Normal bundles of $S^1$ in a Klein bottle Consider the Klein bottle $K$ obtained from a square by gluing the corresponding edges in the blow figure.

Then we get two obvious embedding of $S^1$ to Klein bottle $i_a: S^1\to K$ and $i_b:S^1\to K$, that is, embedded into the circle $a$ and the circle $b$ respectively. I want to identify the normal bundles with respect to those two embedding.
My guess: the first normal bundle is trivial bundle over $S^1$ and the second one is the unbounded Möbius band on $S^1$. But I have trouble in showing this explicitly.
My attempt starts with writing a smooth atlas of the Klein bottle and I hope this can help me do calculation. However, by definition, the normal bundle $N_{K/S^1}$ is the quotient bundle $((i_a)^*TK)/TS^1$, that is,
$$0\to TS^1 \to ((i_a)^*TK)/TS^1 \to N_{K/S^1} \to 0$$
However, I don’t know how to write the first map in local chart so that I can calculate the bundle explicitly.
Any reference, hints, and answers are welcome!
 A: IMO, you have the wrong philosophical understanding of bundles. It seems like you want to show that two bundles are Isomorphic by, say, constructing an open cover and trivializations of both over this open cover so that the transition maps are literally equal, or writing down an explicit formula for a vector bundle Isomorphism. This is possible, especially in this simple situation, but it's not how one usually thinks about such problems.
Vector bundles are topological objects. They're floppy! There aren't many of them! You can often explicitly classify them by more easily accessible information. For instance:
Let $E \to S^1$ be a vector bundle. If $\exp: [0,1] \to S^1$ is the usual quotient map with $\exp(0)=\exp(1) = 1$, then $\exp^* E$ is a trivializable vector bundle over the interval, as every bundle over the interval is contractible. Choose a trivialization $\phi: \exp^* E \to \Bbb R^n \times [0,1]$. Then the composite $\Bbb R^n \xrightarrow{\phi_1^{-1}} (\exp^* E)_1 = E_{\exp(1)} = E_{\exp(0)} = (\exp^* E)_0\xrightarrow{\phi_0} \Bbb R^n$ is given by an element of $GL_n$; call it $A$.
Then you can recover $E$ from $A$ (take the trivial bundle over the interval and glue together the fibers over the endpoints by A), and this only depends on the element $[A] \in \pi_0 GL_n \cong \Bbb Z/2$ (if you homotop A to a different element, you can write down an explicit bundle isomorphism between the resulting bundles.)
So there are exactly two vector bundles over the circle of each dimension and you can tell which one you are by determining whether you flip orientation when you go around once.
Now apply this to your example. There are natural trivialization of the normal bundle restricted to the intervals $a$ and $b$. When you compare the trivializations at the endpoints, you see that $a$'s trivialization gets flipped and $b$'s is left the same. Your desired results follow.
