Why do all incidence theorems follow from Pappus' theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:20:12Z http://mathoverflow.net/feeds/question/65421 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65421/why-do-all-incidence-theorems-follow-from-pappus-theorem Why do all incidence theorems follow from Pappus' theorem? aaron 2011-05-19T11:50:11Z 2011-05-19T20:59:22Z <p>In Hilbert and Cohn-Vossen's Geometry and the Imagination," they state in the last paragraph of Chapter 20 that "Any theorems concerned solely with incidence relations in the [Euclidean projective] plane can be derived from [Pappus' Theorem]."</p> <p>Does anyone know a reference for more details about this claim?</p> <p>Does anyone have an idea of how to prove it?</p> <p>I have seen two books by John Stillwell: Four Pillars of Geometry" and "Yearning for the Impossible." These books discuss the relation between Pappus' theorem in a projective plane over some algebraic structure and commutativity of the multiplication in that algebraic structure. In particular Stillwell describes how all the laws of algebra follow from Pappus' theorem.</p> <p>Presumably one then argues that any incidence theorem can be proved using algebra. I would be happy, for instance, with a precise statement like this, including in particular a definition of incidence theorem."</p> http://mathoverflow.net/questions/65421/why-do-all-incidence-theorems-follow-from-pappus-theorem/65426#65426 Answer by David Speyer for Why do all incidence theorems follow from Pappus' theorem? David Speyer 2011-05-19T12:36:53Z 2011-05-19T15:47:00Z <p>I disagree with this statement. Consider $9$ points in the plane, called $x(i,j)$ where $(i,j)$ is in ${ 0,1,2 }$. There are $12$ triples $((i_1, j_1), (i_2, j_2), (i_3, j_3))$ such that $i_1 + i_2 + i_3 \equiv 0 \mod 3$ and $j_1 + j_2 + j_3 \equiv 0 \mod 3$. Let $L$ be the set of such triples.</p> <p>Consider the following statement:</p> <blockquote> <p>Suppose that, for all $((i_1, j_1), (i_2, j_2), (i_3, j_3)) \in L$, the points $x(i_1, j_1)$, $x(i_2, j_2)$ and $x(i_3, j_3)$ are colinear. Then either all of the $x(i,j)$ are colinear, or else two of the $x(i,j)$ are equal to each other. </p> </blockquote> <p>It seems to me that this statement is an incidence theorem. It is true in $\mathbb{RP}^2$ and false in $\mathbb{CP}^2$, both of which obey Pappus theorem. I learned this example from Kiran Kedlaya.</p> <p>To see this over $\mathbb{R}$, note that a counterexample to this claim is also a counterexample to the <a href="http://en.wikipedia.org/wiki/Sylvester%25E2%2580%2593Gallai_theorem" rel="nofollow">Sylvester-Gallai theorem</a>. Over $\mathbb{C}$, the flexes of any cubic curve form a counter-example, as discussed on the above linked Wikipedia page. More precisely, I believe that the claim is true in $K\mathbb{P}^2$ if and only if $K$ does not contain a root of $x^2+x+1$. </p> <hr> <p>More generally, I would consider an incidence theorem to be a first order statement about points and lines in $\mathbb{RP}^2$ where what we are allowed to say is that a given point does or does not lie on a given line. We can easily turn such a statement into an algebraic claim about $\mathbb{R}$. </p> <p>By a <a href="http://en.wikipedia.org/wiki/Real_closed_field#Model_Theory%3A_decidability_and_quantifier_elimination" rel="nofollow">result of Tarski</a>, any true statement of this form follows from (1) the field axioms (2) the axioms that $\mathbb{R}$ has an ordering $\leq$ obeying the standard properties and (3) the "polynomial intermediate value theorem": for any polynomial $f \in \mathbb{R}[t]$, if $f(a)&lt;0$ and $f(b)>0$, then there exists $c \in (a,b)$ such that $f(c)=0$.</p> <p>For example, to prove that $x^2+x+1$ has no roots in $\mathbb{R}$, just note that $x^2+x+1 = (x+1/2)^2+3/4 \geq 3/4$. Here we have used the field axioms (many times) and the basic properties of $\leq$.</p> <p>Pappus theorem encodes the commutativity of multiplication, and I will believe you that the other field axioms can be deduced from it as well. However, it certainly doesn't include the properties related to inequalities. So, if I rig up an algebraic statement (like the above) whose proof requires the order properties of $\mathbb{R}$, you won't be able to prove it from Pappus theorem. I haven't actually worked this out, but presumably if you apply Tarski's algorithm to the above claim, it will give you a proof which, at some point, involves dividing by $x^2+x+1$ for some unknown quantity $x$.</p> <p>I'll mention that the axioms of an oriented matroid are an attempt to systematize the properties of $\mathbb{RP}^2$ deducible from the order properties of $\mathbb{R}$. It might be true that every true incidence theorem in $\mathbb{RP}^2$ is deducible from Pappus theorem, plus the axiom that the set of points of the plane can be equipped with the structure of an oriented matroid of rank $3$, where the bases are the noncolinear triples.</p> http://mathoverflow.net/questions/65421/why-do-all-incidence-theorems-follow-from-pappus-theorem/65475#65475 Answer by Fedor Petrov for Why do all incidence theorems follow from Pappus' theorem? Fedor Petrov 2011-05-19T20:59:22Z 2011-05-19T20:59:22Z <p>All incidence theorems, which hold over any field, follow from Pappus, because Pappus allows to sum up and multiple points on a line, i.e. implies that your plane is over field (Desargues implies that the plane is over skew field). But some specific properties of the field may be encoded as incidence theorems (I think, any algebraic property like $x^2+1=0$ has no solutions'' may be encoded, just use the geometric construction for addition and multiplication), and hold not for all fields, see spectacular examples in David's answer.</p>