Why are the eigenvectors of the Laplacian of a ring graph sinusoidal?

I think the following is what Spielman is referring to. Let $z(u)$ be the point $(x_k(u), y_k(u))$ in $\mathbb{R}^2$. Consider the vector $z(u-1) - 2 z(u) + z(u+1)$. By the reflection symmetry of the picture, it is parallel to $z(u)$; let $$z(u-1) - 2 z(u) + z(u+1) = \lambda z(u)$$ for some scalar $\lambda$. Moreover, by rotational symmetry, the constant $\lambda$ is independent of $u$.

Taking the first coordinates of this equation, $$x_{k}(u-1) - 2 x_k(u) + x_{k}(u+1) = \lambda x_k(u)$$ This exactly says that the vector $x_k$ is an eigenvector for $D$, with eigenvalue $\lambda$. Looking at the second coordinate, we get the same conclusion for $y_k$.

The ring graph with $n$ vertices has a dihedral symmetry by $D_n$, so the action of $D_n$ commutes with the Laplacian. It follows that the eigenspaces of the Laplacian divide up into irreducible representations of $D_n$, which include direct sums of irreducible representations of $C_n$ and their duals. Laplacian eigenvectors in these representations are given by sinusoids. The whole picture behaves exactly the way you'd expect in the $n \to \infty$ limit.

I'm not sure exactly what Spielman is referring to, though. There is a connection between nice embeddings of graphs into $\mathbb{R}^n$ and eigenvectors but I wasn't aware that it goes in the direction that Spielman claims.