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I am interested in the affine version of the well-known Hessenberg's Theorem (saying that Pappian projective planes are Desarguesian).

First I introduce all necessary definitions.

Definition L. A linear space is a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ satisfying three axioms:

(L1) for any distinct points $x,y\in X$ there exists a unique line $L\in\mathcal L$ containing $x$ and $y$;

(L2) every set $L\in\mathcal L$ contains at least three points;

(L3) $X\notin\mathcal L$.

Elements of the family $\mathcal L$ are called lines.

For any distinct points $x,y\in X$ of a linear space $(X,\mathcal L)$, the unique line $L\in\mathcal L$ containing the points $x,y$ will be denoted by $\overline{xy}$.

Lines $L_1,\dots,L_n$ in a linear space are called concurrent if $\bigcap_{i=1}^n L_i$ is a singleton.

Definition. A linear space $(X,\mathcal L)$ is called

$\bullet$ a projective plane if any distinct lines in $X$ are concurrent;

$\bullet$ an affine plane if for every line $L\in\mathcal L$ and point $x\in X\setminus L$ there exists a unique line $L'\in\mathcal L$ such that $x\in L'$ and $L'\cap L=\emptyset$.

Definition P. A projective plane $(X,\mathcal L)$ is called

$\bullet$ Desarguesian if for every concurrent lines $A,B,C\in\mathcal L$ and points $a,a'\in A\setminus(B\cup C)$, $b,b'\in B\setminus(A\cup C)$, $c,c'\in C\setminus(A\cup B)$, the set $(\overline{ab}\cap\overline{a'b'})\cup(\overline{ac}\cap\overline{a'c'})\cup(\overline{bc} \cap \overline{b'c'})$ is contained in some line;

$\bullet$ Pappian if for every lines $L,L'\in\mathcal L$ and distinct points $a,b,c\in L\setminus L'$ and $a',b',c'\in L'\setminus L$, the set $(\overline{ab'}\cap\overline{a'b})\cup(\overline{ac'}\cap\overline{a'c})\cap(\overline{bc'}\cap\overline{b'c})$ is contained in some line.

By the famous Hessenberg's Theorem, every Pappian projective plane is Desarguesian.

Now let us define affine counterparts of Desarguesian and Pappian projective planes. Given two lines $L,L'\in\mathcal L$ we write $L\parallel L'$ if $L=L'$ or $L\cap L'=\emptyset$.

Definition A. An affine plane $(X,\mathcal L)$ is called

$\bullet$ Desarguesian if for every concurrent lines $A,B,C\in\mathcal L$ and points $a,a'\in A\setminus(B\cup C)$, $b,b'\in B\setminus(A\cup C)$, $c,c'\in C\setminus(A\cup B)$, if $\overline{ab}\parallel\overline{a'b'}$ and $\overline{bc}\parallel\overline{b'c'}$, then $\overline{ac}\parallel \overline{a'c'}$;

$\bullet$ Pappian if for every lines $L,L'\in\mathcal L$ and distinct points $a,b,c\in L\setminus L'$ and $a',b',c'\in L'\setminus L$, if $\overline{ab'}\parallel \overline{a'b}$ and $\overline{bc'}\parallel\overline{b'c}$, then $\overline{ac'}\parallel\overline{a'c}$.

Problem. Is every Pappian affine plane Desarguesian?

Remark. In the lecture notes of Adrien Deloro "Affine and Projective Geometry"
I found Theorem 6.2.1 which contains the proof of Hessenberg's Theorem and also mentions (unfortunately without a convincing proof) that the affine counterpart of the Hessenberg's Theorem also is true. So, what I would like to see is a correct and complete proof of the fact that Pappian affine planes are indeed Desarguesian. Is it published anywhere? If yes, then where exactly?

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  • $\begingroup$ What exactly is it that you find unconvincing about his proof? It looks completely accurate to me: just pass to the projectivization of the affine plane. There is indeed the tricky aspect that for non-Desarguesian projective planes, the removal of different lines could result in non-isomorphic affine planes, but this is not relevant for this proof. $\endgroup$ Commented Oct 25, 2023 at 14:28
  • $\begingroup$ @TomDeMedts The problem with this proof is that when you look at the Affine Pappus Axiom as a property of a projective plane, then it means that you can only apply it to some selected line (at infinity). Without the Desargues Axiom, we have no homogeneity of the projective completion, so cannot move that selected line to other places. But Hessenberg's Theorem applies the Pappus Axiom for three different lines. So, I do not see how it can be applied. Or I do not see something obvious? $\endgroup$ Commented Oct 25, 2023 at 16:50
  • $\begingroup$ @TomDeMedts If indeed everything can be done with the projectivization trick, then it also should be done without that trich just in the affine plane. And it indeed can be done, but such a modified proof requires some different form of the Pappus Axiom (not that one, equivalent to the commutativity of the division ring at the presence of the Desargues Axiom). $\endgroup$ Commented Oct 25, 2023 at 16:55

2 Answers 2

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There is German book Plane geometry (2007) which contains equivalent adapted Hessenberg Theorem for affine planes.

It's too long to translate and adapt proof to your terminology, so here are screenshots of formulations and proof.

enter image description here enter image description here enter image description here enter image description here

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  • $\begingroup$ I would like to thank @LutzMattner for a valuable information about the proof of the affine version of the Hessenberg Theorem, which led to finding the book "Plane Geometry" of Max Koecher and Aloys Krieg, mentioned in the above answer. $\endgroup$ Commented Dec 14, 2023 at 19:37
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It seems that the proof from the book posted by @ihromant follows the lines of the original Hessenberg's proof from his paper in Mathematische Annalen of 1905:

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