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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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(1) How can your $D$ act on ambient tensors? (2) How $D$ can act in the normal direction? –  Yuri Vyatkin Sep 18 '12 at 9:52
@Yuri (1) By restricting $\text{Ric}$ to $\Sigma$. (This involves composition with the immersion map.) (2) $D$ induces a connection in the normal bundle, and this acts "in the normal direction". –  Glen Wheeler Sep 18 '12 at 10:19
You have said in the question that $D$ is the induced connection on $\Sigma$. I don't know how it can induce a connection in the normal bundle. Actually, $\nabla$ on $N$ is responsible for that, not $D$. –  Yuri Vyatkin Sep 18 '12 at 10:22
Also, what do you precisely mean by restricting on $\Sigma$: will $Ric$ act on $T\Sigma$ or on the pullback bundle? In either case the expression $D_{X}\text{Ric}(\nu,\nu)$ is not well defined to me. –  Yuri Vyatkin Sep 18 '12 at 10:31
@Yuri I mean for $\text{Ric}$ to act on $T\Sigma$, although I'm not sure if it turns out to be better for it to act on the pullback bundle. Can you explain why you don't think the expression is well-defined either way? Let's just take $X\in\{e_1,e_2\}$. What I mean by "connection in the normal bundle" is just the projection of $\nabla$ onto $(T\Sigma)^\perp$. Sorry that I am being confusing. –  Glen Wheeler Sep 18 '12 at 10:38

1 Answer 1

Note that $$ e_3=\nu,$$

$$ i=1,\ 2,\ D_{e_i} Ric(e_3,e_3)=e_i( Ric_{33} ) - 2Ric_{k3}\Gamma_{i3}^k =- 2Ric_{k3}\Gamma_{i3}^k = 2Ric_{k3}\Gamma_{ik}^3$$

Note that on some frame $\{ f_i\}$, $$ Ric=cI $$ so that wrt $\{e_i=A_{ij}f_j\}$ we have $$ Ric_{ij} = Ric(A_{ik} f_k, A_{jl} f_l) =A_{ik} A_{jk}c =\delta_{ij}c $$

Hence $$ 2Ric_{k3}\Gamma_{ik}^3 = = 2Ric_{33}\Gamma_{i3}^3=0 $$

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