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I'm following an example about how to determine the signature of a metric induced on a submanifold of some space using the normal vectors to the submanifold but don't understand how the full process works. Could someone help?

I'm trying to compute the signature of the metric induced on the hypersurface:

$t^2-x^2-y^2-z^2 = 1$

in $\mathbb{R}^{1,3}$ (where $ds^2 = dt^2-dx^2-dy^2-dz^2$)

To do this, the normal vector is calculated as the gradient of the function of the hypersurface:

$n = 2t\frac{\partial}{\partial t}-2x\frac{\partial}{\partial x}-2y\frac{\partial}{\partial y}-2z\frac{\partial}{\partial z}$

The metric pairing it then found (using the given metric $\mathbb{R}^{1,3}$) as:

$\langle n,n\rangle = 4(t^2-x^2-y^2-z^2) = 4$ (by definition of the hypersurface)

It is then stated that, since $\langle n,n\rangle>0$, we subtract one + from the signature which means the new signature is:

(-,-,-,-) (from (+,-,-,-))

Could someone explain:

  1. How to calculate a normal vector generally
  2. How the sign of $\langle n,n\rangle$ informs the change in signature generally

Thank you

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1 Answer 1

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Your gradient is computed incorrectly since you didn’t take the Lorentzian signature into account. It should be \begin{align} n &= 2t\frac{\partial}{\partial t}+2x\frac{\partial}{\partial x}+2y\frac{\partial}{\partial y}+2z\frac{\partial}{\partial z}. \end{align} Of course, this still implies $g(n,n)=4(t^2-x^2-y^2-z^2)=4>0$, which means the normal vector is timelike, and hence the hypersurface is spacelike, which according to your convention means it has signature $(-,-,-)$.

For your first question, if $(M,g)$ is a Lorentzian manifold and $f:M\to\Bbb{R}$ is a smooth function with $a\in\Bbb{R}$ a regular value, then the level set $\mathcal{H}:=f^{-1}(\{a\})$ is a smooth hypersurface with tangent spaces given by the kernel of $df$; hence the vector field given by the metric-gradient of the function, $\text{grad}_g(f)=g^{\sharp}(df)$ (which in terms of a local coordinate chart $(U,x)$ intersecting $\mathcal{H}$, is equal to $g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$) is nowhere vanishing on $\mathcal{H}$ and normal to it.

Next, for Lorentzian metrics $g$ with signature $(+,\underbrace{-,\dots, -}_{\text{$d$ times}})$, if $\mathcal{H}$ is any smooth hypersurface (whether or not it is described as the level set of a smooth function) with $n$ as a nowhere-vanishing normal vector field, then we have:

  • if $g(n,n)>0$ (a timelike normal), then $\mathcal{H}$ has signature $(\underbrace{-,\dots, -}_{\text{$d$ times}})$ (a spacelike hypersurface… i.e the pullback of $g$ to $\mathcal{H}$ is minus a Riemannian metric)
  • if $g(n,n)=0$ (a null normal), then $n$ is in fact tangent to the hypersurface, so $\mathcal{H}$ is null (in particular the pullback of the metric to $\mathcal{H}$ is degenerate). This is probably the most unintuitive fact because it is a feature of Lorentzian geometry alone, absent in Euclidean/Riemannian geometry.
  • if $g(n,n)<0$ (i.e a spacelike normal), then $\mathcal{H}$ has signature $(+,\underbrace{-,\dots, -}_{\text{$d-1$ times}})$ (i.e a timelike hypersurface).
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