Let $G$ be a complex algebraic group, $K$ a closed subgroup so that $X=G/K$ is a homogeneous space.

Let $\mathcal{D}(X)$ denote the algebra of differential operators on $X$. The group $G$ acts on $\mathcal{D}(X)$, and so we may consider its invariants under this action, which we write as $\mathcal{D}(X)^G$.

For each closed point $a\in X$, we have a natural restriction map $res_a:\mathcal{D}(X)\to Dist(G/K,a)$, where $Dist(G/K,a)$ are the collection of $\mathbb{C}$-linear maps $D:\mathcal{O}_{X,a}\to\mathbb{C}$ such that $D(\mathfrak{m_a}^n)=0$ for some $n>0$, where $\mathfrak{m}_a$ is the maximal ideal of the closed point in the local ring $\mathcal{O}_{X,a}$.

The map is defined by $res_a(D)(f)=D(f)(a)$.

Now it is not hard to check that for $D\in\mathcal{D}(X)^G$ we have $res_a(D)\in Dist(G/K,a)^{G_a}$, where $G_a$ is the isotropy group of $a$ in $G$. My question is if the converse holds: if $D\in\mathcal{D}(X)$ has that $res_a(D)\in Dist(G/K,a)^{G_a}$ for all $a\in X$, does that imply that $D$ is $G$-invariant?

Also, the above question could be asked for any $G$-variety $X$, and if the above is true I would wonder if the corresponding statement for an arbitrary variety also holds.

Any references/ideas/thoughts would be greatly appreciated!

  • 2
    $\begingroup$ It is not true. For example multiplication by a non-constant function is $G_a$-invariant for every $a$ because it factors through the quotient by $m_a$ and the action on the quotient is trivial of course (it is just $\mathbb{C}$). $\endgroup$ – S. carmeli May 30 '18 at 19:59

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