# Ricci flow on surfaces : step in proof

I am trying to realize the paper of richard hamilton's ricci flow on surfaces from the book of benett chow's Ricci flow : An Introduction.Here Hamilton denoted the trace free part of the Hessian of the potential $f$ of the curvature by $$\ M = \nabla \nabla f - \frac 12 \Delta f . g$$ Next taking divergence of $M$ we have $$(div M)_i = \nabla ^j M_{ji}=\nabla _j\nabla_i\nabla^jf-\frac 12 \nabla_i\nabla_j\nabla^jf=R_{ik}\nabla^kf+\frac 12\nabla_i\Delta f=\frac 12(R\nabla_if+\nabla_iR)$$....But in this calculation I can not find what the term $\nabla^j$ means.$\nabla_j$ is covariant derivative...But what about $\nabla^j$...Please anyone help me to understand the calculations...one more here $R_{ik}=\frac R2 g_{ik}$...

• $\nabla^j = g^{ij} \nabla_i$ (sum over $i$ of course).
– user99914
Commented May 19, 2015 at 6:32
• It's just the usual notation for raising and lowering indexes via the metric as already mentioned Commented May 19, 2015 at 6:54

This is a convention in differential geometry: $$\nabla^{j} = g^{ij}\nabla_i=\sum_{i=1}^{n}g^{ij}\nabla_i$$ where n is the dimension of the manifold.

Recall the following identities : $$\nabla^jM_{ji}:=g^{kj}\nabla_k M_{ji}$$ $$\nabla g=0$$ $$\nabla_i\nabla_jf=\nabla_j\nabla_i f$$

$$\nabla_a\nabla_b\nabla c f-\nabla_b\nabla_a\nabla_c f =-R_{abc}^k\nabla_k f$$

Hence \begin{align} \Delta \nabla_if -\frac{1}{2}\nabla_i\Delta f&=g^{ab}(\nabla_a\nabla_i\nabla_b f)-\nabla_i\Delta f \\&=g^{ab} (\nabla_i\nabla_a\nabla_b f - R_{aib}^k \nabla_k f )-\frac{1}{2}\nabla_i\Delta f \\& =\frac{1}{2}\nabla_i\Delta f +R_i^k\nabla_kf \end{align}