# eigenvalues of a Möbius strip

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?

-
In Windows, if your keyboard doesn't have a ö you can copy and paste one from a web page, or type Alt-0246 using the numeric keypad. –  Douglas Zare Feb 22 '11 at 9:26
Or, you can try this handy tool from Andrew Stacey: type in the TeX and just copy and paste. math.ntnu.no/~stacey/code/latexToUTF/utf.php –  Willie Wong Feb 22 '11 at 10:48
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. –  Michael Beeson Feb 22 '11 at 15:53
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: $\text{K\"ahler}$ –  Michael Hardy Feb 22 '11 at 17:58
Didn't work. Maybe the version of TeX used here is incomplete? –  Michael Hardy Feb 22 '11 at 17:59

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.
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? –  Willie Wong Feb 22 '11 at 21:24
and perhaps you meant $f(x,y) = \sin(\pi a x) e^{i\pi b y}$? –  Willie Wong Feb 22 '11 at 21:26