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.

  • $\begingroup$ Comment to the post (v2): Do you mean ${\rm SO}(4, \mathbb{R})$ rather than ${\rm SO}(4, \mathbb{C})$? $\endgroup$
    – Qmechanic
    Jun 8, 2020 at 6:06
  • $\begingroup$ ...but the entries of ${\rm SO(4)}$ will not all be real if we define $x^\mu\equiv (ict, {\bf x})$. Right? @Qmechanic $\endgroup$
    – SRS
    Jun 8, 2020 at 6:23
  • $\begingroup$ @SRS You can act on complex-valued vectors with real-valued matrices. $\endgroup$
    – J. Murray
    Jun 8, 2020 at 6:45
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Jun 8, 2020 at 6:52
  • $\begingroup$ @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. $\endgroup$
    – J. Murray
    Jun 8, 2020 at 7:11

2 Answers 2


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

  • $\begingroup$ 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 $\endgroup$
    – SRS
    Jun 8, 2020 at 7:00
  • 1
    $\begingroup$ 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 $x^2+y^2-z^2-t^2$. $\endgroup$
    – G. Smith
    Jun 8, 2020 at 7:12
  • $\begingroup$ 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 $\endgroup$
    – SRS
    Jun 8, 2020 at 7:39
  • $\begingroup$ No, I didn’t say that. $\endgroup$
    – G. Smith
    Jun 8, 2020 at 16:23

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

Reasons for using ict:

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

Reasons NOT to use ict:

  1. A vector is a very different geometric object from a one-form.
  2. The Euclidean angle is periodic, whereas the Minkowski-angle, better known as the rapidity ("velocity parameter"), increases monotonically without bound.
  3. Hiding the Lorentzian signature (- + + +) hides the light-cones that encode the causal structure.
  4. 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.


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