Is the Lie derivative along the normal well-defined? This question is cross-posted at https://math.stackexchange.com/q/3274757/247251
Let $(\Sigma, q)$ be a non-degenerate submanifold of a Lorentzian manifold $(M,g)$. Let $N$ be the section of $T\Sigma ^g$. Physicists often talk about the evolution of $q$ along $N$ as $\mathcal{L}_Nq$. But this expression makes no sense as $N$ does not belong to $\mathfrak{X}(\Sigma)$. As such, Lie derivatives are defined using flows of vector fields; I don't see any natural way of extending it to arbitrary vector bundles$^{[1]}$. 
What is happening here? What do physicists mean when they construct quantities like these$^{[2]}$? Even a link to a reference that treats this on a mathematically justifiable level is welcome.
[1] Naively, I would even expect that one would require some sort of a connection on the vector bundle to make this question tractable.
[2] The only argument I can think of is that the operation is actually being performed on the ambient manifold. Say $tan:\mathfrak{X}(M)\to \mathfrak{X}(\Sigma)$ is the canonical projection operation associated to the embedding ($tan:=q^{\sharp}\circ\iota \circ g^\flat$). Now, $tan^*(q)\in \Omega^2(M)$, so $\mathcal{L}_N(tan^*(q))\in\Omega^2(M)$ is well defined. But this feels like an incomplete picture, and possibly even wrong.
 A: To be properly rigorous, let us fix the manifolds involved. In this answer $M$ shall denote an $n$ dimensional spacetime manifold, $\Sigma$ an $n-1$ dimensional manifold, and $\phi:\Sigma\rightarrow M$ an embedding.
OP's inquiries are twofold. First, they are interested in how to represent a geometric object defined on $\Sigma$ as a geometric object defined on $M$, secondly, they are interested how are differential operators acting on fields along $\Sigma$ are defined rigorously.
For the sake of clarity, if $\pi:E\rightarrow M$ is a vector bundle, then a field along $\Sigma$ is defined as a smooth map $\psi:\Sigma\rightarrow E$ such that $\pi\circ\psi=\phi$. This is often called a field along $\phi$ in the mathematical literature, and fields along $\Sigma$ are in 1-to-1 correspondance with sections of the pullback bundle $\phi^\ast E$.

Let us furthermore assume that $M$ has a Lorentzian metric $g$, and that $q=\phi^\ast g$ is the induced metric on $\Sigma$, and let us assume that $q$ is either Riemannian or Lorentzian. In physics terminology, this means that the hypersurface $\Sigma$ is timelike or spacelike everywhere, and has no null points at all.
There is a "God given" tangent map $T\phi:T\Sigma\rightarrow TM$, which can be used to push forward vectors defined on $\Sigma$ to vectors defined on $M$. Vectors of $M$ that are in the image of $T\phi$ are called tangential (to $\Sigma$).
If the condition of $q$ being nowhere degenerate is met, then we also have the following structure: Let us denote as $T^0\Sigma$ the following construction: $$T^0\Sigma=\{\text{The annihilator of }\text{Im}(T\phi)\},$$ eg. $T^0\Sigma$ is the space of all "normal 1-forms" to $\Sigma$. Its metric dual is $N\Sigma$, the space of all normal vectors. I will consider all these manifolds to be fibred over $\Sigma$, rather than the image $\phi(\Sigma)$ (eg. I am taking the pullbacks of these bundles).
Then we have the orthogonal decomposition $$ \phi^\ast TM=T\Sigma\oplus N\Sigma, $$ and to this orthogonal decomposition there is an orthogonal projection $$ P:\phi^\ast TM\rightarrow T\Sigma. $$
This projection can be used to transport covectors defined on $\Sigma$ to covectors defined on $M$ via the dual map $$ P^\ast:T^\ast\Sigma\rightarrow \phi^\ast T^\ast M\subset T^\ast M. $$
Note that:


*

*An arbitrary contravariant tensor of order $k$ can be transported from $\Sigma$ to $M$ by the tangent map $T\phi$ by taking tensor products as $T^k\phi=T\phi\otimes...\otimes T\phi$.

*An arbitrary covariant tensor of order $k$ can be transported from $\Sigma$ to $M$ b the dual of the orthogonal projection by taking tensor products as $P^{\ast k}=P^\ast\otimes...\otimes P^\ast$.

*If these maps are applied to tensor fields defined on $\Sigma$, rather than single tensors defined at points, the end result will not be a tensor field of $M$, but it will be a tensor field of $M$ along $\Sigma$.
To use OP's example, the Lie derivative of $q$ appears. What is meant under $q$ in this case is actually $\tilde q=P^{\ast 2}q$, where $q=\phi^\ast g$. The object $q$ is a tensor field on $\Sigma$, but $\bar q$ is a tensor field along $\Sigma$ on $M$.

This concludes answering the first inquiry. On the second inquiry, whenever there is a hypersurface $\Sigma$ with normal field $N$, we usually assume that $M$ is at least locally foliated by a family of hypersurfaces, and $\Sigma$ is only a leaf of the foliation.
We further take all tensor fields along $\Sigma$ and extend them smoothly to tensor fields defined on open sets contained within the foliation. Since we can assume that $\phi$ is an embedding, the image of $\Sigma$ will not intersect itself, hence these smooth extensions are always possible.
Differential operators such as $\mathcal L$ and $\nabla$ are then applied to these smooth extensions. We usually get a physically sensible result only if the result is independent of the extension.
To use OP's example, we interpret $\mathcal L_N q$. Here by default $q$ is the induced metric on $\Sigma$, eg. it is a section of $T^\ast \Sigma\otimes T^\ast \Sigma$, and $N$ is a vector field along $\Sigma$, a section of $N\Sigma$ (remember, this is fibred over $\Sigma$).
We then replace $N$ with a smooth extension $\bar N$, which is now defined on an open set, and replace $q$ with $\tilde q=P^{\ast 2}q$, which is now a tensor field along $\Sigma$, and then further replace $\tilde q$ with $\bar q$, a smooth extension of $\tilde q$ to an open set.
Now, the expression $\mathcal L_{\bar N}\bar q$ is meaningful, albeit we still need to see if it is independent of the extensions.
I do not want to check it right now, but I think this expression is independent of the extension of $\tilde q$ if $N$ had been extended geodesically. If we do not wish to rely on this geodesic extension, then we can consider the pullbackn $\phi^\ast(\mathcal L_{\bar N}\bar q)$, which is now completely independent of the extensions $\bar q$ and $\bar N$.
Indeed, one possible way to express the extrinsic curvature/second fundamental form of $\Sigma$ is via $$ K=\frac{1}{2}\phi^\ast(\mathcal L_{\bar N}\bar q). $$
