Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the flat torus $T^2=\frac{\mathbb{R}^2}{l_1\mathbb{Z}\oplus l_2\mathbb{Z}}$. It is easy to see that the eigenvalues of the Laplacian on torus, $-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}$, are $\lambda_{m_1,m_2}=(2\pi)^2(\frac{m_1^2}{l_1^2}+\frac{m_2^2}{l_2^2})$ with the associated eigenfunction $$f_{(m_1,m_2)}(x,y)=e^{2\pi i(\frac{m_1}{l_1}x+\frac{m_2}{l_2}y)}.$$ where $m_1,m_2\in \mathbb{Z}$. Furthermore, The closed geodesics of $T^2$ parametrized by the arc length, are $$ \gamma_{(n_1,n_2)}(t)=\frac{1}{l}(n_1l_1t,n_2l_2t)$$ where $n_1,n_2\in \mathbb{Z}$ and $l=\sqrt{n_1^2l_1^2+n_2^2l_2^2}$. A simple computation shows that an eigenfunction, say $f_{(m_1,m_2)}$, restricted on a closed geodesic, $\gamma_{(n_1,n_2)}$, gives
$$f_{(m_1,m_2)}\circ \gamma_{(n_1,n_2)}(t)=e^{2\pi i(\frac{m_1n_1+m_2n_2}{l})t}$$ Which is an eigenfunction on the circle $\mathbb{R}/l\mathbb{Z}$ with the eigenvalue $\tilde{\lambda}=\left( \frac{2\pi}{l}(m_1n_1+m_2n_2)\right)^2$.

Now my question is: Is this true in the general cases? More precisely;

Let $\gamma:[0,l]\to M$ be a closed geodesics on the Riemannian manifold $(M,g)$ which is parametrized by the arc length. If $f\in C^\infty(M)$ is an eigenfunction for the Laplacian on $M$, i.e. $$\Delta(f)=\lambda f$$ Then

Question 1) Is $f\circ \gamma$ an eigenfunction on the circle $S^1=\mathbb{R}/l\mathbb{Z}$? Or, Is it in the form of $$f\circ \gamma(t)=c e^{2\pi i \tilde{\lambda}t}.$$

Question 2) If so, how does $\tilde{\lambda}$ depend on $\gamma$ and $\lambda$?

Thanks.

share|improve this question
    
This is just a complement to Robert's answer: restrictions of eigenfunctions to geodesics (with a view to $L_p$ estimates) are treated here: Restrictions of the Laplace-Beltrami eigenfunctions to submanifolds N. Burq, P. Gérard, and N. Tzvetkov Source: Duke Math. J. Volume 138, Number 3 (2007), 445-486. –  alvarezpaiva Jan 21 '13 at 8:42

1 Answer 1

up vote 11 down vote accepted

The answer is 'no', as you can see by taking the case of $M$ being the unit $2$-sphere in $\mathbb{R}^3$ and the geodesic $\gamma$ being a great circle, say, the horizontal great circle given by $z=0$. If you consider the harmonic polynomials of degree $2$ in $x,y,z$ restricted to the $2$-sphere, these are eigenfunctions of the Laplacian on the $2$-sphere, but their restrictions to the horizontal great circle aren't usually eigenfunctions of the Laplacian on the circle.

More generally, you take the $k$-th eigenspace of the Laplacian on the $2$-sphere for $k>1$, you'll find that the restriction of these functions to each great circle projects into a sum of a finite number of eigenspaces of the Laplacian on the great circle (I think it's about $\tfrac12(k{+}2)$ of them), but not into a single one of these eigenspaces.

share|improve this answer
    
Thanks. You are right. –  Asghar Ghorbanpour Jan 21 '13 at 15:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.