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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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Normal bundles are associated to immersions $i : X \hookrightarrow Y$. Both $X$ and $Y$ are manifolds, and have tangent bundles. The normal bundle is, very informally, what is missing from $TY$ when we look at the image of $TX$ under the differential $i_* : TX \to TY.$

Consider a point $x \in X$ and the tangent space $T_xX$. The image of the differential $(i_*)_x : T_xX \to T_{i(x)}Y$ is a linear subspace. We can consider the quotient of $T_{i(x)}Y$ by the image of $T_xX$. The fibre of $NY$ at $i(x) \in Y$, denoted by $N_{i(x)}Y$, is the quotient space $T_{i(x)}Y / [(i_*)_x(T_xX)]$

If $\dim(X) = p$ and $\dim(Y) = p+q$ then $NY$ is a rank-$q$ vector bundle over $X$.

If $X = \mathbb{R}^2$ and $Y = \mathbb{R}^3$, then the normal bundle is just a line bundle over $\mathbb{R}^2$.

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    So the normal bundle of a submanifold $X \subset Y$ is the disjoint union (over $i(x) \in Y$) of $T_{i(x)}Y / [(i_\ast)_x T_x X]$?2012-09-09
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    As a set: yes. In reality: not quite. It's a vector bundle and has more structure than a simple disjoint union of vector space.2012-09-09
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    It seems a bit painful to define. If we take a Riemannian manifold (just looked it up, Riemannian manifolds come with an inner product on the tangent spaces) then we can take the normal bundle to be the union of all vectors $v$ that are orthogonal to every vector in the submanifold. That's a bit simpler. Also in terms of seeing what it looks like.2012-09-09
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    Ok, now I have one more piece missing in my puzzle: framings. : )2012-09-09
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    You're right about the Riemannian case. In fact, people often use $TX^{\perp}$ to denote $NX$ in this case. I'm pleased that you understand that example; that is the intuition to have. Remember that the normal bundle is more abstract and doesn't need a metric to be defined. The normal bundle is more than a disjoint union of vector spaces. It's a smooth manifold, and has more structure. I know you're working over the reals, but take a look at https://en.wikipedia.org/wiki/Chern_class2012-09-09
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    Thank you : ) I'm pleased, too. And thanks for pointing this out!2012-09-09
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Again, here's a comment (on your answer) that's too long to fit so I'm posting it as an answer.

First of all, these definitions are definitely in books. I'd strongly recommend checking out Bott & Tu's "Differential Forms in Algebraic Topology". (I always recommend Bott & Tu, it's fantastic. It introduces de Rham cohomology and the basics of the algebraic topology of smooth manifolds, but then it also introduces some exciting and relatively advanced algebraic topology, too.)

Next, Jason is right, but there is more: you must require that on each fiber, the maps preserve the relevant structure. (This means, for example, that a map of vector bundles is supposed to be linear on each fiber.)

Then, if $f:S \rightarrow M$ is an immersion of a smooth manifold into a Riemannian manifold, the normal bundle $NS \rightarrow S$ consists of those pairs $(s,v)$ such that $s\in S$, $v \in T_{f(s)}M$, and $v \perp (df)_s(T_sS)$. (In particular, it makes no sense to ask that $v$ be perpendicular to all $s\in S$; vectors can't be perpendicular points, and anyways you only can compare tangent vectors that are based at the same point.)

In the more general case that $M$ doesn't have a Riemannian structure, the normal bundle can still be defined as the set of pairs $(s,\overline{v})$, where $s\in S$ and $\overline{v} \in T_{f(s)}M / (df)_s(T_sS)$. That is, $v\in T_{f(s)}M$ is a representative of an equivalence class of vectors in a quotient vector space, where we're quotienting by the relation that we "can't see" vectors that are tangent to $S\subset M$ at $s$. It might be helpful to think of the example of the standard inclusion $f:\mathbb{R} \rightarrow \mathbb{R}^2$.

I'd recommend that you work out an explicit formula for the normal bundle of the standard inclusion $f:S^1 \rightarrow \mathbb{R}^2$ (with the usual Riemannian structure on $\mathbb{R}^2$!). Then, you can try fancier immersions.

Bonus: Can you see why we're demanding that our map be an immersion?


Added, in response to a question on the comments: Let $F$ be some space that'll be our fiber, and choose a "structure group" $\mbox{Aut}(F)$. For instance, if $F=\mathbb{R}^n$, you could take

  • $\mbox{Aut}(F)=O(n)$ to get "vector bundles with a metric",
  • $\mbox{Aut}(F)=GL_n(\mathbb{R})$ to get "vector bundles",
  • $\mbox{Aut}(F)=\mbox{Diffeo}(\mathbb{R}^n)$ to get ... well, these don't have a name. But you could take this, anyways.

Then, it turns out that $F$-bundles with transition functions in $\mbox{Aut}(F)$ are equivalent to bundles of $\mbox{Aut}(F)$-torsors. (A torsor for a group $G$ is just a copy of $G$ where we've forgotten the basepoint, but we remember how $G$ acts by (say, left) multiplication. So for instance, an abstract copy of the real line is a torsor for the group $\mathbb{R}$.) From this framework, you can see that in fact it's not just that a morphism of $F$-bundles needs to be fiberwise: the individual maps on fibers are also required to respect the $\mbox{Aut}(F)$-action (i.e., they need to be equivariant). This is the general setup for any kind of fiber bundle, of which vector bundles are just a special case. The important thing here is that it's the structure group that matters, not the actual fiber.

By the way, you ask for definitions of trivializations of (i) fiber bundles, (ii) vector bundles, and (iii) normal bundles. But (iii) is just a special case of (ii). I still think you should work out the example of the normal bundle for the standard embedding $S^1 \rightarrow \mathbb{R}^2$.

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    Dear Aaron. I have a copy of Bott and Tu handy. Right now I'm looking at the index. "Trivialization" is indexed for pages 54, 21 and 80. The first two are about local trivializations. The last, page 80, is about locally constant trivialization. I don't think any of these give me the definition of trivialisation of any of the three I want: (i) fibre bundle, (ii) vector bundle, (iii) normal bundle. Of course, they might not be indexed hence: if you know the page number(s) of definitions, I'd be very grateful if you could point me to the right pages.2012-09-11
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    I will read your answer later, I have to go afk right now.2012-09-11
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    @Matt: Well, the important thing to nail down is the correct notion of a morphism of vector bundles. Then, a trivialization is just an isomorphism with a trivial bundle. I'll add more regarding arbitrary fiber bundles.2012-09-11
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    Aaron, sure, by now I figured as much : ) But it's still puzzling why I cannot find the definition saying this explicitly anywhere.2012-09-11
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    Yeah, I don't know. But the word "trivialization" is used throughout math to mean "an isomorphism with a trivial object in the same category"; it's only in the word "trivial" that there's any ambiguity in this definition, but certainly we know what a trivial vector bundle is in any case.2012-09-11
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    I'll do the normal bundle of $S^1$ in a separate thread.2012-09-11
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Ok, let's try and sum this up. First some definitions (not sure why but could not find these in any of my books nor anywhere on the internet, so I'll just make up my own)

($\tt{Def \ 1}$) Let $(E, B, F, \pi: E \to B)$ be a fibre bundle. Then we call it trivial if there exists a homeomorphism $\varphi : E \to B \times F$ such that $p \circ \varphi = \pi$ where $\pi$ is the continuous surjection coming with the bundle and $p: B \times F \to B$ is the projection $(b,f) \mapsto b$. We call $\varphi$ a $\textit{trivialisation}$ of the fibre bundle.

Similarly,

($\tt{Def \ 2}$) Let $(E, B, V, \pi: E \to B)$ be a vector bundle. Then we call it trivial if there exists a linear homeomorphism $\varphi : E \to B \times V$ such that $p \circ \varphi = \pi$ and we call $\varphi$ a $\textit{trivialisation}$ of the vector bundle.

Now most importantly,

($\tt{Def \ 3}$) Let $S$ be a submanifold of dimension $k \leq n$ of $\mathbb R^n$ (I'm restricting to $\mathbb R^n$ for now to keep it simpler). Let $i: S \hookrightarrow \mathbb R^n$ be an embedding. Let $NS$ denote the normal bundle, that is, the set of all paris $(s,v)$ with $s$ in $S$ and $v$ orthogonal to all $s^\prime$ in $T_s S$. Then $NS$ is trivial if there exists a linear homeomorphism $\varphi : NS \to S \times \mathbb R^{n-k}$. We call $\varphi$ a $\textit{trivialisation of } NS$.

Example (as pointed out in the comments) Let $S= S^1$ , $M = \mathbb R^2$ with the inclusion $i: S^1 \to \mathbb R^2$, $\theta \mapsto (\sin \theta, \cos \theta)$. The normal bundle looks like the disjoint union of normals at each point of the circle. Now we rotate them so that the whole thing looks like $S^1 \times \mathbb R$, hence $NS$ is trivial since rotation is a linear homeomorphism.

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    What does $(x,y) \mapsto (x^2,y^2)$ mean? You use the immersion when you write "normals at each point to the circle": what would that mean without identification of $S^1$ with a circle in $\mathbb{R}^2$?2012-09-10
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    @t.b. It means I map the point $(x,y) \in \mathbb R^2$ to the point $(x \cdot x, y \cdot y) \in \mathbb R^2$. I am not sure: an inclusion would work I think. But that's not an immersion. I get normals to a circle because I have an inner product in each tangnet space (that's why I want a Riemannian manifold). But my invented definition of what it means for the normal bundle to be trivial doesn't make sense to me.2012-09-10
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    What puzzles me most though is that there don't seem to be any definitions of this.2012-09-10
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    The inclusion of $S^1$ in $\mathbb{R}^2$ is even better than an immersion, it's an embedding... For arbitrary $n$ identify $S^{n-1}$ with $\{x \in \mathbb{R}^n:\lvert x\rvert = 1\}$. Then the tangent space at $x \in S^n$ is $T_x S^n = x^\bot$ and the normal bundle has fibers $N_x S^n = \operatorname{span}x$. In general, for an embedded submanifold $M$ of $\mathbb{R}^n$ you can realize $TM$ as $\{(x,v) \in \mathbb{R}^n \times \mathbb{R}^n\,:\,x\in M, v \in T_xM\}$ and the normal bundle can be written as $NM = \{(x,n)\,:\,x \in M, n \perp T_x M\}$.2012-09-10
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    The normal bundle of a codimension $k$ submanifold is trivial if and only if there are $k$ sections $X_1,\dots,X_k\colon M \to N M$ such that $\operatorname{span}\{X_1(x),\dots,X_k(x)\} = N_xM$ for all $x \in M$.2012-09-10
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    @t.b. You're my saviour, now I just need to look up what a section is et voilà.2012-09-10
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    A section of a bundle $\pi \colon E \to M$ is a map $X \colon M \to E$ such that $\pi \circ X = \operatorname{id}_M$. Just a map that assigns to each $m \in M$ a point $X(m)$ in the fiber $E_m = \pi^{-1}(m)$ over $m$. For the tangent bundle a section is the same as a vector field. The notion of a section generalizes this to arbitrary bundles. The $k$ sections is what Aaron referred to as "frames". Note: A frame of a bundle is the same as a choice of a basis in each fiber.2012-09-10
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    Your def 2 is wrong. It's not enough for there to be some abstract homeomorphism between $E$ and $B\times V$, the homeomorphism is required to commute with projections.2012-09-10
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    @JasonDeVito Thank you. (For some reason your comment didn't ping me.) Would then def.1 not also be wrong? I mean, a vector bundle is a special case of a fibre bundle, no?2012-09-10
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    @Matt: You're right - I actually read your post in reverse order for some reason. Sorry about not pinging you - I really thought it would automatically. Incidentally, I know of examples of general fiber bundles there the total space is homeomorphic (diffeo, in fact) to a product, but not trivial as a bundle. I don't know of examples in the special case of vector bundles.2012-09-10
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    Dear Matt, Also, in Def2, you don't just want a homeomorphism, you want it to respect the linear (i.e. vector space) structure in the fibres; in short, you want an isomorphism of vector bundles. This is what *trivialization* means: by definition, it is an isomorphism of your given bundle with the trivial bundle of the same dimension. Regards,2012-09-11
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    Dear @JasonDeVito Thank you. Again your comment didn't ping me. It makes me wonder how many comments addressed to me on SE I have missed. I will improve my post!2012-09-11
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    Dear @MattE Thank you. I will edit my post.2012-09-11
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    @t.b. I tried to make up a definition of triviality of the normal bundle that doesn't involve sections. Does this work?2012-09-11
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    This works if by *linear* you mean *fiberwise linear* (also in Def. 3).2012-09-11
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    To see the link between the two definitions of triviality we have now: if you fix a basis $e_1,\dots,e_n$ of $V$ then $X_i(b)$ is simply $\varphi^{-1}(b,e_i)$. In the other direction, given $x \in E$ then $x$ can be considered as a vector in the *fiber* $E_{\pi(x)} = \pi^{-1}(\pi(x))$ and thus be uniquely written as $x = a_1 X_1(\pi(x)) + \dots + a_n X(\pi(x))$ and then $\varphi(x) = (\pi(x),(a_1,\dots,a_n))$.2012-09-11
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    @t.b. Thank you. If by fibre you mean $N_x S \in NS$ then yes and also thanks for pointing it out since I think I didn't think it through properly when I wrote it.2012-09-11
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    Sorry: I don't understand what you're confirming... The *fiber* of a vector bundle $\pi \colon E \to B$ *over a point* $b \in B$ is the vector space $E_b = \pi^{-1}(b)$ "sitting over $b$".2012-09-11
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    @t.b. But I'm talking about the trivialisation of a normal bundle. There is no vector bundle in definition 3. Or is there?2012-09-11
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    @t.b. Sorry. Ignore, I'm re-reading your previous comments.2012-09-11
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    @t.b. Yes, I seem to get it: "fibres $N_x S^n$...". Now I just replace $S^n$ with $S$ et voilà.2012-09-11
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    You still have this bizarre condition "orthogonal to all $s'$ in $S$". That's not what the normal bundle is about. The *fiber of the normal bundle* over $s$ consists of the vectors orthogonal to the tangent space $T_s S$. The fiber is $N_s S = \{v \in \mathbb{R}^n\,:\,v \perp T_sS \}$ and $NS = \{(s,v)\,:\,s \in S,\,v \perp T_sS\}$. The bundle projection is simply $NS \to S$ given by $(s,v) \to s$. // yes to the last comment2012-09-11
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/4805/discussion-between-t-b-and-matt)2012-09-12