Let $\Omega\subset\mathbb{R}^d$ be a compact domain such that its first Sobolev space $H^1(\Omega)$ embeds compactly into $L^2(\Omega)$. (It is sufficient that $\partial\Omega$ be Lipschitz.)
For $\omega\in\mathbb{R}^d$, define $e_\omega:\Omega\to\mathbb{R}$ by $e_\omega(x) = \exp( i\omega\cdot x )$. Notice that $e_\omega\in H^1(\Omega)$, with gradient $\nabla e_\omega = i\omega e_\omega$.
Question
What is the dimension of the subspace $S(\mu) = \operatorname{span} \{ e_\omega\ |\ |\omega|^2 = \mu\}$ in $H^1(\Omega)$?
Thoughts
The Neumann Laplacian has discrete spectrum $0 = \mu_1\leq\mu_2\leq\cdots $ (counted with multiplicity). Set $N(\mu) = \# \{k\ |\ \mu_k \leq \mu\}$.
These eigenvalues satisfy the min-max property: $$ \mu_k = \min_E \max_{u\in E} \frac{\int |\nabla u|^2\ dx}{\int |u|^2\ dx}$$ where $E$ ranges over all $k$-dimensional subspaces of $H^1(\Omega)$. The ratio $R(u) = \int |\nabla u|^2 / \int |u|^2$ is known as the Rayleigh quotient.
In particular any subspace of $H^1$ with dimension greater than $k$ has at least one element with Rayleigh quotient no less than $\mu_{k+1}$. Equivalently if every element of a subspace has Rayleigh quotient at most $\mu_k$, then the dimension of the subspace is no greater than $k$.
As $R(e_\omega) = |\omega|^2$, we must have $\operatorname{dim} S(\mu) \leq N(\mu)$. So there must be an inordinate amount of linear dependencies among the $e_\omega$.
But at the same time, suppose a linear combination of the $e_\omega$ vanish, say $\sum c_k e_{\omega_k} = 0$ on $\Omega$. Because the $e_\omega$ are all bounded, the sum $\sum c_k \exp(i\omega_k x)$ is defined on all of $\mathbb{R}^d$, and as an analytic function equal to zero on an open set, must have all $c_k = 0$.
This seems like a contradiction. What am I missing? Are there any known approaches to the question?