Hodge * Operator

 Basics of wedge products
  In the description below, we are assuming that our underlying space is an orientable $C^\infty$ manifold.

We know vector space has a dual space that consists of functionals. Similarly, if we have a tangent plane there is a dual to it. Typically tangent space is spanned by ${ \frac{\partial }{\partial x^i} }$ where $i=1,2,\cdots,n$ for an n-dimensional basis for tangent space. Since this is a vector space, there is dual to it called ``cotangent space''. Just like vector space duals are functions to real numbers, the corresponding vector space dual is consists of differentials $dx_i$ and these are maps to real. These are called ``covectors''. 

If you consider Riemman integral $\int f dx$, here we consider $dx$ to be limit of $\delta x$ which is considered as small strip in $x-y$ plane. This concept can guide thinking on $dx_i$.
 

The differentials $dx_i$ have an operation called ``wedge'' operation, given by
$$\begin{equation} dx \wedge dy = - dy \wedge dx \end{equation}$$

In regular integration $dxdy$ is considered area element and $dxdydz$ volume element. Similarly, wedge product gives oriented area or more generally oriented volume.

Wedge operation maps to reals. Then $dx \wedge dx = - dx \wedge dx$ implies $dx \wedge dx=0$.
For a function $f(x,y,z)$ in $\mathcal{R}^3$, total differential leads to the following equation.
$$\begin{equation}   df = \frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy + \frac{\partial{f}}{\partial{z}}dz \end{equation}$$
Here $dx,dy, dz$ are covectors or duals. If as in above each expression in addition, $df$ involves single differential or covector $dx,dy,dz$, these are called $1$-form. If you wedge two $1$-forms we get $2$($dx\wedge dy$) form etc.,$0$ forms are simply functions.

Collection of $k$ covectors is denoted by $\Lambda^k_n(M)$, where $M$ is $n$ dimensional manifold or space under consideration.

From the above equation it is evident that the $d$ operator acts on functions - that is $0-form$ and produces $1$ form. Using this as an example, we can write $d:\Lambda^k_n(M) \rightarrow \Lambda^{k+1}_{n+1}(M)$ which means that if we apply $d$ to a $k$ form, we get $k+1$ form.

From equation (2) we can also infer the dual space has basis consists of $dx,dy,dz$ or $3$-dimensional. Similarly for $n=3$ space we take $2$-forms, then the dimension is also $3$. In general, for  $\Lambda^k_n(M)$ the dimension of vector space is given by $^nC_k$. Since $^nC_k=^nC_{n-k}$, the dimensions of $k$ covectors and $n-k$ covector space is same.

Hodge *
Let $M$ be an $n$-dimensional $C^\infty$ Manifold. For a given integer $0 \leq k \leq n$, $\Lambda^k T^*_nM$ and $\Lambda^{n-k}T^{*}_nM$ have same dimension as vector space and they are isomorphic. Here $T^*M$ is dual to Tangent space $TM$. If $M$ has a Riemanian metric and oriented, for each point $p \in M$, there is a natural isomorphism
$$\begin{equation}   *:\Lambda^k T^*_nM \equiv \Lambda^{n-k}T^{*}_nM \end{equation}$$
By varying $p$ we get linear isomorphism
$$\begin{equation}   *:\mathcal{A}^k(M) \rightarrow \mathcal{A}^{n-k}(M) \end{equation}$$
This operator is called ``Hodge'' star operator where $\mathcal{A}^k$ represents vector space of $k$ forms etc.,

A bit of explanation for above definition. $C^\infty$ Manifold means a smooth manifold.
Riemannian metric is a positive definite inner product defined defined on the tangent space at each point of the Manifold $g_p: T_pM \times T_p M \rightarrow \mathcal{R}$ in such a way $g_p$ is $C^\infty$ at point $p$. Very standard way to express this is $ds^2 = \sum_{i,j=1}^n g_{ij}dx_i dx_j$.

Going back to Hodge star, if we write $V$ for $T_p M$, and use $V^*$ for $T_p^* M$, we get,
$$\begin{equation}   *: \Lambda^k V^* \rightarrow \Lambda^{n-k} V^* \end{equation}$$

This linear map can be defined by setting
$$\begin{equation}   *(\theta_1 \wedge \theta_2 \wedge \cdots \wedge \theta_k) = \theta_{k+1}\wedge \cdots \theta_n \end{equation}$$
In particular
$$\begin{equation}   *1 = \theta_1 \wedge \theta_2 \wedge \cdots \wedge \theta_n = 1 \end{equation}$$

Let $\omega$ be a one form for $X \in \Xi(M)$. For example, $\omega = f_x dx + f_y dy + f_z dz$ where $f_x = \frac{\partial f}{\partial x}$ etc., Then,
$$\begin{equation}   div X = *d*\omega \end{equation}$$

For example, for $\omega$ given in above example, we can work this out as follows:
$$\begin{align*}   *\omega = *(f_x dx + f_y dy + f_z dz) = f_x*dx+f_y*dy+f_z*dz \\   = f_x dy \wedge dz + f_y dz \wedge dx + f_z dx \wedge dy \\   d*\omega = f_{xx} dx \wedge dy \wedge dz + f_{yy}  dx \wedge dy \wedge dz + f_{zz}  dx \wedge dy \wedge dz \\   *d*\omega = f_{xx}+f_{yy}+f_{zz} = \frac{\partial^2f}{\partial x^2}+ \frac{\partial^2f}{\partial y^2}+ \frac{\partial^2f}{\partial z^2} = div X \end{align*}$$