Consider the Möbius strip as the unit square with two opposite sides identified (with opposite directions). Consider the eigenvalue equation $\Delta u = \lambda u$ with boundary condition $u=0$. Unlike for orientable manifolds, the least eigenfunction will not be all of one sign; there will be a nodal line. My question generally concerns the behavior of eigenfunctions and eigenvalues in the nonorientable case, but to ask some specific questions: (1) is the eigenspace of the first eigenvalue still onedimensional? (2) does there have to be just ONE nodal line? (3) does any nodal line have to meet the boundary in two points?
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Pass to the double cover, which is a cylinder: $[0,1] \times [0,2]$ with the end points of the second interval identified. On the cylinder, the eigenfunctions are $f(x,y) = \sin (ax \pi) e^{i b y \pi}$ with $a$ and $b$ integers and $\lambda =  a^2  b^2$. An eigenfunction descends to the Mobius strip if and only if $f(x,y) = f(1x, y+1)$ which means that $a+b$ is odd. For the Mobius strip example, your other questions should be straightforward from there. 

