Transformation law in curved space-time

[jfy4]

All diffeomorphisms preserve the norm. That is kind of the point of using tensors.

Is this true? The orthogonal group is a subgroup of the diffeomorphism group, so that group preserves the norm, but do all diffeomorphisms preserve the norm? it doesn't seem the converse is true. Tensors maintain the invariance of coordinates, not the norm right?

 

[atyy]

In a curved space, the metric does not act on "position" ie. coordinates. The metric acts on velocity.

 

[jfy4]

In a curved space, the metric does not act on "position" ie. coordinates. The metric acts on velocity.

In what context are you saying this? clearly this is a valid operation in GR, and a regular one:

$$ds^2=g_{ab}dx^{a}dx^{b}$$

This is the line element which is crucial in defining constant distance in space-time along with the notions of time-like, space-like, and null intervals. Here the metric is being used as the coefficients of the scalar product in a particular basis, and the product can certainly be between coordinates no?

 

[atyy]

In what context are you saying this? clearly this is a valid operation in GR, and a regular one:

$$ds^2=g_{ab}dx^{a}dx^{b}$$

This is the line element which is crucial in defining constant distance in space-time along with the notions of time-like, space-like, and null intervals. Here the metric is being used as the coefficients of the scalar product in a particular basis, and the product can certainly be between coordinates no?

The metric doesn't act directly on the coordinates to produce distance. Roughly speaking, it acts on the velocity vectors at each point to give their magnitude, and the distance is obtained by integrating the velocities over coordinates. It is the metric acting on the velocity vectors to give their magnitude that is coordinate-independent.

So the linear space is a tangent space at each point of the manifold. But different points have different linear spaces.

 

[jfy4]

Ok,

After a bit more reading here is what I have to report on discovering the group. We will take $$G\equiv g_{ab}$$. Now we are interested in the situation

$$B^{\top}GB=G$$

where $$B$$ is our transformation. These transformations form a group, and it has a name and a notation!: the orthogonal group with the "metric ground form", $$\mathcal{O}_{g}(p,q)$$. Now I quote a Lemma from Weyl's book:

"A non-exceptional transformation, B, [of the group mentioned above] may be written in the form

$$B=(E-T)(E+T)^{-1}$$

where T satisfies this condition

$$GT+T^{\top}G=0$$"

That is, T is anti-symmetric. I claim T is anti-symmetric since the indicies are raised and lowered by the metric, and E is the Identity.

Then our group of transformations are of the form, B.