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Tangent vector as derivation question

  1. Jul 8, 2015 #1
    I have a question concerning the tangent space.

    Consider a manifold Mn and take Mn to be ℝn with the Euclidean metric for the purposes of this question.

    The directional derivative of a function in the direction of a vector v is

    (a) vf = ∑ vi(∂f/∂xi)

    where the sum runs from 1 to n. The vector v is then given as

    (b) v=∑ vi(∂/∂xi).

    Furthermore the claim is that the space of derivations at p is isomorphic to the space of geometric vectors at p. Thus, we can make the identifications:

    (c) (∂/∂xi)p <-----------> ei

    And I can see this to be true if we let the operator in (c) operate on the coordinate functions, and indeed that is what the books do. But to me, there seems to be a slight of hand going on here because in (a) the operator operated on an arbitrary function then suddenly in (b) and (c) the assumption is made that the operator operates on coordinate functions and not arbitrary functions. All books that I have read make this change from the vector operating on a arbitrary function to operating on the coordinate functions without justifying it.

    So my question is, what is the justification for specifying a specific set of functions in (b) and (c)?
     
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  3. Jul 8, 2015 #2

    Fredrik

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    The operator in (c) takes an arbitrary smooth ##f:M\to\mathbb R^n## as input. If it didn't, the f on the left in (a) would be different from the f on the right in (a). Maybe you're not aware of the definition of the notation in (c). If ##x:U\to\mathbb R^n## is a coordinate system and ##U\subseteq M##, then ##\big(\frac{\partial}{\partial x^i}\big)_p## is defined for each ##p\in U## and each ##i\in\{1,\dots,n\}## by
    $$\left(\frac{\partial}{\partial x^i}\right)_p f =(f\circ x^{-1})_{,i}(x(p))$$ for all smooth ##f:M\to\mathbb R^n##. (Here ##(f\circ x^{-1})_{,i}## denotes the partial derivative of ##f\circ x^{-1}:x(U)\to\mathbb R## with respect to the ##i##th variable slot).

    If you meant that ##\frac{\partial}{\partial x^i}## (with no specific point specified) acts on functions from a different set, that's incorrect. ##\frac{\partial}{\partial x^i}## (where x is a coordinate system with domain U) denotes the vector field with domain U that takes each ##q\in U## to ##\big(\frac{\partial}{\partial x^i}\big)_q##.

    Also, if v is a tangent vector at p, rather than a vector field with domain U, there should be a ##p## in (a) and (b) too.
     
    Last edited: Jul 8, 2015
  4. Jul 8, 2015 #3
    It's my fault that I didn't specify that these operators act at a particular p. Yes, the vector in (a) acts on an arbitrary smooth function at a point p. That I understand. And that's kind of my point.

    To see what I mean, let me refer you to a typical book that I use, namely J.M. Lee's Introduction to Smooth Manifolds. In the proof to Proposition 3.2 on page 53 (isomorphism between the derivations at p to the geometrical tangent space), he writes:

    "Writing Va = viei|a in terms of the standard basis, and taking f to be the jth coordinate function xj: ℝn→ℝ, thought of as a smooth function on ℝn, we obtain: ..." (emphasis mine).

    My question is why is he able to specify f in order to get the standard basis. In other words, how is it that (∂/∂xi)p is identified ei when the partial derivative operator operates on arbitrary functions (and not just on the coordinate functions).
     
  5. Jul 8, 2015 #4

    Fredrik

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    That theorem is on page 64 in my book. There's something that I didn't expect going on there. He's talking about tangent spaces of an (n-1)-sphere. But the functions he's talking about have domain ##\mathbb R^n##, not ##S^{n-1}##. In particular, the coordinate system ##x## has domain ##\mathbb R^n##. And the ##\frac{\partial}{\partial x^i}## aren't the vector fields discussed above. They are plain old partial derivatives.

    Definition: ##\mathbb R^n_a=\{a\}\times\mathbb R^n##.

    Definition: For each ##v\in\mathbb R^n_a##, we define a map ##D_v|_a## by
    $$D_v|_a f=\frac{d}{dt}\bigg|_0 f(a+tv)$$ for all smooth ##f:\mathbb R^n\to\mathbb R##. (Lee doesn't actually specify the domain of ##D_v|_a##. I'm inferring it from how he uses the chain rule a few lines later).

    What he's doing in the part you're asking about is to prove the following: If ##D_v|_a## takes every function in its domain to 0, then ##v=0##.

    If it takes every function to 0, then for all ##i##, it takes ##x^i## to 0, so for all ##i##, we have
    $$0=D_v|_a x^i =\frac{d}{dt}\bigg|_0 x^i(a+tv)= x^i{}_{,j}(a) v^j =\delta^i_j v^j =v^i.$$ This implies that ##v=0##.
     
  6. Jul 8, 2015 #5
    You have a completely different edition than I do because my book says nothing about an (n-1)-sphere. But the result and conclusion are the same. Ok, there he is using the coordinate functions for a particular purpose. I will give you another example. In Riemannian Geometry by Do Carmo, on page 8 (of my book at least), Do Carmo states,

    "Observe that (∂/∂xi)0 is the tangent vector at p of the 'coordinate curve' ...."


    My question really boils down to understanding the isomorphism between ei and (∂/∂xi)p when the partial derivative operator operates on arbitrary curves and not just coordinate curves. I wish I could express it more clearly than that.
     
  7. Jul 8, 2015 #6

    Fredrik

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    I found a pdf version of the second edition online. Proposition 3.2 is on page 53 in that one. The (n-1)-sphere is mentioned in the intro for the section titled "Tangent vectors" and the intro for the subsection titled "geometric tangent vectors", both on page 51. (That's page 61 in the printed copy of the first edition). What I said in post #4 is still the answer to what you asked in post #3.

    do Carmo is talking about tangent vectors of (curves in) arbitrary manifolds. Lee is talking about tangent vectors of ##\mathbb R^n## (or ##S^{n-1}##).
     
  8. Jul 10, 2015 #7

    Fredrik

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    Did that solve the problem? Do you understand that part of the proof now? Do you understand the rest of it?
     
  9. Jul 10, 2015 #8
    Yeah, thanks, that clears up that part of the proof, but I'm still struggling to understand my general question. I'll try asking it a different way because I think it has gotten clouded by referring to specific passages in books. What I'm trying to get at is how do we make the association (∂/∂xi)p ↔ ei if (∂/∂xi)p acts on arbitrary functions and not just on the coordinate functions?

    The reason why I refer to books is that they tend to choose the partial derivatives operating on the coordinate curves when it is convenient, but these operators are defined as acting on arbitrary functions. I have found other books that do the same thing even when it's not part of a proof. (See, eg, Crampin and Pirani Applicable Differential Geometry)

    To be honest, since I posted this question, I've read so much about it that I think I have absorbed something by osmosis even though I still struggle with justifying it to myself sometimes.

    Thank you for taking time to talk with me on this.
     
  10. Jul 10, 2015 #9

    Fredrik

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    I don't understand the concern about coordinate functions. I think you're seeing a problem where there isn't one. I will elaborate on that below.

    What you need to understand the association is Lee's proposition 3.2(b) and the chain rule.The isomorphism from ##\mathbb R^n_a## to the vector space of derivations at ##a## is the map ##v\mapsto D_v|_a##. The domain of each ##D_v|_a## is the set of smooth real-valued functions on ##\mathbb R^n## (not just "coordinate functions"). This isomorphism takes ##e_i## to ##D_{e_i}|_a##. For all smooth ##f:\mathbb R^n\to\mathbb R##, we have
    $$(D_{e_i}|_a)(f) =\frac{d}{dt}\bigg|_0 f(a+te_i) =f_{,j}(a)\frac{d}{dt}\bigg|_0 (a+te_i)^j =f_{,j}(a)\frac{d}{dt}\bigg|_0 (a^j+t\delta^j_i) =f_{,j}(a)\delta^j_i =f_{,i}(a).$$ This implies that
    $$D_{e_i}|_a =\frac{\partial}{\partial x^i}\bigg|_a$$ where the partial derivative is now denoted by ##\frac{\partial}{\partial x^i}## instead of ##{}_{,i}##. So the isomorphism takes ##e_i## to ##\frac{\partial}{\partial x^i}\big|_a##. Note that this is just the kind of partial derivative that we're familiar with from calculus. It's not the vector field I mentioned in my first post.

    Consider e.g. the function ##f:\mathbb R\to\mathbb R## defined by ##f(x)=x^2## for all ##x\in\mathbb R##. Would you get suspicious if someone says that ##f(2)=4##? We can input integers into this function, because its domain is the set of real numbers, and integers are real numbers. Similarly, we can input a coordinate function into a derivation, because its domain is the set of smooth functions, and coordinate functions are smooth functions.
     
    Last edited: Jul 10, 2015
  11. Jul 10, 2015 #10
    Thanks for explaining the isomorphism to me. That makes a lot of sense.



    This I think is what is bothering me. I will have to think on this more in the next few days to be clearer. Either it will become clearer to me or I will have a clearer question to ask.

    Thanks again.
     
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