# Phorgy Phynance

## Discrete stochastic calculus and commutators

This post is in response to a question from Tim van Beek over at the Azimuth Project blog hosted by my friend John Baez regarding my paper

The basic fact needed to address the question is that we have a set 0-dimensional objects $\mathbf{e}^\kappa$ and 1-dimensional objects $\mathbf{e}^{\kappa\lambda}$ that obey the following geometrically-motivated multiplication rules:

1. $\mathbf{e}^\kappa \mathbf{e}^\lambda = \delta_{\kappa,\lambda} \mathbf{e}^\kappa$
2. $\mathbf{e}^{\kappa\lambda} \mathbf{e}^\mu = \delta_{\lambda,\mu} \mathbf{e}^{\kappa\lambda}$
3. $\mathbf{e}^\mu \mathbf{e}^{\kappa\lambda} = \delta_{\mu,\kappa} \mathbf{e}^{\kappa\lambda}$

To see the geometrical meaning of these multiplications, it might help to consider discrete 0-forms

$f = \sum_\kappa f(\kappa) \mathbf{e}^\kappa.$

and

$g = \sum_\lambda g(\lambda) \mathbf{e}^\lambda$

Multiplication 1.) above implies

$f g = \sum_\kappa f(\kappa) g(\kappa) \mathbf{e}^\kappa,$

which is completely reminiscent of multiplication of functions where we think of $f(\kappa)$ as the value of the function at the “node” $\mathbf{e}^\kappa.$

Multiplications 2.) and 3.) introduce new concepts, but whose geometrical interpretation is quite intuitive. To see this consider

$f \mathbf{e}^{\kappa\lambda} = f(\kappa) \mathbf{e}^{\kappa\lambda}$

and

$\mathbf{e}^{\kappa\lambda} f = f(\lambda) \mathbf{e}^{\kappa\lambda}.$

Multiplying the function $f$ on the left of the “directed edge” $\mathbf{e}^{\kappa\lambda}$ projects out the value of the function at the beginning of the edge and multiplying on the right projects out the value of the function at the end. Hence, the product of functions and edges do not commute.

In the paper, it was shown that the exterior derivative of a discrete 0-form is given by

$df = \sum_{\kappa,\lambda} \left[f(\lambda) - f(\kappa)\right] \mathbf{e}^{\kappa\lambda}.$

Here, we show that this may be expressed as the commutator with the “graph operator”

$\mathbf{G} = \sum_{\kappa,\lambda} \mathbf{e}^{\kappa\lambda}.$

The result is quite simple and follows directly from $\mathbf{G}$ and the multiplication rules, i.e.

$f\mathbf{G} = \sum_{\kappa,\lambda} f(\kappa) \mathbf{e}^{\kappa\lambda}$

and

$\mathbf{G} f = \sum_{\kappa,\lambda} f(\lambda) \mathbf{e}^{\kappa\lambda}$

so that

$[\mathbf{G},f] = \sum_{\kappa,\lambda} \left[f(\lambda) - f(\kappa)\right] \mathbf{e}^{\kappa\lambda}$

or

$df = [\mathbf{G},f].$

Written by Eric

October 27, 2010 at 3:47 pm