# Uniform Position in Characteristic p

Let $X$ be an irreducible degree $d$ curve in $\mathbb{C} \mathbb{P}^r$. Some hyperplanes will be tangent to $X$, or pass through singularities of $X$, or even perhaps contain $X$. But most hyperplanes won’t do any of these things. Let $U$ be the space of hyperplanes which do not exhibit any of these bad behaviors; these hyperplanes are said to be transverse to $X$. Every hyperplane $H$ in $U$ meets $X$ in $d$ points. As we move $H$ through $U$, these $d$ points move around $X$. If $H$ navigates a loop in $U$, before returning to itself, then these $d$ points are permuted. Last week, in his class on Algebraic Curves, Joe Harris proved the following theorem.

The Uniform Position Principle: Let $G$ and $H$ be two hyperplanes in $U$, with $G \cap X = \{ x_1, \ldots, x_d \}$ and $H \cap X = \{ y_1, \ldots, y_d \}$. Then, for any permutation $\sigma \in S_d$, there is a path from $G$ to $H$ within $U$ such that traveling along this path takes $x_i$ to $y_{\sigma(i)}$.

In this proof, it was crucial that we were working with a field of characteristic zero. The key lemma was that there was a little loop in $U$ such that traveling around that loop swapped exactly two points. The proof, in sketch, is to find a hyperplane $H_0$ which is not transverse to $U$, but just barely; so that $H_0$ is tangent to $X$ at one point and meets $X$ transversely at $d-2$ other points. If we then wiggle $H_0$ in a little disc, then the tangency point of $X$ and $H_0$ will separate into two distinct points, while the other $d-2$ points will wiggle around a little. The boundary of that disc will be a loop in $U$ and (exercise!) as you travel around that loop, the two perturbations of the tangency point switch with each other.

In characteristic $p$, there are curves for which there are no such tangent planes $H_0$. As I’ll show you soon, there exist curves $X$ where every single point $x \in X$ is a flex, meaning that the tangent line to $X$ at $x$ touches $X$ with degree $\geq 3$. So the proof falls apart. We got into a discussion in Harris’ class about whether the result also falls apart. I’ve done some computations and the answer is “yes”. Moreover, the monodromy groups we get are very pretty.

Let’s understand what a curve where every point is a flex “looks like”. The standard example of a flex is the point $(0,0)$ on $y=x^3$, back in characteristic zero. At the point $(t,t^3)$, this curve has slope $3 t^2$. As $t$ increases, starting at $-\infty$, the slope of the curve decreases until we hit $t=0$. There, the rate of change of the slope is momentarily zero, after which the slope starts increasing. Note that the slope need not be zero at a flex: The curve $y=x+x^3$ also has a flex at zero. The slope of this curve at $(0,0)$ is $1$; what matters is that this is the point where the rate of change of the slope is zero.

Now, let’s switch to characteristic $p$. Look at the curve $y=x^p$. At every single point, this curve has slope zero! So the slope is never changing, and every point is a flex. Another example is $y=x^{p+1}$. Now the slope is $x^p$. So the slope is changing, but the rate of change of the slope is everywhere zero so, again, every point is a flex. (This is the fundamental difference between characteristic $p$ and characteristic zero. In characteristic $p$, a function with derivative zero need not be constant. The monomial $x^p$ sneaky: you’ll never catch it growing, yet grow it does!)Let’s take our first example: the curve $y=x^p$ in the plane. Let’s intersect this with the line $y=mx+b$, so we get the equation $x^p=mx+b$. When $m \neq 0$, this equation has $p$ distinct roots (exercise!). When $m=0$, all $p$ roots collide to give a $p$-fold root at $b^{1/p}$. So, as $(m,b)$ moves around, with $m \neq 0$, we want to know how these $p$ points move. Of course, if we are going to be rigorous, we can’t actually move points around, we need to talk about the etale fundamental group. But those who know what that means should be able to make what I’m saying rigorous and those who don’t will understand this.

Let $r_0$ be one of the roots of $x^p=mx+b$. Then the other roots are $r_1:=r_0+m^{1/(p-1})$, $r_2:=r_0+2m^{1/(p-1)}$, … $r_{p-1}:=r_0+(p-1) m^{1/(p-1)}$ (exercise!). So, if $a-b \equiv c-d \mod p$ then $r_a-r_b=r_c-r_d$. If our monodromy carries $r_i$ to $r_{\sigma(i)}$, then $\sigma$ must have the property that $a-b \equiv c-d \mod p$ implies $\sigma(a)-\sigma(b)=\sigma(c)-\sigma(d)$. In other words, $\sigma$ must be of the form $i \mapsto gi+h$, for some constants $g$ and $h$. More geometrically, if we move $b$ while keeping $m$ fixed, we shift $r_i$ to $r_{i+h}$, with $h$ depending on which “path” we move $b$ along. If we let $m$ vary as well, then the different $(p-1)$-st roots of $m$ are interchanged, in a cyclic cover branched over $m=0$.

That was pretty, but here is the really fun example. Look at the genus zero, degree $p+1$ curve $X$ in projective three space which is parameterized by $(w:x:y:z)=(1:t:t^p:t^{p+1})$. What do you think the monodromy group is? Feel free to leave an argument in the comments section; I’ll post the answer sometime next week. Alternatively, if that is too easy, what about the curve $(1:t:t^p:t^{p+1}:t^{2p}:t^{2p+1})$? In that case, I don’t even have a guess yet.