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About basis vectors

  1. May 25, 2008 #1
    This is still rather new to me so please pardon my ignorance. My introduction to tensor differentiation involved only manifolds that were embedded in higher-dimensional Euclidean spaces. To describe them, I was instructed to find basis vectors using partial derivatives, as in

    [tex]
    e_\theta = <\frac {\partial x}{\partial \theta},\frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta}>
    [/tex]

    then to use those and the dot product to find both the metric tensor and the Christoffel symbols.

    [tex]g_\phi_\theta =e_\phi \cdot e_\theta[/tex]

    [tex] g_\theta_\theta =e_\theta \cdot e_\theta[/tex]
    .
    .

    [tex]
    \Gamma ^\theta _\theta _\phi = e^\theta \cdot \frac {\partial e_\theta}{\partial \phi}

    [/tex]

    etc.

    Since then I've noticed that GR doesn't seem to bother with higher-dimensional flat spaces, and instead begins with the metric tensor, using that alone to compute the Christoffel symbols. That's fine with me, but what about basis vectors? Can those somehow also be computed using the metric tensor? or should I regard them simply as

    [tex]
    e_0 = <1,0,0,0>[/tex]
    [tex] e_1 =<0,1,0,0>[/tex]
    [tex] e_2 =<0,0,1,0>[/tex]
    [tex] e_3 = <0,0,0,1>[/tex]

    and move on? Does it matter?
     
  2. jcsd
  3. May 25, 2008 #2

    Hurkyl

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    Manifolds don't come with a standard basis. However, if you are using a coordinate chart, then the standard basis on R^n does give you a basis for the vector fields defined on the range of your coordinates.
     
  4. May 25, 2008 #3

    Fredrik

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    If M is a manifold, U is an open subset of M, p is a point in U, and [itex]x:U\rightarrow \mathbb{R}^n[/itex] is a coordinate system, then the partial derivative operators

    [tex]\frac{\partial}{\partial x^\mu}\bigg|_p[/tex]

    are basis vectors of the tangent space TpM of M at p.

    These operators are defined by their action on functions [itex]f:M\rightarrow\mathbb{R}[/itex].

    [tex]\frac{\partial}{\partial x^\mu}\bigg|_p f=(f\circ x^{-1}),_\mu(x(p))[/tex]

    where [itex],_\mu[/itex] denotes the partial derivate of the function, with respect to the [itex]\mu[/itex]th variable.

    So, no need to involve the metric. The components [itex]g_{\mu\nu}(p)[/itex] of the metric at p, are given by

    [tex]g_{\mu\nu}(p)=g\bigg(\frac{\partial}{\partial x^\mu}\bigg|_p,\frac{\partial}{\partial x^\nu}\bigg|_p\bigg)[/tex]

    When the manifold is [itex]\mathbb{R}^n[/itex], it's convenient to take the coordinate system to be the identity map I on [itex]\mathbb{R}^n[/itex], defined by I(p)=p for all p in [itex]\mathbb{R}^n[/itex].
     
  5. May 25, 2008 #4
    Thanks guys, I appreciate that.

    Fredrik, can you provide (or direct me to) an example of this procedure? I don't understand why

    [tex]\frac{\partial}{\partial x^\mu}\bigg|_p f=(f\circ x^{-1}),_\mu(x(p))[/tex]

    instead of

    [tex]\frac{\partial}{\partial x^\mu}\bigg|_p f=(f\circ x),_\mu(x(p))[/tex]

    and it might help if I see it in action.
     
  6. May 26, 2008 #5

    Fredrik

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    The reason for that is that x (the coordinate system) maps a subset of the manifold onto a subset of Rn, and f maps the manifold to R. So [itex]f\circ x^{-1}[/itex] is a real-valued function of n real variables, and we know how to take partial derivatives of those. [itex]f\circ x[/itex] doesn't make sense, since the range of x is not a subset of the domain of f.

    I'm not sure what kind of example I should show you. Here's the relationship between the partial derivative operators associated with two different coordinate systems, x and y:

    [tex]\frac{\partial}{\partial y^\mu}\bigg|_p f=(f\circ y^{-1}),_\mu(y(p)) =(f\circ x^{-1}\circ x\circ y^{-1}),_\mu(y(p))[/tex]

    [tex]=(f\circ x^{-1}),_\nu(x\circ y^{-1}(y(p)))\cdot(x\circ y^{-1})^\nu{},_{\mu}(y(p))[/tex]

    [tex]=(f\circ x^{-1}),_\nu(x(p))\cdot (x\circ y^{-1})^\nu{},_{\mu}(y(p))[/tex]

    [tex]=(x\circ y^{-1})^\nu{},_{\mu}(y(p)) \frac{\partial}{\partial x^\nu}\bigg|_p f[/tex]

    You probably won't see it right away, but all I did to get from the first line to the second was to use the chain rule. (I'm also using the Einstein summation convention. We don't write any sigma symbols for summation, because we know that we're supposed to take a sum when we see the same index appear twice).

    So the relationship between these two sets of basis vectors is

    [tex]\frac{\partial}{\partial y^\mu}\bigg|_p = \Lambda^\nu{}_\mu\frac{\partial}{\partial x^\nu}\bigg|_p[/tex]

    where the [itex]\Lambda^\nu{}_\mu[/itex] are the components of the Jacobian matrix of [itex]x\circ y^{-1}[/itex]
     
    Last edited: May 26, 2008
  7. May 26, 2008 #6
    I just wanted to point out that something important about what Fredrick said-- noticed his coordinate system or chart was from U to R^n and not M to R^n. That's no coincidence, you usually need multiple charts to cover a manifold (i.e. there is no global coordinate system), the manifold in general can only be locally mapped to R^n.

    I also don't like that the exercise seems to teach a false lesson that basis vectors need to be (a) constructed from coordinate systems and (b) live in the manifold (as opposed to the tangent space where they belong). There is nothing technically wrong with the exercise but working out something that is particular to E^n (n dimensional Euclidean space) doesn't help build intuition for the more general treatment.
     
  8. May 26, 2008 #7

    Fredrik

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    This is a good point. The tangent space can be defined as the vector space of all operators that take functions to numbers and are i) linear, and ii) satisfy the Leibnitz rule: v(fg)=v(f)g(p)+f(p)v(g). The result is the same though, because we can show that the partial derivative operators I defined are basis vectors of this vectors space.

    I actually prefer the coordinate dependent method in this particular case, even though I prefer coordinate independence in all other definitions.

    There's nothing in the calculation that depends on the manifold being Euclidean. It makes no reference to a metric, and it doesn't suggest that the tangent vectors live in the manifold.
     
    Last edited: May 26, 2008
  9. May 26, 2008 #8
    I misread the problem, I thought it said that it was Euclidean, and now I see that the OP was saying that his previous exposure was only Euclidean. Sorry.
     
  10. May 26, 2008 #9
    I'd like to ask a slight variation on my original question (one which will probably seem trivial at this point), if that's OK.

    Suppose I have a three-dimensional manifold covered by coordinate system (a,b,c) with metric

    [tex]
    ds^2 = g_{aa}da^2 + g_{ab}da db + g_{ac}da dc..[/tex]

    and I know nothing else about it. Is it alright to define my basis vectors

    [tex]
    e_a = <1,0,0>, e_b = <0,1,0>, e_c = <0,0,1>[/tex]

    so long as I define my basis covectors as

    [tex]
    \omega ^a = g^{ap}e_p, \omega ^b = g^{bp}e_p, \omega ^c = g^{cp}e_p[/tex]?
     
    Last edited: May 26, 2008
  11. May 26, 2008 #10
    If you embed your manifold in some higher R^n, then once you pick your basis vectors and your "induced metric", which is the pullback of the Euclidean R^n metric onto the manifold, then any calculation you can do with this manifold can be done solely with that metric and the basis vectors. The fact that it was embedded can be forgotten, and you can talk purely about intrinsic quantities. Then, next time you want to consider a curved manifold, forget about embedding it (which can always be done, btw), and just write down a metric and some coordinates =)
     
  12. May 27, 2008 #11
    Thanks, Ibrits (lbrits?). Unfortunately, I don’t know how to embed a curved manifold in a higher-dimensional Euclidean space in the first place. Is there a standard technique, or is it a matter of trial-and-error?
     
  13. May 27, 2008 #12
    If it isn't obvious how to, don't try, since you don't need to. You're either given an intrinsic metric, or need to calculate it (e.g., Einstein's equations).

    If you want some practice, though, you can parameterize various objects like 2-spheres and cylinders and cones and whatnot in 3-space and find their induced metrics. You can also do the same for 3-spheres and embed them in 4 dimensions. But the general thing goes along Nash's theorem and I think it assumes you already have some geometric structure, like a metric.
     
  14. May 28, 2008 #13
    OK; I'll see if I can find a proof of Nash's theorem out there.
     
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