Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

My question is regarding the definition of $\mathcal{D}$-modules of level $m$ given in this paper. As an example, let $X=\mathbb{A}^1$ over $S=\text{Spec }\overline{\mathbb{F}_p}$; I was told that a $\mathcal{D}$-module of level $m$ is a module over $\mathbb{k} \langle x, \partial_x, \frac{{\partial_x}^p}{p!}, \cdots, \frac{{\partial_x}^{p^m}}{(p^m)!} \rangle$; I was wondering how to work this out from first principles, using Definition $2.5$ given in that paper.

Consider the immersion $X \rightarrow X \times_S X$. Definition $2.1$ from that paper defines what a divided power structure of level $m$ on an immersion is; Definition $2.3$ (and $2.4$) states constructs the divided power envelope of level $m$, $P_{X,m}(Y)$. Subsequently, on pg $5$ they define $\mathcal{P}_{X,m}^n(Y)$; the sheaf of differential operators of level $m$ is defined to be union of the duals of this family of sheaves (as $n$ varies). Most of the details/proofs are done in this other paper.

I'm having trouble properly understanding these definitions and working out what they are in the case of $X=\mathbb{A}^1, S=\text{Spec } \overline{\mathbb{F}_p}$. So, my question is what the above objects look like in this particular example, and how to explicitly calculate everything in this example.

share|improve this question
add comment

1 Answer

up vote 9 down vote accepted

I find it helpful to first work through the definition of multiplication on $\mathcal{D}^{(m)}$ when $m = \infty$, in which case it reduces to the "classical" ring of differential operators in the sense of Grothendieck; read sections 16.7 and 16.8 of EGA 4, Quatrième partie.

So let $A$ be a commutative base ring, let $S = Spec(A)$ and let $X = \mathbb{A}^1_S$ so that $B = A[t] = \Gamma(X, \mathcal{O})$. We want to work out $\mathcal{D}^{(\infty)}(X)$. Let $Y = X \times_S X$ and $m = \infty$.

Let $n \geq 0$. First we have to work out the global sections of the sheaf $\mathcal{P}^n_{Xm}(Y)$, considered as a $B$-module. This will turn out to be dual (as a $B$-module) to the $B$-module of all Grothendieck differential operators of order at most $n$.

Now $\mathcal{O}(Y) = B \otimes_A B$ is isomorphic as an $A$-algebra to the polynomial ring $A[t,t']$ where $t \mapsto t \otimes 1$ and $t' \mapsto 1 \otimes t$. The diagonal immersion $X \hookrightarrow Y$ corresponds to the algebra surjection $B \otimes_A B \to B$ which is just the multiplication map. So the ideal of the diagonal, namely the kernel of this map, is generated as an ideal by the element

$$\tau := t \otimes 1 - 1 \otimes t.$$

Let's view $\mathcal{O}(Y)$ as a $B$-algebra via the map $b \mapsto b \otimes 1$; then $\mathcal{O}(Y) \cong B[\tau]$. By definition (EGA IV,, the global sections of $\mathcal{P}^n_{X\infty}(Y)$ are just

$$P^n := \mathcal{O}(Y) / (\tau^{n+1})$$

--- this is the algebra of functions on the $n$-th infinitesimal neighbourhood of the diagonal $\tau = 0$ inside $\mathcal{O}(Y)$ (hence the $n+1$ in the exponent). So in particular it is a free $B$-module of rank $n+1$ with generators (the images of) $\tau^i$ for $0 \leq i \leq n$. By definition,

$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(P_n, B) =: D_n $$

which is again a free $B$-module of rank $n+1$; let $\{ \partial^{[i]}, i=0, \ldots, n\}$ be the dual basis for this module.

Now the multiplication map $D_r \times D_s \to D_{r+s}$ is the $B$-module dual of a map $\delta : P^{r+s} \to P^r \otimes P^s$ which is constructed in EGA IV, Lemma Morally $\delta$ sends $a \otimes b$ to $a \otimes 1 \otimes 1 \otimes b$, as Gros/Le Stum/Quirros mention. This turns out to be a $B$-algebra homomorphism, and tts key property is that

$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$

(it is a "primitive element" in an appropriate bialgebra --- see EGA IV.4, Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):

$$(\partial^{[i]} \cdot \partial^{[j]})(\tau^k) = (\partial^{[i]} \otimes \partial^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{a + b = k} \binom{k}{a} \partial^{[i]}(\tau^a) \partial^{[j]}(\tau^b)$$

which is just $\binom{i+j}{i}\delta_{k,i+j}$. Since $\binom{i+j}{i} \partial^{[i+j]}$ has the same effect on each $\tau^k$, we deduce that

$$ \partial^{[i]} \cdot \partial^{[j]} = \binom{i+j}{i} \partial^{[i+j]}$$

which is hopefully the familiar rule for multiplying divided powers (morally $\partial^{[i]} = \partial^i/i!$).

The point of the Berthelot construction is that it is possible to vary the divided-power structure on the diagonal, and thereby control just how many divided powers one gets in $\mathcal{D}^{(m)}$. For example, if $m = 0$ then you instead allow all divided powers on the ideal of the diagonal (algebraically this means you consider the divided power algebra of the ideal $(\tau)$ in $B[\tau]$ to get $\oplus_{n=0}^\infty B \tau^{[n]}$), and when you take the $B$-dual, these divided powers in $\tau$ "remove" the divided powers in $\partial$ and you end up with $\mathcal{D}^{(0)}(X) = B[\partial]$, the ring of crystalline differential operators (no divided powers).

Now to answer your question, let the level $m \geq 0$ be fixed. Then as Gros/Le Stum/Quirros explain just before Definition 2.5,

$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^n B \tau^{ \{ a \} } $$

where $\tau^{ \{ a \} }$ is a symbol that "behaves like $\tau^a / q_a!$" (where $q_a$ is the integer part of $a / p^m$: thus $a = q_a p^m + r_a$ say).

To understand the multiplication of the dual vectors to these $\tau^{ \{ a \} }$, namely the $\partial^{ \langle a \rangle }$, we need to understand how to comultiply the $\tau^{ \{ a \} }$. So we compute (again dropping bars for convenience)

$$ \delta( \tau^{ \{ a \} }) = \frac{1}{q_a!} \delta(\tau)^a = \sum_{i+j = a} \frac{1}{q_a!} \binom{i+j}{i} \tau^i \otimes \tau^j = \sum_{i+j = a} \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \tau^{ \{ i \} } \otimes \tau^{ \{ j \} }$$

and the same computation as above in the case $m=\infty$ shows that

$$ \partial^{\langle i \rangle} \cdot \partial^{\langle j \rangle} = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \partial^{\langle i + j \rangle}.$$

The interesting thing is that this structure constant $\frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i}$ is always a $p$-integral rational number (see Lemma 1.1.3(i) in Berthelot's paper), so it makes sense whenever $A$ is an algebra over $\mathbb{Z}_{(p)}$, say, and in particular if $A$ had characteristic $p$. Note that if $A$ was a $\mathbb{Q}$-algebra, then there would be a ring homomorphism from $\mathcal{D}^{(m)}$ to $A[t; \partial]$ which sends

$$ \partial^{\langle i \rangle} \mapsto \frac{q_i!\partial^i}{i!} $$


$$ \left(\frac{q_i! \partial^i}{i!}\right) \cdot \left(\frac{q_j! \partial^j}{j!}\right) = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \left( \frac{q_{i+j}! \partial^{i+j}}{(i+j)!}\right).$$

Thus morally $\partial^{\langle i \rangle}$ should be thought of as "modified divided powers" $q_i! \partial^i / i!$.

Finally, you don't need all of the $\partial^{\langle i \rangle}$ to generate $\mathcal{D}^{(m)}$. As is well-known, the full ring of Grothendieck differential operators in characteristic $p$ can be generated by the divided powers $\partial^{[p^a]}$ (for all $a \geq 0$). Since $q_i = 0$ for $i < p^m$ and $q_{p^m} = 1$, the modified divided powers $\partial^{\langle p^i \rangle}$ are equal to the "true" divided powers $\partial^{[p^i]}$ for $0 \leq i \leq m$. If $a > m$ then since $q_{p^a} = p^{a-m}$,

$$\partial^{\langle p^a \rangle} = \frac{ p^{a-m}! }{ p^a! } \partial^{p^a} = \left(\frac{ p^{a-m}! (p^m!)^{p^{a-m}} }{p^a!} \right) (\partial^{\langle p^m \rangle})^{p^{a-m}}$$

shows that $\partial^{\langle p^a \rangle}$ is a $p$-adic unit times a power of $\partial^{\langle p^m \rangle}$ for $a \geq m$ since the $p$-adic valuation of that big fraction is

$$\frac{p^{a-m}-1}{p-1} + p^{a-m} \frac{p^m-1}{p-1} - \frac{p^a-1}{p-1} = 0.$$

So we see that $\mathcal{D}^{(m)}(\mathbb{A}^1_S)$ in this case is the $A$-algebra generated by $t$ and the divided powers $\partial^{[p^0]}, \partial^{[p^1]}, \ldots, \partial^{[p^m]}$, subject to appropriate natural relations.

Edit: To see what the map $X \to P_{Xm}(Y)$ looks like in the case $X = \mathbb{A}^1_S$, it is enough to describe the corresponding map $C := \mathcal{O}(P_{Xm}(Y)) \to B = \mathcal{O}(X)$ on functions, because everything in sight is affine. $C$ is a $B$-algebra, generated by symbols $\tau^{ \{a \} }$ for all $a \geq 1$ subject to the relations

$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \frac{q_{a+b}!}{q_a!q_b!} \tau^{ \{ a + b \} }$$

(note that the structure constant $\frac{q_{a+b}!}{q_a!q_b!}$ is actually an integer, this again follows from Lemma 1.1.3(i) in Berthelot's paper.) The map $C \to B$ sends all of the generators $\tau^{ \{a \} }$ to zero, and the map $C \to P^n$ which corresponds to the closed subscheme $P^n_{Xm}(Y)$ of $P_{Xm}(Y)$ sends all of the $\tau^{ \{ a \} }$ to zero for $a \geq n+1$.

This description makes it easy to see that the algebra of functions $C$ on $P_{Xm}(Y)$ is isomorphic to the polynomial ring $B[\tau]$ when $m = \infty$ (since we can take $q_a$ to be always zero in this case), and to the "free" divided-power algebra $B[\tau^{[n]} : n\geq 1]$ that Gros/Le Stum/Quirros call $\Gamma_\bullet(B \tau)$ when $m = 0$. This is because $q_a = a$ in this case, so the defining relations between the $\tau^{ \{a \}}$ reduce to

$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \binom{a+b}{a} \tau^{ \{ a + b \} }.$$

share|improve this answer
Thanks! Just one question - what do the left adjoint $X→P_{X,m}(Y)$ (from Prop $2.3$ in Gros/Le Stum/Quirros), and the ideals $\mathcal{I}^{n+1}_{X,m}(Y)$ (mentioned at the start of pg 5) look like here? (Above you explained in detail what the quotient $P^n_{X,m}(Y)=P_{X,m}(Y)/I^{n+1}_{X,m}(Y)$ looks like.) –  Vinoth Sep 15 '12 at 23:26
I've added an answer to your question above, and also fixed the $n/n+1$ issue with the definition of $P^n$. –  Konstantin Ardakov Sep 16 '12 at 8:28
Thanks, that was really helpful. –  Vinoth Sep 16 '12 at 13:02
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.