0

I have a confusion: First we define the Lorentzian manifold.

A Lorentzian manifold is pair $(M^{n+1}, g)$ where $M^{n+1}$ is a smooth manifold of dimension $n+1$ and $g$ is a Lorentz metric, which assigns for each point $p\in M$ a non degenerate symmetric bilinear form of index $(n,1)$ on $T_pM$.

Now my question is: So, we get a non-degenerate symmetric bilinear form on $T_pM$ of index $(n,1)$. My question is, can I write the matrix of the bilinear form as $$B = \begin{pmatrix} 1 &\\ & 1&\\ & &\ddots &\\ & & & -1\end{pmatrix}?$$ I think that I can write above by the law of inertia. But I am not sure. Can anyone please help me?

  • Your question is unclear: Are you asking if this can be done at one point $p\in M$ by choosing an appropriate basis in $T_pM$? Then this is just linear algebra. Or, maybe, you are asking about a trivialization of the tangent bundle of the entire manifold, i.e. $n+1$ smooth vector fields $X_0,...,X_n$ on $M$ such that at every point $p\in M$ the metric $g$ has the above form with respect to the basis $X_0(p),...X_n(p)$ in $T_pM$. This would be impossible. – Moishe Kohan Nov 20 '24 at 08:57
  • @MoisheKohan Sorry for the cofusion. I asked the first question you mentioned. –  Nov 20 '24 at 09:23
  • Take a look at the answer and references here. – Moishe Kohan Nov 20 '24 at 12:42

0 Answers0