Problem with notation: Laplacian on a manifold In the Aubin's book "Nonlinear analysis on manifolds" the Laplacian operator on functions on some smooth manifold is defined by the formula
$$
     \Delta = -\nabla^\gamma\nabla_\gamma,
$$
where $\nabla_\gamma$ is the covariant derivative. The author didn't write what does mean $\nabla^\gamma$. I've found in the internet that the notation $\nabla^\gamma$ is sometimes used for the so-called contravariant derivative that is given by $\nabla^\gamma = g^{\gamma i} \nabla_i$, where $g_{ij}$ is a metric tensor. If we use this definition then for a smooth function $f$ on a manifold we will have:
$$
   \Delta f = - g^{\gamma i} \nabla_i \nabla_\gamma f = g^{\gamma i}\nabla_i \left( \frac{\partial f}{\partial x^\gamma}\right) = g^{\gamma i} \frac{\partial^2 f}{\partial x^i \partial x^\gamma},
$$
but this formula doesn't coincide with the formula for the Laplace-Beltrami operator. On the other hand I have found that $\Delta = -\nabla_\gamma^* \nabla_\gamma$, where $\nabla_\gamma^*$ is the formal conjugate to the covariant derivative with respect to scalar product $(f,g) = \int f g \, \Omega$, but in the book the operator $\nabla^\gamma$ appears before the definition of scalar product, so it can't be just a notation for the formal conjugate operator to $\nabla_\gamma$.
 A: I think the main confusion come from the notation $\nabla_\gamma \nabla_i f \neq \nabla_\gamma \bigg(\frac{\partial f}{\partial x^i}\bigg)$. Indeed, 
$$\nabla_\gamma \nabla_i f = \big(\nabla ^2 f)_{i\gamma}$$
Just like what you said, $\Delta = - \nabla^* \nabla$, where $\nabla^*$ is the formal adjoint of $\nabla$ defined by 
$$\int_M \langle X ,\nabla f \rangle dA = \int_M \nabla^*X fdA\ ,$$
where $X$ is any smooth vector fields and $f$ is any smooth function on $M$. To calculate everything in local coordinate, we choose $f$ has compact support in that coordinate, by integration by part, write $G = \det(g_{ij})$ and summing repeating indices,
$$\int_{\mathbb R^n} X^i f_i dA = \int_{\mathbb{R}^n} X^i f_i \sqrt{G}dx^1\cdots dx^n = -\int_{\mathbb R^n} f \frac{1}{\sqrt{G}} \big( X^i \sqrt{G}\big)_i dA\ ,$$
hence 
$$(*)\ \ \ \  \nabla^*X =-\frac{1}{\sqrt{G}} \big( X^i \sqrt{G}\big)_i =-\big( X^i_{\ ,i} + \frac{1}{2}X^i(\log G)_i \big)$$
using the formula $(\log \det A)' = tr(A^{-1} A')$ and defintion of $\Gamma_{ij}^k$, we have
$$\nabla^* X = -\big( X^i_{\ ,i} + X^i \Gamma_{il}^l\big) =- (\nabla X)^i_{\ i}\ .$$
As a result, 
$$\Delta f = -\nabla^* \nabla f = \big( \nabla \nabla f\big)^i_{\ i} = \nabla^i \nabla_i f$$
(The last equality is really just notation). If you use $(*)$, 
$$\Delta f = -\nabla^* \nabla f = \frac{1}{\sqrt G} \big(\sqrt G (\nabla f)^i \big)_i = 
\frac{1}{\sqrt G} \big(\sqrt G g^{ij} f_j \big)_i\ ,$$
which might be more familiar. I agree that the notation you are using is confusing, yet it is standard.
