I'm trying to follow a calculation of the curvature tensor of a rotationally symmetric metric, and there's this step I can't justify to myself. The discussion can be found on Petersen's "Riemannian Geometry", section 4.2.3.
The setting is as follows. We have a metric $$g =dr^2 +\rho(r)^2ds^2_{n-1} $$ and we set $$ g_r = \rho^2 ds^2_{n-1} $$
We've proved that $$ Hess(r)=\frac{\partial_r \rho}{\rho} g_r $$
The author then calculates as follows: $$ \begin{align} \nabla_{\partial_r}Hess(r) &= \nabla_{\partial_r}(\frac{\partial_r \rho}{\rho} g_r) \\ &= \nabla_{\partial_r} (\frac{\partial_r \rho}{\rho}) g_r + \frac{\partial_r \rho}{\rho} \nabla_{\partial_r} (g_r)\\ \# &= \frac{(\partial_r^2 \rho)\rho - (\partial_r \rho)^2}{\rho^2}g_r \\ &= \frac{(\partial_r^2 \rho)}{\rho}g_r -Hess^2(r) \end{align} $$
I've marked the step I'm confused about with a #. I think it implies $\nabla_{\partial_r} g_r =0$, but when calculating, I have: $$ \nabla_{\partial_r} (g_r) = \nabla_{\partial_r}(\rho^2 ds^2_{n-1} ) = 2\rho \partial_r (\rho)ds^2_{n-1} $$
What did I get wrong?
You need to define the covariant derivative of a symmetric bilinear form $h\in\mathcal{T}^0_2$. For a vectorfield $X$, we define $(\nabla_X h)(Y,Z):= X(h(Y,Z))+h(\nabla_X Y, Z)+ h(Y,\nabla_X Z)$, where the first term on the right hand side corresponds to the derivative of $h(Y,Z)$ in direction $X$. The product rule of every Levi-Civita connection implies $\nabla_X g=0$. Furthermore, it is straightforward to check that $\nabla_{\partial_r} dr^2=0$. Thus, $\nabla_{\partial_r}g_r=\nabla_{\partial_r}g-\nabla_{\partial_r}dr^2=0$.