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I am looking for an explicit description of the algebra of $SO(n)$- or, better, $O(n)$-invariant differential operators on the real Grassmann manifolds of $k$-dimensional linear subspaces in the Euclidean space $\mathbb{R}^n$.

I was told that for general symmetric spaces there is a Harish-Chandra type theorem saying roughly that the above algebra is isomorphic to the algebra of polynomials invariant under appropriate Weyl group. However in the case of Grassmannians I would like to have a more explicit description not referring to this theorem.

To give an example of what I am looking for, let me mention that the algebra of $SO(n)$- (or $O(n)$-) invariant operators on the sphere $S^{n-1}$ ($n>2$) is precisely equal to the algebra of polynomials in the Laplace operator.

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    $\begingroup$ Hi Semyon, I think either Fulton Gonzalez or Eric Grinberg may help you with that. $\endgroup$ – alvarezpaiva Aug 5 '13 at 16:02
  • $\begingroup$ Thanks, Juan-Carlos, for a good idea. I will write them. $\endgroup$ – MKO Aug 6 '13 at 13:32
  • $\begingroup$ Just operators on functions? The apparatus of BGG gives a description of the invariant operators on sections of invariant vector bundles; see the book of Baston and Eastwood. $\endgroup$ – Ben McKay Aug 6 '13 at 13:58
  • $\begingroup$ @semyon: Your last claim on the $SO(n)$-invariant operators on the sphere is very interesting, I was not aware of that. Do you have an easy reference for it? $\endgroup$ – Renato G. Bettiol Aug 9 '13 at 15:01
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    $\begingroup$ For the sphere, you start by looking for the symbol of such an operator, which must then be invariant under rotations fixing the north pole, so must be the metric. Subtracting off a suitable multiple of the Laplace operator, you reduce order; induction. For the Grassmannian, you need the invariant polynomials on $p \times q$ matrices, up to similarity, which you get from classical invariant theory: $\sigma(A)=\sum_{ij} a_{ij} a_{ij}$. But then you have to find a differential operator with that symbol. $\endgroup$ – Ben McKay Aug 9 '13 at 16:12
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In this particular case, the invariant theory is pretty simple, so there is an explicit description.

Recall that, in general, for a homogeneous space $M=G/H$, a $G$-invariant, $m$-th order linear differential operator $L:C^\infty(M)\to C^\infty(M)$ is specified by its value on the $m$-jets at the identity coset $eH$ of functions in $C^\infty(M)$. In particular, the vector space of such operators is dual to the vector space of $\mathrm{Ad}(H)$-invariant elements in $J^m_{eH}(M,\mathbb{R})$. Taking the inverse limit as $m$ goes to infinity, one sees that the vector space $\mathcal{L}$ of $G$-invariant, linear differential differential operators from functions to functions is dual to the vector space of $\mathrm{Ad}(H)$-invariant elements of the symmetric algebra generated by ${\frak{m}}^*$, where ${\frak{m}}=T_{eH}M$. In other words, as a vector space, $\mathcal{L}$ is isomorphic to the $\mathrm{Ad}(H)$-invariant polynomials on ${\frak{m}}^*$.

In the case of the real Grassmannian $\mathrm{O}(n)/\bigl(\mathrm{O}(k)\times\mathrm{O}(n{-}k)\bigr)$, you are asking for the $\mathrm{O}(k)\times\mathrm{O}(n{-}k)$-invariant polynomials on the $k$-by-$(n{-}k)$ matrices, and this follows from the singular value decomposition for such matrices. Supposing, as one can, that $k\le n{-}k$, if one lets $A$ be such a matrix, then one can define polynomials $p_j(A)$ (of degree $2j$) by the rule $$ \det(A A^T - t\ I_k) = p_k(A) - p_{k-1}(A) t + p_{k-2}(A) t^2 - \cdots + (-1)^kp_0(A)\ t^k. $$ Then these $k{+}1$ polynomials generate the ring of $\mathrm{Ad}(H)$-invariant polynomials. (Of course, $p_0(A) \equiv 1$.) Thus, $p_j$ corresponds to an invariant linear differential operator $L_{2j}$ of order $2j$, and these generate the ring of invariant operators on the Grassmannian.

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    $\begingroup$ Thanks for the beautiful description. I think however that this does describe the invariant operators as a vector space, but not as a ring. For example it seems to follow from the first part of your argument that $\mathcal{L}$ is a commutative ring for any $G/H$. But for non-symmetric pair this may not be true. Indeed take $H=\{e\}$. Then $\mathcal{L}$ is the universal enveloping algebra of $Lie(G)$. $\endgroup$ – MKO Aug 25 '13 at 18:02
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    $\begingroup$ Actually, I didn't say anything about the ring structure in the general case; I just gave the identification as vector spaces. The ring structure of the invariant in the general case is indeed more subtle, as you noted in the case when $H$ is the identity. However, in the case of irreducible Riemannian symmetric spaces (which covers the Grassmannian case you were interested in), things are simpler. $\endgroup$ – Robert Bryant Aug 25 '13 at 20:29
  • $\begingroup$ Then I misunderstood. Sorry. $\endgroup$ – MKO Aug 26 '13 at 3:25
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The answer to my original question of description of the algebra of $O(n)$-invariant operators on real Grassmannians is explicitly contained in Proposition 3.2 of the paper by Fulton B. Gonzalez, Tomoyuki Kakehi; "Pfaffian systems and Radon transforms on affine Grassmann manifolds", Mathematische Annalen 326 (2003).

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