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I am a physics student confused with the notion of null hypersurface, so sorry if this question is very simple.

Given a manifold $M$ and a hypersurface $H$ defined on it, we can always take the hypersurface to be locally equal to a level set of a scalar function $f$ (I think). Given a certain point $P$, we can thus consider the vector $g^{ab}\nabla_{b} f$ and notice it is orthogonal to the subspace $T_{P}H$ of the tangent space $T_{P}M$. $H$ is said to be a null hypersurface if and only if $g^{ab}\nabla_{b} f$ is tangent to $H$.

Now, I can understand that, if $g^{ab}\nabla_{b} f$ is tangent to $H$, then, since $g^{ab}\nabla_{b} f$ is orthogonal to $T_{P}H$, $g^{ab}\nabla_{b} f$ has zero length. Thus, the metric on $T_{P}H$ is degenerate and the name "null hypersurface" is appropriate.

What is not clear to me is that even if $g^{ab}\nabla_{b} f$ has non null length, it might be possible to engineer a situation in which the metric on $T_{P}H$ is still degenerate. It would make sense to call $H$ a null hypersurface in that case. So, is this situation possible?

LeastSquare
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1 Answers1

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This is not possible. If $L$ is a null hypersurfce of a Lorentzian manifold $M$, then $TL^{\perp}$ is a one-dimensional null distribution contained in $TL$ ($TL$ is the tangent bundle).

If $f$ is a function such that (al least locally) $f^{-1}(c)=L$, then $g(v,\nabla f)=0$ for all $v\in TL$ and thus $\nabla f\in TL^{\perp}$ and $\nabla f$ is a null vector field.

benji
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  • Thanks for the answer and sorry for the delay. I am not very familiar with bundles, but isn't your first statement precisely what I'm asking to prove? – LeastSquare Aug 16 '14 at 11:49
  • In retrospect, my question was really simple and this answer is right. Thanks. – LeastSquare Sep 18 '14 at 14:23