# A question on Levi-Civita connection and a fixed hyper surface

Suppose $$(M,g)$$ is a three dimensional smooth compact simply connected Riemannian manifold with boundary and suppose that $$\Sigma$$ is a smooth simply connected hypersurface in $$M$$ with a smooth boundary $$\partial \Sigma \subset \partial M$$. Let $$D_X Y$$ denote the Levi-Civita connection on $$(M,g)$$. Let $$Z$$ be a smooth vector field on $$\Sigma$$ and let $$f:\Sigma \to \mathbb R$$ be a smooth function on $$\Sigma$$. Does there exist a smooth function $$\phi$$ in $$M$$ such that $$\Sigma=\{\phi=0\}$$ and such that $$(D_{\nabla \phi} \nabla \phi)\big|_{\Sigma} = f\, Z.$$ Note that the left hand side is just the restriction of $$D_{\nabla \phi}\nabla \phi$$ to $$\Sigma$$ and that $$\nabla \phi$$ is the gradient of $$\phi$$ with respect to $$g$$.

If the answer is no, would it make a difference if the latter condition is replaced with $$(D_{\nabla \phi} \nabla \phi)\big|_{\Sigma} - f\, Z \in \textrm{Span}\{(\nabla \phi)|_{\Sigma}\}$$

• The Whitney extension theorem could be relevant. (Also, do you need $f$ at all?) Commented Jul 12, 2021 at 20:13
• Indeed $f$ is redundant :)
– Ali
Commented Jul 12, 2021 at 20:18
• Why is this question tagged with lorentzian and semi-riemannian geometry? All the objects in sight seems to be Riemannian. Is there some connection to semi-Riemannian stuff? If so, can you make it explicit? Commented Jul 13, 2021 at 0:26
• Additionally, when $Z$ is said to be a smooth vector field on $\Sigma$, do you mean $Z$ is tangent to $\Sigma$, or a vector field of $M$ restricted to $\Sigma$? And when you say "restriction of $D_{\nabla \phi} \nabla \phi$ to $\Sigma$" do you mean the projection, to $T\Sigma$ or just the restriction? Commented Jul 13, 2021 at 0:28
• $Z$ can be thought of as a vector field on $M$ that when restricted to $\Sigma$, is tangent to $\Sigma$. By the restriction of $D_{\nabla \phi}\nabla \phi$ to $\Sigma$ I mean the restriction, so the equation should be considered on $\Sigma$ only.
– Ali
Commented Jul 13, 2021 at 0:55

Take the simplest case where $$M = \mathbb{R}^3$$ (the boundary is unimportant for the discussion here) and $$\Sigma$$ is the $$x$$-$$y$$ plane.
If $$\phi$$ is a defining function of $$\Sigma$$, then restricted to $$\Sigma$$ we have $$\nabla \phi|_{\Sigma} = \eta \partial_z$$ for some function $$\eta: \Sigma \to \mathbb{R}$$.
Symmetry of the Hessian requires then $$\nabla^2 \phi|_{\Sigma} = \begin{pmatrix} 0 & 0 & \partial_x \eta \\ 0 & 0 & \partial_y \eta \\ \partial_x \eta & \partial_y \eta & * \end{pmatrix}$$ This shows that $$D_{\nabla\phi} \nabla\phi = \eta (\partial_x \eta, \partial_y \eta, *)$$
So a necessary condition for your equation to hold (setting $$f \equiv 1$$ since as you agreed it is unimportant) in the flat case is that $$Z$$ is the gradient of some scalar function on $$\Sigma$$, and that allowing there to be normal components (as in your modified question) has no impact.
The conclusion is unchanged in the Lorentzian setup, if you require $$\Sigma$$ to be non-degenerate. In the case where $$\Sigma$$ is null, however, since $$\nabla\phi$$ now lies in the tangent space of $$\Sigma$$ and $$\Sigma$$ is two dimensional, it may in fact be possible for the second variant to hold (but I don't have the energy to run through the analysis now).