Why is it problematic to regard the Lorentz group as ${\rm SO}(4, \mathbb{C})$?

Question

If the four-vector $x^\mu$ is defined as $x^\mu\equiv(ict,{\bf x})$, instead of $x^\mu\equiv (ct,{\bf x})$, the Lorentz group will be the compact(?) ${\rm SO}(4, \mathbb{C})$ group. But the Lorentz group is regarded as the noncompact group ${\rm SO}(3,1)$. But I could never figure out what is the real problem of using $ict$ instead of $ct$? In short, what will go wrong if I choose to work with $ict$? Does it pose a problem from the point of view of representation theory? I am only interested in flat spacetime.

Comment to the post (v2): Do you mean ${\rm SO}(4, \mathbb{R})$ rather than ${\rm SO}(4, \mathbb{C})$? — Qmechanic, Jun 08 '20 at 06:06
...but the entries of ${\rm SO(4)}$ will not all be real if we define $x^\mu\equiv (ict, {\bf x})$. Right? @Qmechanic — SRS, Jun 08 '20 at 06:23
@SRS You can act on complex-valued vectors with real-valued matrices. — J. Murray, Jun 08 '20 at 06:45
But why do you think, in this case, the ${\rm SO(4)}$ matrices will be real-valued? For example, the boost along x-axis will be $\begin{pmatrix}\gamma & i\gamma\beta & 0 & 0\i\gamma\beta & \gamma & 0 & 0\0 & 0 & 1 & 0\0 & 0 & 0 & 1\end{pmatrix}$. @J.Murray — SRS, Jun 08 '20 at 06:52
@SRS I meant only to say that $SO(4,\mathbb R)$ can act on complex-valued matrices. However, note that the matrix you posted in your comment is not in $SO(4,\mathbb C)$ - for that you'd need a minus sign in the (1,0) entry or the (0,1) entry. — J. Murray, Jun 08 '20 at 07:11
Does this answer your question? Special relativity and imaginary coefficient of the time coordinate — J. Murray, Jun 08 '20 at 07:13

score 1 · Answer 1 · answered Jun 08 '20 at 06:09

1

${\rm SO}(4, \mathbb{C})$ allows all four coordinates of spacetime to be complex. It doesn’t just allow time to be imaginary.

answered Jun 08 '20 at 06:09

G. Smith

51,534

I don't understand the answer. What stops you defining $x^\mu\equiv (ct, {\bf x})$? Then the matrices acting on it which preserve $ds^2=-({\bf x}^2+i^2c^2t^2)$, are $4\times 4$ complex orthogonal. Right? @G.Smith – SRS Jun 08 '20 at 07:00
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Right. My point is that this group is too large, so its representations have no physical significance. For example, you don’t want to allow transformations that preserve $x^2+y^2-z^2-t^2$. – G. Smith Jun 08 '20 at 07:12
Are you saying that all ${\rm SO(4, \mathbb{C})}$ matrices do not preserve $x^2+y^2+z^2-c^2t^2$ while all ${\rm SO(3,1)}$ matrices do? @G.Smith – SRS Jun 08 '20 at 07:39
No, I didn’t say that. – G. Smith Jun 08 '20 at 16:23

score 1 · Answer 2 · answered Jun 08 '20 at 07:38

Misner, Thorne, & Wheeler (MTW) offer arguments in
“Farewell to ict" on Gravitation, p.51.

Reasons for using ict:

It makes spacetime geometry look like Euclidean geometry.
It make a Lorentz transformation look like a rotation.
It allows one to avoid distinguishing components of a vector from its metric-dual one-form.

Reasons NOT to use ict:

A vector is a very different geometric object from a one-form.
The Euclidean angle is periodic, whereas the Minkowski-angle, better known as the rapidity ("velocity parameter"), increases monotonically without bound.
Hiding the Lorentzian signature (- + + +) hides the light-cones that encode the causal structure.
No one has discovered a way to use this in general relativity for a general curved spacetime manifold.
and thus conclude:
"If '$x^4 = ict$' cannot be used there, it will not be used here" [in this book Gravitation].

See page 19 of this 20-page excerpt at http://laplace.physics.ubc.ca/000-People-matt/200/gravitation.pdf.

In my opinion, disadvantage #3 concerning the causal structure is the most important reason not to use it.

Why is it problematic to regard the Lorentz group as ${\rm SO}(4, \mathbb{C})$?

2 Answers2