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

fiber-bundlesklein-bottlevector-bundles

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)

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.