Vector product in a 4-dimensional Minkowski spacetime I'm studying relativity and I lost track of interpretation along the mathematical formalism.
What does vector product mean as an event? I mean, how must one interpret the result of the vector product of two (say time-like for closeness to the common sense) vectors in a 4-dim Minkowski spacetime?
 A: In ordinary vector spaces, the dot product $\cdot$ is a binary operator which takes a pair of vectors $(A,B)$ in the space to the field over which the space is defined. Formally, for a vector space $V$ over a field $K$, the dot product $(\ \ , \ )$ is a bilinear map
$$(\ \ , \ ): V \times V \to K.$$
The inner product only has assumes the standard meaning in certain vector spaces. In the case of Minkowski spacetime, the dot (or inner) product between two four-vectors $A$ and $B$ is
$$(A,B) = A^T \eta B,$$
where $\eta$ is the standard metric with signature $(-, +, +, +)$ or $(+, -, -, -)$. In conventional Einstein summation notation, this is written as
$$(A, B) = \eta_{\mu \nu}A^\mu B^\nu$$
How do we interpret this operation? Well, we cannot use the standard Euclidean notions of distance or direction since we are dealing with hyperbolic space. Instead, it is better to view the product as a Lorentz-invariant quantity that describes the (hyperbolic) geometric relationship between two vectors. That is, one that does not change under a Lorentz transformation $\Lambda \in SO(1,3).$
A: I know that it is maybe too late to respond but I found the question on cross-product of four-vectors in Minkowski spacetime is very interesting.
The answer is actually given in "Lecture Notes on General Relativity" by Sean M. Carroll, on p 23. The generalization of the cross-product for spacetime could be viewed in terms of the Hodge dual of the wedge product. It gives the product that involves the Levi-Civita's symbol (not the metric tensor as per the answer above)
$$
*(A \wedge B)_i=\epsilon_i^{jk} A_jB_k
$$
Carroll concludes: "This is why the cross product only exists in three dimensions  — because only in three dimensions do we have an interesting map from two dual vectors to a third dual vector. If you wanted to you could define a map from n − 1 one-forms to a single one-form,
but I’m not sure it would be of any use."
PS. The situation is slightly different in pure math where we have Clifford algebra, Pin(p,q) and Spin(p,q) groups, and where vector norm can be $\pm 1$.
