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.