What is the image of $\Phi$ ?

The answer is obtained by considering derivations.

LEMMA 8.2   Let $k$ be a field of characteristic $p$ . For each $i=1,2,3,\dots,2 n$ , let $\nabla^{(0)}_i$ be a derivation on $M_{p^n}(k[t_1,t_2,t_3,\dots,t_{2n}])$ defined by

$$\frac{\partial}{\partial t_i}- \sum_j\overline{h}_{i j} \operatorname{ad}(\mu_j).$$

Then:

1. We have

   $$\nabla^{(0)}_i(x)=0 \quad (i=1,2,3,\dots,2n)$$ (*)

   
   

   for any element $x \in \operatorname{Image}(\Phi)$ .

2. Conversely, any element $x$ of $M_p^n(k[t_1,t_2,t_3,\dots,t_{2 n}])$ which satisfy the equations (*) belongs to the image $\operatorname{Image}(\Phi)$ .

PROOF.. (1) Since $\nabla^{(0)}_i$ is a derivation, we see that the set of elements in $M_p^n(k[t_1,t_2,t_3,\dots,t_{2 n}])$ which satisfy the equations (*) above form a $k$ -algebra. It is also easy that $\nabla^{(0)}_i(\Phi(\gamma_j))=0$ for all $i,j$ .

(2) Any element $x$ of $M_p^n(k[t_1,t_2,t_3,\dots,t_{2 n}])$ may be written uniquely as

$$\sum_I t^I f_I(\Phi(\gamma))$$

(where sum is taken over indices $I\subset \{0,1,2,3,\dots,p-1\}^n$ ) for some polynomial $f_I$ .

$$\sum_I \partial_i(t^I) f_I(\Phi(\gamma))=0$$

We may easily deduce that this happens only when $f_I=0$ for all $I\neq 0$ .

$\qedsymbol$

DEFINITION 8.3   For any vector field $v=\sum_j v_j(t)\partial_j$ on $\mathbb{A}^{2 n}$ , We define

$$\nabla^{(0)}_v=\sum_j v_j \nabla^{(0)}_j$$

LEMMA 8.4   $\nabla^{(0)}$ is a connection. That means,

1. $\nabla^{(0)}$ is bi-additive
   $$\nabla^{(0)}_{v_1+v_2}= \nabla^{(0)}_{v_1}+\nabla^{(0)}_{v_2},\quad \nabla^{(0)}_{v}(f+g)= \nabla^{(0)}_{v}(f)+\nabla^{(0)}_{v}(g).$$

2. For each $v$ , $\nabla^{(0)}_v$ is a first order differential operator. Namely, we have
   $$\nabla^{(0)}_v (f m)=(v.f)m +f \nabla^{(0)}_v.m \qquad (\forall f\in k[t], \quad \forall m \in M_{p^n}(k[t])).$$

PROOF.. Easy. $\qedsymbol$

LEMMA 8.5   $\nabla^{(0)}_v$ is the only first order differential operator on $M_{p^n}(k[t])$ such that its principal symbol is $v$ and

$$\nabla^{(0)}_v(\operatorname{Image}(\Phi))=0$$

holds.

PROOF.. Let $D$ be another first order differential operator with the same property. Then we see that the difference

$$P=\nabla^{(0)}_v -D$$

is a $k[t]$ -linear map from $M_{p^n}(k[t])$ to itself, and that $P$ is zero when restricted to the image $\operatorname{Image}(\Phi)$ of $\Phi$ . Since $\operatorname{Image}(\Phi)$ generates $M_{p^n}(k[t])$ as a $k[t]$ -module, we see immediately that $P$ is equal to zero.

$\qedsymbol$

We could go further and describe fully the result obtained in the author's papers in terms of algebras (that means, ``global'' things.)

But the author thinks it unnatural to do so without even mentioning geometric interpretation.

So let us close the part I of this talk and proceed to a more sophisticated world of schemes.