---
tag: ABJ88GR4Z0
kind: definition
name: Dunkl operator
msc: [33C52, "20F55", 33C80]
defines: GTVCECGH40
uses: [06BRDF3F0E, G6FND7X69J, 06KWGHDN2J]
sources:
  - url: https://en.wikipedia.org/wiki/Dunkl_operator
---

## Definition

Let $\R^n$ have the standard inner product $\langle \cdot, \cdot \rangle$.
For $\alpha \in \R^n \setminus \set{0}$, let $\sigma_\alpha(x) = x - 2 \frac{\langle \alpha, x \rangle}{\langle \alpha, \alpha \rangle} \alpha$ be the [[reflection|06BRDF3F0E]] in the hyperplane orthogonal to $\alpha$.
Let $R \subset \R^n \setminus \set{0}$ be a root system: a finite set such that $\sigma_\alpha(R) = R$ and $R \cap \R \alpha = \set{\alpha, -\alpha}$ for every $\alpha \in R$.
Let $W$ be the finite [[reflection group|G6FND7X69J]] generated by the $\sigma_\alpha$, and let $R_+ \subset R$ contain exactly one of $\alpha, -\alpha$ for each $\alpha \in R$.
Let $k \colon R \to \C$ be a multiplicity function, that is, $k(w \alpha) = k(\alpha)$ for all $w \in W$ and $\alpha \in R$.
For $\xi \in \R^n$, the *Dunkl operator* $T_\xi$ acts on smooth functions $f$ on $\R^n$ by
$$T_\xi f(x) = \partial_\xi f(x) + \sum_{\alpha \in R_+} k(\alpha)\, \langle \alpha, \xi \rangle\, \frac{f(x) - f(\sigma_\alpha x)}{\langle \alpha, x \rangle},$$
where $\partial_\xi$ is the [[directional derivative|06KWGHDN2J]] along $\xi$.

## Remarks

Each term of the sum is the same for $\alpha$ and $-\alpha$, so $T_\xi$ does not depend on the choice of $R_+$.
The difference quotient extends smoothly across the hyperplane $\langle \alpha, x \rangle = 0$, so $T_\xi$ maps homogeneous polynomials of degree $d$ to homogeneous polynomials of degree $d - 1$.
Dunkl's theorem states that $T_\xi T_\eta = T_\eta T_\xi$ for all $\xi, \eta \in \R^n$.
For $k = 0$, $T_\xi = \partial_\xi$.

Example: $n = 1$, $R = \set{1, -1}$, $W = \set{x \mapsto x, x \mapsto -x}$ and $k(\pm 1) = k$.
Then $T f(x) = f'(x) + k \frac{f(x) - f(-x)}{x}$.
For $f(x) = x^m$, the quotient is $(1 - (-1)^m) x^{m-1}$, so $T x^m = m x^{m-1}$ for even $m$ and $T x^m = (m + 2k) x^{m-1}$ for odd $m$.

Non-example: the operator $f \mapsto f'(x) + k \frac{f(x)}{x}$ on $\R$ is not a Dunkl operator for $k \ne 0$.
It lacks the reflected term $f(-x)$, and it does not even map the constant $1$ to a polynomial.

The Dunkl Laplacian $\sum_{i=1}^n T_{e_i}^2$ reduces to the Laplacian when $k = 0$.