Understanding the trivialisation of a normal bundle

Question

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.

Answer 1

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 $\mathbb{R}^2$.

Answer 2

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.

Answer 3

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?

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