line bundles over the circle

I will try to sum up the existing comments and talk about the linking number.

Isomorphism vs. Isotopy

So basically there are two things interplaying here when you talk about line bundles over $S^1$: isomorphisms and isotopies. Let's look at the definitions:

• Two line bundles $E, E' \xrightarrow{p, p'} S^1$ are said to be isomorphic if there exists a map $f : E \to E'$ that makes the triangle commute ($p' \circ f = p$), such that the induced map on fibers is linear (these two conditions make $f$ into a vector bundle morphism), and moreover $f$ has an inverse which is also a vector bundle morphism.

• Let two line bundles $E, E' \xrightarrow{p, p'} S^1$ and embeddings $E, E', S^1 \subset \mathbb{R}^3$. Then an isotopy between $E$ and $E'$ is a map $H : [0,1] \times \mathbb{R}^3 \to \mathbb{R}^3$ such that:

  • $H(0, -)$ is the identity of $\mathbb{R}^3$;
  • every $H(t, -)$ is a homeomorphism $\mathbb{R}^3 \to \mathbb{R}^3$;
  • $H(t, -)$ is the identity when restricted on $S^1$;
  • $H(1, E) = E'$ and $p' \circ H(1, -) = p$.

It's easy to see that if $E$ and $E'$ are isotopic in $\mathbb{R}^3$, then they are isomorphic. The converse is not true, as you noted: as band twisted twice is isomorphic to the trivial line bundle, but it is not isotopic in $\mathbb{R}^3$ to the cylinder. The isotopy class depends on the specific embedding in $\mathbb{R}^3$, but the isomorphism class does not.

Classification

Isomorphism classes: Stiefel-Whitned

In fact there are only two isomorphism classes, the trivial one and the one corresponding to the Möbius strip; see this question. Basically (this is a different version of the reasoning in the answers I linked), isomorphism classes of line bundles on $X$ are classified by the first Stiefel-Whitney class^(*) $w_1(X) \in H^1(X; \mathbb{Z}/2\mathbb{Z})$, and $H^1(S^1; \mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.

Isotopy classes: Linking number

Now it is actually possible to classify (smooth?) isotopy classes too.

Let $E \to S^1$ be a line bundle embedded in $\mathbb{R}^3$, and let's assume that it is smooth for simplicity (the continuous case is more complicated). Then you can construct two curves out of it:

• The first one $\gamma_0$ corresponds to the zero section of the line bundle.

• For the second one, take a tubular neighborhood of the zero section in $E$ and take $\gamma_1$ to be one of the boundary components.

Finally consider the linking number: $$\operatorname{lk}(\gamma_0, \gamma_1),$$ which is well defined up to sign. This number basically corresponds to your intuitive notion of "number of twists": it's zero for the cylinder, one for the Möbius strip, two if you add one more twist... And it classifies smooth isotopy classes of smooth line bundles embedded in $\mathbb{R}^3$.

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^(*) The reason for that is one of my favorite facts in mathematics (the complex case one is just as amazing): $$\operatorname{Vect}^\mathbb{R}_1(X) = [X, \mathrm{Gr}_1(\mathbb{R}^\infty) ] = [X, \mathbb{RP}^\infty] = [X, K(\mathbb{Z}/2\mathbb{Z}, 1)] = H^1(X; \mathbb{Z}/2\mathbb{Z}).$$