Is a Vector Space Always a Manifold in Special Relativity?

[atyy]

I'm asking because I think of Minkowski space as a manifold with a Riemannian metric. However, I've also seen treatments in which an event in spacetime is chosen as origin, and special relativity treated as a vector space given the choice of origin.

Does this mean that a vector space is a manifold? Or is the vector space treatment of special relativity somewhat different from the Riemannian metric treatment?

 

[stevendaryl]

I'm asking because I think of Minkowski space as a manifold with a Riemannian metric. However, I've also seen treatments in which an event in spacetime is chosen as origin, and special relativity treated as a vector space given the choice of origin.

Does this mean that a vector space is a manifold? Or is the vector space treatment of special relativity somewhat different from the Riemannian metric treatment?

The term "vector space" is used to mean something more general than what you are talking about: matrices and quantum states can be described as a vector space. What you're talking about is a particular kind of vector space, which is n-tuples of real numbers. Yes, those vectors form a manifold.

The other way around doesn't work in General Relativity. A curved manifold is not a vector space, because there is no notion of "adding" two positions to get another position. In Special Relativity, as you said, you can uniquely associate a point in spacetime with a vector, and so it sort of makes sense to add two such vectors, but in curved spacetime, points are not associated with vectors.

Within a small enough region of curved spacetime, it is possible to map the region to a vector space, but not for large regions of spacetime.

 

[WannabeNewton]

Every finite dimension vector space is a smooth manifold. See the examples in Lee chapter 1. For infinite dimensional look up Banach manifolds or see Lang's differential geometry text.

 

[espen180]

Yes, any finite-dimensional vector space admits a smooth manifold structure.
The two formulations are equivalent, since Minkowski space, viewed as a manifold, admits a global chart with linear coordinates.

 

[Fredrik]

It doesn't matter much if we choose to define Minkowski spacetime as a vector space, an affine space, or a manifold. We get the same predictions about results of experiments either way. These versions of SR differ only in the details of how the assumptions that define the version are stated.

I'm not sure what this tells us about what manifolds and vector spaces have in common.

Every finite dimensional vector space is isomorphic to the vector space $\mathbb R^n$. The set $\mathbb R^n$ has both a natural vector space structure and a natural manifold structure.

 

[WannabeNewton]

Remember that Minkowski spacetime is just $\mathbb{R}^{4}$ with a specific indefinite quadratic form. The pseudo-riemannian metric is the same for all geometric tangent spaces of Euclidean 4-space so it's just an indefinite quadratic form on the underlying $\mathbb{R}^{4}$.

 

[Fredrik]

Not sure if "semi-inner product" is the appropriate term. Conway (A course in functional analysis) defines it as a form that satisfies all the conditions on an inner product except that <x,x>=0 implies x=0 (so in particular <x,x> ≥ 0 for all x). I don't know if there is a suitable term. Maybe "pseudo-inner product" would be better, but I would guess that there's no consensus about the terminology (not even a semi-consensus or pseudo-consensus).

Edit:. Ah, I see that you edited out the word "semi-inner" while I was typing this.

 

[WannabeNewton]

I'm not sure what this tells us about what manifolds and vector spaces have in common.

I don't really think it says anything other than the fact that vector spaces are a special class of smooth manifolds. A more interesting class of algebraic objects to study that also happen to be smooth manifolds would be lie groups. Now that is a beautiful area of study :!)

 

[WannabeNewton]

Edit:. Ah, I see that you edited out the word "semi-inner" while I was typing this.

I originally read it in Roman but he was talking about a generalization of norms it seems and then I actually did look at Conway and saw he defined it completely differently so yeah lol best to just keep it as indefinite quadratic form.

 

[atyy]

Thanks to all, that's very helpful!

@WannabeNewton: a Banach manifold isn't a smooth manifold?

 

[micromass]

@WannabeNewton: a Banach manifold isn't a smooth manifold?

A smooth manifold is usually defined as something that is locally diffeomorphic to $\mathbb{R}^n$. (this is a very loose definition, but you get the point). A Banach manifold is something way more general and is something that is locally diffeomorphic to a Banach space. This Banach space can be infinite dimensional.

For example, a Banach space like $L^2$ would be naturally a Banach manifold, but has no canonical structure as a smooth manifold.

 

[atyy]

What is a Banach manifold (say something more complicated than $L^2$) used for?

Is there an infinite dimensional analogue of a Riemannian manifold, where the tangent spaces are, say, Hilbert spaces?

 

[WannabeNewton]

What is a Banach manifold (say something more complicated than $L^2$) used for?

Is there an infinite dimensional analogue of a Riemannian manifold, where the tangent spaces are, say, Hilbert spaces?

http://math.stackexchange.com/questions/8322/a-banach-manifold-with-a-riemannian-metric

Note that the notion of tangent spaces has nothing to do with a Riemannian structure. Tangent spaces make perfect sense for any manifold with nothing but a smooth structure but also the concept of a tangent space carries over to Hilbert manifolds. Wether or not Banach manifolds admit any vague notion of a Riemannian metric is discussed in the link.

 

[micromass]

Since Banach spaces such as $L^2$ show up in QM, and since smooth manifolds show up in GR, I think it is a fair guess to show that the combination of those (= Banach manifolds) show up naturally in some theory that combines QM and GR. But I don't know anything about physics, so it's just a guess.

 

[martinbn]

Just a pedantic comment. A vector space is always over a field, and the field need not be the real numbers (or the complex). A finite dimensional vector space over the rational numbers is not a manifold.

 

[atyy]

martinbnb, that's helpful too. Together with WanabeNewton's comment on infinite dimensional vector spaces, these exceptions indicate why vector spaces are not necessarily manifolds, although some are, as in special relativity.

micromass, the gauge/gravity duality is conjectured to be a complete theory of quantum gravity, so I guess one might check there to see if your intuition is borne out!