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

In french wikipedia of 4-vector :

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

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)) :

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

where the metric tensor $$g_{ij}$$ has been introduced. $$ = \delta^i_j$$

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 : $$ = \sum_{i,j=0}^{3} u_i.v^j.\delta^i_j$$

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

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)

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 : $$$$u_i v^j \delta^i_j = g_{ik}u^k v^j \delta^i_j = u^k v^j \times (g_{ik}\delta^i_j) = u^k v^j \times g_{ij}$$$$