Confusion regarding normal vector fields

Question

Let $(\mathcal{M},g)$ be a Lorentzian manifold of dimension $d$ and $(\Sigma,h)$ a Riemannian hypersurface. It is well-known that we can split the tangent bundle as $$T\mathcal{M}\vert_{\Sigma}=T\Sigma\oplus N\Sigma$$ Now, a unit normal time-like vector of $\Sigma$ is by definition a section of the normal bundle $N\Sigma$, such that $$g(N,N)=-1.$$ Since $\Sigma$ is a hypersurface, the rank of $N\Sigma$ is one. Now, I am a litle bit confused about two points:

1. How is $g(N,N)$ defined? $N$ is a vector field on $\Sigma$ and not on $\mathcal{M}$.

Secondly, in physics literature, one often encounters the situation, in which the metric takes the form

$$g_{\mu\nu}\mathrm{d}x^{\mu}\otimes\mathrm{d}x^{\nu}=(-n^{2}+X_{\lambda}X^{\lambda})\mathrm{d}t^{2}+2X_{\lambda}\mathrm{d}t\otimes\mathrm{d}x^{\lambda}+h_{ij}\mathrm{d}x^{i}\mathrm{d}x^{j}$$ where $n\in C^{\infty}(\Sigma)$ and $X\in\mathfrak{X}(\Sigma)$. Then, it is often stated that the normal vector is given by $$N=\frac{1}{N}\partial_{t}+\frac{1}{N}X^{i}\partial_{i}$$

2. How does this make sense from a mathematical point of view? $N$ is the section of a rank $1$-bundle, so why does it have four coordinates??

Answer 1

1. $g(N,N)$ is the mapping $p\mapsto g_p(N_p,N_p)$, defined for $p\in\Sigma$. This is a smooth function on $\Sigma$. Note that all these fields are just some smooth sections of certain bundles over submanifolds of $M$. The metric tensor is a certain section of the $(0,2)$ tensor bundle over $M$, which is why given $p\in \Sigma\subset M$, we have a $(0,2)$ tensor $g_p$ on $T_pM$. Next, $N$ being a normal vector field means given $p\in \Sigma$, we have $N_p\in N_p\Sigma\subset T_pM$. Hence, we can make the evaluation $g_p(N_p,N_p)$.

2. $(t,x^1,\dots, x^{d-1})$ is NOT a coordinate system for the submanifold $\Sigma$; it is a coordinate system for the ambient manifold $M$, which is why $\frac{\partial}{\partial t},\frac{\partial}{\partial x^1},\dots, \frac{\partial}{\partial x^{d-1}}$ is a local basis for the tangent spaces of $M$. If you want to only see “one component”, then you need to use a local basis for the rank 1 vector bundle $N\Sigma\subset TM$. Take a look at Understanding vector field(s) on $\Bbb{S}^3$ for a similar misunderstanding. At a simpler level, think about this: in the vector space $\Bbb{R}^2$, the vector $v=(-7,-7)$ lies in the 1-dimensional subspace \begin{align} E=\{(x,y)\in\Bbb{R}^2\,:\, x=y\}=\text{span}\{(1,1)\}. \end{align} However, if you use only the standard basis $e_1=(1,0),e_2=(0,1)$ of $\Bbb{R}^2$, then we have $v=-7e_1-7e_2$, which requires the use of both vectors (this is not an issue specific to $e_1,e_2$… even if you used another basis, say $\{w_1=(2,3),w_2=(1,9)\}$, you’d still need to use both vectors). This is simply because the bases of $\Bbb{R}^2$ used are such that no subset of it forms a basis for the subspace $E$. In this sense, the basis is not adapted to the subspace.