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Let $\mathbf{F}_p$ be the field with $p$ elements, and $X = (\mathbf{P}^1_{\overline{\mathbf{F}}_p})^\text{perf}$ the inverse perfection of the projective line over $\mathbf{F}_p$.

Let $\Gamma$ be the graph of the inverse of geometric Frobenius $F:=\phi\otimes\overline{\mathbf{F}}_p$, with $\phi$ the $p$-th power map on $\mathbf{P}^1_{\mathbf{F}_p}$.

The intersection $\Gamma\cap \Delta$, with $\Delta$ the diagonal, is $X(\mathbf{F}_p)$, although intuitively each of the elements of $X(\mathbf{F}_p)$ should be counted with multiplicity $1/p$.

Is this intuition correct and is there a way to make it precise?

For example, if one could talk about cycles on $X$, perhaps one could argue that $$(1\times F)^*(\Gamma\cap\Delta)=\Delta\cap\Gamma_F$$ and since $(1\times F)$ multiplies cycles by $p$ (hopefully), then $\Gamma\cap\Delta = (1/p)(\Gamma_F\cap\Delta)$, with $\Gamma_F$ the graph of $F$. Now $\Gamma_F\cap\Delta$ is supported on $X(\mathbf{F}_p)$ with its elements counted with multiplicity one.

I would be interested in a reference, if any.

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  • $\begingroup$ The question seems out of order—surely an intuition can't be correct or incorrect until it is made precise? $\endgroup$
    – LSpice
    Commented Oct 29, 2023 at 23:14
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    $\begingroup$ @LSpice I disagree. There are ideas based on intuition that are "morally" correct, and fruitful, without necessarily being precise. (I take by "out of order" you don't mean "against the rules", but "in the wrong order"?) $\endgroup$
    – David Roberts
    Commented Oct 30, 2023 at 0:40
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    $\begingroup$ I have confusions about your notation. In the perfection, you wrote $\bar{\mathbb F}_p$ in the perfection, but is $X$ over $\bar{\mathbb F}_p$ or $\mathbb F_p$? In the former case, $X(\mathbb F_p)$ does not make much sense. $\endgroup$
    – Z. M
    Commented Oct 30, 2023 at 13:09

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