User talk:Chris

Hi Marty!

I've been having some issues with the theory and some of the problems in Chapter 2....

First off, I was looking at Exercise 2.8.8 and I have what I think is a counterexample... I'm not sure if I've got mixed up someplace or if there's a problem with the question. My example was in the case $m=4$. Consider the usual rank two geometry of the square over points and lines. Add to this two more lines that are incident to opposite vertices (I'm thinking of a square with a cross in the center) and let this be $\Gamma$. Then, at least the way I've understood the definitions, this satisfies the hypothesis but is not a generalized 4-gon since $g=3$, $d_p=3$ and $d_l=4$... It also seems to contradict the hint...
 * I don't think that your counterexample satisfies condition (a) of the problem. Consider the pair of elements, consisting of a corner $c$ of the square, and a diagonal $d$ of the square which is not incident to $c$.  Then, there are two chains of length $3$ joining $c$ to $d$.  Marty 08:52, 5 February 2009 (PST)

The remarks in the text concerning Example 2.1.3 (necessary for Exercise 2.8.1) were also confusing. At one point the example states `The digon diagram of $\Gamma$ is a quadrangle in which $b$ and $w$ represent opposite vertices and $g$ and $r$ likewise' which I took to mean that neither $b$ and $w$ nor $g$ and $r$ were `joined' (definition 2.1.1). However, I'm under the impression that $b$ and $w$ are joined since (using the notation and statement of definition 2.2.1) taking the flag $F$ to be any element of type $r$ (so that typ$(F)$ belongs to $I\backslash\{b,w\}$), the residue $\Gamma_F$ will contain, amongst other things, all the vertices of that cell/cube. Therefore, considering any two vertices that are opposite in that cube, one will be of type $b$, the other type $w$ but the two are not incident. Doesn't this show $b$ and $w$ are joined?
 * The digon diagram of $\Gamma$ is a square, with $b$ and $w$ at opposite corners and $g$ and $r$ at opposite corners, as Cohen suggests. In fact, $b$ and $w$ are not joined, and neither are $g$ and $r$.  Here's the reason that $b$ and $w$ are not joined: any flag $F$ of type $\{ g,r \}$ (you have to consider type $\{ g, r \}$ and not just $\{ r \}$ as you suggest), consists of two adjancent cubes.  The residual geometry $\Gamma_F$ consists of the "square" shared by these two cubes.  Namely, $\Gamma_F$ consists of four points, two of type $b$ and two of type $w$.  Any point of type $b$ is adjancent to any point of type $w$.  I.e., there exists no pair of nonincident points in $\Gamma_F$ of types $b$ and $w$.  So $b$ and $w$ are not joined.  The argument for $g$ and $r$ not to be joined is the same (in fact, there is a symmetry switching the sets $\{ b,w \}$ and $\{g, r \}$). Marty 08:52, 5 February 2009 (PST)

Finally, the second part of that same exercise 2.8.1 was giving me some trouble since I'm having a hard time understanding the incidence given by `principal of maximal intersection' (cf. chapter one, Exercise 1.7.23). The definition of the maximum intersection geometry is given in terms of a set P, a family of subsets $B$ and a map typ$: B \to I$. The exercise in question relates back to example 2.1.4 so when you have time, do you think you could just briefly explain what $P$ is (my guess is all elements of types $0,1,g,r$?), what the family of subsets $B$ are (the power set of the set of elements all of one specific type?!) and give an example of an incidence between elements in relation to the geometry of Example 2.1.4 (or alternatively a example of incidence in Exercise 1.7.23 if its easier to explain)?... I think that would clear it up.
 * This is confusing to me too. Here would be my guess:  the set $P$ is $\ZZ^3$ (the set of all vertices in the tesselation of three-space).  The subsets of type $0$ are just the elements of $P$.  The subsets of type $1$ are the sets $(p,q)$ where $p,q$ are joined by an edge in the tesselation.  The subsets of type $r$ are octuples of points, forming a "red cube".  The subsets of type $g$ are octuples of points, forming a "green cube".  Adjacency is given by touching in the tesselation, but the principle of maximal intersection implies the following:  a cube is only adjacent to an edge, if they are "fully adjacent" -- the edge is contained in the (closure of the) cube.  A cube is not adjacent to an edge, if they only have a vertex in common (non-maximal intersection). Marty 09:05, 5 February 2009 (PST)

Thank you, Chris