$p$ -curvature

In this section, we always assume $p$ to be a prime number.

DEFINITION 7.1   Let $A$ be a commutative ring of characteristic $p$ . Let $B$ be a commutative $A$ -algebra. Let $M$ be a $B$ -module. Let

$$\nabla: M\to \Omega^1_{B/A}\otimes_B M$$

be a connection with zero curvature. Then the $p$ -curvature of $\nabla$ is defined as

$$\operatorname{curv}_p(\nabla)(X)=(\nabla_X)^p -\nabla_{(X^p)}$$

From the argument of the previous section we see that the p-curvature is $p$ -linear.

A fairly good account on $p$ -curvatures is given in [2]. Our treatment here is a bit different. It is not so general, but is easy using only arguments on rings and modules.

DEFINITION 7.2   Let $X$ be a separated scheme over a scheme $S$ . Let $\mathcal{F}$ be a quasi coherent sheaf on $X$ .

$$\nabla: \mathcal{F}\to \Omega^1_{X/S}\otimes_{\mathcal{O}_X} \mathcal{F}$$

be a connection on $\mathcal{F}$ . Then a local section $s$ of $\mathcal{F}$ is said to be ($\nabla$ -)parallel If $\nabla s=0$ . $\mathcal{F}$ is said to be generated by parallel sections if $\mathcal{F}$ is generated by the set

$$\{ s \in \mathcal{F}(X); \nabla s=0\}$$

of parallel sections as a $\mathcal{O}_X$ -module.

$\mathcal{F}$ is said to be locally generated by parallel sections if there exists an open covering $\{U_\lambda\}$ of $X$ such that $\mathcal{F}\vert U_\lambda$ is generated by parallel sections.

PROPOSITION 7.3   Let $S$ be a scheme on which $p=0$ . Let $X$ be a smooth scheme over a scheme $S$ of relative dimension $n$ . Let $\mathcal{F}$ be a quasi coherent sheaf on $X$ .

$$\nabla: \mathcal{F}\to \Omega^1_{X/S}\otimes_{\mathcal{O}_X} \mathcal{F}$$

be a connection on $\mathcal{F}$ .

Then the following conditions are equivalent.

1. $\mathcal{F}$ is locally generated by parallel sections.
2. The curvature and the $p$ -curvature of $\nabla$ are both zero.

Since the question is local, we may reduce the proposition to an lemma which we describe later. Before we do that, we need some preparation. First we note that when $X$ is smooth of relative dimension $n$ , we may locally choose a set of elements $\{x_1,x_2,\dots,x_n\}$ (``coordinates'') such that $\Omega^1_{X/S}$ is freely generated by $\{d x_1, d x_2 \dots, d x_n\}$ over $\mathcal{O}_X$ . Let $U$ be an affine open subset of $X$ on which such a local coordinate system exists. Then there exists vector fields

$$\partial/\partial x_1,\quad \partial/\partial x_2,\quad \partial/\partial x_3,\quad \dots ,\quad \partial/\partial x_n$$

on $U$ which are dual to $\{d x_j\}_j$ . The derivations may also be characterized by the following formula.

$$(\partial/\partial x_j). x_i =\delta_{i j}.$$

Note also that from this observation we see

$$(\partial/\partial x_j)^p=0.$$

Next, let us set some more notation. We employ the graded lexicographic order on an index set

$$\Lambda=\{0,1,2,3,\dots,{p-1}\}^n.$$

We define an order reversing map $\Lambda \in K \mapsto \bar K \in \Lambda$ as follows

$$\overline{(j_1,j_2,\dots, j_n)} =(p-1-j_1,p-1-j_2,\dots, p-1-j_n)$$

$\Lambda$ contains a distinguished element $0=(0,0,\dots,0)$ , elementary vectors $e_i$ and $\bar 0=(p-1,p-1,\dots,p-1)$ .

Finally, for any $I=(i_1,i_2,\dots,i_n)\in \Lambda$ , we define $I!$ by

$$I!=i_1! i_2! i_3! \dots i_n!$$

LEMMA 7.4   Let $A$ be a commutative ring of characteristic $p$ . Let $B$ be a commutative $A$ -algebra. We assume that $B$ contains elements $\{x_1,x_2,\dots,x_n\}$ (``coordinates'') such that $\Omega^1_{B/A}$ is freely generated by $\{d x_1, d x_2 \dots, d x_n\}$ over $B$ . Let $M$ be a $B$ -module. Let

$$\nabla: M\to \Omega^1_{B/A}\otimes_B M$$

be a connection on $M$ . Then the following conditions are equivalent.

1. $M$ is generated by parallel sections.
2. The curvature and the $p$ -curvature of $\nabla$ are both zero.
3. For each element $m\in M$ there exists $\{m_I; I\in \Lambda\}$ such that
   $$m=\sum_{I\in \Lambda} x^I m_I$$
   holds.

PROOF.. Let $M_0$ be the set of parallel sections of $M$ .

$$M_0=\{m\in M; \nabla m=0\}.$$

$(1)\implies (2)$ : We have

$$\nabla \nabla M_0=0$$

$$((\nabla_{D})^p-\nabla_{D^p}) M_0=0 \qquad (\forall D\in \operatorname{Der}_A(B)).$$

Since we know that curvature is $B$ -linear and $p$ -curvature is $p$ -linear over $B$ , we deduce that the curvature and the p-curvature are zero.

$(3)\implies (1)$ : obvious.

$(2)\implies (3)$ : For any $I=(i_1,i_2,\dots,i_n)\in \Lambda$ and for any $v \in M$ , we put

$$\partial_I v= (\nabla_{\partial/\partial x_1})^{i_1} (\nabla_{\pa... ..._{\partial/\partial x_3})^{i_3}\dots (\nabla_{\partial/\partial x_n})^{i_n} v.$$

Note that the condition (1) tells us that

$$\partial_I \partial_J v = \begin{cases} \partial_{I+J} v & \text{if } I+J \in \Lambda \\ 0 & \text{otherwise}. \end{cases}$$

Now we claim :

CLAIM 7.5   For any $J\in \Lambda$ , there exists elements $\{m_I\}_{I\leq J}$ of $M_0$ such that

$$\partial_{\bar K} (m-\sum_{I\leq J} x^I m_I) =0 \qquad (\forall K \leq J).$$

Let us prove the claim above by induction on $J\in \Lambda$ . For $J=0$ , it is easy to see that

$$\partial_{\bar 0}m \in M_0.$$

So it is enough to put

$$m_0=\partial_{\bar 0}m.$$

Assume now that the claim holds for all $J'<J$ . Since $\Lambda$ is well-ordered set, there exists an index $J_0$ which is just before $J$ . (That means, $J_0$ is the largest index which is smaller than $J$ .)

$$\partial_{\overline {K}} (m-\sum_{I\leq J_0} x^I m_I) =0 \qquad(\forall K<J_0)$$

Let us put

$$\tilde m=\partial_{\overline {J}} (m-\sum_{I\leq J_0} x^I m_I) .$$

Then for any $j$ , $J-e_j$ is smaller than $J$ so that we have

$$\partial_j \tilde m=\partial_{\overline {J-e_j}} (m-\sum_{I\leq J_0} x^I m_I)=0 .$$

Thus we have $\tilde m\in M_0$ .

Then we put

$$m_J=(J!)^{-1} \tilde m.$$

We may easily see that

$$\partial_{\overline{K}} (x^J m_J)= \begin{cases} 0 & \text{ if } K< J\\ \tilde{m} & \text{ if } K=J \end{cases}$$

holds and thus the claim holds for $J$ . $\qedsymbol$

It is worthwhile to note that the coefficients $\{m_J\}_J$ in the condition (3) of the Lemma above is unique. Namely,

COROLLARY 7.6   Under the same assumption as the Lemma, assume one (hence, all) of the condition of the Lemma holds. Then a map $\psi$ defined by

$$\psi: \bigoplus_{J\in \Lambda} M_0 \ni (m_J) \mapsto \sum_{I \in \Lambda} x^I m_I\in M$$

is bijective.

PROOF.. We have already shown that $\psi$ is surjective. Let us prove that $\psi$ is injective. Assume on the contrary that there exists a non-zero $(m_I) \in \operatorname{Ker}(\psi)$ . Let $J_0$ be the maximal index such that $m_{J_0}\neq 0$ . Then we have

$$0=\partial_{J_0}(\psi((m_J)))=\partial_{J_0} (\sum_{I\in \Lambda} x^I m_I)=m_{J_0}.$$

This contradicts the assumption.

$\qedsymbol$

COROLLARY 7.7   Let $A$ be a commutative ring of characteristic $p$ . Let $B$ be a commutative $A$ -algebra. We assume that $B$ contains elements $\{x_1,x_2,\dots,x_n\}$ (``coordinates'') such that $\Omega^1_{B/A}$ is freely generated by $\{d x_1, d x_2 \dots, d x_n\}$ over $B$ . Let $B_0$ be the subalgebra of $B$ defined by

$$B_0=\{b \in B; d b=0\}.$$

Then every element $b$ of $B$ is written uniquely as

$$b=\sum_{ 0 \leq j_1,j_2,\dots,j_n \leq p-1} b_{j_1 j_2 j_3\dots j... ..._1} x_2^{j_2} x_3^{j_3}\dots x_n^{j_n} \qquad (b_{j_1 j_2 \dots j_n} \in B_0).$$

REMARK 7.8   The proof of Corollary 7.6 may be replaced by a direct computation.

To obtain that, first we recall Taylor expansion

$$f(s)=\sum_I \frac{1}{I!}f^{(I)}(t)\cdot (s-t)^I$$

of a polynomial $f$ over (say) $\mathbb{C}$ . putting $s=0$ , we obtain

$$f(0)=\sum_I \frac{1}{I!}f^{(I)}(t)\cdot (-t)^I.$$

By analogy we put

$$m_J=\sum_{I \in \Lambda} \frac{1}{I!} (-x)^I (\partial^{I+J} m)$$

Then we see that $m_J$ is parallel.

Indeed, we have the following lemma.

LEMMA 7.9   We employ the same assumption as the Corollary 7.6. Then for any $m\in M$ , an element

$$\sum_{I\in \Lambda} \frac{1}{I!}(-x)^I \partial ^I m$$

is parallel.

PROOF..

$$\partial_j \sum_{I\in \Lambda} \frac{1}{I!}(-x)^I (\partial ^I m)$$

$$= - \sum_{I\in \Lambda} \frac{i_j}{I!} (-x)^{I-e_j} (\partial ^I m)$$

$$+ \sum_{I\in \Lambda} \frac{1}{I!} (-x)^I (\partial ^{I+e_j} m)$$

$$= - \sum_{I\in \Lambda, I\geq e_j}\frac{1}{(I-e_j)!} (-x)^{I-e_j} (\partial ^I m)$$

$$+ \sum_{I\in \Lambda+e_j} \frac{1}{(I-e_j)!} (-x)^{I-e_j} (\partial ^{I} m)$$

If $I=(i_1,i_2,\dots,i_n)\in \Lambda +e_j\setminus \Lambda$ , then we have $i_j=p$ so that $\partial^{I} m=0$ . (We use the assumption $\operatorname{curv}_p(\nabla)=0$ here.) So we finally have

$$\partial_j \sum_{I\in \Lambda} \frac{1}{I!}(-x)^I (\partial ^I m)=0$$

as required.

$\qedsymbol$

Finally we have

PROPOSITION 7.10   We put

$$m_J=\sum_{I \in \Lambda} \frac{1}{I!} (-x)^I (\partial^{I+J} m).$$

Then:

$$m=\sum_{J \in \Lambda} \frac{1}{J !} x^J m_J.$$

PROOF..

$$\sum_{J \in \Lambda} \frac{1}{J !} x^J m_J$$

$$= \sum_{J \in \Lambda} \frac{1}{J !} x^J \sum_{I \in \Lambda} \frac{1}{I!} (-x)^I (\partial^{I+J} m)$$

$$= \sum_{I,J \in \Lambda} \frac{(-1)^{\vert I\vert}}{I! J !} x^{I+J} (\partial^{I+J} m)$$

$$= \sum_{K\in \Lambda+\Lambda }\sum_{I,J \in \Lambda,I+J=K} \frac{(-1)^{\vert I\vert}}{I! J !} x^{K} (\partial^{K} m)$$

Since for $K\in \Lambda+\Lambda$ , $(\partial_K m)=0$ unless $K\in \Lambda$ , the last line is equal to

$$\sum_{K\in \Lambda}\sum_{I,J \in \Lambda,I+J=K} \frac{(-1)^{\vert I\vert}}{I! J !} x^{K} (\partial^{K} m)$$

$$= \sum_{K\in \Lambda}\sum_{I,J \in \Lambda,I+J=K} \frac{K! (-1)^{\vert I\vert}}{I! J !} \frac{1}{K !}x^{K} (\partial^{K} m)$$

So the problem is reduced to computing the following combinatorial number:

$$c_K=\sum_{I,J \in \Lambda,I+J=K} \frac{K! (-1)^{\vert I\vert}}{I! J !}$$

From ordinary binomial theorem, we see easily that

$$\sum_{I,J \in \Lambda,I+J=K} \frac{K! x^I}{I! J !} = \left((1+x_1)^{k_1}(1+x_2)^{k_2}\dots (1+x_n)^{k_n}\right).$$

So we have

$$c_K = \left((1+x_1)^{k_1}(1+x_2)^{k_2}\dots (1+x_n)^{k_n}\right) \vert _{x_1=-1,x_2=-1,\dots,x_n=-1}$$

$$= \begin{align*}\begin{cases}1 & \text{if } K=0 0 &\text{otherwise.} \end{cases}\end{align*}$$

This clearly gives the desired result. $\qedsymbol$