Consider a linear parabolic partial differential equation:
$$t \partial_{tt}\varphi_t(x)=\pm x\partial_x \varphi_t(x)$$
which (essentially) takes the form of the backwards heat equation (minus sign) and heat equation (plus sign) in one dimension (we're in one dimension so time and space can be used interchangeably).
With a Wick rotation $t \mapsto -it$ you can essentially get the free Schrödinger equation:
$$it \partial_{tt}\varphi_t(x)=x\partial_x \varphi_t(x) $$
So my question is why does a Cauchy foliation $\mathcal{F}$ of $\zeta^{1,1} \simeq \Bbb M^{1,1}$ satisfy the first equation? Is there any geometric reason for this?
The diffeomorphism is given by $f: \Bbb M^{1,1} \to \zeta^{1,1}$ with $f(x,y)=(e^x,e^y)$ and $\Bbb M^{1,1}$ is in Dirac coordinates. The foliation of $\zeta^{1,1}$ is $\mathcal{F}= \big\lbrace e^{\frac{\mp t}{\log x}}: t > 0 \big\rbrace. $ And $\varphi_t(x)=e^{\frac{\mp t}{\log x}}$ solves the equation.
I just don't understand why a Cauchy foliation of a certain Lorentzian manifold that's equivalent to the Minkowski plane should obey the backwards heat equation.
I went ahead and partitioned $\Bbb M^{1,1}$ with $\mathcal G=\big\lbrace t/x: t \in \Bbb R \big\rbrace $ and it does not obey the heat equation nor the backwards one.
I thought: "Is this something to do with the metric?"
I believe the crux of the matter is this: The heat equation in the above form assumes Euclidean space with the usual Euclidean metric; therefore in some sense restricting $\mathcal{F}$ to the usual Euclidean metric implies a solution to the heat equation. $\mathcal{F}$ is only a Cauchy foliation of the corresponding Lorentzian manifold $\zeta^{1,1}$ when the metric is chosen to be $g=\frac{dxdy}{xy}$ (this is obtained through the diffeomorphism). Forcing the metric to be $ds^2=dx^2+dy^2$ somehow allows $\mathcal{F}$ to solve the heat equation although I don't have a good reason why. Maybe there is no satisfactory geometrical reason. In any event once the metric is chosen to be the Euclidean $ds^2=dx^2+dy^2$ there's no point talking about a Lorentzian manifold anymore as it fails to carry that structure.