# Carleson's Theorem on Manifolds

Let $M$ be an oriented, compact, differentiable manifold with some Riemmanian metric $g$, so that $(M,g)$ has a nice volume form and one can define $L^2(M,g)$ as the completion of $C^\infty(M)$ under the inner product induced by the volume form (one can similarly define $L^2$ spaces of arbitrary forms). Since $M$ is compact, all of the $L^2(M,g)$ are essentially the same for any metric $g$. The Laplacian is defined naturally as $\Delta = -*d*d$, and the Hodge theorem says that its eigenfunctions form a basis for $L^2(M,g)$.

Now if $M = S^1$ and $g$ is the natural metric induced by the covering of $\mathbb{R}$ then $\Delta = -\frac{d}{dx^2}$ and the eigenfunctions are $e^{in\cdot}$. Carleson's theorem states that, unlike an arbitrary basis for $L^2(S^1)$, the convergence of the trigonometric polynomials is pointwise convergence almost everywhere.

Is it suspected that this pointwise almost everywhere convergence of $\Delta$ eigenfunction expansions holds over $L^2(M,g)$ for any compact manifold $(M,g)$ with Laplacian $\Delta$? What about for $L^2$ spaces of forms? Are there any known counter-examples?

• Note that the answer will depend on how you order the eigenfunctions. Ordering by eigenvalue is perhaps the most natural choice. In the case of the the $2$-d torus this gives spherical summation for $2$ dimensional Fourier series. Determining if pointwise convergence holds in this setting is a longstanding open problem. – Mark Lewko Jan 15 '14 at 19:14
• This isn't directly an answer, but an interesting difference between $S^1$ and $S^3$ is that pointwise convergence can fail at a smooth point of the function being represented. Take $f$ to be $1$ on the northern hemisphere and $-1$ on the southern hemisphere. The eigenfunctions which are $SO(3)$ invariant are naturally indexed by the representations of $SU(2)$. Ordering them in the obvious way, the sum does not converge at the north and poles, even though $f$ is locally constant there. See sbseminar.wordpress.com/2011/02/18/a-peter-weyl-counter-example – David E Speyer Jan 15 '14 at 19:43
• In that case it would at least be interesting to know if there are any counterexamples. – Greg Zitelli Jan 21 '14 at 16:13