Total and exterior derivative of a 1-form

[mma]

Suppose we have a one-form \alpha having a skew-symmetric total derivative matrix. I mean something like \alpha(x,y) = -y dx + x dy, that is, in canonical (x, y, \xi, \eta) coordinates of the cotangent bundle, \alpha(x,y) = (x, y, -y , x ).
The "total derivate matrix" I mean

D\alpha = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)
​

because the tangent mapping of \alpha is

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}​

and the projection of this vector on the \{\frac{\partial}{\partial x}, \frac{\partial}{\partial y} \} plane is always

a \frac{\partial}{\partial x} + b\frac{\partial}{\partial y}​

itself (independently of \alpha), while the projection on the \{ \frac{\partial}{\partial \xi}, \frac{\partial}{\partial \eta} \} plane is

-b \frac{\partial}{\partial \xi} + a \frac{\partial}{\partial \eta},
​

that is, in column vector representation :
\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)

A bit more generally, if our 1-form is \alpha(x,y) = \xi(x,y) dx + \eta(x,y) dy then the "total derivative matrix" is 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).
This matrix can always decompose into the sum of its symmetric and antisymmetric parts, where the symmetric and antisymmetric parts are
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) and 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) respectivelly. If D\alpha is itself symmetric then D\alpha = S, while when it is antisymmetric (as in our previous example), then D\alpha = A.

The exterior derivative of \alpha is 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.
This is just the twice of the negative of the bilinear form represented by matrix A. In the special case when D\alpha is antisymmetric, then this holds for D\alpha, 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 S also have any meaning in the world of forms?

 

[mma]

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?

 

[Hurkyl]

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

Are you're asking if the derivative of this function, \omega_* : TM \to TT^* M, has any relationship to the 2-form d\omega?

 

[mma]

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

Are you're asking if the derivative of this function, \omega_* : TM \to TT^* M, has any relationship to the 2-form d\omega?

Exactly.