# From affine space to a manifold?

+ 2 like - 0 dislike
980 views

One of the several definitions of an affine space goes like this. Let $M$ be an arbitrary set whose elements are called points, let $\mathcal{V}$ be a vector space of dimension $n$, and let $\lambda:\mathcal{M}\times\mathcal{M}\to\mathcal{V}$ have the following properties:

1. For each $p$ in $M$ and each $\vec{v}$ in $\mathcal{V}$ there is a unique $q$ in $\mathcal{V}$ such that $\lambda(p,q)=\vec{v}$

2. $\lambda(p,0)=p$ for each $p$ in $M$

3. $\lambda(p,r)+\lambda(r,q)=\lambda(p,q)$

For classical and special relativitistic physics, an affine space seems to model the physical facts nicely, but not for general relativity. For the latter, we jump to manifolds with an enormous jump in complexity and variability from one author to another.

My question is this: Where does the definition of affine space fall short? I strongly suspect it is in Axiom 3 above, which is a kind of linearity assumption. Unfortunately, no author seems to tackle the transition; all launch full bore into manifold theory. Could someone provide a reference that does-one which discusses the manifold axioms from the point of view of physical phenomena?

Furthermore, can we truly separate the issue of affine versus nonaffine from that of the metric?

Added Later: Please try to understand what I am asking here. There are many treatments available which cover the formalism, but they merely postulate local coordinates. What I am looking for is a discussion of the physical reason for limiting them to local neighborhoods. If our basic set is affine we can establish global coordinates relative to an arbitrarily chosen origin. Thus, given a vector $\vec{v}$ for instance, we can describe a straight line by $\vec{x}=\vec{x}_0+\vec{v}t$ extending to infinity in both directions. Can we thus describe a straight line in a general manifold? If not, then why not?

This post imported from StackExchange Physics at 2014-12-12 22:30 (UTC), posted by SE-user Heaviside
$\lambda$ is introduced as $\mathcal{R}\times\mathcal{R}\to\mathcal{V}$, then it becomes $M\times\mathcal{V}\to\mathcal{V}$ in property 1, then it is $M\times(\mathcal{V}?)\to M$ in property 2.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Chris White
for geometric or topological reasons (eg curvature, singularities), distance parallelism and single-valuedness may fail in GR and gauge theories like electromagnetism; see also holonomy and monodromy

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Christoph
My guess is the intended definition is $\lambda : \mathcal{M} \times \mathcal{M} \to \mathcal {V}$ gets the displacement vector between two points. Then axiom 1 should read "$q$ in $\mathcal{M}$", and axiom 2 should read $\lambda(p, p) = 0$.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Nathan Reed
Nathan: True, thanks for pointing out my glitch. Correction made.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Heaviside
Whoops! And Chris!

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Heaviside

+ 1 like - 0 dislike

Affine spaces use vectors to model displacements between points. This fails in a curved space because displacements no longer add according to the parallelogram law, so it makes no sense to model them as vectors. It's not so much that any particular one of those axioms fails, as that the whole definition is no longer reasonable.

In a curved space, displacements have to be along curves that live in the space. Vectors play the role of tangents to these curves, i.e. elements of tangent spaces. Each point of the manifold is to be thought of as having its own distinct tangent space, and vectors that live in different tangent spaces can't be added or compared directly to each other. (This is all formalized by the notion of tangent bundle.) In order to work with vectors at different points we have to first parallel-transport them to the same point, a process which is path-dependent in curved space.

Re: "can we truly separate the issue of affine versus nonaffine from that of the metric", I am not sure exactly what you're asking, but if the question is how to distinguish between flat and curved spaces (i.e. affine and nonaffine) by looking at the metric, then we can do that by examining the Riemann tensor, which will be zero in flat spaces even if they are described in a weird curved coordinate system (and therefore have a metric that is not obviously the flat space metric).

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Nathan Reed
answered Dec 10, 2014 by (10 points)
Thanks. Your outline of manifold theory is helpful and cogent, but I wonder about your assertion that "the whole definition is no longer reasonable." Here's what motivates my question. If i accept the definition of a manifold at the outset, I am only assured of coordinate mappings "in patches." But my intuition says that each point in physical space (or spacetime) can be associated with a unique point in $R^n$. So apply a nonlinear coordinate system to account for the curvature of the manifold. In other words, why can't one embed in $R^n$? (Not talking about Nash theorems here!)

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Heaviside
@Heaviside: the topology of spacetime might be incompatible with $\mathbb R^n$

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Christoph
@Heaviside It's true that a single coordinate patch can cover the entire spacetime in many cases of interest, so you can map the whole thing into $\mathbb{R}^n$. And you can make a "straight line" $x_0 + vt$ in $\mathbb{R}^n$. But this will not in general be a straight line or a geodesic in the spacetime...it will just be some arbitrary coordinate-dependent curve. If the spacetime has intrinsic curvature it's not possible for straight coordinate lines to be geodesics everywhere. So the affine structure on $\mathbb{R}^n$ isn't physically meaningful or useful.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Nathan Reed
Thanks, Nathan. So-if I begin with the affine axioms and follow a straight line-how do I determine that there is "intrinsic curvature". Note that we haven't (as yet) the mathematical tools to discuss the Riemann tensor.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Heaviside
@Heaviside Well, you have to have the metric as that's what lets you talk about physical distances and angles. Without that, you just have an abstract space with no connection to reality. Intrinsic curvature is encoded in the metric; the Riemann tensor is just made from a bunch of derivatives of the metric.

This post imported from StackExchange Physics at 2014-12-12 22:31 (UTC), posted by SE-user Nathan Reed

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverfl$\varnothing$wThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.