davi2686 said:
My friend you REALLY help me a LOT, with your explanations several books that i have for some time make much more sense to me now, i really thank you.
From what I've read, i may have been wrong but i think that notation \displaystyle x'î=\frac{\partial x'î}{\partial x_j}x_j is something about cartan spaces and generalisation of euclidean spaces, that's make sense or i completely wrong about this?
I don't really know about "Cartan spaces" very much, but the tensor transformation law is simply the definition of what a co variant or contra variant vector on any differentiable manifold is. As I mentioned in my last post, it is a practical definition that obscures the geometric meaning of a vector.
Let's look at how a vector is defined in the "new" way (this is a contra-variant vector). Basically a vector on a manifold is the tangent vector to curves that pass through a point P. Because the vector is tangent to curves, they lie in a space called the tangent space at the point P (##T_P##). Let's look at how it is constructed:
Let's consider a curve ##\Lambda:\mathbb{R}\rightarrow \mathcal{M}## which is a map of (an interval of) the real line into the manifold which passes through a point ##P## on the manifold. In some local coordinates, this curve has the parametric equations ##\{x^i(\lambda)\}##. We define an arbitrary differentiable function ##f:\mathcal{M}\rightarrow \mathbb{R}## which maps points on the manifold to some real number (assigns a real number to each point on the manifold, this should be intuitive). This function will take on the values ##f(x^i(\lambda))## along our curve. At the point ##P## we can find the derivative of a function using the chain rule:
$$\left.\frac{df}{d\lambda}\right|_P=\left.\frac{dx^i}{d\lambda} \frac{\partial f}{\partial x^i}\right|_P$$
Since this is true for all arbitrary differentiable ##f## we can writ the derivative as:
$$\left.\frac{d}{d\lambda}\right|_P=\left.\frac{dx^i}{d\lambda} \frac{\partial }{\partial x^i}\right|_P$$
We can show that these derivative operators ##\left.\frac{d}{d\lambda}\right|_P## form a vector space because they obey all of the axioms of a vector space. I will not prove rigorously here that they do, but I will motivate an intuitive explanation. Consider that these operators are in 1-1 correspondence with curves that pass through ##P## where a curve with a different parametrization is considered a different curve. Changing the length of these vectors is equivalent to changing the parametrization, changing the direction is equivalent to changing the curves, so you would intuitively expect that these differential operators form a vector space. The only non-trivial property of vector spaces that we should prove is the property of closure under linear combinations. Consider a second curve ##\text{M} :\mathbb{R}\rightarrow \mathcal{M}## which have coordinates ##\{\tilde{x}^i(\mu)\}##. The tangent vector to this curve can be expressed in our coordinate system as:
$$\left.\frac{d}{d\mu}\right|_P=\left.\frac{d\tilde{x}^i}{d\mu} \frac{\partial }{\partial \tilde{x}^i}\right|_P=\left. \frac{dx^i}{d\mu} \frac{\partial }{\partial x^i}\right|_P$$
Where the last equality comes from the fact that we are evaluating at point P for which ##x^i(P)=\tilde{x}^i(P)##. So we look at a linear combination of these two vectors (with the evaluation at point ##P## implicit):
$$a\frac{d}{d\lambda}+b\frac{d}{d\mu}=\left(a \frac{dx^i}{d\lambda}+b\frac{dx^i}{d\mu}\right) \frac{\partial}{\partial x^i}$$
Now there must be some curve ##\Phi:\mathbb{R}\rightarrow \mathcal{M}## with parametrization ##\{\bar{x}^i(\phi)\}## passing through the point ##P## such that:
$$\frac{dx^i}{d\phi}=a\frac{dx^i}{d\lambda}+b\frac{dx^i}{d\mu}$$
So that we know therefore:
$$\frac{d}{d\phi}=a\frac{d}{d\lambda}+b\frac{d}{d\mu}$$
Proving closure under linear combinations. We therefore make the claim (definition) that "contra variant vectors are isomorphic (really just are) to derivative operators at point ##P##". We can finally now look at how the coordinates of these contra variant vectors change under a change of coordinates, and we can show that they do indeed transform as contra variant vectors according to the change of basis law. We have a new set of coordinates ##x^{i'}=x^{i'}(x^i)## (don't get this confused with the tilde and bar versions of ##x^i## earlier, those were literally the coordinates of different curves, this is the coordinates of the same curve but in a different coordinate system). Now, in new or old coordinates the vector doesn't change, we can apply the chain rule with this new coordinate system to obtain:
$$\frac{d}{d\lambda}=\frac{dx^i}{d\lambda} \frac{\partial}{\partial x^i}=\frac{dx^{i'}}{d\lambda}\frac{\partial}{\partial x^{i'}}$$
We also have that the chain rule says:
$$\frac{\partial}{\partial x^{i'}}=\frac{\partial x^j}{\partial x^{i'}} \frac{\partial}{\partial x^j}$$
In order for the above equality (involving the ##\lambda##) to hold it must then be that:
$$\frac{dx^{i'}}{d\lambda}=\frac{\partial x^{i'}}{\partial x^j} \frac{dx^j}{d\lambda}$$
Which is exactly the component transformation rule given:
$$A^i\equiv\frac{dx^i}{d\lambda}$$
One can then prove that one forms, which are functions of contra variant vectors into scalars will transform "oppositely". I will not provide the proof here as this post is already getting too long lol.
