $s$ is a $90$ degree rotation of $s$: $t_s = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} s = Rs$. We have $T_s S^1 = \{ \lambda Rs \mid \lambda \in \mathbb R \}$ so that $$ TS^1 = \bigsqcup_{s \in S^1} T_s S^1 = \{ (s, \lambda R s ) \mid s \in S^1, \lambda \in \mathbb R \}$$

Now define $\varphi : S^1 \times \mathbb R \to TS^1, (s, \lambda) \mapsto (s, \lambda R s)$. It is clear that this map is a homeomorphism and that it is fibrewise linear.

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The embedding $S^1 \to \mathbb R^2, s \mapsto s$ is used to identify $S^1$ as a subset of $\mathbb R^2$. I think it's not used anywhere else. Perhaps it would be interesting to do the same example with a different embedding. –  Matt N. Sep 12 '12 at 20:53
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It is used of course when identifying $N_s S^1 \perp T_{s} S^1 = (\operatorname{span}{s})^{\bot}$. I'm missing the bundle projection $\pi\colon NS^{1} \to S^{1}$. And to be completely explicit: the vector space structure on $N_s S^{1}$ is given by $(s,\lambda s) + (s,\mu s) = (s,(\lambda + \mu)s)$ and $\alpha(s,\lambda s) = (s,\alpha s)$ (first coordinate fixed; $\varphi$ is linear in the second component - the fibers - but not in the first one). The last definition should read: The normal bundle of a $k$-dimensional submanifold $S$ of $\mathbb R^n$ is trivial if and only if there are $n-k$... –  t.b. Sep 12 '12 at 21:01
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Maybe you want to try to figure out why the normal bundle of a Möbius band is not trivial. –  t.b. Sep 12 '12 at 21:03
    
@t.b. I thought about the Möbius band already. It's because the normals change sign. Yes, that's a good example to work out in detail. –  Matt N. Sep 12 '12 at 21:12
    
And now you... ;-) –  Asaf Karagila Sep 24 '12 at 21:32
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