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My latest comment:

Minkowski did not suggest CxR3. He used one (imaginary) dimension for t, & 3 real dims for space. There is no reason to think this is merely a trick. When a math representation works in physics, it is naïve to think it's just a 'trick'. What works is correct. The history of math & physics is full of ideas which were originally thought just to be a useful trick (e.g., imaginary numbers), but such ideas usually turn out to be profound reps of the truth, if they give correct results. There is no reason to think that space-time can't be a 4-dim subspace of an 8-dim complex space-time, even though we haven't yet been able to attribute meaning to the other 4 dims. Penrose thinks there must be a meaning to them, because complex numbers usually turn out to be useful in modern physics.

I read that Minkowski's model of space-time, in which one of these qties is imaginary so as to make the geometry Euclidean, cannot be extended to G.R. Where can I find a proof of this?

You seem to have understood my question perfectly, and answered it by pointing me to Visser's paper on how to adapt Minkowski's t and ix coordinates to general relativity. (Minkowski's paper used t and ix, unlike the relativists of the following decades, who switched to it and x.) For SR, his notation is fine, and better than it & x, as it makes the imaginary angle of rotation positive. No one seems to have noticed that the it & x notation gives a negative angle. His notation also corresponds to the quaternion representation of space-time, in which the spatial dimensions are imaginary, not the time dimension. Much has been done with the quaternion representation since then in physics. Penrose still believes there should be a use for fully complex (8-dim.) space-time (see Road to Reality). We can't avoid the t & ix notation, as imaginary coordinates come up again in black holes, and the usual treatment there behaves in such a way as to justify Minkowski'.

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closed as unclear what you're asking by ACuriousMind Jan 14 '17 at 14:19

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    $\begingroup$ It is not clear what you are asking. Minkovski spacetime is not $\mathbb{C} \times \mathbb{R}^3$, ie sequences $(it, \mathbf{x})$. It is $\mathbb{R}^4$ endowed with the pseudo-riemannian metric $\eta = \mathrm{diag}(-1,+1,+1,+1)$ so that the inner product of two vectors is $A\cdot B = -A^0B^0 + A^1B^1 + A^2B^2 + A^3B^3$. One can get rid of the minus sign with a `Wick rotation', that is the formal substitution $t \leftrightarrow i\tau$ where both $t$ and $\tau$ are real, getting $\mathbb{R}^4$ with usual euclidean metric. For Wick rotation in GR see physics.stackexchange.com/q/131355 $\endgroup$ – yoric Jan 14 '17 at 12:29
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    $\begingroup$ @yoric: in the past, it was seen as convenient to use $x_0=it$ and $x^0=-it$. Completely nonstandard these days, obviously, but it's possible OP is seeing this notation. $\endgroup$ – Kyle Kanos Jan 14 '17 at 16:43
  • $\begingroup$ @KyleKanos Fine with that, I was aware of it, but I do not still get the question. $\endgroup$ – yoric Jan 14 '17 at 16:47
  • $\begingroup$ @yoric I concur, the question makes little sense. $\endgroup$ – Kyle Kanos Jan 14 '17 at 16:49
  • $\begingroup$ You seem to have understood my question perfectly, and answered it by pointing me to Visser's paper on how to adapt Minkowski's t and ix coordinates to general relativity. (Minkowski's paper used t and ix, unlike the relativists of the following decades, who switched to it and x. For SR, his notation is fine, and better than it & x, as it makes the imaginary angle of rotation positive. No one seems to have noticed that the it & x notation gives a negative angle. $\endgroup$ – murray denofsky Jan 16 '17 at 2:42