When is the restriction of a Lorentzian metric to a regular submanifold degenerate everywhere?

Question

Let $M$ be a $C^\infty$ manifold, $N\subset M$ be a regular submanifold and $g$ be a Lorentzian metric on $M$. I would like to find $M$, $N$, $g$ such that the restriction of $g$ to the tangent bundle $TN$ is degenerated everywhere.

I would be most grateful if you help me solve this problem.

Answer 1

This is too long for a comment . . .

Let $p$ be a point in $M$ and consider $v \in T_{p}M$. Since $M$ is equipped with with a Lorentzian metric $g$, then there are three possibilities for $g(v, v)$:

1. $g(v, v) > 0$ (We say that the vector $v$ is spacelike)
2. $g(v, v) = 0$ (We say that the vector $v$ is null or lightlike provided $v\ne 0$)
3. $g(v, v) < 0$ (We say that the vector $v$ is timelike).

(Remark: By definition , the zero vector is spacelike.)

The null cone at $T_{p}M$ is defined to be $NC_p = \left\{ v \in T_pM \Big \vert g(v, v) = 0\right\}$.

A parametrized curve $\alpha: I\subseteq \mathbb{R} \to M$ is said to be a null curve if $\frac{d\alpha}{dt}$ is a null vector (i.e. $g(\frac{d\alpha}{dt}, \frac{d\alpha}{dt})= 0$) for all $t$. In this case, the induced metric on the curve $\alpha$ is degenerate.

Example: Take $M = \mathbb{R}_{2, 1}$ (i.e. $\mathbb{R}^{3}$ with metric $ds^2 = dx^2 + dy^2 - dz^2$). The null cone at the origin is the cone defined by $x^2 + y^2 - z^2 = 0$. The curve $\alpha(t) = (t, 0, t)$ is a null curve. In fact, any line through the origin whose direction vector lies in the null cone is a null curve and the restriction of the metric to such a curve will be degenerate.

Now, for a surface $S$ in $\mathbb{R}_{2, 1}$ such that $g$ restricts to a degenerate metric on $S$, we are looking for a surface $S$ such that at every point $p \in S$, the tangent plane contains a unique null line. As mentioned in the comment, the plane spanned by the vectors $v = (1, 0, 1)$ and $w = (0, 1, 0)$ (here we are taking advantage of the linear structure of $\mathbb{R}_{2,1}$) will do the trick. Observe the following:

• $g(v, v) = 0$ (i.e. $v$ is a null vector),
• $g(w, w) = 1$ (i.e. $w$ is spacelike),
• $g(v, w) = 0$ (i.e. $v$ and $w$ are $g$-orthogonal).

The fact that the metric is degenerate follows from the fact that any vector $u$ that belongs to the span of $v, w$ can be expressed as $u = av + bw$ for some constants $a$ and $b$. Now observe that $v$ $g$-orthogonal to all such $u$: \begin{equation} g(v, u) = g(v, av + bw) = a\cdot g(v, v) + b\cdot g(v, w) = 0. \end{equation}

Alternatively, you can parametrize the the plane described above by $f : \mathbb{R}^{2}_{(r, s)} \to \mathbb{R}_{2, 1}$, where $f(r, s) = (r, s, r)$. Finding the $E = g\left(f_{*}\left(\frac{\partial}{\partial r}\right), f_{*}\left(\frac{\partial}{\partial r}\right)\right)$, $F = g\left(f_{*}\left(\frac{\partial}{\partial r}\right), f_{*}\left(\frac{\partial}{\partial s}\right)\right)$ and $G = g\left(f_{*}\left(\frac{\partial}{\partial s}\right), f_{*}\left(\frac{\partial}{\partial s}\right)\right)$ of the metric will reveal that the metric is degenerate (i.e. $EG- F^2 = 0$).

Now, with all of this being said, the key to finding more interesting surfaces in $\mathbb{R}_{2,1}$ where the restriction of $g$ is degenerate is for each tangent space of $S$ to have a basis satisfying the properties that the vectors $v$ and $w$ have above.

You can also carry this out for more general Lorentzian manifolds by making the appropriate adjustments.