Suppose the differential relation $ds=\alpha dx+\beta dy$. Squaring each side of the relation, we obtain:
$$ (ds)^2=\alpha^2(dx)^2+\beta^2(dy)^2+\alpha\beta dxdy+\beta\alpha dydx $$
The structure of the square of the relation is reminiscent of a metric, defining the distance between two points $p_1$, $p_2$ of $\mathbb{R}\times \mathbb{R}$. In this case, it is the Taxi-cab metric.
Comparing this relation to the Pythagorean metric $(ds)^2=(dx)^2+(dy)^2$, we notice that the Taxi-cab metric contains "correlation" terms in the form of $\alpha\beta dxdy+\beta\alpha dydx$.
In the case where $\alpha$ and $\beta$ are elements of the real, the correlation terms cannot be eliminated. But, if $\alpha$ and $\beta$ are matrices with suitable commutative properties, namely:
$$ \alpha^2=I\\ \beta^2=I\\ \alpha\beta+\beta\alpha=0 $$
Interestingly the Pauli matrices have the required properties. We give one solution:
$$ \alpha=\pmatrix{0&1\\1&0}\\ \beta=\pmatrix{1&0\\0&-1} $$
Therefore, the Pythagorean metric is equal to a linear matrix-form, as follows:
$$ \alpha dx+ \beta dy=I \sqrt{(dx)^2+(dy)^2} $$
I am trying to "visualize" what is happening when the Pythagorean metric is of this linear matrix-form, and compare it to the usual case.
In the case of the usual formulation of the Pythagorean metric $\sqrt{(dx)^2+(dy)^2}$ I can visualize an integral over the path whose result is the distance between the points along the path. If the path $C$ between $p_1$ and $p_2$ follows orthogonal axis $\Delta x \cup \Delta y$, one can simply use a double definite integral over $dxdx$ and $dydy$ and add them up. Explicitly, we write:
$$ \int_{\Delta s}\left( \int_{\Delta s} ds \right)ds = \int_{\Delta x}\left( \int_{\Delta x}dx \right)dx +\int_{\Delta y} \left( \int_{\Delta y}dy \right) dy\\ \implies \Delta s\int_{\Delta s}ds = \Delta x \int_{\Delta x} dx + \Delta y \int_{\Delta y} dy\\ \implies (\Delta s)^2=(\Delta x)^2+(\Delta y)^2 $$
In the case of the matrix formulation, it is harder to visualize what is happening. For instance, to find the distance over the path, I would integrate over matrixes as follows:
$$ I\int_{\Delta s} ds =\pmatrix{0&1\\1&0} \int_{\Delta x} dx + \pmatrix{1&0\\0&-1} \int_{\Delta y} dy\\ \implies I \Delta s = \pmatrix{\Delta y & \Delta x \\ \Delta x & -\Delta y} $$
My question is how can I visualize $\Delta s = \pmatrix{\Delta y & \Delta x \\ \Delta x & -\Delta y}$ as equivalent to the Pythagorean theorem? The connection with the Pauli matrices indicates the relevant of Lie group $SU(2)$. One can, of course, produce the steps in reverse and go from the matrix representation to the sum representation:
$$ I (\Delta s)^2 = \pmatrix{\Delta y & \Delta x \\ \Delta x & -\Delta y}\pmatrix{\Delta y & \Delta x \\ \Delta x & -\Delta y}=\pmatrix{(\Delta x)^2 + (\Delta y)^2 & 0 \\ 0 & (\Delta x)^2 + (\Delta y)^2} $$
but I would like to understand it without going back to the usual Pythagorean form. What happens, illustratively, when I integrate over a matrix between two points on the cartesian plane?