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If we take a spacetime with one spatial dimension, we can write a vector as $A^\mu=(t, x)$. This is a contravariant vector, and we can calculate the covariant vector by multiplying it with the Minkowski metric: $$ A_\mu= \begin{bmatrix} 1& 0\\0&-1 \end{bmatrix} A^\mu =\begin{bmatrix} t\\-x \end{bmatrix} $$

(I'm not sure if this is notationally valid, since I might be mixing up tensor notation and matrix notation, but you get the idea. )

We can then calculate the proper distance using the einstein summation convention: $A^\mu A_\mu$.

We can see that coindidentally, the conjugate of a complex number transforms in the same way. If we have a complex number written as a $2\times 1$ matrix $\alpha = \begin{bmatrix} a\\bi \end{bmatrix}$

Then the complex conjugate of $\alpha$ is found by

$$ \alpha^*= \begin{bmatrix} 1& 0\\0&-1 \end{bmatrix} \alpha $$

So we see that the relation between contravariant and covariant vectors in spacetime is similar to the relation between complex conjugates.

Is there a deeper connection between the two, or is this just a mathematical coincidence with no significance?

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  • $\begingroup$ The essence of Minkowski metric is the invariance of the speed of light. How to connect this to the conjugate of a complex number it is arduous. $\endgroup$
    – Michele Grosso
    Mar 20, 2018 at 13:28
  • $\begingroup$ I think they're formally identical. Both $A^\mu$ and $\alpha*$ can be thought of as covectors, i.e., as linear operators that act on a vector and give a real result. Isn't this equivalent to the old-fashioned custom of writing four-vectors as $(ict,x,y,z)$? BTW, you also get some expressions when you deal with quaternions that look a lot like the metric and inner products in 3+1 dimensions. $\endgroup$
    – user4552
    Mar 20, 2018 at 17:32
  • $\begingroup$ I think you have some typos in your post. You seem to have rediscovered the old fashioned way of describing the "proper distance" without using a metric (or with 4-dim Euclidean inner product). Not sure if there is a "deeper meaning" to this. We abandoned that method 100 years ago or so. $\endgroup$
    – user196418
    May 25, 2018 at 1:57

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