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Let $M$ be a hypersurface embedded in $\mathbb{R}^n$. It is known that if the norm squared of the second fundamental form of $M$ is bounded, then we can find a uniform lower bound for the radius $R>0$ of a geodesic ball $B(p,R)$ such that for any point $p \in M$, the ball can be expressed a graph over its tangent plane $p\in M$.

Can this fact be generalised when $\mathbb{R}^n$ is now any Riemannian manifold, i.e. can we be sure that $M$ is uniform graphical when it has bounded second fundamental form? What does it even mean for a hypersurface to be a graph over the tangent plane in this situation? Does one need to use Fermi coordinates on the normal exponential map for this to make sense?

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    $\begingroup$ "What does it even mean for a hypersurface to be a graph over the tangent plane in this situation?" is the main question. One possible formulation is this: given $N$ a complete Riemannian manifold and $\Sigma$ a hypersurface. Let $M$ be a hypersurface that is tangent to $\Sigma$ at some $p\in N$, find conditions that guarantee $M$ can be expressed as a section of the normal bundle (or tubular nbhd?) over $\Sigma$. But I bet there are other equally reasonable formulations. $\endgroup$
    – Willie Wong
    Feb 14, 2021 at 14:58

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