Background
I've heard that it is possible to construct a Penrose triangle in the 3D geometry Nil. And I wondered: Can we build a Penrose triangle in the real world if spacetime is appropriately curved? It can be written as: Given the distributions of matter and fields properly, is it possible that a metric tensor $g_{\mu \nu}$ formed according to Einstein's equations exhibits a $3$-dimensional Nil geometry?
(I just want a 'possible' metric in reality according to Einstein's equations, so I assume that the distribution of matter, $T_{\mu \nu}$, is artificially manipulable 'if realistic'. In other words, the effect of the gravitational field on the distribution of matter again, or the geodesic equation is not taken into account.)
Since the time axis is arbitrarily added, it's not easy to know what the Nil geometry should look like in spacetime. So, I first thought of the following metric tensor: $$N_{\mu \nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -x^2 - 1 & x \\ 0 & 0 & x & -1 \end{pmatrix}$$ where, $(x^0, x^1, x^2, x^3) = (ct, x, y, z)$ (Cartesian coordinates). That's because, in the $(x, y, z)$ coordinate system, the metric tensor and line element of the Nil geometry is respectively given as $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & x^2 + 1 & -x \\ 0 & -x & 1 \end{pmatrix},$$ $$ds^2 = dx^2 + dy^2 + (-x dy + dz)^2.$$ Since the time axis of this spacetime is not 'curved' ($N_{0 \nu} = \eta_{0 \nu}$), I didn't know if this spacetime was possible, but I assumed it was possible anyway. Then, according to Einstein's equation $$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R - \Lambda g_{\mu \nu} = \kappa T_{\mu \nu} \ \ \ \ \ \ (\kappa = \frac{8 \pi G}{c^4})$$ (The sign of the $\Lambda g_{\mu \nu}$ term is $-$ because I use $+---$ metric signature.), we get: $$G_{\mu \nu} = \begin{pmatrix} -\frac{1}{4} & 0 & 0 & 0 \\ 0 & -\frac{1}{4} & 0 & 0 \\ 0 & 0 & \frac{1}{4} (3x^2 - 1) & -\frac{3}{4} x \\ 0 & 0 & -\frac{3}{4} x & \frac{3}{4} \end{pmatrix},$$ $$T_{\mu \nu} = \frac{1}{\kappa} \begin{pmatrix} -\Lambda - \frac{1}{4} & 0 & 0 & 0 \\ 0 & \Lambda - \frac{1}{4} & 0 & 0 \\ 0 & 0 & (\Lambda + \frac{3}{4}) x^2 + (\Lambda - \frac{1}{4}) & -(\Lambda + \frac{3}{4}) x \\ 0 & 0 & -(\Lambda + \frac{3}{4}) x & \Lambda + \frac{3}{4} \end{pmatrix}.$$ (Note that if we set $\Lambda = -\frac{3}{4}$, we get the following clean result: $$T_{\mu \nu} = \frac{1}{\kappa} \begin{pmatrix} \frac{1}{2} & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ Although in our universe $\Lambda > 0$, so we cannot make $\Lambda = -\frac{3}{4}$ at any scale.)
The problem is, I didn't know if this energy-momentum tensor is 'possible'. I have heard of the energy conditions and thought $T_{00} \ge 0$ should hold according to the energy conditions, but I didn't know if that is the only condition for the energy-momentum tensor to be an allowed value. So, I tried to create the desired $T_{\mu \nu}$ by actually constructing a matter distribution such as fluid or the electromagnetic field, but I did not succeed.
(The degrees of freedom of the energy-momentum tensor made by the matter distribution is too low to produce the desired $T_{\mu \nu}$. For example, I tried using the electromagnetic field $F_{\mu \nu}$, which is calculated from $A^\mu$. After fixing the gauge, the degrees of freedom of $A^\mu$ become (roughly) $4 - 1 = 3$, but this is far deficient to make the above $T_{\mu \nu}$.)
Even, I thought of a hypothetical rank $2$ tensor matter field $X_{\mu \nu}$ with the following Lagrangian density: $$\mathcal{L}_{matter} = X.$$ (I'll denote $T = g^{\mu \nu} T_{\mu \nu}$ and $X = g^{\mu \nu} X_{\mu \nu}$.) Then, from the formula of $T_{\mu \nu}$, $$T_{\mu \nu} = -2 \frac{\partial \mathcal{L}_{matter}}{\partial g^{\mu \nu}} + g_{\mu \nu} \mathcal{L}_{matter} \\ = -2 X_{\mu \nu} + g_{\mu \nu} X$$ holds, and contracting with $g^{\mu \nu}$ gives $$T = -2X + 4X = 2X \\ X = \frac{1}{2} T,$$ so we obtain $$X_{\mu \nu} = -\frac{1}{2} T_{\mu \nu} + \frac{1}{4} g_{\mu \nu} T \\ = -\frac{1}{2 \kappa} (R_{\mu \nu} + \Lambda g_{\mu \nu}). \ \ \ \ \ \ (\mathrm{by \ the \ Einstein \ field \ equation})$$ Hence, if there exists a matter field $X_{\mu \nu}$ with a Lagrangian density equal to $\mathcal{L}_{matter} = X$, there are enough degrees of freedom and we can always find $X_{\mu \nu}$ to yield desired $T_{\mu \nu}$ or $g_{\mu \nu}$ through the above formulas. However, there is no free matter field of the form $X_{\mu \nu}$ among the matters used in general relativity or in reality, so this is not a solution.
Questions
I will pose my questions. Let $\mathbb{T}$ be the set of energy-momentum tensors resulting from 'physically possible matter distributions'. (I think the definition of 'physically possible matter distributions' would be ambiguous.
- It can be non-quantum and ideal matters, such as point masses, fluids, the electromagnetic field, etc., used only within the framework of general relativity. (I think parameters in general relativity, i.e. $\Lambda$, or even $\kappa$, can be arbitrary fixed values.)
- It can be quantum matters (gluons, electrons, etc.) that appear when quantum mechanics, such as the Standard Model, is assumed. (In this case, there are new options as the matter fields in addition to the tensor fields: the spinor fields.) (In this microscopic perspective, the concepts like 'point mass' or 'continuous ideal fluid' will disappear.)
- It can be true, experimentally observed real-world matters. (I guess, the cosmological constant $\Lambda$ would be fixed in this case.)
If it is necessary to distinguish which of these 'physically possible matter distributions' are used, the set $\mathbb{T}$ will be denoted as $\mathbb{T}_1$, $\mathbb{T}_2$, and $\mathbb{T}_3$, respectively.) And according to the Einstein field equation, the set of metric tensors formed by energy-momentum tensors in $\mathbb{T}$ is denoted $\mathbb{G}$. In other words, $$\mathbb{G} = \left\{ g_{\mu \nu} \ | \ \frac{1}{\kappa} (R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R - \Lambda g_{\mu \nu}) \in \mathbb{T} \right\}.$$ Also, for convenience, we define a function $f$ that connects corresponding $g_{\mu \nu}$ and $T_{\mu \nu}$ as follows. $$f: \mathbb{G} \to \mathbb{T} \\ f: g_{\mu \nu} \mapsto \frac{1}{\kappa} (R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R - \Lambda g_{\mu \nu})$$ Then, my questions are the following.
(1) Is 'Nil metric' $N_{\mu \nu}$ an element of $\mathbb{G}$?
(2) If (1) fails, is there any metric in $\mathbb{G}$ that we can 'build a Penrose triangle' in spacetime? It's a bit ambiguous question; specifically, I'm curious about both the metrics that we can build at least one Penrose triangle in the whole spacetime / the metrics that we can build Penrose triangles at every time (or, in every spacelike hypersurface).
(3) (This is a generalization of (1)) Among the elements of $\mathbb{G}$, are there metrics other than $\eta_{\mu \nu}$ that has 'the flat time axis', that is, $$g_{0 \nu} = \eta_{0 \nu}$$ (where the $0$th coordinate is a timelike coordinate) ? Which energy-momentum tensor provides a metric with curved space but not time?
And then, we can pose more general questions:
(4) What is the structure of $\mathbb{T}$? Or more precisely, what are the equivalent conditions for any $T_{\mu \nu}$ to belong to $\mathbb{T}$? Maybe the energy conditions provide necessary conditions for $T_{\mu \nu}$ to be an element of $\mathbb{T}$; But, are those sufficient?
(5) (This is a stronger version of (3)) What is the structure of $\mathbb{G}$? What are the equivalent conditions for $g_{\mu \nu}$ to belong to $\mathbb{G}$? This question can be interpreted as 'What is the structure of spacetimes possible in reality?'.
And, while not related to the Nil universe, I have a bit more question about the correspondence (the function $f$) between $\mathbb{G}$ and $\mathbb{T}$.
(6) Is $f$ surjective? That is, for any $T_{\mu \nu}$ belonging to $\mathbb{T}$, does $g_{\mu \nu}$ that satisfies $f(g_{\mu \nu}) = T_{\mu \nu}$ always exist? (Probably, for physical reasons, this must hold, but that doesn't prove that it's true.)
(7) Is $f$ injective? That is, for any $T_{\mu \nu}$ belonging to $\mathbb{T}$, is there always unique $g_{\mu \nu}$ (if it exists) such that $f(g_{\mu \nu}) = T_{\mu \nu}$? (If not, what happens..? How does the universe determine one $g_{\mu \nu}$ from a given $T_{\mu \nu}$?)
Thanks for reading this long question! Any help will be appreciated.
