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Take Wick rotatability being as the way defined in the following article by Helleland and Hervik:

Is the Gödel universe globally and smoothly Wick rotatable according to this definition?

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    $\begingroup$ Note for logicians: The Gödel universe mentioned here is not Gödel's constructible universe $L$, but rather his model of the Einstein equations in physics, and GIT does not refer to the Gödel incompleteness theorem, but to geometric invariant theory. $\endgroup$
    – Joel David Hamkins
    Commented Jan 23 at 16:37
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    $\begingroup$ Now I want to read a Greg Egan novel where scientists discover at the same time that we live in Gödel relativistic universe and that our sets are in L. $\endgroup$
    – PseudoNeo
    Commented Jan 23 at 21:00
  • $\begingroup$ Hi @PseudoNeo, I think that if we lived in an "only slightly perturbed universe", as that you propose, our (idealized) brains would be well modeled by neural networks, thus have the computational power of TMs. Other dynamics too. All usual formal systems would be mutually representable -computationally equivalent to TMs. We could use the same mathematical theories and would need only very weak ones to do physics; customary foundations would be as debatable as here because Gödel's 2nd theorem would still hold, physics would not help discriminate, and our sociology would probably pick only ZFC. $\endgroup$
    – plm
    Commented Jan 25 at 16:33
  • $\begingroup$ Is this me who feels schizophrenic outta these comments or sth worse? $\endgroup$
    – Bastam Tajik
    Commented Jan 25 at 16:38
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    $\begingroup$ Closely related crosspost: physics.stackexchange.com/q/798544 (it is considered good practice to mention this in your question to avoid duplicate work) $\endgroup$
    – Jules Lamers
    Commented Feb 5 at 11:24

2 Answers 2

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$\DeclareMathOperator\SL{SL}$Clearly, as Robert Bryant indicates, it is Wick-rotatable to a different Lorentzian space. However, it is also Wick-rotatable to a Riemannian space, albeit negative definite space. In our paper these are considered equivalent due to the anti-symmetry $g\mapsto -g$.

The key here is to realise that the 3-dim space $z=\mathrm{constant}$ restricts to a Lorentzian metric on $\SL(2,\mathbb{R})$. Explicitly, first we do the coordinate changes $x=-\ln X$, $y=\sqrt{2}Y$, to transform the Gödel metric to: $$ g=\frac{1}{2\omega^2}\left(-\left(dt+\frac{\sqrt{2}dY}{X}\right)^2+\frac{1}{X^2}(dX^2+dY^2)+dz^2\right). $$ This is one of the standard forms of a metric on the LRS $\SL(2,\mathbb{R})$ space. I can't remember the the explicit coordinate transformation but this can again be rewritten (up to determining some appropriate constants $a,b$): $$ g=\frac{1}{2\omega^2}\left[-\left(dt+a\sinh(x_1)d x_2\right)^2+b^2((dx_1)^2+\cosh^2(x_1)(dx_2)^2)+dz^2\right]. $$ Now, the Wick rotation $(t,x_1,x_2,z)\mapsto (t,ix_1,ix_2,iz)$ turns it into: $$ g=\frac{1}{2\omega^2}\left[-\left(dt-a\sin(x_1)d x_2\right)^2-b^2((dx_1)^2+\cos^2(x_1)(dx_2)^2)-dz^2\right]. $$ Note that this is the negative-definite version of a (Berger sphere)${}\times \mathbb{R}$, which indeed has 5 Killing vectors, same as Gödel.

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  • $\begingroup$ What does it mean for a Riemannian metric to be negative definite? I thought Riemannian metrics are by definition positive definite, aren't they? $\endgroup$
    – Bastam Tajik
    Commented Feb 7 at 6:43
  • $\begingroup$ Sure, usually we define the Riemannian metric to be positive definite. But we could as well defined it to be negative-definite. This is just a convention. Especially, in Lorentizian geometry, both sign conventions are common: (+---) or (-+++). In neutral signature this is even more mixed up as they are (--++) or (++--). In any case, considering Wick rotations as we did in the paper, they are considered identical as they may be mixed up under a Wick rotation anyway. $\endgroup$
    – Sigbjørn Hervik
    Commented Feb 7 at 11:03
  • $\begingroup$ Thanks. Well I would ageee with you that the problem can be reduced to a matter of convention, if only there did not exist any positive-definite Riemannian slice. But if there exists at least one, then one faces the problem another way around and cannot escape from it. Do you think it's possible to prove such argument? $\endgroup$
    – Bastam Tajik
    Commented Feb 7 at 12:15
  • $\begingroup$ Moreover, along with my personal reason of interest in your article, I want to know why exactly (in each chosen and fixed frame of reference), time can not be complexified to result in a sign-definite(so to speak) Riemannian geometry, and apart from time (if at all!) also space should be complexified. What is exactly the nature of the obstruction for time to do the job alone? Is it geometric or topologic and manifests through geometry? To my guess the topological mismatch between the Alexandrov topology and the manifold topology might have some role though it's merely a guess. $\endgroup$
    – Bastam Tajik
    Commented Feb 7 at 14:44
  • $\begingroup$ If you consider a specific observer following a worldline with proper time t, then you'd get directly constraints on the metric functions. In particular, if the worldlines are non-twisting, you can foliate using the proper time t. Then again, if the metric is symmetric under the time-reversion, t--> -t, then it seems to be possible to Wick-rotate \tau=it. $\endgroup$
    – Sigbjørn Hervik
    Commented Feb 7 at 16:10
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I may be misreading the sources that you list for the definition of Wick-rotatable, but, I believe that the following construction does fit that definition: According to the Wikipedia page that the OP cites, the Gödel metric can be written on $\mathbb{R}^4$ with coordinates $(t,x,y,z)$ in the form $$ g = \frac1{2\omega^2}\left(-(\mathrm{d}t+ e^x\,\mathrm{d}y)^2+\mathrm{d}x^2 + \tfrac12(e^x\,\mathrm{d}y)^2+\mathrm{d}z^2\right). $$ Then writing $t=i\,\tau$ and $y = i\,\eta$, one gets a globally and smoothly Wick-rotated Lorentzian metric $$ \tilde g = \frac1{2\omega^2}\left((\mathrm{d}\tau+ e^x\,\mathrm{d}\eta)^2+\mathrm{d}x^2 - \tfrac12(e^x\,\mathrm{d}\eta)^2+\mathrm{d}z^2\right). $$ on $\mathbb{R}^4$ with coordinates $(\tau,x,\eta,z)$.

Note that the source cited does not require that one of the Wick-related metrics be positive definite. Perhaps more computation could determine whether there is a positive definite Wick-rotation of Gödel's metric.

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  • $\begingroup$ Right. But at the very least, one can say in this case not only time but also space needs to be complexified. Right? $\endgroup$
    – Bastam Tajik
    Commented Jan 27 at 13:04
  • $\begingroup$ @BastamTajik: Yes, but, remember that in the paper that you cite, the authors complexify everything in the sense that they first extend the metric to a holomorphic metric on $\mathbb{C}^4$ and then consider 'real slices' of dimension 4. $\endgroup$
    – Robert Bryant
    Commented Jan 27 at 13:42
  • $\begingroup$ True. Now I'm thinking of how to limit the paper approach to complex time and real space. Also your point on the positive definitness of the metric is quite pertinent. I'm super curious. Thanks $\endgroup$
    – Bastam Tajik
    Commented Jan 27 at 14:09
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    $\begingroup$ @BastamTajik: Note, however, that the whole point of Lorentzian geometry is that there is no natural separation of time and space. As Minkowski famously said, "Henceforth, space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality." $\endgroup$
    – Robert Bryant
    Commented Jan 27 at 15:09
  • $\begingroup$ Yes actually I noted this. The only question left for me for the moment is the positive-definitness of the Wick rotated metric. $\endgroup$
    – Bastam Tajik
    Commented Jan 27 at 15:11

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