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In french wikipedia of 4-vector :

https://fr.wikipedia.org/wiki/Quadrivecteur#De_la_base_covariante_aux_quadrivecteurs_covariants

The description starts with an explanation, in context of special relativity, for the dot product : (let's call this part (A)) : " $<u;v> = \sum_{i,j=0}^{3} u_i.v^j <e^i,e_j> = \sum_{i,j=0}^{3} u_i.v^j.\delta^i_j = \sum_{i=0}^{3} u_i.v^i = u_i.v^i$ "

Then, they describe the case of general relativity and they state that it is the same method. They write (let's call this part (B)) :

" $<u;v> = \sum_{i,j=0}^{3} g_{ij}.u^i.v^j = g_{ij}.u^i.v^j$

where the metric tensor $g_{ij}$ has been introduced. $<e^i;e_j> = \delta^i_j$

They state also $e^i = g^{ij}e_j$ and $u_i = g_{ij}u^j$

"

Now let's analyse : let's start again from A which states : $<u;v> = \sum_{i,j=0}^{3} u_i.v^j.\delta^i_j$

This is different to (B) : $<u;v> = \sum_{i,j=0}^{3} g_{ij}.u^i.v^j$

since $g_{ij}$ is not equal to $\delta^i_j$.

Is there a contradiction in the description of the 4-vector french page of wikipedia ?

What is missing to make the agreement between the two parts ? (the metric tensor do exist in special relativity as well)

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You say that the two are different because $g_{ij} \neq \delta^i_j$, but this is because they are different that $A$ and $B$ are the same : $u_i = g_{ik}u^k$. Then : \begin{equation} u_i v^j \delta^i_j = g_{ik}u^k v^j \delta^i_j = u^k v^j \times (g_{ik}\delta^i_j) = u^k v^j \times g_{ij} \end{equation}

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  • $\begingroup$ A huge thanks for your great help $\endgroup$ – Mathieu Krisztian May 15 at 19:05

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