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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 non-orientable case, but to ask some specific questions: (1) is the eigenspace of the first eigenvalue still one-dimensional? (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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  • $\begingroup$ I guess for the particular Mobius strip I mentioned, the eigenspace is not one-dimensional, as everything is invariant under a translation. That wouldn't be generally true, but I think it does show that the eigenspace isn't always one-dimensional. But the other two questions still need an answer. $\endgroup$ Commented Feb 22, 2011 at 15:53
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    $\begingroup$ It's also acceptable to write "oe" instead of "ö". Thus "Möbius" and "Moebius" are essentially the same spelling, whereas "Mobius" is different. Likewise "Kähler" and "Kaehler" are the same, whereas "Kahler" is different. In TeX, I think \"o and \"a work. And \"u And if you can't instantly locate an example to copy and paste, just go to dict.leo.org and find an example. I entered "aendern" and got the dictionary entry for "ändern" and pasted the "ä" above. Let's try a TeXperiment: $\text{K\"ahler}$ $\endgroup$ Commented Feb 22, 2011 at 17:58
  • $\begingroup$ Can you show an example of an element in the first eigenspace? $\endgroup$ Commented Feb 22, 2011 at 22:01
  • $\begingroup$ It seems to me that the first eigenfunction is positive (no nodal line), given by $f(x,y)=\sin\pi x$. It satisfies the boundary conditions $f(0,y)=f(1,y)=0$ and the skew-periodicity condition $f(1-x,y)=f(x,y+1)$. $\endgroup$ Commented Feb 23, 2011 at 10:30
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    $\begingroup$ Möbius... just use a German keyboard ;-) $\endgroup$ Commented Oct 1, 2014 at 14:58

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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 $f(x,y) = f(1-x, y+1)$ which means that $a+b$ is odd.

For the Mobius strip example, your other questions should be straightforward from there.

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    $\begingroup$ um, why $f(x,y) = f(-x, y+1)$? Should it be $f(x,y) = f(1-x, y+1)$? In particular, is $a = 1, b = 0$ not an eigenfunction on the Moebius strip? $\endgroup$ Commented Feb 22, 2011 at 21:24
  • $\begingroup$ and perhaps you meant $f(x,y) = \sin(\pi a x) e^{i\pi b y}$? $\endgroup$ Commented Feb 22, 2011 at 21:26
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    $\begingroup$ There appears to be a slight problem with the final part of David Speyer's answer: Pass to the double cover, which is a cylinder: [0,1]×[0,2] with the end points of the second interval identified. On the cylinder, the eigenfunctions are f(x,y)=sin(axπ)e^{ibyπ} with a and b integers and λ=−a^2−b^2. An eigenfunction descends to the Mobius strip if and only if f(x,y)=f(1−x,y+1) which means that a+b is odd. It makes sense (to me) that the conditions on the x/y directions (i.e. on a and b) should be coupled, otherwise I assume the double cover would be unnecessary... $\endgroup$
    – AndyF
    Commented Oct 1, 2014 at 14:10

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