# Understanding the trivialisation of a normal bundle

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.

• 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 $\mathbb{R}^3$. Sep 9 '12 at 13:48
• 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 $2$). Sep 9 '12 at 13:53
• 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.
– Matt
Sep 9 '12 at 14:06
• 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.
– Matt
Sep 9 '12 at 14:22
• You can also define the normal bundle of $f:M \hookrightarrow N$ as the quotient bundle $f^*TN / TM$ (using the pushforward $TM \rightarrow f^*TN$). Sep 9 '12 at 23:03

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 $\mathbb{R}^2$.

• So the normal bundle of a submanifold $X \subset Y$ is the disjoint union (over $i(x) \in Y$) of $T_{i(x)}Y / [(i_\ast)_x T_x X]$? Sep 9 '12 at 19:15
• 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. Sep 9 '12 at 19:36
• 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. Sep 9 '12 at 19:38
• Ok, now I have one more piece missing in my puzzle: framings. : ) Sep 9 '12 at 19:39
• 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 en.wikipedia.org/wiki/Chern_class Sep 9 '12 at 19:52

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 $S^1 \rightarrow \mathbb{R}^2$.

• 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. Sep 11 '12 at 5:56
• I will read your answer later, I have to go afk right now. Sep 11 '12 at 5:58
• @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. Sep 11 '12 at 11:20
• Aaron, sure, by now I figured as much : ) But it's still puzzling why I cannot find the definition saying this explicitly anywhere. Sep 11 '12 at 11:56
• 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. Sep 11 '12 at 13:47

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.

• 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 $\mathbb{R}^2$?
– t.b.
Sep 10 '12 at 16:08
• 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 $NM = \{(x,n)\,:\,x \in M, n \perp T_x M\}$.
– t.b.
Sep 10 '12 at 16:27
• 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 $x \in M$.
– t.b.
Sep 10 '12 at 16:28
• 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.
– t.b.
Sep 10 '12 at 16:32
• 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. Sep 10 '12 at 17:44