I come form General Relativity background and have a question about differential geometry.
Assume a pseudo-Riemannian manifold $M$ equipped with a metric tensor field $g$ inducing inner product on the tangent bundle. Assume a codimension 1 submanifold in $M$ called $\Sigma$ with unit normal field $n$ and pullback of metric $g$ onto it called $h$.
The extrinsic curvature tensor of $\Sigma$ is defined as (giving two options, I am more used to the left case suggesting tensor structure by indices, but I understand that it is not the standard mathematical notation)
$$K_{ab}=h_{a}^c\nabla_c n_b\text{ or equivalently }K=h(\nabla,n)$$
where $\nabla$ is the Levi-Civita connection of the metric $g$.
Assume furthermore that $M$ and $\Sigma$ are maximally symmetric manifolds, i.e. the Riemann tensors are
$$R(M)_{abcd} = \pm L^{-2}(g_{ac}g_{bd}-g_{ad}g_{bc})$$ $$R(\Sigma)_{abcd} = \pm \ell^{-2}(h_{ac}h_{bd}-h_{ad}h_{bc})$$
for some constants $L$, $\ell$.
Assume furthermore that $\Sigma$ is of codimension 1. The Gauss-Codazzi equation than provides
$$\frac{Tr(K)^2-Tr(K\cdot{}K)}{(m-1)(m-2)}=\pm \ell^{-2} \pm L^{-2}$$
where $m=dim(M)$, $Tr$ denotes trace.
I believe that in such settings I should be able to prove that $\Sigma$ needs to be umbilical, i.e.
$$K = \frac{Tr(K)}{m-1}h$$
and $Tr(K)$ is constant.
While I am not sure this is useful, one finds that $K = \frac{Tr(K)}{m-1}h$ is equivalent to $Tr(K\cdot K) = \frac{Tr(K)^2}{m-1}h$ since
$$Tr(K\cdot K) - \frac{Tr(K)^2}{m-1}h=Tr((K-\frac{Tr(K)}{m-1}h)(K-\frac{Tr(K)}{m-1}h)).$$
Can I somehow prove $\Sigma$ is totally umbilical without computing $K$ directly, i.e. without having to compute $h\cdot\nabla n$?
No foliation of $M$ is assumed. I hope that from the knowledge of Riemann tensors I should be somehow able to deduce $Tr(K)$ or $Tr(K\cdot K)$, but I do not seem to be able to do that. I hoped to maybe use some other projection of Riemann tensors, but to no avail. These hopes stem from the intuition that I am basically taking a "ball" in a constant curvature manifold, so it should be also of constant exrinsic curvature.
In my usecase I assume $m=3$ so please if this was actually solvable only for that case, do not hesitate to provide answer for $m=3$ only.
EDIT:
I realized that for codimension 1 submanifold of const. curvature in space of const. curvature I can even deduce that the trace of extrinsic curvature $Tr(K)$ is constant along $\Sigma$. It seems that I have all the ingedients to show that $K = \frac{Tr(K)}{dim(M)-1}$. I reviewed several papers again and there are many similar results, but for different dimension, compact positive curved submanifolds, riemannian (not pseudoriemanian) sapceforms etc.