Harmonic forms from a variational perspective? In short: Are harmonic forms critical points of some generalized Dirichlet functional?
Let $(M,g)$ be a smooth Riemannian manifold, denote by $\Omega^k(M)$ the space of $k$-valued forms on $M$. Let $d:\Omega^k(M) \to \Omega^{k+1}(M),\delta:\Omega^{k+1}(M) \to \Omega^k(M)$ be the exterior derivative and its adjoint, respectively.
A $k$-form $\sigma$ on $M$ is called harmonic if $d\sigma=\delta \sigma=0$ (or equivalently if $\Delta \sigma=0$ where $\Delta=d\delta +\delta d$).
$0$-harmonic forms (also known as harmonic functions) are critical points of the Dirichlet integral: $E(\sigma)=\int_M |d\sigma|_{g}^2 \operatorname{Vol}_g$. Is there an analogous variational realization for forms of higher degree? 
(Perhaps $E(\sigma)=\int_M |d\sigma|_{g}^2 + |\delta \sigma|_{g}^2 \operatorname{Vol}_g$ is the right choice?)
Also, what about vector-valued forms?
Any reference for a variational treatment would be appreciated.
 A: Suppose $M$ is closed.
Define
$$ E:\Omega^k(M) \to \mathbb{R} \, \, \,, \, \, \,E(\sigma)=\frac{1}{2}\int_M |d\sigma|^2 + |\delta \sigma|^2 \operatorname{Vol}_M.$$
Let $\sigma \in \Omega^k(M)$, and let $\sigma_t$ be a smooth perturbation of $\sigma$, i.e. $\sigma_0=\sigma$. Denote $V:=\frac{\partial \sigma_t}{\partial t}|_{t=0} \in  \Omega^k(M)$.
Then,
$$\frac{d}{dt} E(\sigma_t)=\int_M \langle d\sigma_t, \frac{\partial}{\partial t} d\sigma_t  \rangle  + \langle \delta \sigma_t,  \frac{\partial}{\partial t}\delta \sigma_t \rangle \operatorname{Vol}_M.$$
Now, we note* that $\frac{\partial}{\partial t} d\sigma_t=d\frac{\partial}{\partial t} \sigma_t$, so in particular $\frac{\partial}{\partial t} d\sigma_t|_{t=0}=d(\frac{\partial \sigma_t}{\partial t}|_{t=0})=dV$. 
Similarly, we have $\frac{\partial}{\partial t} \delta\sigma_t|_{t=0}=\delta(\frac{\partial \sigma_t}{\partial t}|_{t=0})=\delta V$. 
This implies 
$$\frac{d}{dt} E(\sigma_t)|_{t=0}=\int_M \langle d\sigma, dV  \rangle  + \langle \delta \sigma,\delta V \rangle \operatorname{Vol}_M=\int_M \langle \delta d\sigma, V  \rangle  + \langle d\delta \sigma, V \rangle \operatorname{Vol}_M$$
$$ =\int_M \langle \delta d\sigma+d\delta \sigma, V  \rangle  \operatorname{Vol}_M=\int_M \langle \Delta\sigma, V  \rangle  \operatorname{Vol}_M.$$
So, the Euler-Lagrange equation of $E$ is exactly $\Delta\sigma=0$, hence the critical points are precisely the harmonic forms.
The same proof carries through to the setting of vector-valued forms.

*The equality $\frac{\partial}{\partial t} d\sigma_t=d\frac{\partial}{\partial t} \sigma_t$ is immediate from the coordinate expression for $d$, and the fact that $\sigma_t$ is a smooth family.
