When can we factor out the time dimension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:38:17Z http://mathoverflow.net/feeds/question/32055 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32055/when-can-we-factor-out-the-time-dimension When can we factor out the time dimension? Daniel Litt 2010-07-15T19:26:45Z 2010-07-15T23:13:53Z <p>First of all, my knowledge of both GR and differential geometry is quite weak, so forgive me if the physics here doesn't make much sense.</p> <p>Let $(M, g)$ be a smooth, connected Lorentzian manifold of dimension $n$. Let $f: \mathbb{R}\to M$ be a smooth curve such that the pullback of $g$ through $f$ is everywhere negative (where we've chosen an orientation on $\mathbb{R}$); we say that $f$ is time-like. Say that we can "factor" $f$ out of $M$ if there exists a manifold $S$ of dimension $n-1$ and an isomorphism $M\simeq S\times \mathbb{R}$ so that the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to S$ is constant and the map $\mathbb{R}\to M\simeq S\times \mathbb{R}\to \mathbb{R}$ is the identity. Intuitively, this factorization exhibits $f$ as "time" in some reference frame, and $S$ as space. My question is:</p> <blockquote> <p>For which $(M, g)$ can every time-like path be factored out?</p> </blockquote> <p>Minkowski space seems like an obvious example unless I'm missing something; it seems one can take a tangent vector to $f$ at any point and consider a perpendicular subspace to that vector as $S$. I'd accept as an answer a characterization of all such $(M, g)$ in dimension $4$, or some nice sufficient condition on $M$ for factorization to always work.</p> <p>If the motivation isn't obvious already, this is supposed to codify the intuition that in my reference frame, I seem to be standing still -- and that the same is true for everyone else, even if they seem to me to be moving. My apologies if I've overloaded terms, or used them incorrectly.</p> <hr> <p>Added: Note that this condition is much stronger than stable causality; indeed, it certainly implies stable causality, as choosing any timelike path $f$ and then considering the given projection to $\mathbb{R}$ gives a global time function. However, I am asking for (1) a product structure on $M$ for each path $f$ and (2) in order to formalize the notion that I seem to be standing still (to myself), the projection of $f$ to $S$ must be constant. </p> <hr> <p>Added: I don't think global hyperbolicity suffices either. The theorem of Geroch (it and other splitting theorems are discussed <a href="http://arxiv.org/abs/0712.1933" rel="nofollow">here</a>, for example) does indeed give a decomposition of $M$ as $\mathbb{R}\times S$. But I don't think this is enough. In particular, I am asking for the following---for <strong>every</strong> timelike path $f: \mathbb{R}\to M$, there is a product structure $M\simeq \mathbb{R}\times S$ such that the projection to $\mathbb{R}$ is a section of $f$, and that $f$ is constant upon projection to $S$. This is much stronger than Geroch's splitting theorem, as far as I can tell.</p> <hr> <p>Added: As the accepted answerer rightly points out in the comments to his question, I was wrong to claim that my condition is stronger than global hyperbolicity. They are in fact equivalent.</p> http://mathoverflow.net/questions/32055/when-can-we-factor-out-the-time-dimension/32058#32058 Answer by JosÃ© Figueroa-O'Farrill for When can we factor out the time dimension? JosÃ© Figueroa-O'Farrill 2010-07-15T19:52:51Z 2010-07-15T21:08:43Z <p><strong>Edit</strong></p> <p>The answer below is not correct. Upon further reflection, I believe that the correct causality condition is indeed <strong>global hyperbolicity</strong> and not the weaker stable causality.</p> <hr> <p>I believe this is a causality condition known as <strong>stably causal</strong>. You can read about this in Wald's "General Relativity" Chapter 8, in particular his Theorem 8.2.2.</p> <p>A spacetime $(M,g)$ is <strong>stably causal</strong> if there exists a continuous nowhere vanishing timelike vector $t$ such that the spacetime $(M,\tilde g)$ with $$\tilde g = g - t^\flat \otimes t^\flat,$$ where $t^\flat$ is the dual one-form to $t$ relative to $g$, has no closed timelike curves.</p> <p>Wald's Theorem 8.2.2 says that this is equivalent to the existence of a global time function on $M$.</p> <p><em>Theorem 8.2.2</em> A spacetime $(M,g)$ is stably causal if and only if there is a differentiable function $f$ on $M$ such that its gradient is a past directed timelike vector field.</p> <p>There is a whole hierarchy of causality conditions in GR. <em>Global hyperbolicity</em> implies stable causality and this in turn implies <em>strong causality</em>. <del>Global hyperbolicity might be too strong.</del></p> http://mathoverflow.net/questions/32055/when-can-we-factor-out-the-time-dimension/32059#32059 Answer by userN for When can we factor out the time dimension? userN 2010-07-15T20:01:17Z 2010-07-15T20:01:17Z <p>This sounds to me like you're asking that your spacetime admit a family of <a href="http://en.wikipedia.org/wiki/Cauchy_surface" rel="nofollow">Cauchy surfaces</a> (modulo annoyances like having $f$ be closed and acausal). There's a theorem of Geroch which guarantees that this is equivalent to <a href="http://en.wikipedia.org/wiki/Globally_hyperbolic" rel="nofollow">global hyperbolicity</a>. </p> <p>I don't think that globally hyperbolic and stably causal are equivalent conditions, but I'm not an expert on causality conditions in GR, so take this claim with a grain of salt.</p>