MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Dunkl intertwinig operator $V_k$ on $C(\mathbb{R}^d)$ is defined by:

$$V_k f(x)=\int_{\mathbb{R}^d}f(y)d\mu_x(y),$$

where $d\mu_x$ is a probability measure on $\mathbb{R}^d$ with support in the closed ball $B(0,||x||)$ of center $0$ and radius $||x||$. For all $x\in\mathbb{R}^d$, $z\in\mathbb{C}^d$, we have:


Let $^tV_k$ the operator on $D(\mathbb{R}^d)$ satisfying for all $f\in D(\mathbb{R}^d)$ and $g\in C(\mathbb{R}^d)$,

$$\int_{\mathbb{R}^d} {^tV_k(f)(y)g(y)dy}=\int_{\mathbb{R}^d}V_k(g)f(x)w_k(x)dx.$$

Then there exists a positive measure $\nu_y$ on $\mathbb{R}^d$ with support in the set $\{ x\in \mathbb{R}^d, ||x||\geq ||y||\} $ for which


This operator $^tV_k$ is called the dual Dunkl intertwining operator.

The operators $V_k$ and $^tV_k$ satisfy the flolowing properties:

For all $f\in \xi(\mathbb{R}^d)$ $T_iV_k(f)=V_k(\frac{\partial}{\partial y_j}f(x))$, and for all $f\in D(\mathbb{R}^d)$, $^tV_k(T_jf)(y)=\frac{\partial}{\partial y_j} {^tV_k}(f)(y).$

Where $T_j$ the Dunkl operators, $\xi(\mathbb{R}^d)$ the space of $C^{\infty}$-functions on $\mathbb{R}^d$ and $D(\mathbb{R}^d)$ the space of $C^{\infty}$-functions on $\mathbb{R}^d$ with compact support, from this two properties cames the importance of the Dunk intertwining operator.

My question is, there is an explicit formula for Dunk intertwining operator, in general ? if the answer in no, then there an explicit formula for some special cases ?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.