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].(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Total and exterior derivative of a 1-form

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