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 $\nabla_\gamma$.

  • $\begingroup$ @Neal the definition from this book: the covariant derivative on functions is given by $D_X f = X(f)$. In the case $X = \frac{\partial}{\partial x^\gamma}$ one denote $D_X = \nabla_\gamma$, so $\nabla_\gamma f = \frac{\partial f}{\partial x^\gamma}$ $\endgroup$
    – Appliqué
    Commented Nov 13, 2013 at 19:50
  • $\begingroup$ Yes, my bad. I misremembered that fact. $\endgroup$
    – Neal
    Commented Nov 13, 2013 at 20:26

1 Answer 1


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.

  • $\begingroup$ Thank you for the great answer! But could you please give me a reference to a book where one can find the notation $\nabla^i \nabla_i$ for the Laplacian (not the Aubin's book)? $\endgroup$
    – Appliqué
    Commented Nov 14, 2013 at 17:52
  • 2
    $\begingroup$ This notation is standard in geometric analysis. I think you can find this in some book on (e.g.) Ricci flow. The first chapter of the book "Ricci flow and Poincare Conjecture" or the book "Hamilton's Ricci flow" would be good (I suppose). $\endgroup$
    – user99914
    Commented Nov 14, 2013 at 21:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .