User:Polnasam

Well, I signed in because I saw a clear case of vandalism in one page that, apparently, had been around for several months. As a regular user of WP, I felt the need to start contributing back, whence, signing in and removing that edit.

Any way, I think now I might stay around and see if I can really contribute with anything useful.

Metric tensor
This is a coordinate-free, algebraic characterization of a (pseudo-) metric tensor.

Dual space and forms
Let $$\mathbb{E}$$ denote a finite-dimensional linear space over a field $$\mathbf{F}$$ (e.g., $$\mathbb{R},\,\mathbb{C}$$), with vectors $$\mathbf{v}\,=\,v^\alpha\mathbf{e}_\alpha$$, where summation over repeated indices (Einstein convention) is assumed and the set $$\{\mathbf{e}_\alpha\}$$ is a basis of $$\mathbb{E}$$. The dual of the space, denoted as $$\mathbb{E^*}$$, is the vectorial space of Linear functionals or forms (see also one-form), denoted as $$\mathbf{f}\,=\,f_\alpha\mathbf{e}^\alpha$$, that map $$\mathbb{E}$$ into $$\mathbf{F}$$, i.e.,

$$\forall\,\mathbf{f}\,\in\,\mathbb{E^*}$$
 * $$\mathbf{f}\,:\,\mathbb{E}\longrightarrow \mathbf{F}$$
 * $$\mathbf{v}\mapsto \langle\,\mathbf{f}\,,\,\mathbf{v}\rangle_\mathbb{E}\,\equiv\,f_\alpha\,v^\alpha\,\in\,\mathbf{F}$$

where $$\langle\;,\;\rangle_\mathbb{E}$$ will be called the duality product in $$\mathbb{E}$$. $$\{\mathbf{e}^\alpha\}$$ is the basis of $$\mathbb{E^*}$$ called the dual basis of $$\{\mathbf{e}_\alpha\}$$ if it satisfies that


 * $$\langle\mathbf{e}^\alpha\,,\,\mathbf{e}_\beta\rangle\,=\,\delta^\alpha_\beta$$.

Polnasam (talk) 21:11, 24 June 2011 (UTC)