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I seem to have trouble with an elementary computation, and figure it may help others if faced with a similar situation. The basic question is as follows: if I have a tensor field $T$ on some Riemannian manifold $(M,h)$, and I know that $T$ is constant for some choice of local orthonormal frame, does it hold that $ DT \equiv 0\,, $ where $D$ is the covariant derivative along a smooth isometrically immersed submanifold of $M$?

I think it is only polite to explain the precise situation I am considering, so let me do so here.

Let $(N,h)$ be a Riemannian 3-manifold and consider an immersion $f:\Sigma\rightarrow N$ of the surface $\Sigma$. Equip $\Sigma$ with the pullback of the metric $h$ so that $(\Sigma, f^*h) = (\Sigma, g)$ is a Riemannian 2-manifold. Let us denote by $\nabla$ the connection on $N$ compatible with $h$ and by $D$ the connection on $\Sigma$ compatible with $g$. The Christoffel symbols of $\nabla$ we denote by $\Gamma$ and those of $D$ we denote by $\Lambda$.

Now assume $\Sigma$ is orientable and let $\nu:\Sigma\rightarrow (T\Sigma)^\perp$ be a choice of unit normal. Further let us assume that the curvature of $N$ is constant; since $N$ is a 3-manifold, this is the same as assuming that the Ricci curvature tensor $\text{Ric}$ of $N$ is constant.

Let us perform our computations in an orthonormal frame $\{\nu,e_1,e_2\}$ which is adapted to the submanifold $\Sigma$. By definition $ \nabla_{e_k} \text{Ric}(e_i,e_j) = \partial_k\text{Ric}(e_i,e_j) - \Gamma_{ki}^l\text{Ric}(e_l,e_j) - \Gamma_{kj}^l\text{Ric}(e_i,e_l)\,. $ As far as I can tell, there is no reason that $\Gamma \equiv 0$ in the orthonormal frame we have. So it seems to me that although the Ricci curvature of $N$ is constant, it's covariant derivative does not vanish in general. This seems bizarre to me, and is certainly confusing, but not my main question.

Restricting the tensor $\text{Ric}$ to $\Sigma$, we can consider $D\text{Ric}$. In particular, my main question is: $ \text{Does it hold that $D_{X}\text{Ric}(\nu,\nu) = 0$ for $X\in\{\nu,e_1,e_2\}$}? $

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    I upvoted because it was good to discuss these things. I don't have time for an answer, sorry. It would be good if you answer it yourself.2012-09-18

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