# Going from four to five dimensions can we overlook the entirety of time? [closed]

On youtube, I saw N.D.Tyson explain (in connection with the Interstellar movie) how we can in the fifth dimension oversee the whole of all events. But how can that be? If you go into the fifth dimension, you go one space dimension higher. You can see all sides of a three-dimensional object at once (Picasso liked the fourth space dimension in the time that Einstein and others were playing with it and tried to depict this in his paintings in which we can see the strange women heads) like you can oversee the whole of two-dimensional space if you go to three dimensions.

Tyson tells us that you can move from the time when you were born to the time near the beginning of the universe. All events exist at once. But that means that you move in a new time dimension, implying that going one dimension higher isn't in the space realm.

But increasing spacetime from four to five dimensions does in my humble opinion mean that you increase the number of space dimensions from three to four. The number of time dimensions always stays one. If not, you still move in three space dimensions, and you have access to the events laid out before you from ? to ? (in the movie it is said that time has become a physical dimension).

So my question: is it possible that the time dimension becomes a physical one? It may be obvious that I don't think so.

What Tyson is describing is known as an embedding. This is a well defined mathematical procedure, though in general it's far more complicated than Tyson suggests. Indeed it isn't always possible to embed a Lorentzian manifold (i.e. spacetime) into a higher dimensional manifold,and when it is you may need a startlingly large number of additional dimensions to do it.

Anyhow Tyson is being somewhat vague in his statements, which is perhaps excusable in a popular science TV programme. It doesn't mean that time becomes something you can literally see. The higher dimensional manifold would need at least one timelike dimension to embed spacetime within it i.e. you cannot embed a pseudo-Riemannian manifold in a Riemannian manifold.

• I'm pretty sure the Nash embedding theorem has a pseudo-Riemannian manifolds analogue too. See here: mathoverflow.net/questions/127734/… although I haven't as yet been through the references cited there (at least not thoroughly), but the Campbell's theorem argument seems reasonable, even though some of those mathematicians seemed a little unsure. Do you have a conflicting reference for this - if so, it would be great to take a new question to MathOverflow. – Selene Routley Dec 15 '16 at 7:23
• @WetSavannaAnimalakaRodVance my recollection is that you can't prove embedding is possible if the Lorentzian manifold allows CTCs. It is known that every Lorentzian manifold without CTCs can be embedded. But this is a vague recollection so it could be otherwise. – John Rennie Dec 15 '16 at 7:48
• My equally vague recollection is that CTCs invalidate the Nash Twist, which is an argument central to his theorem, but there are other routes to like, but weaker results - weaker in the sense that now one cannot come up with bounds on the dimension of the embedding. I'm going to have to look into this again. I thought I'd licked this: I even thought I understood the proof of Nash's theorem at the time I looked into all of this (just), but now it looks as ununderstandable as when I began to try to understand it. I guess the few years since have wiped that particular part of my brain state. – Selene Routley Dec 15 '16 at 8:01
• @WetSavannaAnimalakaRodVance (and JohnRennie, but you own the post) There is an embedding theorem for general pseudo-Riemannian manifold called the Clarke embedding theorem, see e.g. theorem 2.8 here. However, it does not embed all Lorentzian manifolds into Minkowski space, it generically raises the number of temporal dimensions by one. The existence of an embedding into Lorentzian space is indeed obstructed by the existence of closed timelike curves, and guaranteed for globally hyperbolic manifolds. – ACuriousMind Dec 16 '16 at 0:10
• @ACuriousMind Thanks heaps. That looks like a beautiful document. Certainly CTCs destroy the Nash argument: I think now that "guaranteed for globally hyperbolic manifolds" is what I am thinking - I may well have been mistaken thinking other routes can get weaker results. When you say, "obstructed by the existence..." do you mean "as far as the Clarke theorem is concerned", or are you aware of a proof that an impossible to embed Lorentzian manifold with CTC can always be found (i.e. that there is no hope of an embedding theorem of any kind, with or without dimension bounds)? – Selene Routley Dec 16 '16 at 0:30

Here is a piecemeal post record some follow up to John Rennie's answer.

There are general embedding theorems for manifolds, which show that there is always a way to think of any manifold as an embedding in a higher dimensional, flat manifold. This is essentially what NDT is talking about. Although such embeddings are always in principle possible, there are limitations to these theorems that mean we sometimes cannot do them in a way that is useful for physics.

The most general and intuitive theorem is the Strong Whitney Embedding theorem which shows that any real, smooth, Hausdorff and second countable m-dimensional manifold manifold can be embedded in $\mathbb{R}^n$, where $n$ is at most $2\,m$.

Historically, this was important, because it showed that the modern conception of a manifold (as a geometrical object charted on overlapping patches) was the same notion as the dominant early conception of a manifold as simply some hypersurface in $\mathbb{R}^n$.

But this is not all we want for differential geometry General Relativity, of course. We are concerned with Riemannian and (for GR) Lorentzian manifolds, i.e. manifolds whose tangent spaces are kitted with a nondegenerate symmetric billinear form (inner product), whence one defines the concept of a metric (or pseudometric, in the case of Lorentzian manifolds).

So we want embeddings to preserve this structure on the tangent space. We want them to preserve length. Thus we are concerned with the much more specialized isometric embeddings. The Whitney theorem is not concerned with this further structure to work, so it is no surprise that the embeddings whose existence it guarantees may not be isometric ones.

Enter now the Nash Embedding Theorem. This guarantees an isometric embedding of any Riemannian manifold (i.e. if the symmetric billinear form is positive definite - which is NOT what we are concerned with in GR) into a higher dimension Euclidean space. If the manifold is compact, and of dimension $m$, it needs at most a Euclidean space of dimension $\frac{m}{2}(3\,m +11)$ to be embedded into. Things are even worse for a noncompact space: the bound is $\frac{m}{2}(m+1)(3\,m +11)$.

But this result does not work for a Lorentzian manifold, where the billinear form is not positive definite and there are null vectors and light cones. In particular, closed timelike curves destroy Nash's central argument.

There are weaker embedding theorems for pseudo-Riemannian manifolds, with nonpositive definite billinear form. See this Math Overflow Thread for a discussion. They still guarantee an isometric embedding, but these theorems are weaker in the sense that, unlike the Nash theorem, there are no bounds on dimensions - they give no guarantee as to how many dimensions you need to embed in.

There is one critical point that also prevents them from being generally useful in physics.

For a Lorentzian manifold, the embedding theorems above tell you nothing about the number of timelike dimensions. Sure, you can always embed the Lorentzian model, but, in the words of ACuriousMind:

An isometric embedding in particular preserves the time/light/spacelike character of a curve, but there are no closed timelike curves in Minkowski space, so you simply can't have such an embedding without raising the number of temporal dimensions on the target space.

That is, if you isometrically embed our signature $(1,\,3)$ manifold into a flat higher dimensional one, and if our manifold has closed timelike curves, then the embedding is going to be of the form $(n,\,m)$ where $m\geq3$ and $n$ is strictly greater than 1. So we shall end up with more than one timelike dimension, and that is not generally useful for physics. Not the least because there are then always isometries on the metric that both belong to the identity component of the group of isometries AND reverse the time direction of a vector. So it becomes very hard, if not impossible, to do any physics with a notion of causality in such an embedding.

If, however, the manifold is globally hyperbolic, then the Clarke embedding theorem guarantees an isometric embedding with one timelike dimension, see e.g. theorem 2.8 here.