# Radial part for O(n) invariant differential operators acting on symmetric matrices ?

Consider the vector space of symmetric matrices. O(n) acts on it by conjugation $gMg^t$.

Question informally: consider invariant differential operators what is known about their radial parts ? (Is there something nice like Harish-Chandra's homomorphism (this MO quest.))

More precisely: a) Consider O(n) invariant differential operator with CONSTANT COEFFICIENTS D. It can be identified with some symmetric polynomial. Can we write down its radial part in terms of this polynom ? (In Harish-Chandra's case the answer is this polynom conjugated by the Vandermonde (this MO quest.)).

b) Very particularly take $D=det(\partial_{ij})$ what is its radial part ? (This operator enters Capelly and Cayley identities).

c) If it is not with constant coefficients what can be said ?

• In (a), you mean your differential operator to be $O(n)$-invariant, right? – macbeth Mar 24 '12 at 21:22
• By the way, I don't know the answer to your general question, but by my calculation the radial part of $\det(\partial_{ij})$ (aka the Monge-Ampère operator) is given by: the operator sends $f(|\cdot|^2)$ to $[2f'(|\cdot|^2)]^{n-1}[4f''(|\cdot|^2)|\cdot|^2+2f'(|\cdot|^2)]$. Proof: The endomorphism induced from the Hessian is $4f''(|\cdot|^2)|\cdot|^2\text{Proj}_{\cdot}+2f'(|\cdot|^2)]I$. – macbeth Mar 25 '12 at 2:02
• @Macbeth Thank you. Yes in a) I mean O(n) invariant. But I do not understand yours second comment - actually your notations is not clear for me - what do you mean by f( ) - is it the function on which radial part of D acts ? Then something is wrong - what you write is non-linear expression. While radial part is linear operator. May be you can write more details as answer, not as a comment ? Also you mention "Monge-Ampere" is there some relation between det(d/dx_ij) and "Monge-Ampere" ? – Alexander Chervov Mar 25 '12 at 7:22
• Ah -- maybe I've misread your question? My interpretation was: operator is Monge-Ampère operator $\det(\frac{\partial^2}{\partial x_i\partial x_j})$, on the space $\mathcal{C}^\infty(\mathbb{R}^n)$ (say). Is your operator actually $\det(\frac{\partial}{\partial x_{ij}})$, on the space $\mathcal{C}^\infty(Mat_n(\mathbb{R}))$? – macbeth Mar 25 '12 at 12:47
• @macbeth Yes my operator is on $Mat_n(R)$ it is \$D=det(d/dx_{ij}) (with symmetry ij=ji). Such operators are used in Capelli and Cayley identities. Symmetric version of Capelli identity is due to Turnbull - see en.wikipedia.org/wiki/Capelli's_identity . – Alexander Chervov Mar 25 '12 at 16:29