How to visualize the Frobenius endomorphism?

Question

As the question title asks for, how do others "visualize" the Frobenius endomorphism? I asked some people in real life and they said they didn't know and that I could go and ask on MO and possibly get miseducated. So that's what I am doing. Bonus points for pictures.

Answer 1

Probably the best way to think about the Frobenius morphism, and more generally purely inseparable finite morphisms of schemes, is as some sort of foliation. For example, Ekedahl proved the following result:

Proposition [Ek, Prop. 2.4]. Let $Y$ be a smooth variety over a perfect field $k$ of characteristic $p > 0$. Then there is a $1$-$1$-correspondence between finite flat height $1$ purely inseparable morphisms $Y \to X$ to a smooth variety $X$ and subbundles $\mathscr E \subseteq \mathcal T_{Y/k}$ stable under Lie brackets and $p$-th powers, associating to $Y \to X$ the subbundle $\mathcal T_{Y/X} \subseteq \mathcal T_{Y/k}$.

Here, a purely inseparable morphism $Y \to X$ has height n¹ if and only if there exists a morphism $X \to Y^{(p^n)}$ such that the composition $Y \to X \to Y^{(p^n)}$ is the $p^n$-th power relative Frobenius.

Among height $1$ morphisms, the Frobenius $Y \to Y^{(p)}$ is clearly terminal, hence corresponds to the subbundle $\mathcal T_{Y/k}$ (this is also clear from the description of the correspondence). If $Y$ is smooth of dimension $n$ over a perfect field $k$, then the fibres of relative Frobenius $F_{Y/k} \colon Y \to Y^{(p)}$ are isomorphic to the 'thick point' $\operatorname{Spec} k[x_1,\ldots,x_n]/(x_1^p,\ldots,x_n^p)$.

What does this look like?

So we are to think of purely inseparable morphisms as 'infinitesimal foliations': they define a subbundle of the tangent bundle, and 'integrating' this subbundle should give the 'leaves' of the foliation. Here is a very crude picture:

We see that the tangent space of the leaves is the entire space, and they form some sort of thick covering of $Y$, whose fibres are Artin local schemes of length $p^n$ (as above).

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¹This seems to be standard terminology. If $Y \to X$ has height $n$, it also has height $m$ for any $m \geq n$. Probably a better definition would be that $Y \to X$ has height $\leq n$.

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References.

[Ek] Ekedahl, Torsten, Foliations and inseparable morphisms. In: Algebraic geometry, Proc. Summer Res. Inst., Brunswick/Maine 1985, Part 2. Proc. Symp. Pure Math. 46, No. 2, 139-149 (1987). ZBL0659.14018.