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

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

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