Phorgy Phynance

Discrete stochastic calculus and commutators

leave a comment »

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].

Advertisements

Written by Eric

October 27, 2010 at 3:47 pm

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: