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

Total and exterior derivative of a 1-form

  1. Dec 22, 2009 #1

    mma

    User Avatar

    Suppose we have a one-form [tex] \alpha[/tex] having a skew-symmetric total derivative matrix. I mean something like [tex] \alpha(x,y) = -y dx + x dy[/tex], that is, in canonical [tex] (x, y, \xi, \eta)[/tex] coordinates of the cotangent bundle, [tex] \alpha(x,y) = (x, y, -y , x )[/tex].
    The "total derivate matrix" I mean
    [tex] D\alpha = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)[/tex]
    because the tangent mapping of [tex] \alpha[/tex] is
    [tex] a \frac{\partial}{\partial x} + b\frac{\partial}{\partial y} \mapsto a \frac{\partial}{\partial x} + b \frac{\partial}{\partial y} -b \frac{\partial}{\partial \xi} + a \frac{\partial}{\partial \eta}[/tex] ​
    and the projection of this vector on the [tex] \{\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \}[/tex] plane is always
    [tex] a \frac{\partial}{\partial x} + b\frac{\partial}{\partial y}[/tex] ​
    itself (independently of [tex] \alpha[/tex]), while the projection on the [tex] \{ \frac{\partial}{\partial \xi}, \frac{\partial}{\partial \eta} \}[/tex] plane is
    [tex] -b \frac{\partial}{\partial \xi} + a \frac{\partial}{\partial \eta}[/tex],
    that is, in column vector representation :
    [tex] \left(\begin{array}{c} -b \\ a \end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) \left(\begin{array}{c} a \\ b \end{array}\right)[/tex]

    A bit more generally, if our 1-form is [tex] \alpha(x,y) = \xi(x,y) dx + \eta(x,y) dy[/tex] then the "total derivative matrix" is [tex] D\alpha = \left( \begin{array}{cc} \frac{\partial \xi}{\partial x} & \frac{\partial \xi}{\partial y} \\ \frac{\partial \eta}{\partial x} & \frac{\partial \eta}{\partial y} \end{array} \right)[/tex].
    This matrix can always decompose into the sum of its symmetric and antisymmetric parts, where the symmetric and antisymmetric parts are
    [tex] S = \frac{1}{2} \left( \begin{array}{cc} 2\frac{\partial \xi}{\partial x} & \frac{\partial \xi}{\partial y} + \frac{\partial \eta}{\partial x} \\ \frac{\partial \eta}{\partial x} + \frac{\partial \xi}{\partial y} & 2\frac{\partial \eta}{\partial y} \end{array} \right)[/tex] and [tex] A = \frac{1}{2} \left( \begin{array}{cc} 0 & \frac{\partial \xi}{\partial y} - \frac{\partial \eta}{\partial x} \\ \frac{\partial \eta}{\partial x} - \frac{\partial \xi}{\partial y} & 0 \end{array} \right)[/tex] respectivelly. If [tex] D\alpha[/tex] is itself symmetric then [tex] D\alpha = S[/tex], while when it is antisymmetric (as in our previous example), then [tex] D\alpha = A[/tex].

    The exterior derivative of [tex] \alpha[/tex] is [tex] d \alpha= -\frac{\partial \xi}{\partial y} dx\wedge dy + \frac{\partial \eta}{\partial x} dx\wedge dy = (-\frac{\partial \xi}{\partial y} + \frac{\partial \eta}{\partial x}) dx\wedge dy[/tex].
    This is just the twice of the negative of the bilinear form represented by matrix [tex] A[/tex]. In the special case when [tex] D\alpha[/tex] is antisymmetric, then this holds for [tex] D\alpha[/tex], i.e for the total derivative matrix itself.

    My questions:
    1. Is this accidental, or there is a deeper geometrical interconnection between the total derivative matrix and the exterior derivative?
    2. Does the matrix [tex] S[/tex] also have any meaning in the world of forms?
     
    Last edited: Dec 22, 2009
  2. jcsd
  3. Dec 23, 2009 #2

    mma

    User Avatar

    Perhaps it isn't quite clear what do I mean. I try to explain it by an analogy in one lower degree, and on vector spaces instead of manifolds.

    Take a 0-form, i.e. a function f on a vector space. The "total derivative" of this function (in some point of the vector space) is a vector that on the one hand can be regarded as the 1 by n matrix of the linear approximation of the function, while on the other hand it defines the same linear functional (via the scalar product) as the exterior derivative df of f does (really, this is the definition of the gradient vector).

    In the original (one higher) degree, the "total derivative" of a vector-vector function (after all, the 1-form on a vector space is also a vector-vector function) is a matrix that defines a linear approximation of our vector-vector function, while on the other hand, its antisymmetric part defines (up to constant, via what?) the same bilinear form as the exterior derivative of our 1 form does.

    Does this have any real sense? And what can we do with the symmetric part?
     
  4. Dec 23, 2009 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're looking at a one-form as a function [itex]\omega : M \to T^*M[/itex] from the manifold to the cotangent bundle.

    Are you're asking if the derivative of this function, [itex]\omega_* : TM \to TT^* M[/itex], has any relationship to the 2-form [itex]d\omega[/itex]?
     
  5. Dec 23, 2009 #4

    mma

    User Avatar

    Exactly.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook