I recently encountered the following nice fact, and I'm wondering if it's part of a more general story.

Let $x\in \mathbb{C}^n$ satisfy

$$x^2:=\sum_i x_i^2 = 0,$$

and consider functions $f(x)$ which are homogeneous of degree $2-\frac n 2$ in $x$ ($f$ is allowed to have singularities, but it should be locally analytic). Since $x$ is constrained to lie on a subvariety, formal derivatives with respect to $x$ don't make sense in general. However, in the case at hand, the Laplacian $\Delta=\sum_i \frac{\partial}{\partial x_i}\frac{\partial}{\partial x_i}$ is actually a well-defined operation acting on $f(x)$. This is because it maps the ideal generated by $x^2$ to itself. Indeed, suppose $f(x)=x^2 g(x)$, where $g(x)$ is homogeneous of degree $-\frac n2$. Then

$$ \Delta (x^2 g(x)) = \left(2n + 4\frac {-n} 2\right)g(x) + x^2 \Delta g(x)=x^2 \Delta g(x) $$

where we've used that $\sum_i x_i \partial_i g(x) = -\frac n 2 g(x)$.

So there exists a natural $\mathrm{SO}(n)$-invariant second-order differential operator mapping sections of a particular line bundle to sections of a different line bundle over the quadric $x^2=0$ in $\mathbb{P}^{n-1}$. Is there some general reason I should have expected this to be the case?

Edit: Note that $\Delta$ is distinct from the Casimir operator $C=\frac 1 2\sum_{i,j} L_{ij}L_{ij}$, with $$ L_{ij}=x_i \frac{\partial}{\partial x_j}-x_j \frac{\partial}{\partial x_i}. $$ For example, $C$ preserves degree, while $\Delta$ reduces the degree by 2. In fact, we can compute $$ \Delta = \frac{1}{x^2}(C-D(2-n-D)) $$ where $D=\sum_i x_i\partial_i$ measures the degree in $x$. This expression makes it clear that existence of $C$ is not enough to define $\Delta$ on the quadric $x^2=0$.

  • 2
    $\begingroup$ Whenever you have an action of $SO(n)$, you should have a $SO(n)$-invariant second-order differential operator, the Casimir operator. $\endgroup$
    – Will Sawin
    Feb 16 '14 at 4:27
  • 2
    $\begingroup$ In general (in the affine, algebraic context), a differential operator $L$ belonging to the algebra $\mathcal D$ of differential operators on affine space $\mathbb A^n$ restricts to a subvariety $X\subseteq\mathbb A^n$ cut by an ideal $I$ if it belongs to the idealizer of $I\mathcal D$ in $\mathcal D$ (and the ring of differential operators on $X$ is the quotient of that idealizer by $I\mathcal D$) This is explained, for example, in the last chapter of the book by McConnell and Robson. $\endgroup$ Feb 16 '14 at 5:13
  • $\begingroup$ @WillSawin: $\Delta$ is not the Casimir operator (though it's probably closely related) -- see my edit. $\endgroup$
    – David S-D
    Feb 16 '14 at 23:49
  • $\begingroup$ @MarianoSuárez-Alvarez: Thanks. I think you're saying precisely what I checked explicitly above, although I guess you are directly answering the question title. $\endgroup$
    – David S-D
    Feb 16 '14 at 23:52
  • $\begingroup$ I have edited the title of the question to make it more specific. It was previously "When does a differential operator restrict to a subvariety" (hence Mariano's answer). $\endgroup$
    – David S-D
    Feb 17 '14 at 19:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.