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Given a four dimensional Lorentzian manifold $\mathcal{M}$ (a manifold with a metric $g_{\mu\nu}$ in the tangent bundle with signature (-1, 1, 1, 1)), we define a global spatial foliation by a time-like vector field ($n^\mu{}n^\nu{}g_{\mu\nu} = -1$). Is it true that given another global foliation defined by a time-like vector field $v^\mu$ there is (at least one) a diffeomorphism $\Upsilon:\mathcal{M}\rightarrow\mathcal{M}$ generated by a vector field $A^\mu$ such that $v^\mu = \exp(\mathcal{L}_A)n^\mu$, where $\mathcal{L}$ represents the Lie derivative and $\exp(\mathcal{L}_A)$ the exponential map generated by it?

If it is true, is there a good reference for such result?

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  • $\begingroup$ If there were such a vector field, how would you relate it to the (certainly existing) diffeomorphism between the 2 foliations? $\endgroup$
    – magma
    May 12, 2014 at 12:41
  • $\begingroup$ @magma Suppose $p \in \mathcal{M}$, we can find the integral submanifolds using the vectors orthogonal to $n^\mu$ or $v^\mu$. Hence, assuming $n^\mu \neq v^\mu$ we have two different hypersurfaces which coincide at least in $p$. I expect that $A^\mu$ would be the tangent vector field of the flow which connects the two hypersurfaces. $\endgroup$ May 12, 2014 at 12:51
  • $\begingroup$ In this level of generality, the statement is false. You should at least require that both foliations on $M$ correspond to product decompositions $M=N_i\times R$, $i=1,2$, where $N_i\times \{t\}$ are leaves of foliations. This assumption is quite common and physically reasonable. I am still quite sure that the statement is false even in this case; maybe you need to assume compactness of $N_i$'s. Then you have a chance. $\endgroup$ May 12, 2014 at 17:08
  • $\begingroup$ In that case the answer is no, there is no such $A^\mu$ lying at the intersection of the 2 hypersurfaces: take 1+2 flat spacetime with 2 inertial frames (primed and unprimed) moving along the common x axis. Then the intersection is the common y axis. Now $e_{t'} = a e_t + b e_x$ (it has a x component). Vector $A = \alpha (x,y,t) e_y$. $\mathcal{L}_A e_t$ does not have a x component and neither do its $\mathcal{L}_A$ iterates. So $\exp(\mathcal{L}_A)e_t$ does not have a x component and we cannot get $e_{t'}$ $\endgroup$
    – magma
    May 12, 2014 at 18:03
  • $\begingroup$ To give $A^\mu$ a chance to exist, it should be on the plane spanned by $n'$ and $n$ $\endgroup$
    – magma
    May 12, 2014 at 18:07

1 Answer 1

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Here are details for my comment.

The answer to your question (without further assumptions) is negative. Here is an example.

Consider $R^{3,1}$ with the (standard) flat Lorenztian metric $x_1^2+x_2^2+x_3^2-x_4^2$. Let $\Gamma$ be the discrete isometry group of $R^{3,1}$ consisting of translations by vectors with integer coordinates. The quotient $M=R^{3,1}/\Gamma$ is the 4-dimensional torus with flat Lorenztian metric. Next, let $\tilde \nu$ be the (parallel) vector field on $R^{3,1}$ consisting of vectors parallel to the vector $(0,0,0,1)$ and let $\tilde n$ be the parallel vector field defined similarly, using the vector $(t,t,t,1)$, where $0<t<1/\sqrt{3}$ is a small irrational number (it is then also time-like). Then take $\nu$ and $n$ to be projections of the vector fields $\tilde\nu, \tilde n$ to the torus $M$. The normal foliation to $\tilde \nu$ is by horizontal hyperplanes. Each leaf projects to a 3-dimensional torus in $M$, which gives us a foliation of $M$ normal to $\nu$. On the other hand, each hyperplane normal to $n$ projects to a hypersurface which is dense in $M$ (this is an irrational foliation of $M$ normal to $n$). Now, it is clear that there is no homeomorphism $M\to M$ which carries one foliation to the other one.

My guess is that your question has positive answer (but the diffeomorphism might be more complex than the one you are attempting to construct) in the following setting: Both $\nu$ and $n$ are normal to product foliations, defined below.

Definition. A foliation $F$ on $M$ is a product foliation if there exists a diffeomorphism $h: M\to X\times {\mathbb R}$, where $X$ is a compact 3-dimensional manifold, such that $h$ sends leaves of $F$ to submanifolds of the form $X\times \{t\}, t\in {\mathbb R}$.

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  • $\begingroup$ Thanks for your answer. Yes, I should have been more specific, the manifold $\mathcal{M}$ is globally hyperbolic (which gives the topology of $\mathcal{M}$ as of $R\times\Sigma$ where $\Sigma$ is a Cauchy surface of the manifold). I was also referring to product foliations. Should I edit my original question to add these requirements? $\endgroup$ May 13, 2014 at 7:38
  • $\begingroup$ @SandroVitenti: You definitely should. As a general rule, you should always ask the question that you mean to ask in order to get a good answer. $\endgroup$ May 13, 2014 at 16:44

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