Let $C \subset {\bf CP}^2$ be an irreducible algebraic smooth (projectively) planar curve over the complex numbers of degree $d$ (we allow finitely many points to be deleted from $C$ to make it smooth). Then a generic complex line in ${\bf CP}^2$ intersects $C$ in $d$ distinct points $z\_1,\ldots,z\_d$.



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> **Question 1:** for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, is it possible to continuously deform the $d$ points $z\_1,\ldots,z\_d$ to $z\_{\sigma(1)},\ldots,z\_{\sigma(d)}$ while keeping all the points collinear, distinct, and on $C$ at all times? In other words, does there exist continuous maps $z\_i: [0,1] \to C$ with $z\_i(0) = z\_i$, $z\_i(1) = z\_{\sigma(i)}$, and the $z\_1(t),\ldots,z\_d(t)$ distinct and collinear for all $t\in [0,1]$?
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I believe the answer to this question is yes, because one can transpose any two of the $z\_i,z\_j$ while keeping the other $z\_k$ unchanged by bringing them close together near a generic point of $C$ (and making the collinear line close to the line of tangency of this point to $C$, which generically will not be tangent to anywhere else in $C$), performing the transposition, and then returning back to the original position. One can also phrase the question as follows: if we let $S \subset C \times Gr(2,1)$ be the set of incidences $(p,\ell)$ with $p \in C$ and $\ell \in Gr(2,1)$ a line through $p$, then (after passing to an open dense subset of $Gr(2,1)$), $S$ is a covering space over (most of) $Gr(2,1)$ whose fibre has $d$ points, and the question asserts that the fundamental group of the base acts completely transitively on the fibre (in that every permutation of the fibre shows up); equivalently, the question asserts that the $d^{th}$ exterior power of $S$ over $Gr(2,1)$ is irreducible.


However, I'm having more trouble with proving the following generalisation of Question 1. Keep the same setup as before, but now suppose we also have another irreducible curve $C'$ covering $C$ with fibres of cardinality $d'$. Thus, above a generic $z\_i$ in $C$ we have $d'$ points $w\_{i,1},\ldots,w\_{i,d'}$ in $C'$.



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> **Question 2:** for any permutation $\sigma: \{1,\ldots,d\} \to \{1,\ldots,d\}$, and any $j\_1,\ldots,j\_d,k\_1,\ldots,k\_d \in \{1,\ldots,d'\}$ is it possible to continuously deform the $d$ points $w\_{1,j\_1},\ldots,w\_{d,j\_d}$ to $w\_{\sigma(1),k\_1},\ldots,w\_{\sigma(d),k\_d}$ while keeping all the points on $C'$, and their projections onto $C$ collinear and distinct?
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Like Question 1, one can phrase this question in terms of the transitivity properties of the action of the fundamental group of a suitable open dense subset of $Gr(2,1)$ on some fibre, but this formulation becomes rather complicated to state and I won't give it here.


I believe I can resolve this problem in the affirmative if $C$ has a smooth closure, in which case I can use ad hoc arguments to move around each of the $j\_i$ independently, but encountered a problem if $C$ has a singular point in its closure, and $C'$ ramifies above this point, as then the points $z\_i$ appear to become entangled with each other near this point in a manner that I was not able to analyse by ad hoc topological arguments.


[My motivation for this problem was to understand arcs in a finite plane ${\bf F}\_p^2$, that is to say sets that do not contain any collinear triples, and to classify when these arcs can be constructed as the ${\bf F}\_p$-points of a low-degree algebraic curve $C$ (or as the projection of the ${\bf F}\_p$-points of a low-degree algebraic curve $C'$). Using a Lefschetz principle argument, one can transfer the problem (in the large $p$ limit) to a problem in the complex plane related to the questions above.]



Permuting collinear points on a curve
The answer to question 1 is yes, and is known as Harris's Uniform Position Lemma. It was proved in Harris's 1980 paper [Galois groups of enumerative problems](http://www.ams.org/mathscinet-getitem?mr=552521).
You can find a nice exposition in Chapter 9 of [Solving Polynomial Equations](http://books.google.com/books?id=aIlSmBV3yf8C&lpg=PA367&ots=G9ldcJ8-gQ&dq=uniform%20position%20theorem%20harris&pg=PA407#v=onepage&q=uniform%20position%20theorem%20harris&f=false), by Bronstein, Dickenstein and Emiris.


You should be warned that the analogous statement is not true in characteristic $p$, which you say is your motivation. For example, consider the curve $y=x^p$ in characteristic $p$. Any line $y=mx+b$ meets this curve at $p$ points, whose $x$ coordinates are the roots of $x^p=mx+b$. Note that the roots of this polynomial form an arithmetic progression: If $r$ and $s$ are roots, so is $r+k(s-r)$ for every $k \in \mathbb{Z}/p$. That means that the monodromy, as you move the line, must preserve this structure, so it is a subgroup of $\mathbb{Z}/p^{\ast} \ltimes \mathbb{Z}/p$. 




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**UPDATE** I think question 2 is false! Let $C$ be a smooth cubic curve in $\mathbb{P}^2$, and choose a flex on $C$ to be the origin, so $C$ has a group structure. Take $C'$ to be equal to $C$, but with the map $C' \to C$ being the doubling map in the group law. So $d'=4$.


If $L$ is a line in $\mathbb{P}^2$, and $L \cap C = \{ x,y,z \}$, then $x+y+z=0$ in the group law. If $2 x'=x$, $2y' =y$ and $2 z'=z$, then $x'+y'+z'$ is a $2$-torsion point of $C'$. If we move $(L, x',y',z')$ continuously, we can't change which $2$-torsion point $x'+y'+z'$ is. So there are (at least) four separate connected components in the space of $(x',y',z')$ triples.


