For a $C^1$ function $f:U\to\mathbb{C}$, where $D\subset U\subseteq \mathbb{C}$ with piecewise $C^1$ boundary $\partial D$, we have the following generalized Cauchy integral formula:

$$ f(\zeta) = \frac1{2\pi i}\int_{\partial D}\frac{f(z)}{z-\zeta}dz \ + \ \frac1{2\pi i}\int_D \frac{\partial f}{\partial \overline{z}}(z) \frac1{z-\zeta}dz\wedge d\overline{z} $$

If $f$ is holomorphic then it has certain rigid properties imposed by the Cauchy-Riemann equations, like being infinitely differentiable and having the maximum-modulus principle hold. The result of this rigidity is that $f$ can be determined from its values on the boundary $\partial D$, and so the equation reduces to the first term.

I was recently interested to find that there exists an analogous result in Clifford analysis, where functions from $\mathbb{R}^n$ to $C\ell_n$ are considered. Like holomorphic functions, elements of the kernel of the left and right Dirac operators $\sum_{j=1}^n e_j \frac{\partial}{\partial x_j}$, $\sum_{j=1}^n \frac{\partial}{\partial x_j} e_j$ can be recovered from expressions involving their boundary behavior, notably using the Clifford-Cauchy integral formulas

$$ f(y) = \frac1{\omega_n}\int_S G(x-y) \ n(x) \ f(x) \ d\mu(x) $$

$$ g(y) = \frac1{\omega_n}\int_S g(x) \ n(x) \ G(x-y) \ d\mu(x) $$

where $f,g$ are $C^1$ functions from a subset of $\mathbb{R}^n$ to $C\ell_n$, $S$ is the nice boundary of some domain in $\mathbb{R}^n$ where these things are defined, $n(x)$ is the outward normal to $S$, $\omega_n$ is the area of the unit sphere in $\mathbb{R}^n$, and $G(x) = \frac{x}{\|x\|^n}$. In this example $f$ satisfies $\sum_{j=1}^n e_j \frac{\partial f}{\partial x_j}\equiv 0$ (called left monogenic or left regular), and $g$ satisfies $\sum_{j=1}^n \frac{\partial g}{\partial x_j}e_j \equiv 0$ (right monogenic).

Like complex holomorphic functions, left and right monogenic functions inherit a variety of properties from this expression, like a Taylor series and a Cauchy theorem. It also becomes possible to develop Hardy spaces for Clifford algebras using these ideas.

The impetus for these properties is the fact that such functions (including complex holomorphic functions) can be thought of as lying in the kernel of one of these Dirac operators, and so my question is what is the natural setting to study such operators, Cauchy-type integral formulas, and generalized Hardy spaces? Are Clifford algebras the appropriate way to approach such a topic, or are they simply one kind of structure where such operators are easy to express?

  • $\begingroup$ what is 'monogenic'? $\endgroup$
    – JHM
    Jun 28 '12 at 15:44
  • $\begingroup$ I phrased it poorly. When I say that f is left monogenic such that the sum is zero, what I really meant is that left monogenic functions are functions such that the sum of the e_j partial_j is always 0, which the function f is assumed to be (sim. right monogenic and g). $\endgroup$ Jun 28 '12 at 16:56
  • $\begingroup$ Just an idea - but you get similar integral identities for diffusions, where $f$ is in the kernel of the generator. This is just by writing $f(x)$ as the expected value of $f(X)$ at the first time at which $X$ hits the boundary, where $X$ is the diffusion started from $x$. On the appropriate Clifford algebra, the generator splits into a product of first order differential operators. So, it is enough for $f$ to be in the kernel of such a linear term in order for it to also be in the kernel of the generator. $\endgroup$ Jun 28 '12 at 18:27
  • $\begingroup$ The Cauchy-Riemann equations are an elliptic system. Their solutions have properties that are special cases of results for somewhat more general elliptic systems: analyticity -- elliptic regularity, analytic continuation -- unique continuation, Cauchy integral formula -- Green function identity for Dirichlet boundary value property. $\endgroup$ Jun 30 '12 at 9:10
  • $\begingroup$ Igor, do you have any texts to recommend for the subject of general elliptic operators. $\endgroup$ Jul 16 '12 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.