Hodge theory: $\Delta \alpha = 0$ iff $d\alpha = d^* \alpha = 0$ on a noncompact manifold?

Question

Let $M$ be a Riemannian manifold (connected, oriented).

One can define the co-differential $d^* : \Omega^k(M, \mathbb{R}) \to \Omega^{k-1}(M, \mathbb{R})$ even if $M$ is not compact (for example use the definition with the Hodge star, there is also a definition with the covariant derivative). $d$ and $d^*$ are formally self-adjoint, in the sense that $\langle d \alpha, \beta \rangle_{L^2} = \langle \alpha, d^* \beta \rangle_{L^2}$ if $\alpha$ or $\beta$ has compact support.

One can then define the Hodge Laplacian $\Delta = d d^* + d^* d$. If $M$ is compact, then \begin{equation} \Delta \alpha = 0\quad \Leftrightarrow \quad d\alpha = 0 \text{ and } d^* \alpha = 0~. \end{equation}

What if $M$ is not assumed compact? The direction $\Leftarrow$ still obviously holds.

What about $\Rightarrow$?

Thoughts: In the compact case, it is easy to show $\Rightarrow$: one just writes $\langle \Delta \alpha, \alpha \rangle_{L^2} = \langle d \alpha, d\alpha \rangle_{L^2} + \langle d^* \alpha, d^* \alpha \rangle_{L^2}$ and quickly concludes. I still the argument might still work in the noncompact case by evaluating $\langle \Delta \alpha, \varphi \alpha \rangle_{L^2} $ for a wisely chosen bump function $\varphi$, but I can't quite make it work. Otherwise maybe it can be proven by direct computation somehow, after all $\Delta \alpha$, $d \alpha$, and $d^* \alpha$ are all local quantities.

Answer 1

Consider $M = \mathbb{R}^n$ equipped with the standard metric. For a $0$-form $f$, i.e. a real-valued function, we have $d^*f = 0$ for degree reasons, and

$$df = \sum_{i=1}^n\frac{\partial f}{\partial x_i}dx^i$$

which is zero if and only if $f$ is constant. While

\begin{align*} \Delta f &= dd^*f + d^*df\\ &= -*d*df\\ &= -*d*\left(\sum_{i=1}^n\frac{\partial f}{\partial x_i}dx^i\right)\\ &= -*d\left(\sum_{i=1}^n\frac{\partial f}{\partial x_i}*(dx^i)\right)\\ &= -*d\left(\sum_{i=1}^n\frac{\partial f}{\partial x_i} (-1)^{i-1}dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^n\right)\\ &= -*\left(\sum_{i=1}^n\sum_{j=1}^n\frac{\partial^2 f}{\partial x_j\partial x_i} (-1)^{i-1}dx^j\wedge dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^n\right)\\ &= -*\left(\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2} (-1)^{i-1}dx^i\wedge dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^{i+1}\wedge\dots\wedge dx^n\right)\\ &= -*\left(\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2} dx^1\wedge\dots\wedge dx^{i-1}\wedge dx^i\wedge dx^{i+1}\wedge\dots\wedge dx^n\right)\\ &= -*\left(\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2} dx^1\wedge\dots\wedge dx^n\right)\\ &= -\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2}*(dx^1\wedge\dots\wedge dx^n)\\ &= -\sum_{i=1}^n\frac{\partial^2 f}{\partial x_i^2} \end{align*}

which is zero if and only if $f$ is harmonic.

So your question becomes:

Is every harmonic function on $\mathbb{R}^n$ constant?

The answer is no. For example, $f(x_1, \dots, x_n) = x_1$ is a harmonic function which is not constant, so $\Delta f = 0$ but $df \neq 0$.