I am studying differential forms and I am trying to characterize exterior derivatives.
This journey keeps taking me back to linear algebra and my most recent insight has been the Singular Value Decomposition where any rectangular matrix can be decomposed into two rotational matrices and a diagonal matrix. There is an excellent YouTube video that illustrates this by Visual Kernel.
The whole video series that precedes the SVD video also characterizes some important fundamental types of matrices, orthogonal, diagonal and symmetric. These are easy to identify. But I suspect that it might be possible to identify an arbitrary matrix as having some rotational component without decomposing it.
I am motivated by the fact that the curl of a vector field is identified in two dimensions by $\frac{df_1}{dx_2} - \frac{df_2}{dx_1}$.** But these are just the diagonally opposing entries of the Jacobian Matrix of the vector field, $$ \begin{matrix} \frac{df_1}{dx_1} & \frac{df_1}{dx_2} \\ \frac{df_2}{dx_1} & \frac{df_2}{dx_2} \\ \end{matrix} $$ And the Jacobean is just a linear approximation of the transformations characterized by the vector field. It is a transformation matrix.
A linear transformation that is strictly a rotation is a skew symmetric matrix, which is to say that the diagonally opposing entries of a m x m matrix are equal and opposite, their difference is always not equal to 0.
The curl of a vector field - in two dimensions anyway - also has difference always greater than 0. If it equals 0, there is no curl.
So, could you say that if you took an arbitrary m x m matrix and took the difference between the diagonally opposing entries, would their difference always be not equal to zero if there was an SVD that had a rotational component?
Notice that if you take the exterior derivative of a differential form, the new partial derivative components have a similar character.
** See this excellent sequence of YouTube videos that motivates this conclusion. The sequence is in a playlist here that starts with that video.