2
$\begingroup$

In the following lecture by prof. Schuller, He tries to define absolute space at time $\tau$ mathematically as,

$$ S_\tau= \{p \in M \mid t(p)= \tau\} $$

Where $t$ is the absolute time function defined on a Newtonian spacetime. Now, He claims that by setting $(dt)_p\neq0, \forall p\in M $, the whole manifold can be seen as a disjoint union of absolute spaces at different times. i.e. $$ M = \dot\cup S_\tau$$

I don't understand how, by setting the gradient of absolute time function to be non zero, one can see the manifold as a disjoint union of all the absolute spaces?

Reference: https://www.youtube.com/watch?v=IBlCu1zgD4Y&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_&index=9

$\endgroup$

1 Answer 1

3
$\begingroup$

As sets, $M$ is the union of the preimages $S_\tau = t^{-1}(\tau)$ because the absolute time map $t : M \to \mathbb{R}$ is a function, and for any function $f: X \to Y$ we have that $X$ is the union of preimages, i.e. $X = \cup_{y\in Y} f^{-1}_y$, simply because every point of $X$ has to lie in one of the preimages.

However, the condition $(\mathrm{d}t)_p \neq 0$ everywhere is the same as the derivative map $D_pt : T_pM \to T_p\mathbb{R}$ being surjective everywhere, also called "having full rank".

This, in turn, means that $t$ is a submersion and e.g. the submersion theorem applies that tells us that all preimages $S_\tau$ also carry a smooth manifold structure.

$\endgroup$
3
  • 3
    $\begingroup$ Not only they carry a smooth structure, but, as a consequence of $dt\neq 0$ everywhere, that structure and the compatible topology are the ones induced by the corresponding structures of $X$. In other words, those preimages are co-dimension $1$ embedded submanifolds of $X$. $\endgroup$ Mar 21, 2020 at 11:33
  • 2
    $\begingroup$ I found the wikipedia page a bit sloppy on these issues. It is fundamental that the $t$-constant submanifolds are embedded and not symply immersed because one wants also to construct the classical spacetime as the Cartesian product of the time axis and the absolute space. This is not possible, in general if the absolute space is only immersed. Also it is necessary to assume that every absolute space carries a structure of 3D Euclidean affine space compatible with the differentiable structure. Obviously there are as many these Cartesian decompositions as many observers exist. $\endgroup$ Mar 21, 2020 at 11:46
  • $\begingroup$ does one have to assume it, or can one prove that absolute space carries a structure of 3D affine space? Why can it be not curved at all from the mathematical POV? $\endgroup$
    – ProphetX
    Feb 4, 2022 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.