Cross product in E_u from a 4-form via metric duality

[aclaret]

I study from Gourgoulhon's text 'special relativity in general frames', I have some difficulty to understanding Chapter 3 Page 84. I already learn that there exist a orthogonal projection mapping $\bot_{u}:E \rightarrow E_u(P)$ from the vector space $E \cong R^4$ to the subspace $E_u(P)$ associated with local rest space $\mathscr{E}_u(P)$ of the observer at event $P$.

Now want to proof the proposition (3.37), that given timelike $u \in E$ and antisymmetric bilinear form $A$, there exist unique form $q = A(., u) \in E^*$ and unique vector $b \in E$ such that $A = u \otimes q - q \otimes u + \epsilon(u, b, \dots)$. During proof author writes "By metric duality, $\epsilon_u$ induces the cross product of two vectors of $E_u$ by $$\forall (v, w) \in {E_u}^2, \quad v \times_u w := \epsilon_u (v, w, \dots) = \epsilon(u, v, w, \dots)$$where $\epsilon_u(v,w \dots)$ stands for vector of $E_u$ associated by $g$-duality to the linear form $E_u \rightarrow R$, $z \mapsto \epsilon(v, w, z)$... [and] $\varepsilon(u,v,w, \dots)$ stands for vector in $E$ that is $g$-dual of the linear form $E \rightarrow R, z \mapsto \epsilon(u,v,w,z)$"

I don't understand this part, please somebody can please explain how exactly this induces a cross product? (I do undertand what author mean by metric duality, that is simply the map $\Phi_g$ associating any $u \in E$ to a one-form $\tilde{u} \in E^*$ such that satisfy $\langle \tilde{u}, v \rangle = g(u,v)$ for all $v \in E$, but I don't understand how it relate to the concept above).

 

[martinbn]

Not sure which part is unclear to you. In 3D space there is one trilinear antisymmetric form (up to a constant multiple). If you feed two vectors into it, you get a one form. By metric duality it gives you a vector.

 

[aclaret]

Thank yes I did now understand, what confuse me is that it look on paper like the object $\epsilon_u(v,w, \dots)$ is a oneform $E_u \rightarrow R$ (LV tensor with a one unfilled slot for a vector), but author instead mean that this object above is g-dual $(\in E_u$) of what I thinking before. so I simply was imagining the isomorphism wrong way round in my brain ;) ;)

apologise for trivial question :), thank @martinbn

 

[martinbn]

No need for apology. It is worded in an unusual way. It is easy to loose track of the notations and not see the forest because of the trees.

 

[Perezsv]

For me In 3D space there is one trilinear antisymmetric form