Let $M$ be a smooth, bounded, oriented Riemannian manifold-with-boundary. Let $\alpha$ be a harmonic differential $p$-form on $M$, subject to the boundary condition $\alpha\wedge\nu^\sharp|\partial M = 0$ or $\iota_\nu \alpha|\partial M = 0$. Here $\nu$ is the normal vectorfield along $\partial M$, $\nu^\sharp$ is its dual $1$-form, and $\iota$ is the interior multiplication. Assume that $\alpha \in W^{1,2}$.

The question is: can we conclude that $\alpha \in C^\infty$?


1 Answer 1


The answer depends on your definition of harmonic; if you mean $\Delta \alpha=0$ then you can not conclude that $\alpha$ is smooth. It is easy to find counter examples on the unit disk in $\mathbb{R}^2$. Indeed a $1$-form on the unit disk which is a solution of the equation $\Delta \alpha=0$ can be written as $$\alpha=f(x,y)dx+g(x,y)dy$$ where $f$ and $g$ are harmonic functions on the disk. Hence $f$ and $g$ are uniquely determined by their boundary values on the circle. Let now $\varphi\colon \partial \mathbb{D}^2\rightarrow \mathbb{R}$ be a continous but not smooth function and let $f$ be the harmonic extension of $y\varphi$ and $g$ be the harmonic extension of $-x\varphi$. Then $\alpha=f(x,y)dx+g(x,y)dy$ solves the equation $\Delta \alpha=0$ and $\left.\iota_{\nu} \alpha\right|_{\partial \mathbb{D}^2}=xf+yg=0$.

If you mean $(d+d^*) \alpha=0$ then $\alpha$ is smooth because this boundary condition is elliptic (see the chapter 5 in the beautiful book of M. Taylor: Partial Differential Equations I - Basic Theory).

  • $\begingroup$ Thank you very much! -- S. $\endgroup$ Commented Feb 1, 2019 at 23:50

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