Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is a manifold?

  1. Jul 31, 2014 #1

    iyz

    User Avatar

    Can somebody explain to me what is a manifold.Also what it means for a space to be curved and how we define curvature.I know that a sphere is a curved 2d object, can a curved 3d object live in 3-dimensional space?
     
  2. jcsd
  3. Jul 31, 2014 #2

    WWGD

    User Avatar
    Science Advisor
    Gold Member

  4. Jul 31, 2014 #3

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    I think you can start off by thinking of a manifold as basically any set of points (elements) which locally looks like flat (Euclidean) space, which means I can draw a map of it locally (although my map might not necessarily be 2-D), and ALSO that I can somehow smoothly stitch together these maps to form an atlas of the whole manifold.

    That's basically it in an intuitive sense. More complicated if you ask for the full math definition.
     
  5. Jul 31, 2014 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Yeah, so an intuitive definition of a manifold is difficult. As S. Weinberger said:

    I guess my favorite definition would be to look at the graphs of smooth functions ##f:\mathbb{R}^n\rightarrow \mathbb{R}^m## or subsets of those. If you glue those together in a smooth way, then you get a a manifold. This can be made rigorous, but you need some topological notions for that.

    It's easier just to list some examples: sphere, circle, smooth curve, torus, basically anything which feels smooth if you touch it and which can't cut you (like a square).

    Now, when you look at the most general definition then you will probably not at all see the connection between what I described above and the definition. However, it can be proven (but is not easy to do so), that it does give the same thing. It is called the Whitney embedding theorem(s).

    Why do they make the definition so complicated? First, because mathematicians like to do things abstractly and like to make things more complicated than they are. Second, because we want to get rid of the ambient space. For example, a sphere is embedded in the ambient space ##\mathbb{R}^3##, but we want to study the sphere independently of its ambient space. This is useful in physics like relativity, because spacetime is supposed to be a manifold, but it is nonphysical to speak of an ambient space (in fact, if one exists, then it has to have close to 100 dimensions!).
     
  6. Jul 31, 2014 #5

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If we put mathematical rigor aside for a while, I think it's easy to explain manifolds, but not curvature. So I will only attempt to explain manifolds. A post like this can still get very long if we go deep into the subject, so I will stop right after we get to tangent spaces.

    The basic idea is this: Consider a set M. Suppose that there's a bunch of injective functions ##x:U\to\mathbb R^n## (all with the same value of n) such that ##U\subseteq M##. These functions are called coordinate systems or charts. We use the notation ##x(p)=(x^1(p),\dots,x^n(p))##, and call the components of this n-tuple the coordinates of the point p in the coordinate system x.

    Suppose also that these coordinate systems "cover" M in the sense that the union of their domains is M. If these conditions are met, as well as a bunch of technical assumptions of course, we call the set M a manifold.

    The claim that a manifold "locally" "looks like" ##\mathbb R^n## means precisely that for each ##p\in M##, there's a coordinate system ##x:U\to\mathbb R^n## such that ##p\in U##.

    The coordinate systems can be used to define partial derivatives of functions ##f:M\to\mathbb R##. The ith partial derivative of f at p is defined by
    $$\frac{\partial}{\partial x^i}\bigg|_p f=(f\circ x^{-1})_{,i}(x(p)),$$ where I have used the notation ##{}_{,i}## for the (usual kind of) partial derivative of a function with a domain that's a subset of ##\mathbb R^n##.

    Linear combinations of the derivations ##\frac{\partial}{\partial x^i}\big|_p## can be defined in an obvious way. The set of all linear combinations turns out to be a vector space. In fact, it turns out to be the same vector space, regardless of what coordinate system we use (as long as its domain includes p). This vector space is called the tangent space of M at p, and is denoted by ##T_pM##.
     
  7. Aug 7, 2014 #6

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    curvature requires a way to define length. then we compare the radius of circles to their circumference to get curvature. i.e. curvature zero means the circumference is 2π times the radius, and positive curvature means the circumference is less than that and negative curvature means the circumference is more than that. thats all, at least on surfaces.
     
  8. Aug 7, 2014 #7

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    Affine curvature is a perfectly valid way of defining curvature.. that only requires a connection between adjacent tangent spaces, with no need for a definition of length on the manifold. Unless I'm missing a subtle point here?
     
  9. Aug 9, 2014 #8
    I think I'll take a stab at this. I'll try to make this fun and step away from the formalities for now.

    Below is a hyperlink to a map of Yellowstone National Park*.

    http://yellowstone.net/maps/files/2012/06/yellowstone-map.jpg

    The map is perfectly flat; you can see it on your computer monitor, after all! But, as everyone knows by now, Earth is not flat. It is (roughly) spherical.

    Is the map wrong? How can something be flat and not flat at the same time?

    Clearly, the map isn't "wrong". Anyone who has navigated using a map knows that maps work very well for finding locations. Resolution comes from the idea of local flatness. This can be made rigorous with a bit of topology, using the language of "homeomorphisms". Again, we will avoid this. The idea is fairly simple, though: if you were walking around on a sphere, then you would not be able to tell the difference between your surroundings and a plane ([itex]\mathbb{R}^2[/itex]).

    Generalizing, the "defining" property of a manifold is the following: if you were walking around on a manifold, then you would not be able to tell the difference between your surroundings and Euclidean space ([itex]\mathbb{R}^n[/itex]).

    Curvature is a little bit harder to describe, especially since there are a LOT of ways to look at it. I'll do my best.

    The tangent space to a point on a manifold is something that we can define if our manifold is really nice. Intuitively, we can think of it as exactly what it sounds like: the space of vectors originating from a point on a manifold such that the vectors are tangent to the manifold. One problem for manifolds is that we cannot add vectors from different tangent spaces without extra structure. After all, if you are pushing on a box, the force you exert will not magically act on another box that is several miles from you; it acts on your box.

    We can resolve this in several ways. One way, from "Riemannian geometry," is to introduce a "connection". Without going into too much detail, connection provides a way of pushing tangent vectors along a curve. Most of the time, we disguise connections as things called "covariant derivatives," which are a kind of generalization of the derivative. Curvature, vaguely speaking, is a way of measuring how much the covariant derivative doesn't act like regular derivatives, in the sense of "equality of mixed partials."



    *For non-American readers, all you need to know is that Yellowstone is a REAAALLY big park.
     
  10. Aug 10, 2014 #9
    You are correct.

    There are three intuitive ways to think about curvature. One approach is that if you understand curvature in 2-dimensions, you can try to use that to generalize it to higher dimensions by considering well-chosen 2-dimensional slices. Just define the curvature at a point by looking at all possible 2-dimensional slices of the manifold near that point (this is a little tricky because these need to be "geodesic" slices to have any meaning). This is the way Riemann thought about it when he came up with the idea. To avoid getting too technical, maybe I won't attempt to flesh this idea out, but it should give you some idea.

    By the way, curvature of a surface is, of course, not some vaguely defined thing. If you have a surface in 3-dimensional space, you can measure its curvature at a point by intersecting it with a plane and looking at the curvature of the resulting curve. The Gaussian curvature is defined to be the maximum curvature you can get that way, times the minimum. One of Gauss's achievements was to prove that this curvature value doesn't change if you you bend the surface without stretching it, so that you can think of this curvature as somehow not depending on exactly how the surface is sitting in 3-dimensional space.

    Another way to think about curvature is in terms of geodesics or curves in the manifold of locally minimum length that act like "straight" lines in the manifold. If you want to go from point A to point B on the globe, you have to travel along a great circle. That's an example of a geodesic. If you imagine 2 ants on a sphere, both starting on the equator, fairly close to each other and have them both go up perpendicular to the equation, what you notice is that the ants will be heading towards the North pole, so they are getting closer together. This getting closer together is exactly what curvature measures. It measures how geodesics get closer together. If you measure it in all different directions (for each 2-dimensional plane passing through a point), you have a kind of curvature that makes sense in higher dimensions.

    Finally, the 3rd intuitive way to think about curvature is in terms of a concept called parallel transport. Basically, you can slide vectors around a little tiny parallelogram, and the curvature measures the difference between. the original vector and the vector that slid all the way around the parallelogram, to some approximation. Again, it's hard to flesh this out here, but hopefully, it gives you some idea. You can read more about it here:

    http://math.ucr.edu/home/baez/einstein/einstein.html
     
  11. Aug 10, 2014 #10
    I should add that curvature can also measure how much geodesics spread apart if they happen to do so (or don't diverge at all, as in the case of parallel straight lines in Euclidean space), not just get closer.
     
  12. Aug 10, 2014 #11
    Also, I should stress that the viewpoints I mentioned can be arrived at purely by studying the Gaussian curvature of surfaces in 3-dimensional Euclidean space. So, the point is when you understand curvature of surfaces really, really well, it becomes clear that, if you take the right point of view, the concept still makes sense in higher dimensions.
     
  13. Aug 10, 2014 #12

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    Thank you for the correction, please replace the word "requires" in the following sentence by the phrase, "can be simply defined in terms of":

    "curvature requires a way to define length."


    My goal was to give the simplest, most intuitive possible interpretation I could think of.
     
  14. Aug 11, 2014 #13

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    Here is the simplest characterization I know of the Gaussian curvature of a surface at a point. Arrange the point to have coordinates (0,0,0) in 3 space and the surface to be the graph of a smooth function z = f(x,y) near (0,0,0). Moreover make the tangent plane at the origin equal to the x axis. Then both the constant term and the linear terms of the taylor expansion are zero, and now and rotate until the quadratic polynomial term is diagonal, i.e. both mixed 2nd partials are zero. then the quadratic term of the taylor expansion looks like AX^2 + BY^2, and Z = AX^2+BY^2 is a quadratic approximation tot he surface near (0,0,0).

    Then the Gauss curvature at the origin is the product AB, i.e. it is the determinant of the matrix of the quadratic taylor polynomial. If it is positive, i.e. the curvature is positive, it tells you the surface looks locally like a paraboloid, and if it is negative it looks locally like a hyperboloid, and if zero, the it looks locally like a cylinder or plane. So the curvature tells you the rough local shape of the surface at that point.

    This relates to what homeomorphic said, because the maximum and minimum curvatures of a slice of the surface are in the directions of the x and y axes here, and those curvatures are A and B, so the Gauss curvature is the product of those "principal curvatures".

    Gauss and Riemann originally thought of it in terms of comparing the extent to which the angles of a small triangle exceed or fail to reach 180 degrees, with the area of the triangle. This is related to the version I gave comparing the circumference of a small circle to its radius.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook