It is a long time that I am looking for an abstract definition of four-vectors. This is the definition that I have reach to so far:

A four-vector is an element of the representation space of the Lorentz group whenever the representation space is chosen to be $\mathbb R^{4}$.

Following this definition, however, every vector in $\mathbb R^{4}$ is to be identified a four-vector. Why? because the group action is a total function from $\Lambda \times \mathbb R^{4}$ to $\mathbb R^{4}$ ($\Lambda$ here denotes the Lorentz group) and a total function covers its domain. Hence, every vector in $\mathbb R^{4}$ is a member of our representation space.

Am I right?


1 Answer 1


95% correct, let me add some elements.

We define the "mathematical Minkowski spacetime" as the vector space $\mathbb R^4$ endowded with a bilinear form we denote by $\eta$ ("mostly minus") with the property that for whatever 4-tuples $x, y\in \mathbb R^4$

$$ \eta(x,y) =: x^0 y^0 - x^1 y^1 -x^2 y^2 - x^3 y^3 $$

We call "a Minkowski 4-vector" a generic element of the "mathematical Minkowski spacetime" on which the fundamental representation of the full Lorentz (1,3) group is defined.

  • $\begingroup$ Hello and thank you very much for your response. I agree with you; however, to my eyes that bilinear form is an additional piece of information. In defining the vector space and the group action we do not depend on the definition of the bilinear form. Hence, four-vectors defined as the elements of the representation space (here $\mathbb R^{4}$), would have to be defined irrespective of this bilinear form. Do you agree? $\endgroup$ Jun 14 at 19:31
  • $\begingroup$ @AliKoohpaee Additional but essential. By 4-vectors, physicists usually mean "Minkowski spacetime 4-vectors", thus the bilinear form is essential. $\endgroup$
    – DanielC
    Jun 16 at 7:38
  • $\begingroup$ Yes I see. It is from the mathematical point of view that this piece of information is additional. $\endgroup$ Jun 16 at 8:43

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