Conflict in expression for Lie derivative of a metric

Question

Let $S \subset M$ be an embedded submanifold. Here, $(M, g)$ is a (pseudo)Riemannian manifold. One can choose a vector field defined on $S$ (not necessarily normal or tangent to $S$), say, $X$. We can decompose this into its normal/tangental components $X = \lambda \nu + X^T$ where $X^T$ is tangent to $S$ and $\nu$ is the normal field to $S$. Here $\lambda$ is a scalar function. Computation shows that $$ L_X g = 2\lambda\theta + \pi, $$ where $\pi_{ab} = \nabla_a X^T_b + \nabla_b X^T_a$ is the deformation tensor of $X^T$ and $\theta$ is the second fundamental form of $S$ in $M$, defined by $\theta(X, Y) = g(\nabla_X \nu, Y)$. Edit: Here, $S$ is codimension 1 and $\nabla$ is the connection on $M$.

My question: The Lie derivative is known to depend not just on the vector $X$ at a point, but the 1-jets; which, in this case, may leave $S$ (they do, as long as $\lambda \neq 0$). However, the right hand side of this equation, to the best of my knowledge, depends only on the values of $X$ on the submanifold $X$.

A possible explanation: At least, this is definitely true for $\theta$ and $\lambda$, as $\theta$ doesn't depend on $X$ at all, and $\lambda$ was defined in a way that only depends on $X|_S$. So perhaps $\pi$ depends on $X$'s values outside of $S$. However, if we restrict the above to tensor fields on $S$ (so they take $V, W$ in $T_p S$ for some $p \in S$), I believe this is false: $$ \pi(V, W) = g(\nabla_V X^T, W) + g(\nabla_W X^T, V) $$ and since $V, W$ are tangent to $S$, this only depends on the values of $X$ (in particular, the values of $X^T$) on $S$. So, some follow-up questions are:

1. Does $L_X g(V, W)$ not depend on the 1-jets of $X$, but only the value at a point, if $V, W$ are in $S$? Or perhaps it still depends on 1-jets of $X$, but only the tangential components? However, this goes against what I know about Lie derivatives.

2. Does $\pi(V, W)$ in general depend on the values of $X$ outside $S$, if the vectors $V, W$ are not assumed to be tangent to $S$?

Answer 1

Let $X = X^{T} + \lambda \nu$ be the above decomposition. Let $U$ and $V$ be vector fields. Then, \begin{align} \left(L_Xg\right)(U,V) &= X\left(g(U,V) \right) - g\left([X,U],V \right) - g\left(U,[X,V]\right) \\ &= g \left(\nabla_XU,V \right) + g\left(U,\nabla_XV \right) -g\left([X,U],V \right) - g\left(U,[X,V]\right) \\ &= g\left(\nabla_XU - [X,U],V \right) + g\left(U,\nabla_XV-[X,V] \right)\\ &= g\left(\nabla_U X,V \right) + g\left(U,\nabla_V X \right) \\ &= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T \right)\right) \\ &~~~+ g\left(\nabla_U\left(\lambda\nu\right),V \right) + g \left(U,\nabla_V(\lambda \nu) \right) \\ &= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T\right) \right) \\ &~~~ + g\left( \mathrm{d}\lambda(U)\nu,V \right) + g\left(\lambda\nabla_U\nu,V \right)\\ &~~~ + g\left(U, \mathrm{d}\lambda(V)\nu\right) + g\left(U,\lambda \nabla_V\nu\right)\\ &= g\left(\nabla_U\left(X^T\right),V \right) + g\left(U, \nabla_V\left(X^T \right)\right) \\ &~~~ + g\left(\mathrm{d}\lambda(U)\nu,V \right) + g\left(U, \mathrm{d}\lambda(V)\nu\right) \\ &~~~+ 2\lambda \theta\left(U,V \right) \end{align} So it seems your formula is valid only if $U$ and $V$ are tangent to $S$, i.e orthogonal to $\nu$. Otherwise, you would have to consider the term $g\left((U\cdot \lambda)\nu,V \right) + g\left(U, (V\cdot \lambda)\nu\right)$, and $\mathrm{d}\lambda$ is a component of the $1$-jet of $X$. Moreover, $\nabla_UX^T$ and $\nabla_V X^T$ depends too on the $1$-jet of $X$, not just on the value of $X$ at a point.

More generally, $\left(\nabla_AB\right)_p$ depends on the value $A_p$ and on the value and the $1$-jet of $B$ at $p$.

So to answer directly your questions:

• $(L_Xg)(U,V)$ depends on the $1$-jet of $X$ by $\mathrm{d}\lambda$ and $\nabla \left(X^T\right)$.
• if $U$ and $V$ are tangent to $S$ (thus, orthogonal to $\nu$), $\lambda$ affects the result only at the $0$-th order. Otherwise, its infinitesimal growth affects the result, thus the growth of $X$ along the integral curves of $\nu$ affects the results too.

Edit If $U$ and $V$ are tangent to $S$, the calculation shows that $(L_Xg)(U,V)$ depends on $X$ this way: it depends on the $1$-jet of $X^T$ and on the $0$-jet of $\lambda$. This can be interpreted like this: $U$ and $V$ are intrinsic object of $S$. If $\lambda = 0$, $X$ is also an intrinsic object of $S$, so is $(L_Xg)(U,V)$. So the result should not depend on how you expand $X$ outside of $S$. As you can expand $X$ by many ways and at many growth outside of $S$ (with any sort of function $\lambda)$, this should only depend on the fact that $\lambda = 0$ on $S$ and not on its derivative.