# What is the meaning of the symbol $\nabla^k$?

In T.Aubin's book, a course in differential geometry, he write the formula $\Delta f=-\nabla^k\nabla_kf$ on a Riemannian manifold, but he never define the symbol $\nabla^k$. It seems that the notation is not the kth covariant derivative. So my question is: Does any one know what is the meaning of the symbol? Thank you very much and any reference is welcome.

• This may not be what it means, but I know in time series theory its means the differencing operator – Kamster Jul 21 '14 at 2:29
• mathworld.wolfram.com/BackwardDifference.html – Kamster Jul 21 '14 at 2:30
• or maybe laplace operator? en.wikipedia.org/wiki/Del en.wikipedia.org/wiki/Laplace_operator – Kamster Jul 21 '14 at 2:36
• It seems to me, that it's intended to simply specify the kth component of the derivative. The Einstein summation rule turns $\nabla^k\nabla_k$ into the $\Delta$ laplace operator. The only odd thing is the minus sign though. – Klaas van Aarsen Jul 21 '14 at 2:37
• @IlikeSerena sometimes a minus is included for the Laplacian. There is not a universal convention on this point. – James S. Cook Jul 21 '14 at 3:40

It looks like here on functions $\nabla^k \nabla_k f$ means $g^{jk}\nabla_j\nabla_k f$ where $\nabla_j\nabla_k f = \nabla_{\partial_j}\nabla_{\partial_k}f = (\nabla \nabla f)_{jk}$. So for an $r$-tensor $\alpha_{i_1 \cdots i_r}$, we could lift an index on the $(r+1)$-tensor $\nabla_k \alpha_{i_1 \cdots i_r} = (\nabla \alpha)_{ki_1 \cdots i_r}$ and write $\nabla^k \alpha_{i_1 \cdots i_r}$ for $g^{jk}\nabla_j \alpha_{i_1 \cdots i_r}$.
• So $g^{jk}$ would be defined in such a way that the result turns out negative? Or let me rephrase: where does the minus sign come from? – Klaas van Aarsen Jul 21 '14 at 3:48