# Expressing the sections of the Möbius bundle on $S^1$ as globally defined real valued functions

Consider the real Möbius bundle over $S^1,$ defined as follows. The bundle is trivial over $U_1 = S^1 \setminus \{i\}$ and $U_2 = S^1 \setminus \{-i\}$ and the transition function $T_{12}$ defined on $U_1 \cap U_2$ is given by $T_{12}(z) = 1$ when $\textbf{Re}\ z > 1$ and $T_{12}(z) = -1$ when $\textbf{Re}\ z < 1.$

Sections of this bundle can be considered as pairs of real valued functions $f_1, f_2$ defined on $U_1, U_2$ respectively such that $f_1(z) = f_2(z)$ when $\textbf{Re}\ z > 1,$ and $f_1(z) = -f_2(z)$ when $\textbf{Re}\ z < 1.$

Can we express the sections naturally as real valued functions defined on all of $S^1$satisfying some further condition? My guess would be that sections are functions satisfying $f(-z) = -f(z)$ or perhaps just $f(-i) = -f(i).$ More generally, can sections of a (real or complex) bundle always be expressed as globally defined real/complex valued functions satisfying some further conditions?

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## 1 Answer

In general, sections of a line bundle on $X$ cannot be identified with a subspace of the functions on $X$.

If $p:\tilde{X}\to X$ is a normal covering space, and the bundle on $X$ pulls back to the trivial bundle on $\tilde{X}$, then we can identify sections of the bundle on $X$ with functions defined on $\tilde{X}$ that satisfy some equivariance property with respect to the group of deck transformations of $p$.

The Möbius bundle pulls back to the trivial bundle along the degree 2 map $S^1\to S^1$, so we can express sections of the Möbius bundle as real valued functions on $S^1$ satisfying some condition, and that condition is yours: $f(-z)=-f(z)$. The thing to be careful about is that the $S^1$ in this case is really a degree 2 covering of the $S^1$ we started with, and it's just a coincidence that the two spaces are abstractly homeomorphic.

Given $f_1, f_2$ on $U_1$, $U_2$, we form the corresponding $f:S^1\to\mathbb{R}$ as follows: $$f(z)=\begin{cases} f_1(z^2) \text{ if }\arg(z)\in (-3\pi/4,\pi/4)\\ f_2(z^2) \text{ if }\arg(z)\in (-\pi/4,3\pi/4)\\ -f_1(z^2) \text{ if }\arg(z)\in (\pi/4,5\pi/4)\\ -f_2(z^2) \text{ if }\arg(z)\in (3\pi/4,7\pi/4)\\ \end{cases}$$

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Great! Could you explain how to take the pair of functions $f_1, f_2$ which represent a section and construct the associated function $f$ satisfying $f(-z) = -f(z)?$ – mck Dec 4 '12 at 18:31
@mck Sure. I've added this to my answer. – Julian Rosen Dec 4 '12 at 18:44
Thank you!! 15 char – mck Dec 4 '12 at 19:04