Let $M$ be a 2-dimensional closed Riemannian manifold and let $$\phi:M\rightarrow M$$ be an isometry with $\phi^2=Id\_M$. Consider the fixed point set $$F:=\lbrace x\in M: \phi(x)=x \rbrace\subset M,$$ and
suppose $F\subset M$ is a closed $1$-dimensional submanifold. Let $u:M\rightarrow \mathbb{R}$ be an eigenfunction on $M$ for the Laplace-Beltrami operator. Suppose further $u$ is symmetric with respect to the isometry $\phi$, i.e.\ $u\circ \phi=u$. 


It follows easily that $\frac{\partial u}{\partial \nu}\_{\lvert p}=0$ for all $p\in F$, where $\frac{\partial}{\partial \nu}\_{\lvert p}\in (T\_p F)^{\perp}$ is a perpendicular vector along $F$. 


I am wondering if there must be at least one point $p^{\*}\in F$ such that $u(p^{\*})\neq 0$? Or is it possible that $u$ vanishes everywhere along $F$?


The easy examples suggest the former, but I could not prove it in general. Does anyone has an idea how to approach this problem?


Best wishes



Zero set of eigenfunction along a sub manifold
This appears to be a result of Dong. See Theorem 3.4.


*Rui-Tao Dong*, [**Nodal sets of eigenfunctions on Riemann surfaces**](http://projecteuclid.org/euclid.jdg/1214448750), *J. Differential Geom.* **36** (1992), no. 2, 493--506.


