# Interpretation of degenerate metrics

I was studying the metric tensor and saw all about degenerate metrics. I would like what is the physical or geometrical intuition of a degenerate metric.

What is the meaning of $$g(v,w) = 0$$ for a fixed $$\vec{v}$$ vector and any $$\vec{w}$$ vector? Because there is a vector orthogonal to all others. I understand that for matrix form, you can study if a metric is degenerate or not with the determinant of its matrix form.

When I studied determinant and linear dependence, I saw that if a determinant was 0, it is because the transformation degenerates the space, and turns it into a lower dimensional space. Since metrics can be used to describe space, what sense does it make for a metric to be degenerate, that is, for its determinant to be 0? I also don't understand what relation exists that the determinant is 0 with $$g(v,w) = 0$$ for a fixed $$\vec{v}$$ vector and any $$\vec{w}$$ vector. Thanks in advance

When the determinant of the components of the metric is zero, then one cannot write its inverse-metric. This makes the metric tensor "degenerate".

With a non-degenerate metric tensor, one can "raise and lower indices". For example, one can take a vector $$v^a$$ and get its metric-dual $$v_b=g_{ab}v^a$$ (reusing the $$v$$ in the standard notation and "lowering the index"). We would use an inverse-metric $$g^{bc}$$ to "raise the index of $$v_b$$" to get back the original vector: $$g^{bc}v_b=g^{bc}g_{ab}v^a=\delta^c{}_a v^a=v^c.$$

If the metric is degenerate, then $$g^{bc}$$ isn't available.

One might have to introduce additional structures to try to do some calculations like those done with a non-degenerate metric. This is done in https://en.wikipedia.org/wiki/Newton%E2%80%93Cartan_theory , which includes "Galilean Spacetime" (the spacetime geometry of PHY 101) as a special case. For example, from the Wikipedia link, in some formulations of Galilean spacetime, one has a pair of degenerate metrics with components $$t_{ab}=\left(\begin{matrix}1 & 0\\ 0 & 0\end{matrix}\right)$$ and $$h^{ab}=\left(\begin{matrix}0 & 0\\ 0 & 1\end{matrix}\right)$$.

My favorite example of the occurrence of degenerate metrics in general relativity is on null surfaces. Take for example Minkowski spacetime with Cartesian coordinates $$(t,x,y,z)$$. Then the plane $$t-x = u$$ (with $$u$$ a constant) is a null plane, and the metric on it is degenerate. To see this, let us define an auxiliary coordinate $$v = t + x$$. Using coordinates $$(u,v,y,z)$$ the Minkowski metric is written as $$\mathrm{d}s^2 = - \mathrm{d}u\mathrm{d}v + \mathrm{d}y^2 + \mathrm{d}z^2.$$ Now set $$u = \text{constant}$$. Then $$\mathrm{d}u = 0$$ and the induced metric is $$\mathrm{d}\sigma^2 = \mathrm{d}y^2 + \mathrm{d}z^2,$$ which is a "two-dimensional metric on a three-dimensional surface".

The reason is that the two spatial directions of the plane are orthogonal to the spatial direction of the null vector, and the remaining direction of the plane (the null direction) is null. This is quite different from a Riemannian manifold, where all directions are spatial, or from a Lorentzian manifold, in which the distinguished direction is timelike.

A manifold with a degenerate metric with a single degenerate direction is called a Carrollian manifold, after Lewis Carroll (from Alice in Wonderland). The reason is that these metrics are essentially the $$c \to 0$$ limit of Lorentzian metrics, meaning no one can go anywhere in space, and this resembles a passage from Through the Looking Glass and what Alice found there. I particularly enjoy some discussions on Carroll manifolds given by Duval and collaborators in arXivs: 1402.0657 and 1402.5894, for example.