# Katz's paper on $p$-curvature – help with proof understanding

I am studying N. Katz's paper "Nilpotent connections and the monodromy theorem: applications of a result of Turrittin" where I found a fairly good account on $$p$$-curvatures.

I don't understand the following proof:

Let :

$$\Psi: \operatorname{Der}(S|T) \to \operatorname{End}_T(\mathcal E)$$

$$D \to (\nabla(D))^p -\nabla(D^p)$$

Where $$\nabla: \operatorname{Der}(S|T) \to \operatorname{End}_T(\mathcal E)$$ such that: $$\nabla(D)(ge) = D(g)e+g\nabla(D)(e)$$, $$e$$, $$g$$ and $$D$$ sections of $$\mathcal E$$, $$\mathcal O_S$$ and $$\operatorname{Der}(S|T)$$ respectively and $$\mathcal E$$ is a vector bundle on $$S$$.

1. To prove $$(5.4.4)$$, we have by $$p$$-linearity and additivity of the p-curvature:

$$\Psi(D)=\sum_i a_i^p \Psi\big( \frac{\partial}{\partial s_i}\big) = \sum_i a_i^p \Big(\nabla\big(\frac{\partial}{\partial^p s_i}\big)\Big)^p -\sum_i a_i^p \nabla\big(\frac{\partial^p}{\partial s_i^p}\big)$$ but the term $$\sum_i a_i^p \nabla\big(\frac{\partial^p}{\partial s_i^p}\big)$$ disappears in the proof and I don't see why?

2. At the end of the proof, it looks like we use the fact that $$\frac{\partial}{\partial s_i}$$ and $$\frac{\partial}{\partial s_j}$$ commute, but why is that true?

• When you write "$\nabla(D)(g e) = D(g)e + g\nabla(D)(e)$", you say what $g$ and $D$ are, but not what $e$ is. Mar 17, 2020 at 17:07
• Thank you, I just edited the question to define $e$. Mar 17, 2020 at 17:12

1. The term in question vanishes because the derivation $$(\partial/\partial s_i)^p$$ is zero, and hence $$\nabla((\partial/\partial s_i)^p) = 0$$. This is because, in characteristic $$p$$, $$(\partial/\partial x)^p(x^n) = 0$$ for all $$n \in \mathbb{N}$$.
2. These derivations do commute, because $$\partial/\partial s_i$$ and $$\partial/\partial s_j$$ commute on polynomials $$k[s_1,\ldots,s_r]$$.
• I don't know what you are referring to. 2. is the algebraic analog of the calculus fact that, for sufficiently smooth functions, mixed partial derivatives commute. It is proven by: $\partial_i \partial_j (s_i^m s_j^n) = m n s_i^{m-1} s_j^{n-1} = \partial_j \partial_i (s_i^m s_j^n)$. Mar 17, 2020 at 20:56