**An Approach to Spacetime Triangulation as the Benchmark towards Gravitational Unification**
It is well known from Pythagoras' theorem that there exists the spacetime inequality ~

$$AB + BC > AC$$

$$AB + AC > BC$$

$$AC + BC > AB$$

Is it possible to apply a spacetime commutator inside of this inequality? Yes I think so! Or at least, this occurred to me.

For a scalar product defind on a vector space the length of vector is determined by

$$|X_a|^2 = X_a \cdot X_a$$

With some invesigation (see references) a spacetime inequality can indeed satisfy the following relationship

$$|X_a| + |X_b| \geq |X_a + X_b|$$

Squaring both sides also yields

$$|X_a + X_b|^2 = |X_a|^2 + |X_b|^2 + 2|X_a| |X_b|$$

$$(X_a + X_b) \geq ...$$

$$|X_a|^2 + |X_b|^2 + 2X_a \cdot X_b \geq |X_a|^2 + |X_b|^2 + 2|X_a| |X_b|$$

from which it follows

$$|X_a \cdot X_b| \geq |X_a||X_b|$$

which is known as the Cauchy Schwartz inequality which can be thought of as a direct interpretation of a spacetime uncertainty. Another important identity whicch further can be identified from the spacetime relationships is

$$|X_a||X_b| \geq \frac{1}{2} |<X_a|X_b> + <X_b|X_a>|$$

If you have trained your eye on all my previous work into gravity, this looks like the structure of commutators!

In a Hilbert space, you can define new vectors

$$\sqrt{|<\Delta X_A^2>< \Delta X_B^2>|} \geq \frac{1}{2} i< \psi|X_AX_B|\psi > + i<\psi|X_BX_A|\psi> = \frac{1}{2} <\psi|[X_A,X_B]|\psi>$$

The left hand side calculates the deviation of the derivative from the mean of the derivative, at least, that is how it would be interpreted in the approach we took to the quantization of gravity. We will use these solutions as a benchmark into how to treat this commutivity in spacetime from the connections we solved.

ref

http://rocs.hu-berlin.de/qm1415/resources/Lecture_Notes_10_11_12.pdf