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I've been looking for a definition of "trivialisation of normal bundle".

I think I understand what a vector bundle, fibre bundle and a local trivialisation of either is. I also know what a tangent bundle is.

  1. I'm not too sure about what a normal bundle is. Let's consider the torus $T = S^1 \times S^1$. This should be a particularly easy example since it's orientable and hence if considered as the total space over the base space $X = S^1$, the fibre bundle is a trivial bundle. What does the normal bundle of $T$ look like? I think I see what the tangent bundle looks like. I assume to keep it simple we want to take the inclusion map $T \hookrightarrow \mathbb R^3$ as our immersion. Is the normal bundle $\{0\}$ at each point? If yes, can someone give me an immersion so that the normal bundle becomes more interesting? Thanks loads.

  2. I quote from ncatlab: A framing is a trivialisation of the normal bundle of a manifold. What is a trivialisation of the normal bundle? It's also not clear to me whether "local trivialisation" and "trivialisation" are used interchangeably.

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    The normal bundle is the orthogonal complement of the tangent bundle of a (Riemannian) manifold embedded in another. It is a generalisation of the normal vector field of a surface embedded in $\mathbb{R}^3$.2012-09-09
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    Thanks, that's very helpful to know.2012-09-09
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    Also, note that the normal bundle of a manifold $M$ is not an object that exists intrinsically, like the tangent bundle. It requires a bigger manifold $N$ inside of which $M$ sits, and the normal bundle will vary depending on $N$. If you consider the circle $S^1$ inside $\mathbb{R}^2$, the normal bundle will consist of *lines* at each point. If you consider $S^1$ inside $\mathbb{R}^3$, say in the $xy$ plane, the normal bundle will consist of *planes* at each point (i.e., will have rank $2$).2012-09-09
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    Thank you! This is also very helpful to know!2012-09-09
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    Given any fiber bundle, it is locally trivial which means about any point you can find a neighborhood on which it is isomorphic to a product of the open set with whatever "fiber structure" you have. So in the case of a rank $2$ vector bundle you'd get $U\times \mathbb{R}^2$. A local trivialization is a choice of such an isomorphism. Thus a trivialization of a vector bundle is a choice of a global such isomorphism, i.e. the total space of the vector bundle over $X$ is $V\simeq X\times \mathbb{R}^n$. You could also rephrase this in terms of global sections of the bundle.2012-09-09
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    @Matt Thank you! And a trivialisation of the normal bundle is an isomorphism between what and what? I can only guess: if the trivialisation of a vector bundle $V$ over $X$ is an isomorphism $V \to X \times \mathbb R^n$ then it could be an isomorphism from the normal bundle $N$ to $X \times \mathbb R^n$.2012-09-09
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    For example, student's comment gives the normal bundle $N$ of $S^1\hookrightarrow \mathbb{R}^2$ as a vector bundle whose fibers are lines, i.e. copies of $\mathbb{R}$. Not every normal bundle has a trivialization, but in this case it does. There is an explicit isomorphism $N\simeq S^1\times \mathbb{R}$. One such map induced by the non-vanishing global section $S^1\to N$ given by $v\mapsto (v,v)$, i.e. you just take the unit normal to the circle everywhere.2012-09-09
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    @ZhenLin: I would suggest you remove the brackets. Orthogonality doesn't make sense for general manifolds. The riemannian condition is central.2012-09-09
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    @FlybyNight He won't be able to edit his comment, they're only editable for a short period of time (something like 2 minutes or so).2012-09-09
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    @Matt: Ahh, yes, quite right.2012-09-09
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    You can also define the normal bundle of $f:M \hookrightarrow N$ as the quotient bundle $f^*TN / TM$ (using the pushforward $TM \rightarrow f^*TN$).2012-09-09
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    Dear @student regarding your sentence that the normal bundle at $S^1 \subset \mathbb R^2$ consists of lines at each point: there is only one line at each point, no? (the line through the point and orthogonal to the tangent line)2012-09-10
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    @Matt Right. Maybe my sentence gave you the impression that I meant many lines at one point, sorry about that. The plural is because there are infinitely many lines in this bundle, but only one at each point, as you said.2012-09-10
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    @Matt Oops, wrong Matt was pinged. Or were we both? Oh, but you were probably pinged due to the comment on your question ...2012-09-10
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    Yes, I was pinged, too : )2012-09-10
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    @Matt Where you write $S^1 \to N, v \mapsto (v,v)$, how does this map induce an isomorphism between $N$ and $S^1 \times \mathbb R$?2012-09-10

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