eigenvalues of a Mobius strip - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T12:13:44Z http://mathoverflow.net/feeds/question/56249 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56249/eigenvalues-of-a-mobius-strip eigenvalues of a Mobius strip Michael Beeson 2011-02-22T06:30:45Z 2011-02-28T15:56:31Z <p>Consider the Mobius 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?</p> <p>Sorry, I couldn't get an umlaut to appear in Mobius; neither TeX nor HTML worked to accomplish that.</p> http://mathoverflow.net/questions/56249/eigenvalues-of-a-mobius-strip/56300#56300 Answer by David Speyer for eigenvalues of a Mobius strip David Speyer 2011-02-22T17:09:55Z 2011-02-28T15:56:31Z <p>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 $b$ is odd.</p> <p>For the Mobius strip example, your other questions should be straightforward from there. </p>