Fano Plane

This list of configurations comes from https://en.wikipedia.org/wiki/Fano_plane#Configurations

$$ Fano = {1 \over 168} ( a_1^7 + 21 a_1^3a_2^2 + 42 a_1 a_2 a_4 + 56 a_1 a_3^2 + 48 a_7 ). $$

There are 7 points

 * There are 7 points, and 24 symmetries fixing any point.

How many points are in a Fano plane ? The cycle index for Fano acting on points is

$$ {1 \over |Fano|} \sum_{\sigma \in Fano } fix(\sigma). a^\sigma $$

$$ {Fano \over X} = {1 \over 24} \left(a_1^6 + 9 a_1^2a_2^2 + 6 a_2 a_4 + 8 a_3^2 \right) = 6T7 $$

42 ordered pairs, 21 unordered pairs
$$ {Fano \over X.X} = a_1.{1 \over 4}(a_1^4 + 3 a_2^2) = X.Klein $$
 * There are 42 ordered pairs of points, and each may be mapped by a symmetry onto any other ordered pair. For any ordered pair there are 4 symmetries fixing it.


 * Correspondingly, there are 21 unordered pairs of points, each of which may be mapped by a symmetry onto any other unordered pair. For any unordered pair there are 8 symmetries fixing it.

How many 2-sets are in a Fano plane ? The cycle index for Fano acting on 2-sets is

$$ {Fano \over { \{ X,X  \} } } = X.Square = 1T1.4T3 $$

21 flags + 84 pointed triangles

 * There are 21 flags consisting of a line and a point on that line. Each flag corresponds to the unordered pair of the other two points on the same line. For each flag, 8 different symmetries keep it fixed.
 * There are 84 ways of specifying a triangle together with one point on that triangle, each of which has two symmetries fixing it.

How many X.2-sets are in a Fano plane''' ? The cycle index for Fano acting on 2-sets is

$$ {1 \over |Fano|} \sum_{\sigma \in Fano } fix(\sigma). a^\sigma $$

$$ {Fano \over {X. \{X,X\} } } = {1 \over 8}(a_1^4 + 3 a_2^2 + 2 a_1^2a_2 + 2 a_4) + a_1^2.{1 \over 2}(a_1^2 + a_2) = Square + X.X.Ens_2 $$

7 quadrangles, (7 lines) + 28 triangles

 * There are 7 lines, and 24 symmetries fixing any line.
 * There are 28 triangles, which correspond one-for-one with the 28 bitangents of a quartic (Manivel 2006). For each triangle there are six symmetries fixing it, one for each permutation of the points within the triangle.
 * There are 7 ways of selecting a quadrangle of four (unordered) points no three of which are collinear, and 24 symmetries that fix any such quadrangle. These four points form the complement of a line, which is the diagonal line of the quadrangle.

How many 3-sets are in a Fano plane ? The cycle index for Fano acting on 3-sets is

$$ {1 \over |Fano|} \sum_{\sigma \in Fano } fix(\sigma). a^\sigma $$

$$ {Fano \over { \{ X,X,X \} } } = {a_1^4 + 3a_2^2 + 6 a_1^2a_2  + 8 a_1 a_3 + 6 a_4 \over 24} + a_1.{a_1^3 + 3 a_1a_2 + 2 a_3 \over 6} = 2{b_1^3 + 3 b_1b_2 + 2 b_3 \over 6} $$

$$= Ens_4 + X.Ens_3 = 4T5 + 1T1.3T2 $$

168 labeled triangles

 * This means that there are 168 labeled triangles fixed only by the identity collineation

$$ {Fano \over { X.X.X }} = Klein + X.X.X.X $$

28 anti-flags
There are 28 ways of selecting a point and a line that are not incident to each other (an anti-flag), and six ways of permuting the Fano plane while keeping an anti-flag fixed. For every non-incident point-line pair (p,l), the three points that are unequal to p and that do not belong to l form a triangle, and for every triangle there is a unique way of grouping the remaining four points into an anti-flag.

$$ {Fano \over {X. \{X,X,X \} } } = { a_1^3 + 3 a_1a_2 + 2 a_3 \over 6} + { a_1^3 + 3 a_1a_2  + 2 a_3 \over 6} + { a_1^3 + a_1 a_2 \over 2} = Ens_3 + Ens_3 + X.Ens_2 $$

28 hexagons
TO DO
 * There are 28 ways of specifying a hexagon in which no three consecutive vertices lie on a line, and six symmetries fixing any such hexagon.

84 pentagons

 * There are 84 ways of specifying a pentagon in which no three consecutive vertices lie on a line, and two symmetries fixing any pentagon.

(X,X,X)
$$ {Fano \over { (X,X,X) } } = {1 \over 12}(a_1^4 + 3 a_2^2 + 8 a_1 a_3) + a_1.{1\over 3}(a_1^3 + 2 a_3) = Alt_4 + X.Alt_3 $$

[X,X,X,X]
$$ {Fano \over { [X,X,X,X] } } = 2 .a_1 .{1 \over 2} (a_1^2 + a_2) = X.Ens_2 + X.Ens_2 $$

Klein
$$ {Fano \over Klein} = 2. a_1^3 = X.X.X + X.X.X $$

(X,X,X,X)
42 cyclic quadrangles

There are 42 ways of selecting a quadrangle of four cyclically ordered points no three of which are collinear, and four symmetries that fix any such ordered quadrangle. For each not-oriented quadrangle there are two cyclic orders.

$$ {Fano \over { (X,X,X,X) } } = a_1{a_1^2 + a_2 \over 2} + a_1^3 = X.Ens_2 + X.X.X $$

$$ = a_1^3b_1^4 + a_1{a_1^2b1^4 + a1^3b_2^2 + 2a_2b_4 \over 4} + = Y.Y.Y.Y + C_4(Y) $$

{X,X,X,X}
$$ {Fano \over \{ X,X,X,X \} } = {1\over 6} (a_1^3 + 3.a_1a_2 + 2.a_3) + {1\over 6} (a_1^3 + 3.a_1a_2 + 2.a_3) = Ens_3 + Ens_3 =3T2+3T2 $$