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This question has a motivation in physics, thus its formulation may not be entirely rigorous.

Let $f$ be a function that takes values on a Moebius strip of fixed length $L$ and maps them to operators in a Hilbert space, that is to say the function satisfies $\begin{equation}f(x+L,y) = f(x,-y)\end{equation}$ for all real numbers $x$ and $y$. In particular this function is periodic in the case that $y=0$, namely $f(x+L,0) = f(x,0)$.

Let's consider the "transformation" given by:

$$\begin{equation} \left( \begin{array}{c} g_{(x,y)} \\ h_{(x,y)} \\ \end{array} \right) = \left( \begin{array}{cc} \frac{1}{\sqrt{2}}\cdot e^{i\cdot\frac{\pi}{L}\cdot x} & -\frac{1}{\sqrt{2}}\cdot e^{i\cdot\frac{\pi}{L}\cdot x} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)\cdot \left( \begin{array}{cc} f_{(x,y)} \\ f_{(x,-y)} \\ \end{array} \right) \end{equation}$$

The inverse transformation is given by: $$\begin{equation} \left( \begin{array}{c} f_{(x,y)} \\ f_{(x,-y)} \\ \end{array} \right) = \left( \begin{array}{cc} \frac{1}{\sqrt{2}}\cdot e^{-i\cdot\frac{\pi}{L}\cdot x} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\cdot e^{-i\cdot\frac{\pi}{L}\cdot x} & \frac{1}{\sqrt{2}} \\ \end{array} \right)\cdot \left( \begin{array}{cc} g_{(x,y)} \\ h_{(x,y)} \\ \end{array} \right) \end{equation}$$

The given matrix is actually unitary and the functions g and h satisfy periodic boundary conditions:

$\begin{equation}g(x+L,y) = g(x,y)\end{equation}$ $\begin{equation}h(x+L,y) = h(x,y)\end{equation}$

When considering the inverse transformation though we have to restrict $y$ to $y>0$ since otherwhise there are to ways of writing $f(x,y)$ in terms on $g(x,y)$ and $h(x,y)$. In other words we use the first line of the inverse transformation for positive y values and the second line for negative y values. This is just a convention that removes the ambiguity that was explained above.

To me this construction seems to have it's roots in the theory of vectorbundles, since in some sense you have the Moebius strip as a base space and you attach the vector space given by all linear combinations of f(x,y) and f(x,-y) to each point (x,y) of the Moebius-Strip.

Especially since the matrix is unitary it kind of looks like a transition matrix between different local trivialization.

It also sort of looks like a construction that was made "on top" of the original vector bundle structure of the Moebius strip, especially since we have to distinguish between y>0 and y<0. (cmp. for example page 3 for the original moebius vector bundle construction)

Since I'm a student in physics I do have some background in differential geometry, though I'm not an expert in this area, that is why I am asking here.

I'm looking forward to some ideas in what direction to think.

Best regards. Stan

share|cite|improve this question the question is: Is this matrix the transition matrix between different local trivializations of some vector bundle? – Matt Sep 9 '12 at 14:16

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