Why the fiber of the normal bundle of the Klein bottle is $\mathbb{R}^2$ instead of $\mathbb{R}$?

Question

What I know:

• If M is an m-dimensional manifold embedded in $\mathbb{R}^{{m+k}}$, the normal bundle $NM$ is with the typical fiber $\mathbb{R}^k$.

My Question:

• When I think about a torus ($T$) or a sphere ($S^2$), they are 2-manifold and embedded in $\mathbb{R}^3$, so the fiber of $NT$ or $NS^2$ is $\mathbb{R}$.
• But when I think about the Klein Bottle ($K$), it is a 2-manifold and embedded in $\mathbb{R^4}$, so its fiber should be $\mathbb{R}^2$. Why do I feel its fiber of $NK$ is still $\mathbb{R}$ (see the figure below)?
• I understand that in the figure I put the Klein bottle in $\mathbb{R}^3$, so the Klein bottle is not really embedded, is it the reason why I feel the fiber is $\mathbb{R}$?
• Could I have an intuitive understanding of the $\mathbb{R}^2$ fiber? Is the $\mathbb{R}^2$ fiber like a 'surface' instead of a 'line' in Euclidean space?
• Why the fiber of $TK$ must be $\mathbb{R}^2$? A line bundle is not enough?

Figure: the normal fiber of the Klein bottle (what I imagine) (Please ignore the arrows)

Answer 1

It seems like you might be confused regarding well-definedness of normal bundles and well-definedness of fiber dimension. So I will write an answer focussing on that issue.

The normal bundle of an $m$-dimensional manifold $M$ is not a well-defined thing. First you have to choose an embedding (or an immersion) of $M$: only then can you talk about the normal bundle of $M$ relative to that embedding (or immersion); and only then can you talk about the dimension of the fibers of that normal bundle.

Maybe $M$ was already given to you as a submanifold of $\mathbb R^n$ so you can certainly talk about the normal bundle of the inclusion map (which is an embedding), but that should not deter you from studying other (noninclusion) embeddings of $M$ and their normal bundles.

For example, if you embed the circle $S^1$ into the plane $\mathbb R^2$ then the normal bundle of that embedding has fibers of dimension $2-1=1$, i.e. lines. But if on the other hand you embed $S^1$ as a knot in $\mathbb R^3$ then the normal bundle of this new embedding has fibers of dimension $3-1=2$, i.e. planes.

Regarding "the" Klein bottle $K$ (scare quotes intentional), which is a 2-dimensional manifold, like any manifold it has many different embeddings. The lowest dimensional Euclidean space into which $K$ embeds is $\mathbb R^4$, and under any such embedding the normal bundle has fibers of dimension $4-2=2$. In fact you can embed $K$ into Euclidean space $\mathbb R^n$ of any dimension $n \ge 4$, and you would get a normal bundle with fibers of dimension $n-2$.

The usual image of a Klein bottle in $\mathbb R^3$ is not an embedding but it is an immersion, and one can still define the normal bundle of an immersion. And the same formula holds: the normal bundle of an $m$-dimensional manifold immersed into $\mathbb R^n$ has fibers of dimension $n-m$. So the normal bundle of a Klein bottle immersed into $\mathbb R^3$ has fibers of dimension $3-2=1$, i.e. lines.

Answer 2

As you noted, your picture does not show an embedding of the Klein bottle. That would require 4 spacial dimensions and the drawing uses only 3 (technically 2 but your mind will interpret the picture as representing something 3-dimensional).

The Klein bottle cannot be embedded into $\mathbb{R}^3$ but this is a global constraint not a local one. A small neighborhood of any point can be embedded into $\mathbb{R}^3$ and using that and glueing the pieces together is what the picture is doing. Hence the normal bundle is only a line at each point but globally this is not an embedding.

There are methods for representing 4-dimensional objects in 2 or 3 dimensions but I don't know how useful these would be to get an idea of what the normal bundle of the Klein bottle looks like. That is what you would need to actually have a 2-d normal bundle on a surface.

If you just want an idea of what a normal bundle with 2-d fibres looks like you could picture the embedding of a knot into $\mathbb{R}^3$. The knot has only 1 dimension, so the fibre of the normal bundle is a plane at each point.