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Tangent space basis

  1. Oct 5, 2011 #1
    I am unable to understand as to how the basis for the tangent space is
    [itex]\frac{\partial}{\partial x_{i}}[/itex]. Can this be proved ,atleast intuitively???
    Bachman's Forms book says that if co-ordinates of a point "p" in plane P are (x,y), then
    [itex]\frac{d(x+t,y)}{dt}=\left\langle 1,0\right\rangle[/itex] [itex]\frac{d(x,y+t)}{dt}=\left\langle 0,1\right\rangle[/itex] denote vectors in [itex]\T_{p}(P)[/itex] , in fact they are the basis.
    Also any point of [itex]T_{p}(P)[/itex] is [itex]dx\left\langle 0,1\right\rangle +dy\left\langle 1,0\right\rangle ;dx,dy\in\mathbb{R}[/itex]
    where does this "t" come from...is it a result of parametrization??

    This topic has propped up a few times in this forum, but after having gone through them I am still confused. I would appreciate any help.
    I hope I am not breaking any forum rules.
  2. jcsd
  3. Oct 5, 2011 #2
    First let me get some notation out of the way: If [itex] x \in \mathbb{R}^k [/itex] then a tangent vector to [itex] \mathbb{R}^k [/itex] at x is denoted by [itex] (x;v) [/itex] for some [itex] v \in \mathbb{R}^k [/itex]. Note that [itex] (x; v) + (x; w) = (x; v + w) [/itex] and if c is a scalar then [itex] c(x;v) = (x; cv) [/itex]

    Now, if M is a k-manifold in [itex] \mathbb{R}^n [/itex] then for any [itex] p \in M [/itex], if [itex] \alpha [/itex] is a coordinate patch about p such that [itex] \alpha(x) = p [/itex] for some [itex] x [/itex] in [itex] \mathbb{R}^k [/itex], then [itex] T_p(M) = \{ (\alpha(x); D\alpha(x) v) : v \in \mathbb{R}^k \} [/itex].

    Now that that's out of the way, choose any tangent vector in [itex] T_p(M) [/itex]. Say we choose [itex] (\alpha(x); D\alpha(x)v') [/itex] Note that [itex] v' = \sum_{i = 1}^k c_i e_i [/itex] for some scalars [itex] c_i [/itex]. So [itex] (\alpha(x); D\alpha(x)v') = (\alpha(x); D\alpha(x) \sum_{i = 1}^k c_i e_i) = \sum_{i=1}^k c_i (\alpha(x); D\alpha(x) e_i) = \sum_{i=1}^k c_i (\alpha(x); \frac{\partial \alpha}{\partial x_i})[/itex].

    Thus we see that [itex] \{ (\alpha(x); \frac{\partial \alpha}{\partial x_i}) : 1 \le i \le k \} [/itex] spans [itex] T_p(M) [/itex]. Linear independence of the set is guaranteed because since [itex] \alpha [/itex] is a coordinate patch about p, [itex] D\alpha(x) [/itex] has rank k. That's how the set is a basis.
  4. Oct 5, 2011 #3

    I think Bachman is trying to use the fact that directional derivatives are just the
    derivatives along the directions of the x-y planes dotted --dot product with--the
    direction of the product; you can see dx, dy respectively as the projections into
    the first, second components of a vector (x,y). In this sense, (x,y)=x(1,0)+y(0,1),
    so that {(1,0),(0,1)} are a basis. There is too, the correspondence between a vector,
    and the directional derivative in the direction of the vector.

    But the notation is confusing--welcome to Differential Geometry.
  5. Oct 5, 2011 #4
    first of all thanks a ton both of u....JG89 and Bacle...
    @JG89....i know very little about patches ...i am using Barrett O'Neill....but the content of ur post is pretty clear....however can this be done without bringing in patches??....
    also..by Bacle's statement...shudnt it be
    [itex]dx\left\langle 1,0\right\rangle +dy\left\langle 0,1\right\rangle ;dx,dy\in\mathbb{R}[/itex]
    and those brackets dont denote inner product by the book...they just denote vectors in
    and yes the 2 angled bracket elements are basis of the tangent space
  6. Oct 5, 2011 #5
    Yes, Paluskar, I think it should be dx(1,0)+dy(0,1).
  7. Oct 6, 2011 #6


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    I like the proof in Isham's book "Modern differential geometry for physicists". I think it's the same as the one in Wald's GR book, so I'm guessing that it appears in lots of books. You will have to read section 2.3.5, starting on page 79. The actual proof is on pages 82-84.
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