# Trying to understand how the metric is formed from basis vectors

In the line element

$ds=\frac{\partial s}{\partial x^1} dx^1+\frac{\partial s}{\partial x^2} dx^2$

(superscripts are indices, not powers)

the basis vectors are defined as $e_1=\frac{\partial s}{\partial x^1}$ and $e_2=\frac{\partial s}{\partial x^2}$

The metric is then said to be obtained by multiplying these basis vectors together in each of their possible combinations. In this case there are two basis vectors so there will be four elements in the metric:

$e_1e_1=g_{11}$

$e_1e_2=g_{12}$

$e_2e_1=g_{21}$

$e_2e_2=g_{22}$

$\implies g=\begin{bmatrix} g_{11} &g_{12} \\ g_{21}& g_{22} \end{bmatrix} = \begin{bmatrix} (e_1)^2 & e_1e_2 \\ e_1e_2& (e_2)^2 \end{bmatrix}$

This much I follow. Now, as I understand it, the metric is a diagonal matrix. Take the Euclidian metric for eg.

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

So in what way does $e_1e_2$ equal $0$?

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I don't understand "in what way does $e_1e_2$ equal $0$?". In what sorts of different ways can things equal $0$? Perhaps you mean "why"? – joriki Nov 24 '11 at 13:44
I mean I don't see how $\frac{\partial s}{\partial x^1}\frac{\partial s}{\partial x^2}=0$, which is what has to happen for the metric to be a diagonal matrix. – ben Nov 24 '11 at 14:00
Why do you insist on writing "how"? Do you really mean "how"? If so, please explain what that means. If you mean "why", that would be clearer. – joriki Nov 24 '11 at 14:04