# 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$$^{}$$.

What is happening here? What do physicists mean when they construct quantities like these$$^{}$$? Even a link to a reference that treats this on a mathematically justifiable level is welcome.

 Naively, I would even expect that one would require some sort of a connection on the vector bundle to make this question tractable.

 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.

• In physicist-speak, the metric on the hypersurface is also a tensor on the whole manifold. The lie derivative you're describing is the second fundamental form of the hypersurface, which is something that doesn't only depend on data from the hypersurface (it is in fact a piece of boundary information for the Cauchy problem) – John Donne Jun 26 at 22:19
• Yes, I know all this. The second fundamental form is not intrinsic to the hypersurface, but it exists on the normal bundle of the hypersurface. The metric- can also be thought of as a metric on the whole foliation of spacelike Cauchy surfaces; so it can also be though of as a bundle metric over the bundle whose integral manifolds are the Cauchy surfaces. The question, though, is how to make sense of all of this in a manner consistent with differential geometry. – Sandesh Jr Jun 27 at 9:25

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 $$K=\frac{1}{2}\phi^\ast(\mathcal L_{\bar N}\bar q).$$

• For a codimension 1 submanifold, $\mathcal{L}_{\bar N} \bar q$ is independent of the extension even if $\bar N$ is not geodesic. For codimension 2 or greater, this expression is not independent of the extension, and depends on the twist tensor for the normal congruence of the extension. Of course, the pullback you mentioned remains independent of the extension. – asperanz Jun 28 at 15:55
• @asperanz That's possible, I wrote this from memory. Then maybe it is independent of the extension, but is "tangential" only if $N$ is geodesic. At least I recall that the pullback can be omitted if $N$ is geodesic. – Bence Racskó Jun 29 at 6:29
• Orthogonal projectors are maps $\mathfrak{X}(M) \rightarrow \mathfrak{X}(H)$, and correspondingly the metric is a section of $T^*H \otimes T^*H$. But apart from this technical point, the argument seems more or less right. – Sandesh Jr Jun 29 at 9:21
• @SandeshJr But in my answer it is not the orthogonal projector acting on the metric but the second tensor power of its dual map, whose domain consists of second order covariant tensors on $\Sigma$. – Bence Racskó Jun 29 at 17:12