Consider a spacetime $(\zeta^{3,1},g)$
where $$g=\frac{dudv}{uv}-\frac{dr^2}{r^2}-\frac{dw^2}{w^2} \quad \forall u,v,w,r \in (0,1)$$ Now this is just Minkowski space in different coordinates (related to Dirac/Light cone coordinates/null coordinates). I'm looking to take a Cauchy foliation of $\zeta^{3,1},$ change the metric back to $g'=dudv-dw^2-dr^2,$ and use the induced measure from $g'$ by means of the volume form to transform the foliated past light cone region onto a new manifold $(\Psi^{~3,1},d).$ I'm not sure how this will work so I will try to construct a $(1+1)$ dimensional example and then define a product manifold to acheive a $(3+1)$ dimensional spacetime. But the idea is similar.
Let's look at the steps involved for the $(1+1)$ dimensional case $(\zeta^{1,1},u)$ where $u=\frac{dxdq}{xq}.$ The Cauchy foliation is simply $\ln(b)\ln(y)=t$ (solving for $y$ gives an explicit representation) which can then be transformed via Mellin-like transform:
$$\Phi_s(t)= \int_{S=(0,1)} \exp {\frac{t s}{\ln b}}~db = \int_0^1 tb^{t-1} \exp \frac{s}{\ln b}~db = 2\sqrt{ts}K_1(2\sqrt{ts})$$
Observe that this is an unnormalized K-distribution and it's inverse transform yields an unnormalized distribution call it the "Zeta-distribution" (our Cauchy foliation). Observe that the Fisher information metric of the Zeta-distribution is $\Phi_s(t)$ up to a factor. Observe that the Zeta-distribution yields solutions to the Killing field $X=\langle x \ln x, -y\ln y \rangle.$ It also provides a distributional solution to the backwards heat equation with diffusivity depending on both space and time:
$$\frac{\partial^2}{\partial t^2}\Phi(t,b)=-\frac{b}{t}\frac{\partial}{\partial b}\Phi(t,b)$$
In short, $\Phi_1(t)=v$ is a Lorentzian metric on the smooth $(1+0)$ manifold $(\Psi^{~1,0}, v).$ We have that $\Psi^{~3,1}=\Psi^{~1,1}\times \Psi^{~1,0} \times \Psi^{~1,0}$ so we can define the product metric on $\Psi^{~3,1}.$
What is the exact form of the metric (line element) for $(\Psi^{~3,1},v)?$ I believe it is of the form $v=v_1 \oplus v_2 \oplus v_3$ and should be expressed in terms of $K_1$ (Bessel functions). I think I am not too far off from getting the precise form. Am I on the right track here?
Best Answer
A couple of things.
I think you had some misconceptions about metrics and transformations when writing your post. The point you overlooked is that the metric does not automatically change along with your transformation of the foliation. Yes the leaves are mapped to a new function space but you could easily just leave the metric as it is. The foliation of course will not preserve this metric anymore but it satisfies your question because you only wanted to transport the foliation to some new manifold.
Let $(M,g')$ be Minkowski space in light cone/Dirac coordinates. Then the foliation inherited from $(\zeta,g)$ is the product foliation:
$$\Omega_s(x,y,z)=\varphi_s(x)\varphi_s(y)\varphi_s(z)\space\space\space\space{s>0}$$
for $x,y,z \in (0,1).$
Then we can embed $\Omega$ into $M$:
$$ e: \Omega \to M $$
And use the measure and volume form on $M$ to transform $\Omega$. Under this transformation the metric does not change, only the leaves of the foliation are being piecewise transformed to the new function space.
Therefore, we have the $3$-tuple $(M,g',\Omega)$ and finally we obtain $(M,g',\chi)$
where $$\Delta: \Omega \to \chi $$
is given by
$$\Delta:=\int_{(0,1)^3} x^{a}y^{b}z^{c}\Omega_s(x,z,y)~\frac{dxdydz}{xyz}=\chi_s(a,b,c).$$
By the way your partial differential equation is pretty irrelevant in the context of this question because it is expressed on $(\Bbb R^2,\mathrm{stan})$ which does not carry a Lorentzian structure.
I wouldn't read much into the fact that $\Omega$ being a foliation of a Lorentzian manifold also satisfies said linear PDE on Euclidean space. There is no important relationship between these two things at all.