Connection between the Hodge laplacian and the Laplace operator Let M a riemannian manifold. How can I show that the hodge-laplace-operator of a function $f$ is the negative of the laplace-operator?
 A: A rather short proof can be found here.
I assume you are interested in the case when $f$ is a scalar function. Otherwise the
Hodge Laplacian differs from the Laplace–Beltrami operator not only by a sign due to the Ricci curvature. See the Weitzenböck identity.
A: For the function case, choose local coordinate $(x_1,\cdots, x_n)$, then $\nabla_i\nabla_j f=Hessf(\frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j})=\frac{\partial^2 f}{\partial x_i\partial x_j}-\Gamma^k_{ij}f_k$, and
\begin{equation} 
\Delta f =g^{ij}\nabla_i\nabla_j f=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x_j}),
\end{equation}
where $g=\det{g_{ij}}$.
Next, calculating $\Box f=d^*d f$. First, for $\alpha\in A^1(M)$,  we calculate $d^*\alpha$.
Let $\alpha=\alpha_i dx^i$, $\beta\in C^{\infty}(M)$, let $d^*\alpha=A$, then,
\begin{equation}
\begin{split}
\int_X<d^*\alpha, \beta>dV&=\int_X<\alpha, d\beta>dV\\
&=\int_X \alpha_i\partial_j\beta g^{ij}dV\\
&=-\int_X\frac{\partial}{\partial x_j}(\alpha_i g^{ij}\sqrt{g}) \sqrt g\beta dV
\end{split}
\end{equation}
So, $d^*\alpha=-g^{-\frac{1}{2}}\frac{\partial}{\partial x_j}(\alpha_i g^{ij} g^{\frac{1}{2}})$, and 
\begin{equation}
\begin{split}
\Box f &=d^*d f=d^*(f_i dx^i)\\
&=-\frac{1}{\sqrt{g}}\frac{\partial}{\partial x_i}(\sqrt{g}g^{ij}\frac{\partial f}{\partial x_j})=-\Delta f.
\end{split}
\end{equation}
