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How do I prove that for bases $g^1,g^2,g^3$ and $g_1,g_2,g_3$ is $g^j=G^{ij}g_i$ and $g_j=G_{ij}g^i$?

$g^1,g^2,g^3$ and $g_1,g_2,g_3$ are the bases of the tensor first tank.

$G_{ij}$ and $G^{ij}$ are the metric coefficients.

Can I prove it with using this rule:

The metric coefficients are symmetrical. $G_{ij} =G_{ji},G^{ij} = G^{ji}$. Vectors with respect to the bases $g_1,g_2,g_3$ and $g^1,g^2,g^3$ according to $x = x^ig_i = x_ig^i$ gives $x_j = x^iG_{ij} $ and $ x^j = x_iG^{ij}$. This is called the raising and lowering of indices in tensor analysis.

If yes, than why are metric coefficients symmetrical? Why is $x = x^ig_i = x_ig^i$?

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Your metric is defined as $G_{ij}= g_i g_j$ it does not matter how you multiply the $g_i g_j =g_j g_i $. On the other had the $g$'s with upper index are covectors. And these co-vectors are also special, because they are the dual of down-index $g$'s and the rule for vector and dual is $g^j g_i = \delta^j_i$ if the basis is of normal length.

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  • $\begingroup$ So if I want to show this I just write $G_{ij}=g_ig_j$, $g_j=G_{ij}g^i=g_ig_jg^i=g_j$ or? $\endgroup$
    – Maica
    Jul 6, 2017 at 14:03
  • $\begingroup$ exactly like this $\endgroup$ Jul 6, 2017 at 17:35

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