My objection is quite nuanced, in that manifolds, as objects of their own right, do not come equipped with coordinates or patches anymore than spacetime, the Euclidean plane or the Earth does. Rather patches and coordinates systems of human invention can be applied to manifolds and properties of the manifold will become apparent when quantities like arc-length and area are computed. If you learn manifolds as part of a class on point-set topology, you might have manifolds defined for you as a collection of all coordinate systems with limited applicability over a "patch" of some common set of points such that where patches overlap the transformation between coordinates systems is smooth and one-to-one. That's excellent for calculation of properties, but kind of hard to apply to found manifolds which don't start out with patches defined, let alone coordinate systems.
Yes they do, it's part of the definition of a manifold Well sure - but if one chooses, as GR does, to describe spacetime as a 4-manifold, then the mathematical baggage comes with it. I simply do not understand your phrase "Rather patches and coordinates systems of human invention can be applied to manifolds". You are assuming the existence of something called a "manifold" but denying the mathematical properties that define it What is a "found manifold" without the mathematical definition of a manifold?
My 2c: If you have a sphere and a plane, how free are the choices of coordinate systems? For instance, embedding a graph, or some geometric figure like a triangle has different consequences in each case, but that's obviously because a sphere is curved and a plane isn't. So how can we define a sphere or plane without invoking graphs? [And if we don't, is either a manifold] Ed: I think what I'm asking comes down to whether intersecting lines exist somewhere. Or maybe, how to find out you're on a sphere rather than a plane (if you don't know and you're 2 dimensional).
So, if you can discover that every point on the surface you're on has a locally Euclidean neighbourhood (by doing Euclidean geometry in that neighbourhood), you know you've satisfied one requirement of manifolds; if you're on a sphere you will need overlapping charts. But: Which explains the phrase "manifold of solutions" in one of Einstein's papers.
Ahem. One way to tell a sphere and a plane apart is to have them intersect. The largest intersection will be a great circle, and intuitively, any parallel planes will not intersect any sphere as all great circles. On the sphere of earth, there is an intersection with the plane of the ecliptic. The lunar orbit is inclined (by about 5 deg.) relative to this plane.
Looking at the triangle inscribed on a sphere... Choose each corner to be North in turn, the angle is then the longitude of the third point taking the second point to be 0. Edit - Thanks to rpenner for pointing out the sphere didn't arrive with any sort of coordinate system defined.
But there's the thing. How did the sphere, or spherical symmetry which the earth has approximately, "arrive". We know the earth isn't a flat plane, but how did we prove it? And, is rpenner referring to these kinds of manifolds, the one the earth and moon are in for instance, as "found". Physically, and before humans recognised the fact, the plane of the ecliptic has always intersected the earth with (roughly) a great circle, pretty much since the earth and moon formed. We "found" this system, and we did something mathematical eventually, which amounts to triangulating the surface of a sphere to predict when lunar phases occur--the manifold of solutions of orbital equations of motion. But not using calculus, as such.
Be careful, arfa. In your Wiki quote, the key word is "resembles", and even then it is not quite right. Look...... the Real plane \(R^2\), say, is called Euclidean if and only if it admits of the Euclidean metric. But a manifold need not have a metric, and even if it does, it need not be the Euclidean metric - the spacetime 4-manifold does not, for example (it has the Minkowski metric). Moreover, not all manifolds are Real I ran into trouble with this before, but I will say it again....... A manifold \(M\) is a topological space equipped with a continuous, invertible, operation-preserving map \( h:M \to R^n\). For simplicity, let's set \(n=4\). Then for the point \(p \in M\) the image point \(h(p)\) is the 4-tuple, say, \((u^0,u^1,u^2,u^3) \in R^n\). But since \(R^4 =\mathbb{R}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R}\) then there exist 4 projections \(\pi_j:R^n \to \mathbb{R}\) i.e.\(\pi_j \circ h(p)= u^j \in \mathbb{R}\). It is customary to refer to the set of composite functions \(\{\pi_j \circ h\}\) as the set of coordinates \(\{x^0,x^1,x^2,x^3\}\). And that is all you need to be a manifold. No metric. Conversely, I cannot convince myself that it makes sense to refer to "something" as a manifold, without taking this definition on board
If M is a topological space then every point p in M is contained in at least one open set of M. \(\forall p \in \textrm{points}(M) \, \exists O \in \textrm{opensets}(M) \; p \in O\) Then every open set of M, because it has no boundary, is itself a topological space. A homeomorphism is a continuous bijection between topological spaces where the inverse function is continuous. Thus it maps points to points and open sets to open sets. If there exists a homeomorphism between two topological spaces then they are in the same equivalence class, as we can write \(A \sim B\), A is homeomorphic to B. The surface of the sphere, \(S^2\) is not homeomorphic to the Euclidean plane, \(E^2\) because you can slice a sphere into circles and any homeomorphism would map oriented circles to oriented closed curves and there would always be one circle that could not be mapped to either orientation. But if we puncture the sphere by removing just one point, a homeomorphism does exist between the open set subspace of the sphere and the Euclidean plane. If n is a fixed positive number and \(E^n\) is the Euclidean topological space of n dimensions and if there is an indexed finite set, \(B_i\), of open sets of M which cover M, meaning every point p is contained in at least one open set, a neighborhood, that is in turn is in at least one of the open sets that cover M and each of those covering opensets is homeomorphic to \(E^n\), then I would say M is a manifold. \(\exists n,m \in \mathbb{N}^{+} \, \exists B \in \textrm{opensets}(M)^m \, \forall p \in \textrm{points}(M) \, \exists O \in \textrm{opensets}(M) \, \exists i \in m \; p \in O \subseteq B_i \in \textrm{opensets}(M) \; \wedge \; B_i \sim E^n\) \(\mathbb{R}^n\) is a model for \(E^n\), useful in algebra because it reduces statements about figures in Euclidean space to statements about relations between ordered n-tuples of numbers, and algebra is happy to work with numbers.
The above definition looks similar to what modern texts have as a "topological manifold" and assuming it would be easy to add a definition of derivative, while QuarkHead is focusing on a "differentiable (or smooth) manifold". For n < 4, based on a skim of some introductory material, I don't think there is a difference. For n=4, it's hard to point to a concrete example of something that falls in one category and not another. Many modern books seem to define \(E^n \equiv \mathbb{R}^n\) and \(\mathbb{R}^n\) already comes with a natural set of coordinates, the Cartesian coordinates. But the \(E^n \) I think of has points, nonzero lengths between pairs of distinct points, angles between ordered triples of distinct points, aka an affine space while \(\mathbb{R}^n\) comes with all these, an origin (which makes it a vector space), and natural axes or directions (which makes it a representation of a vector space). Using \(\mathbb{R}^n\) where I would use \(E^n \) makes talk of differentiation easier, while obscuring that the choice of coordinates remains free. This might be a bias on my part after years of talking to people who don't get the geometry of SR is independent of choice of inertial coordinates.
There's quite a lot of terminology to understand; here Wikipedia lays out a definition of an homogenous space, a particular kind of manifold, "X" is a G-set and its group G acts transitively, meaning that for every pair of elements (symmetries) of G, say x,y, there is a (not necessarily unique) element g in G, such that gx = y. Or of course you can click on the pasted links to get the Wiki definitions. "locally the same" = "homogenous" Given all that, suppose X is the Euclidean plane. X is homogenous because every point has a local neighbourhood that "looks the same". What's an automorphism group acting on X that makes X a single G-orbit?
And the answer is: the group of Euclidean plane isometries (!) If you have some geometric figure, translating it to a different 'neighbourhood' preserves the geometry and the orientation, there are rotations which also preserve orientation, and there are reflections which don't preserve orientation. Finally, scaling (not an isometry!) will preserve the ratios of sides and the orientations of figures. So all translations, rotations, and reflections preserve some geometry and give the whole plane the same G-orbit where G is the isometry group.
arfa, as is often the case here, you are attempting to run before you can walk! Look, the Lie groups are very particular sorts of manifold - a group that is also a manifold, and conversely. In order to understand these guys, one needs a pretty good understanding of topological manifolds and also of group theory. Sure, you can get information from the Wiki, but you won't get an education there. So if you want to know more about manifolds ask away here. If you want to know about the Lie groups after that, ask away. Just don't keep going off half-cocked on subjects you clearly do not fully understand
Please, Quarkhead. I'd love to know a little about manifolds. I suspect my preconceptions are flawed at best...
However, the plane, or \( E^2 \), is a manifold. Or, it is a manifold when isometries are included. I presume that means the "bare space", with no definition of lengths, or angles, is not a manifold. But, I can see a problem with that, along the lines you've described: how do you mathematically define a Euclidean plane without invoking geometry? Moreover, Lie groups correspond to continuous symmetries whereas isometries are generally discrete. How would you correct the Wikipedia claim that the isometry group makes \( E^2 \) a single orbit? Or how would you explain what that means? I did (I think) learn what orbits and stabilizers are, when I studied a bit of group theory.
No -- isometries are not definitions of lengths and angles, but mappings that preserve lengths and angles. What makes the Euclidean plane a topological manifold is that there exists at least one way to cover it by a collection of 1-or-more overlapping chunks with homeomorphisms between each of the chunks and the Euclidean plane. This is trivial. What makes the Euclidean plane a differentiable manifold is that there exists at least one way to cover it by a collection of 1-or-more overlapping chunks with diffeomorphisms between each of the chunks and \(\mathbb{R}^2\). This is a trickier concept than homomorphisms since if you distinguish between the two, the Euclidean plane has no origin or axes while \(\mathbb{R}^2\) comes with the axes and origin of Cartesian coordinates. But the Euclidean plane does have precisely the type of continuity exhibited by \(\mathbb{R}^2\) and the geometric concepts of tangent and curvature of smooth figures in the Euclidean plane have simple expressions in representations of those curves as parametric expressions in Cartesian coordinates. \(0 = \left( x - x(\lambda_0) \right) y'(\lambda_0) - x'(\lambda_0) \left( y - y(\lambda_0) \right) \\ k = \frac{ x'(\lambda_0) y''(\lambda_0) - x''(\lambda_0) y'(\lambda_0) }{ \left( x'(\lambda_0)^2 + y'(\lambda_0)^2 \right)^\frac{3}{2}} \) But can also be expressed in polar coordinates: \( 0 = r(\lambda_0) \theta'(\lambda_0) \left( r \cos(\theta - \theta(\lambda_0) ) - r(\lambda_0) \right) - r r'(\lambda_0) \sin(\theta - \theta(\lambda_0) ) \\ k = \frac{ 2 r'(\lambda_0)^2 \theta'(\lambda_0) + r(\lambda_0)^2 \theta'(\lambda_0)^3 + r(\lambda_0) r'(\lambda_0) \theta''(\lambda_0) - r(\lambda_0) r''(\lambda_0) \theta'(\lambda_0) }{ \left( r'(\lambda_0)^2 + r(\lambda_0)^2 \theta'(\lambda_0)^2 \right)^\frac{3}{2}} \) So the easy diffeomorphism \(E^2 \to \mathbb{R}^2\) is not necessarily the one that is simple with respect to derivatives and straight lines.
But without lengths and angles, there aren't any isometries either. The overlapping chunks you speak of, would they not be sets of points, and so would their elements be described by ordered pairs? Does this pair have to be a pair of numbers?
My experience was that we assume the Euclidean plane comes with distance (between pairs of points) defined on it. Since (in 2d) we can have intersecting lines, these intersections also define single points. Likewise planes (in 3d) can intersect along lines, so lines are defined by intersecting planes. Is a plane surface the intersection of two surfaces with one more dimension, and, is there such a thing as the intersection of two points? And when the topic was plane symmetric figures, these aren't located relative to any coordinates. The Euclidean plane has no boundary, so the discrete symmetries apply everywhere.
So, Euclid defined the plane based on geometric axioms, of which there are five. We have an interesting comment from Wikipedia: " Euclidean geometry is an example of synthetic geometry, in that it proceeds logically from axioms to propositions without the use of coordinates. This is in contrast to analytic geometry, which uses coordinates. " And an example of analytic geometry, I presume, is via the map of the celestial sphere which enables proper astronomy.
The would be the "open sets" of topology. By definition in \(\mathbb{R}^2\). In \(E^2\) it depends on the definition. In my thinking they would just be "geometric points" without any structures involved. I would say it doesn't even have to be a pair. Euclidean geometry existed for thousands of years without orders pairs of numbers.
