Eric Forgy
Noncommutative Stochastic Calculus

Given any first order calculus $(d,{\Omega }^1(A))$ on a commutative algebra $A$, we have

$$d(fg)=d(gf)\Rightarrow [df,g]=[dg,f].$$d(f g) = d(g f)\implies [d f,g] = [d g,f].

According to

• H. C. Baehr, A. Dimakis, F. Müller-Hoissen, Differential Calculi on Commutative Algebras,

this is enough to show that the commutator is a symmetric $A$-bilinear function of $df$ and $dg$, i.e.

$$[df,g]=C(df,dg).$$[d f,g] = C(d f,d g).

The symmetric $A$-bilinear map $C$ extends to a product

$$C:{\Omega }^1(A)\times {\Omega }^1(A)\to {\Omega }^1(A),$$C:\Omega^1(A)\times\Omega^1(A)\to\Omega^1(A),

since ${\Omega }^1(A)$ is generated by elements of the form $a(db)$ for $a,b\in A$. We will denote this product

$$\alpha •\beta ≔C(\alpha ,\beta ).$$\alpha\bullet\beta \coloneqq C(\alpha,\beta).

This allows us to write the product rule in “left component form” as

$$d(fg)=f(dg)+g(df)+df•dg,$$d(f g) = f(d g) + g(d f) + d f\bullet d g,

in “right component form” as

$$d(fg)=(dg)f+(df)g-df•dg,$$d(f g) = (d g)f + (d f)g - d f\bullet d g,

or the “symmetrized form” as

$$d(fg)=\frac{1}{2}[f(dg)+(dg)f]+\frac{1}{2}[g(df)+(df)g].$$d(f g) = \frac{1}{2}\left[f(d g) + (d g)f\right] + \frac{1}{2}\left[g(d f) + (d f)g\right].