Metric-Dual Vector Space

a metric is also used to raise/lower indices.

[itex] g_{\nu \mu } x^{\mu} = x_{\nu} [/itex]

[itex] g^{ \nu \mu} x_{\mu} = x^{\mu} [/itex]



In general a metric [with lower indices] is a map from [itex]V_{(1)} \times V_{(2)} \rightarrow \mathbb{R} [/itex]

whereas the upper indices are the map from [itex] V^{*}_{(3)} \times V^{*}_{(4)} \rightarrow \mathbb{R}[/itex]

I used the subscript to denote the vectors later I'm going to take, and the * means the dual space.

In this case then you have:

[itex]g(x_1, x_2) = s \in \mathbb{R}[/itex]

[itex]g(x_{3}, x_{4}) = s \in \mathbb{R}[/itex] , with [itex]x_{i} \in V_{(i)}^{(*)}[/itex]



Which means that [itex]g_{\mu \nu} x_{1}^{\mu} x_{2}^{\nu} = x_{1 \nu} x_{2}^{\nu}[/itex]

and also [itex] g^{\mu \nu} x_{3\mu} x_{4 \nu} = x^{\nu}_3 x_{4 \nu}[/itex]



Now if these are equal then it means that [itex]g(x_{1})=g(x_4)[/itex] or in other words the metric maps a vector in a vector space to its dual.

But isn't the dual space basis given by derivatives? So if [itex]x^{\mu}[/itex] is a vector, then [itex]x_{\mu}[/itex] should be written in terms of derivatives?