Recent content by isaacdl

Yeah totally understood that part, but what I said in the previous post is that I don't know if g in S is non-degenerated. It's not given in the exercise. PD: Wait, u are totally right, we can use it, it's given in a previous part of the exercise! Thanks a lot.

Yes, I can see it graphically, but my problem is to show it in a free cordinate way.

Ok, I was using proof by contradiction wrong. Ok, if ##g(\mathbf{u},\mathbf{w}) = 0## for all ##\mathbf{w} \in \mathbf{R}^4##, by non-degenacy condition if S is non-degenerated u is a null vector u=0. But I'm not sure if the space S or equiv, g | S is non-degenerate. The only info I have is that...

Mmm, I think I have to use the fact w and u are linearly independent. My first idea was to use the properties of bilinearity of the metric. So by contradiction you mean to consider g(u,v)=0 and get to u is not lightlike?

In a coordinates it's straight forward. If u=(1,1,0,0) and w=(1,0,0,0) timelike or same to ligthlike, it's not difficult to prove they are colinear. But I don't know how to traduce it in a free coordinate way.

I'm trying to prove that there exist always a vector w whose contraction with a lightlike vector u (g(u,u)=0) it's always different from zero: $g(u,v)≠0$I know how to do this with coordinates, but in a free cordinate scheme I'm totally lost. Any help? PD: Both vectors are linearly independent.