Contraction between basis vectors and basis one-forms

Question

Discretion: The title may be misleading, because I am not certain whether the one-forms are actually basis one-forms.

I always thought by definition, $dx^i (e_j) =\delta^i_j $. But, I am confused because of what it says on one of the book I am reading, "Mathematical Methods of Classical Mechanics", by Arnold. In page 178 - 179, he says if we define the metric $g = diag(E_1,E_2,E_3)$, then $dx^i (e_i) = 1/\sqrt E_i$.

I cannot wrap my head around this. Is $dx^i$ the dual basis one-form, or is it something else? Thank you so much for your help.

I think this is probably a better fit for mathstackexchange? — , Apr 16 '14 at 06:03

score 2 · Accepted Answer · answered Apr 16 '14 at 06:34

To any coordinate system on a manifold (Riemannian or not) there is associated a basis for the tangent space usually denoted by $\{ X_j \}$ or $\{ \partial/\partial x^j \}$. For these basis vectors, it's always true that $dx^i(X_j) = \delta^i_j$. However, if you have a metric, it's usually not the case that the vectors $X_j$ form an ON basis with respect to this metric. In the case that Arnold considers, the metric tensor is assumed to be diagonal, so the $X_j$ are pairwise orthogonal, but they are not normalized, and the unit vectors $e_j$ are what you get when you normalize them: $e_j = X_j \,/ \sqrt{E_j}$.

Thank you sincerely for your answer, they make perfect sense! — Quantization, Apr 16 '14 at 06:53

Contraction between basis vectors and basis one-forms

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