18
$\begingroup$

Let $B$ be a commutative $A$-algebra, and let $M$, $N$ be two $B$-modules. We can talk about the set of $A$-linear module homomorphisms $M \to N$, i.e. the set $\text{Hom}_A(M, N)$. Differential operators of order zero should be the $B$-linear maps from $M$ to $N$, i.e. $\text{Hom}_B(M, N)$.

First, note that the commutator $[f, b]$ (where $b \in B$) is a well-defined morphism $M \to N$. Then we make our first definition, the "Weyl Algebra" one.

Definition 1 (Weyl Algebra). Let $\mathcal{D}_A^0(M, N) = \text{Hom}_B(M, N)$. Define $$\mathcal{D}_A^n(M, N) = \{f \in \text{Hom}_A(M, N) \text{ such that }[f, b] \in \mathcal{D}_A^{n-1}(M,N)\}.$$We set $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0}\mathcal{D}_A^n(M, N)$.

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$. We now have our "Crystalline" definition.

Definition 2 (Crystalline). Let $I$ be the kernel of the diagonal map (i.e., the map $B \otimes_A B \to B, \ b \otimes b' \mapsto bb'$). Then $D: M \to N$ is said to be a differential operator of order $\le n$ if $\overline{D}$ annihilates $I^{n+1} \otimes_B M$. Let $\mathcal{D}_A^n(M, N)$ be the $B$-module of differential operators of order $\le n$. We define $\mathcal{D}_A(M, N) = \bigcup_{n \ge 0} \mathcal{D}_A^n(M, N)$.

My question is, what is the easiest way to see that/the intuition behind the definitions of rings of differential operators between modules given above are equivalent?


EDIT: In the comments, Michael Bächtold is asking me to spell out the definition of $\delta_{B/A}$ and $\overline{D}$.

So say we have $B$ a commutative $A$-algebra. We want to formalize the notion of an $A$-linear endomorphism of $B$ which is ``close" to being $B$-linear. Let $D: B \to B$ be an $A$-linear endomorphism of $B$. Using $D$, we obtain a map$$\tilde{D}: B \otimes_A B \to B$$defined by $\tilde{D}: b \otimes b' \mapsto bD(b')$, which can also be viewed as a map$$\overline{D}: B \otimes_A B \otimes_B B \to B,$$where we have identified $B$ and $B \otimes_B B$ and the map is defined by $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$. Let us define $\delta_{B/A} = B \otimes_A B$. Then, we have a map:$$\overline{D}: \delta_{B/A} \otimes_B B \to B.$$

In order to formulate the crystalline definition, we introduce some notation. Let $D: M \to N$ be an $A$-linear map. Then, $D$ induces the map $\overline{D}: \delta_{B/A} \otimes_B M\to N$ defined by the same formula as above (that is, $\overline{D}: b \otimes b' \otimes b'' \mapsto bD(b'b'')$ for $b \in B$, $b' \in B$ and $b'' \in M$). We now have our "Crystalline" definition.

In the quoted text, the inducing is in perfect analogy to what I wrote above.

$\endgroup$
  • 2
    $\begingroup$ Could you spell out the definition of $\delta_{B/A}$ and $\bar{D}$? $\endgroup$ – Michael Bächtold Jul 6 '15 at 6:44
  • $\begingroup$ Dear Kevin, how come you ask such astonishingly advanced questions while still being an undergraduate? Do you self-study much? $\endgroup$ – Georges Elencwajg Nov 23 '15 at 10:15
  • $\begingroup$ Apologies for coming back to this older questions: do you know who first gave this crystalline definition of DO or the reference where you learned about it? $\endgroup$ – Michael Bächtold May 30 '16 at 13:59
6
$\begingroup$

It's straightforward to check that the definitions agree for order $0$ differential operators.

To show that they agree for higher order DOs it suffices to show that $D$ is a crystalline DO of order $\leq n$ iff $[b,D]$ is a crystalline DO of order $\leq n-1$ for all $b\in B$.

Keeping in mind that the kernel $I$ of $B\otimes_A B \to B$ is generated, as $B\otimes B$-module, by elements of the form $b\otimes 1 - 1 \otimes b$, it follows from the next computation, where $c\in I^n$ and $m\in M$ are arbitrary: $$ \overline{D}\left(( (b\otimes 1 - 1\otimes b)\cdot c) \otimes m\right )=\overline{D}((b\cdot c - c \cdot b)\otimes m) = b \overline{D}(c\otimes m) - \overline{D}(c\otimes bm)=\overline{[b, D]}(c\otimes m) .$$

Here $b\cdot c$ means multiplication on the first factor of $c\in B\otimes_A B$ and similarly $c\cdot b$ is multiplication on the second factor.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Another way to catch the common footing of the two definitions is encoded in the following commutative diagram (I shall assume that both the modules $M$ and $N$ are $B$, just to make the point clearer): $$ \require{AMScd} \begin{CD} B @>{D}>> B\\ @V{\mu}VV @| \\ \delta_{B/A} @>{\widetilde{D}}>> B\, , \end{CD} $$ where $\mu(b):=1\otimes b$ and $D\in\mathrm{End}_A(B)$. Indeed, it is easy to check that $D\in\mathcal{D}_A^n(B)$ if and only if $\widetilde{D}$ vanishes when composed with all the $A$-homomorphisms of the form $$ [b_1\cdots [b_{n},[b_{n+1},\mu]]\cdots]\, ,\quad b_1,\ldots,b_{n+1}\in B\, , $$ and that the images of the above homomorphisms span precisely the $\delta_{B/A}$-module $I^{n+1}$ (the proof is by induction, as in Michael Bächtold's answer).

In other words, if $D\in\mathcal{D}_A^n(B)$, the above diagram descends to the commutative diagram $$ \require{AMScd} \begin{CD} B @>{D}>> B\\ @V{\widetilde{\mu}=:j_n}VV @| \\ J^nB:=\frac{\delta_{B/A}}{I^{n+1}} @>{\widetilde{\widetilde{D}}}>> B\, , \end{CD} $$ where, by construction, $j_n\in\mathcal{D}^n_A(B,J^nB)$. Then $$ \mathcal{D}_A^n(B)\ni D\longrightarrow \widetilde{\widetilde{D}}\in\mathrm{Hom}_B(J^nB,B) $$ is a $B$-module isomorphism, capturing the equivalence you were interested about (it all works for projective and finitely generated modules).

| cite | improve this answer | |
$\endgroup$

Your Answer

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