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Basis vectors one-forms

  1. Aug 29, 2012 #1
    Greetings,

    I have just started studying manifolds, and have come across the idea that the basis vectors can be expressed as:

    e[itex]\mu[/itex] = [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex].

    I tried to convince myself of this in 2D Cartesian coordinates using a pretty non-rigorous derivation (the idea being to get a simple intuitive grasp rather than an abstract mathematical one) starting with the position vector, as follows:

    r(x,y) = xe1 + ye2

    dr = [itex]\partial[/itex]r/[itex]\partial[/itex]x dx + [itex]\partial[/itex]r/[itex]\partial[/itex]y dy = [itex]\partial[/itex](xe1+ye2)/[itex]\partial[/itex]x dx + [itex]\partial[/itex](xe1+ye2)/[itex]\partial[/itex]y dy

    Expand with the product rule, everything other than [itex]\partial[/itex]x/[itex]\partial[/itex]x e1dx and [itex]\partial[/itex]y/[itex]\partial[/itex]y e2dy (each partial combination is obviously 1) goes to 0 since the Cartesian basis vectors are constant and dx and dy are independent, so that:

    dr = [itex]\partial[/itex]r/[itex]\partial[/itex]x dx + [itex]\partial[/itex]r/[itex]\partial[/itex]y dy = e1dx + e2dy

    My first question: I would conclude that e[itex]\mu[/itex] = [itex]\partial[/itex]r/[itex]\partial[/itex]x[itex]\mu[/itex], rather than e[itex]\mu[/itex] = [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex]. What did I miss here?

    My second question: Is there an equivalent expression to e[itex]\mu[/itex] = [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex] for basis one-forms? If so, could anyone please provide a quasi-derivation similar to mine above, except correct in the way mine was wrong?

    Thanks for any help you can give.

    -HJ Farnsworth
     
  2. jcsd
  3. Aug 29, 2012 #2

    pervect

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    The short answer is that just as [itex]\partial / \partial x[/itex] can be interpreted as a basis vector, dx can be interpreted one form. Some authors prefer to use the notation dx, boldfacing the d, for this purpose.

    Note that any one form is defined as a map from a vector to a scalar, so only when dx is thought of as such a mapping does it qualify as a one form. Some times dx is just a scalar and doesn't "operate" on anything. So the difference is subtle, the value of dx is always a scalar, but when it's a one-form, dx operates on a vector to return a scalar.

    Note that assuming a cartesian basis, dx maps the basis vector [itex]\partial / \partial x[/itex] to 1 and other basis vectors to zero, which is what one wants from a basis one-form
     
  4. Aug 29, 2012 #3
    Thanks for the response, Pervect.

    Now that you say it, it seems obvious. I knew why [itex]\tilde{d}[/itex]x[itex]\mu[/itex] = [itex]\tilde{\omega}[/itex][itex]\mu[/itex], for some reason it didn't occur to me that [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex] = e[itex]\mu[/itex] was just a mirror expression to that. So thank you for clearing that up.

    Does anyone know the answer to the first question in my original post?

    Thanks again for the help.

    -HJ Farnsworth
     
    Last edited: Aug 29, 2012
  5. Aug 29, 2012 #4
    What makes you think that [itex]d{\bf r} ={\bf e}_1 dx + {\bf e}_2 dy[/itex]?
    (in fact is dr a tangent vector?)
     
  6. Aug 29, 2012 #5
    Thanks for the reply.

    I got that when I expanded my equation with the product rule. I didn't type that out explicitly in my original post, but will do so now (the r's and e's below should be bold and some of the numbers should be subscripted, I wasn't sure how to get Latex to do that within the [itex]\frac tag. Anyone know how?):

    dr = [itex]\frac{\partial r}{\partial x}[/itex]dx+[itex]\frac{\partial r}{\partial y}[/itex]dy = [itex]\frac{\partial (xe1+ye2)}{\partial x}[/itex]dx+[itex]\frac{\partial (xe1+ye2)}{\partial y}[/itex]dy

    Using the product rule, the dx term in the RHS of the above equation is:

    [itex]\frac{\partial (xe1+ye2)}{\partial x}[/itex]dx = [itex]\frac{\partial x}{\partial x}[/itex]e1dx + x[itex]\frac{\partial e1}{\partial x}[/itex]dx + [itex]\frac{\partial y}{\partial x}[/itex]e2dx + y[itex]\frac{\partial e2}{\partial x}[/itex]dx

    And thy dy term is basically the same:

    [itex]\frac{\partial (xe1+ye2)}{\partial y}[/itex]dy = [itex]\frac{\partial x}{\partial y}[/itex]e1dy + x[itex]\frac{\partial e1}{\partial y}[/itex]dy + [itex]\frac{\partial y}{\partial y}[/itex]e2dy + y[itex]\frac{\partial e2}{\partial y}[/itex]dy

    e1 and e2 are constant, ie., they stay the same regardless to the x- and y-coordinates, so [itex]\frac{\partial e1}{\partial x}[/itex] = [itex]\frac{\partial e2}{\partial x}[/itex] = [itex]\frac{\partial e1}{\partial y}[/itex] = [itex]\frac{\partial e2}{\partial y}[/itex] = 0. Additionally, the x- and y-coordinates are independent from one another, so [itex]\frac{\partial x}{\partial y}[/itex] = [itex]\frac{\partial y}{\partial x}[/itex] = 0. So the equation should be:

    dr = [itex]\frac{\partial r}{\partial x}[/itex]dx+[itex]\frac{\partial r}{\partial y}[/itex]dy = [itex]\frac{\partial (xe1+ye2)}{\partial x}[/itex]dx+[itex]\frac{\partial (xe1+ye2)}{\partial y}[/itex]dy = [itex]\frac{\partial x}{\partial x}[/itex]e1dx+[itex]\frac{\partial y}{\partial y}[/itex]e2dy.

    Or:

    dr = e1dx+e2dy
     
    Last edited: Aug 29, 2012
  7. Aug 30, 2012 #6
    I have looked around a little more, and have found a couple of sources that imply that [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex] is in fact short-hand for [itex]\partial[/itex]r/[itex]\partial[/itex]x[itex]\mu[/itex].

    While this makes sense of the math I did in my first post, it seems like a very strange short-hand way of writing something, since the two expressions blatantly appear to mean different things.

    Does this seem reasonable to anyone? Or, is there anyone who knows for a fact whether or not it is indeed a short-hand way of writing the expression? If it is, why are the two expressions equivalent? If it is not, does anyone know why the derivation in my first post was incorrect?

    Thanks again.

    -HJ Farnsworth
     
  8. Aug 30, 2012 #7
    you seem to have missed my point. I was asking how you
    are reasoning about tangent vectors using dr?
    When in doubt go back to definitions. What is your definition of a tangent vector?

    As to your most recent question. [itex]\frac{\partial}{\partial x^\mu}[/itex] is clearly a different
    mathematical entity than [itex] \frac{\partial {\bf r}}{\partial x^\mu}[/itex]. That being
    said, there is an obvious isomorphism between the two.
     
  9. Aug 30, 2012 #8
    Thanks for responding again, qbert. I tried it again from a tangent vector perspective, using curves. This was successful, though there was still one part at the end that I find unintuitive. My reasoning is below.

    Definition of a tangent vector
    A manifold is, by definition, locally Euclidean. Therefore, at a point P on a manifold, we can define a Euclidean space, of equal dimension to the manifold, which contains P. This space is the tangent space to the manifold at P. A tangent vector is any vector that lies within that tangent space. Furthermore, any vector defined at any point on a manifold lies within the tangent space at that point.

    Finding the basis vectors
    We can index all of the points on the manifold with arbitrary coordinates x[itex]\mu[/itex]. A curve on a manifold is a mapping from the real numbers to that manifold - we can define such a curve C as x[itex]\mu[/itex] = x[itex]\mu[/itex](t), where t is a real number and [itex]\mu[/itex] ranges over all the coordinates. We can also define a test scalar function f(x[itex]\mu[/itex]) on the manifold.

    An infinitesimal displacement along the curve is found by changing t infinitesimally. Then, the directional derivative of f along the curve at P will be given by:

    df/dt = [itex]\partial[/itex]f/[itex]\partial[/itex]x[itex]\mu[/itex] dx[itex]\mu[/itex]/dt

    Clearly, each dx[itex]\mu[/itex]/dt is a component of the tangent vector to C at P - because of how we defined C, a vector made of these components will just have to have the same direction as the curve itself. The tangent vector is obviously equal to the sum of its components times the basis vector for each component. Then, if d/dt is the tangent vector to the curve (this is a guess I'm taking because I know it will get me the answer I want, but I don't have an intuitive grasp on why the statement is true), we can get rid of the test function f from the above equation. This would clearly mean that the basis vectors are:

    e[itex]\mu[/itex] = [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex]

    As I mentioned above, I am unclear as to why d/dt is the tangent vector to the curve (at least, it seems like that is necessary for my derivation in this post to reach the right conclusion). I know that the test function f I used was arbitrary, but still don't quite see how d/dt is the tangent vector.

    I'll think about it and see if I can figure out that part. I will also probably follow up on this post later to see if I can figure out why my argument of the first post in this thread was wrong, using what I've learned from this exercise. That being said, any assistance is still always appreciated.

    Thanks again for the help.

    -HJ Farnsworth
     
  10. Aug 30, 2012 #9
    When I learned differential geometry many, many, years ago, I learned to express the coordinate basis vectors as partial of r wrt xu. In recent years, in studying relativity, I realized that modern math had replaced this with just partial wrt xu. They are really the same thing.

    The thing is, in a curved tangent space, it is not possible to draw an absolute position vector from one location in the tangent space to another neighboring location (without going out of the tangent space), unless the curved tangent space is immersed in a higher dimension flat space. However, the same differential position vector expression is still applicable, since the tangent space could actually be immersed in a higher dimension flat space.
     
  11. Aug 30, 2012 #10
    interesting. you have given a definition that i haven't seen before. maybe it's really
    the same as i'm used to, but not directly. It seems to rely on the natural association
    you can make: the tangent space to a vector space is isomorphic to the original vector space.
    And it's that isomorphism that's hiding the true nature of tangent space.
    [tex] M \cong {\mathbb R}^n \text{locally, and } T_x\mathbb{R}^n \cong \mathbb{R}^n.[/tex]
    Where we'd need to start unraveling is then how [itex]T_x\mathbb{R}^n[/itex], works.

    take a look at the formal definitions given on wikipedia http://en.wikipedia.org/wiki/Tangent_space#Formal_definitions

    1. Tangent vectors as velocity vectors to curves.
    2. Tangent vectors as equivalence classes of functions.
    3. Tangent vectors as derivations.

    If you have access to Lee's book (http://books.google.com/books/about/Introduction_to_Smooth_Manifolds.html?id=eqfgZtjQceYC),
    I'd recommend reading it. I think he has an excellent discussion.
     
  12. Aug 30, 2012 #11
    Hey Chestermiller, great to hear from you.

    If that's true, it is certainly reassuring, because as it stands now I've tried my original way of going about things and the way qbert suggested, and reached what seem like two different conclusions that both seem right to me (aside from not yet understanding that d/dt bit at the end of my last post).

    Do you know why the two basis vector definitions are equivalent? It seems like one is an operator, while the other is a vector. Now, I know that vectors are themselves operators since they definitively map one-forms to the real numbers, but for example, if I compare the two expressions:

    [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex] V

    with

    [itex]\partial[/itex]r/[itex]\partial[/itex]x[itex]\mu[/itex] V

    The first is obviously a derivative vector, while the second doesn't really mean anything to me yet.

    I unfortunately don't have access to the Lee book. I briefly looked at that section of the Wikipedia article yesterday, but foolishly ignored it because it had a lot of notation that I had been introduced to but hadn't worked with much, and I didn't feel like diving into it that much yet. But I will give it a more careful look.

    -HJ Farnsworth
     
  13. Aug 30, 2012 #12
    Basically what's going on is that you write [itex]x = x^0 e_0 + x^1 e_1 + x^2 e_2 + x^3 e_3[/itex] and you take a derivative,

    [tex]e_i \equiv \frac{\partial x}{\partial x^i}[/tex]

    Conversely, write [itex]\nabla = e^i \frac{\partial}{\partial x^i}[/itex] and the one forms [itex]e^i[/itex] are extracted by

    [tex]e^i \equiv \nabla x^i[/tex]

    Edit: both of the above are special cases of general identities. Namely, [itex](a \cdot \nabla) x = a[/itex] and [itex]\nabla (x \cdot a) = a[/itex] for arbitrary vector [itex]a[/itex]. See that [itex]e_i \cdot \nabla x = \frac{\partial x}{\partial x^i}[/itex] and that [itex]\nabla (x \cdot e^i) = \nabla x^i[/itex].

    Of course, phrasing it this way seems dangerously circular, but that's why the definitions in terms of partial derivatives are typically given.
     
    Last edited: Aug 30, 2012
  14. Aug 30, 2012 #13
    bah!

    for Rn you have god given basis vectors {e_1= (1,0,0 ...), e_2= (0,1,0, ...), etc. }
    so it makes sense to write [itex] {\bf x} = x^ie_i [/itex], if you are talking about an
    embedded surface in Rn (i.e. a smooth map) [itex] S : (u_1, u_2, \cdots u_n) \mapsto {\bf x}({\bf u}) [/itex]
    Then it makes sense to think of [itex] {\bf e}_i = \frac{\partial {\bf x}}{\partial u^i} [/itex],
    as a vector tangent to the surface. These vectors are tangent in the usual sense.

    For a manifold, M, however, given a point p, and a chart that maps p to Rn,
    [itex] p \stackrel{\phi}{\mapsto} (u^1(p), u^2(p), \ldots, u^n(p))[/itex], then trying to use [itex] {\bf e}_i = \frac{\partial {\bf x}}{\partial u^i} [/itex]
    is not meaningful, what's x(u), here?

    So you back up and try to find another way to define the tangent vectors. Some
    way that's close enough that when you apply it to surfaces in Rn it reproduces the
    "correct" results.

    For surfaces in Rn, for each vector [itex] {\bf e}_i = \frac{\partial {\bf x}}{\partial u^i} [/itex],
    we can always form the directional derivative
    [tex]D_i f = \lim_{t \rightarrow 0} \frac{f({\bf{x}} + t \bf{e}_i) - f(\bf{x})}{t}[/tex],
    Look at what that is in terms of u
    [tex]D_i f = \lim_{t \rightarrow 0} \frac{f({\bf{x(u)}} + t \frac{\partial {\bf x(u)}}{\partial u^i}) - f(\bf{x(u)})}{t} = \frac{\partial f}{\partial x^j}\frac{\partial x^j}{\partial u^i} = \frac{\partial f}{\partial u^i} [/tex]

    So, for each tangent vector we can associate a differential operator, such that
    the directional derivative in the direction of the tangent vector is the differential
    operator. Similarly any tangent vector is the directional derivative of SOME
    vector. So the two spaces tangent vectors and directional derivatives are really
    the same (isomporphic). Since directional derivatives is an idea we can apply
    on manifolds, that's the one we use.
     
  15. Aug 30, 2012 #14

    atyy

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    That's a nice example. A vector is written as d/dt with an empty "slot" because a vector is conceived as something that eats a function and spits out a number (the derivative is evaluated at the point, since the vector space is the tangent space at that point, and different points have different tangent spaces). So the empty slot just shows that the vector can eat many different functions, but for any function it satisfied all the formal requirements of being a vector (and a derivative). For example, f(x,y) could be the temperature, and g(x,y) could be the humidity. Then as you change your position over time (x(t),y(t)), you will experience changes in temperature df/dt and humidity dg/dt. But the change in temperature and humidity at each point are generated by the same underlying d/dt acting on different functions.

    It also makes sense because then in the expression df/dt=dx/dt.df/dx+dy/dt.df/dt, we have the common sense expression for the velocity (dx/dt,dy/dt) as the components of the tangent vector d/dt using the basis vectors d/dx and d/dy.
     
    Last edited: Aug 30, 2012
  16. Aug 30, 2012 #15

    DrGreg

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    Just to get the terminology straight, it's not a "curved tangent space", it's a "manifold" (M). The tangent space at an event x (TxM) is flat. Vectors exist in the tangent space, not the manifold.

    200px-Tangentialvektor.svg.png
     
  17. Aug 30, 2012 #16
    Thank you, atyy, that makes the reasoning for choosing d/dt as the tangent vector much more clear to me.

    I think I understand what you are saying now, qbert, but let me ask you to make sure...

    The reason that I can get either [itex]\partial[/itex]r/[itex]\partial[/itex]x[itex]\mu[/itex] or [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex] for the basis vector e[itex]\mu[/itex] is because I am defining vectors in two different ways, which are not equivalent. In the first way, which results in e[itex]\mu[/itex] = [itex]\partial[/itex]r/[itex]\partial[/itex]x[itex]\mu[/itex], vectors are defined based on the existence of a position vector r, while in the second way, which results in e[itex]\mu[/itex] = [itex]\partial[/itex]/[itex]\partial[/itex]x[itex]\mu[/itex], they are not. If we cannot define a position vector, then the second way still works, while the first way does not. We cannot always define a position vector - therefore, the second way is a more general definition than the first way, and is the one we should use.

    So, the first way is perfectly reasonable for certain, but not all, manifolds. However, the second way is also reasonable for those manifolds, and for those manifolds every vector defined in the first way corresponds to exactly one vector defined in the second way, and vice versa. Relating this to your last post, the first way tangent vector is what is normally thought of as a "traditional" tangent vector, while the second way tangent vector is a differential operator.

    For the other manifolds, only the second way works.

    In other words, the source of my confusion that started this thread was that I was using two different definitions of basis vectors, or tangent vectors, or just vectors, and thinking they should be expressed in the same way. There were two vector spaces, and I thought there was one, which led to a contradiction.

    Is this correct?

    -HJ Farnsworth
     
  18. Aug 30, 2012 #17
    i'd say you've got it.
     
  19. Aug 30, 2012 #18
    Great.

    Thank you very much for the help, everyone.

    -HJ Farnsworth
     
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