$\mathcal{D}$-modules of level m - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:40:25Z http://mathoverflow.net/feeds/question/106740 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106740/mathcald-modules-of-level-m $\mathcal{D}$-modules of level m Vinoth 2012-09-09T17:07:31Z 2012-09-16T08:28:02Z <p>My question is regarding the definition of $\mathcal{D}$-modules of level $m$ given in <a href="http://arxiv.org/abs/0811.1168" rel="nofollow">this</a> 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.</p> <p>Consider the immersion $X \rightarrow X \times_S X$. Definition $2.1$ from that paper defines what a <em><code>divided power structure of level $m$</code></em> on an immersion is; Definition $2.3$ (and $2.4$) states constructs the <em><code>divided power envelope of level $m$</code></em>, $P_{X,m}(Y)$. 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 <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf" rel="nofollow">this other paper</a>.</p> <p>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. </p> <hr> http://mathoverflow.net/questions/106740/mathcald-modules-of-level-m/107131#107131 Answer by Konstantin Ardakov for $\mathcal{D}$-modules of level m Konstantin Ardakov 2012-09-13T22:35:05Z 2012-09-16T08:28:02Z <p>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 <a href="http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_" rel="nofollow">EGA 4, Quatrième partie</a>.</p> <p>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 $m = \infty$.</p> <p>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 <em>dual</em> (as a $B$-module) to the $B$-module of all Grothendieck differential operators of order at most $n$.</p> <p>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 <em>ideal</em> of the diagonal, namely the kernel of this map, is generated as an ideal by the element </p> <p>$$\tau := t \otimes 1 - 1 \otimes t.$$</p> <p>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 (<a href="http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_" rel="nofollow">EGA IV, 16.7.1.1</a>), the global sections of $\mathcal{P}^n_{X\infty}(Y)$ are just</p> <p>$$P^n := \mathcal{O}(Y) / (\tau^{n+1})$$</p> <p>--- 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, </p> <p>$$\mathcal{D}^{(\infty)}_n (X) := Hom_B(P_n, B) =: D_n$$</p> <p>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. </p> <p>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 <a href="http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_" rel="nofollow">EGA IV, Lemma 16.8.9.1</a>. 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</p> <p>$$\delta( \overline{\tau} ) = \overline{ \tau} \otimes 1 + 1 \otimes \overline{\tau}$$</p> <p>(it is a "primitive element" in an appropriate bialgebra --- see <a href="http://www.numdam.org/numdam-bin/feuilleter?id=PMIHES_1967__32_" rel="nofollow">EGA IV.4, 16.8.9.4</a>). Let's now work out how to multiply $\partial^{[i]}$ by $\partial^{[j]}$ (drop the bars for clarity):</p> <p>$$(\partial^{[i]} \cdot \partial^{[j]})(\tau^k) = (\partial^{[i]} \otimes \partial^{[j]})(\tau \otimes 1 + 1 \otimes \tau)^k = \sum_{a + b = k} \binom{k}{a} \partial^{[i]}(\tau^a) \partial^{[j]}(\tau^b)$$</p> <p>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</p> <p>$$\partial^{[i]} \cdot \partial^{[j]} = \binom{i+j}{i} \partial^{[i+j]}$$</p> <p>which is hopefully the familiar rule for multiplying divided powers (morally $\partial^{[i]} = \partial^i/i!$). </p> <p>The point of the Berthelot construction is that it is possible to <em>vary</em> 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 <em>all</em> 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 <em>crystalline differential operators</em> (no divided powers).</p> <p>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,</p> <p>$$\Gamma(Y, \mathcal{P}^n_{Xm}(Y)) = \oplus_{a=0}^n B \tau^{ \{ a \} }$$</p> <p>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).</p> <p>To understand the multiplication of the <em>dual</em> vectors to these $\tau^{ \{ a \} }$, namely the $\partial^{ \langle a \rangle }$, we need to understand how to <em>comultiply</em> the $\tau^{ \{ a \} }$. So we compute (again dropping bars for convenience)</p> <p>$$\delta( \tau^{ \{ a \} }) = \frac{1}{q_a!} \delta(\tau)^a = \sum_{i+j = a} \frac{1}{q_a!} \binom{i+j}{i} \tau^i \otimes \tau^j = \sum_{i+j = a} \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \tau^{ \{ i \} } \otimes \tau^{ \{ j \} }$$ </p> <p>and the same computation as above in the case $m=\infty$ shows that </p> <p>$$\partial^{\langle i \rangle} \cdot \partial^{\langle j \rangle} = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \partial^{\langle i + j \rangle}.$$</p> <p>The interesting thing is that this structure constant $\frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i}$ is always a <em>$p$-integral</em> rational number (see Lemma 1.1.3(i) in <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf" rel="nofollow">Berthelot's paper</a>), 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 </p> <p>$$\partial^{\langle i \rangle} \mapsto \frac{q_i!\partial^i}{i!}$$</p> <p>since</p> <p>$$\left(\frac{q_i! \partial^i}{i!}\right) \cdot \left(\frac{q_j! \partial^j}{j!}\right) = \frac{q_i! q_j!}{q_{i+j}!} \binom{i+j}{i} \left( \frac{q_{i+j}! \partial^{i+j}}{(i+j)!}\right).$$</p> <p>Thus morally $\partial^{\langle i \rangle}$ should be thought of as "modified divided powers" $q_i! \partial^i / i!$. </p> <p>Finally, you don't need <em>all</em> 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 $i &lt; p^m$ 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 $q_{p^a} = p^{a-m}$,</p> <p>$$\partial^{\langle p^a \rangle} = \frac{ p^{a-m}! }{ p^a! } \partial^{p^a} = \left(\frac{ p^{a-m}! (p^m!)^{p^{a-m}} }{p^a!} \right) (\partial^{\langle p^m \rangle})^{p^{a-m}}$$</p> <p>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 </p> <p>$$\frac{p^{a-m}-1}{p-1} + p^{a-m} \frac{p^m-1}{p-1} - \frac{p^a-1}{p-1} = 0.$$</p> <p>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.</p> <p><strong>Edit:</strong> 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 </p> <p>$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \frac{q_{a+b}!}{q_a!q_b!} \tau^{ \{ a + b \} }$$</p> <p>(note that the structure constant $\frac{q_{a+b}!}{q_a!q_b!}$ is actually an <em>integer</em>, this again follows from Lemma 1.1.3(i) in <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_1996_4_29_2/ASENS_1996_4_29_2_185_0/ASENS_1996_4_29_2_185_0.pdf" rel="nofollow">Berthelot's paper</a>.) 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 $a \geq n+1$.</p> <p>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</p> <p>$$\tau^{ \{a \} } \cdot \tau^{ \{b \} } = \binom{a+b}{a} \tau^{ \{ a + b \} }.$$</p>