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## Classification of Flocks for q < 7

In order to solidify the ideas that have already been presented and to give a specific example of a non-linear flock, we shall now classify all flocks of wide cones for q in the range 2q5.

The case of q = 2 has already been dealt with (), the flock of any cone which has a flock is linear.

For q = 3, by examining baselines we can conclude that:

Proposition 9.1: The flocks of all non-flat cones in PG(3,3) that have flocks are linear. The flocks of any cone are star flocks.

To deal with higher order spaces, the following theorem will be useful. Essentially it says that a flock with "too many" planes having a common line must be a linear flock.

Theorem 9.1: If C(V,S) is a wide cone, then any flock of it which contains more than q -[q+2/2] planes that meet in a common line is a linear flock.

We may now show that:

Proposition 9.2: The flocks of all wide cones in PG(3,4) that have flocks are linear. The flocks of any cone with more than two points in its carrier are star flocks.

We define two new types of flocks needed in the next case.

A flock is of Fisher type if q - [q+2/2] of its planes pass through a common line and the remaining planes pass through a common point. Note that there will be two varieties of Fisher type flocks, depending upon whether or not the common point is incident with one of the planes through the common line (The common point can not lie on the common line for wide cones , so these flocks are never star flocks of thick cones.) If the common point is incident with a plane through the common line we will call the flock a Fisher type A flock, otherwise a Fisher type B flock.

A flock is of FTW type if no four of its planes pass through a common point (and hence, no three of its planes pass through a common line). We see that a flock of FTW type can be linear only if q = 2 and it can be a proper star flock only if q = 3 (but, by proposition 9.1 above, the cone must be a flat cone). It is possible that a flock of FTW type can also be of Fisher type, but this can only occur, for thick cones, if q = 5.

In the proof of the last proposition we used the herd space of the flocks in a fairly trivial way, we shall do so again.

Proposition 9.3: The flocks of all wide cones in PG(3,5) that have flocks are linear, or FTW type flocks (which are also Fisher type B flocks). The flocks of any cones having more than 6 points in their carriers are star flocks.

This proposition does not guarantee the existence of a non-linear flock of a wide cone, it just says that if a wide cone has a non-linear flock it must be of FTW type. We shall carry the analysis further to produce an example. So, assume that a wide cone of PG(3,5) has a non-linear flock. Since this must be an FTW type flock, and these are not star flocks since q = 5, the carrier of the cone must have either 5 or 6 points (4 or fewer points would give thin cones). Now, the permutations of GF(5) which fix 0 are the scalar multiples of f1(t) = t, f2(t) = t3, f3(t) = 3t3 + 3t2 + t, f4(t) = 4t3 + 3t2 + 2t, f5(t) = t3 + 3t2 + 3t, and f6(t) = 2t3 + 3t2 + 4t. A little calculation shows that the only linear combinations of two of these that results in another one must start with two functions which are scalar multiples of each other. In terms of the herd space of this flock, this implies that no more than two permutations can be associated to the points on a line unless the flock is a star flock. So, in our situation, no more than two points of the carrier can be collinear, i.e. the points of the carrier form a 5-arc or a 6-arc. A 6-arc is an oval and since the characteristic of the field is odd, it is a conic (Segre's Theorem). Any 5-arc in this plane extends uniquely to a 6-arc, so every 5-arc is obtained by removing a point from a conic. Thus, to obtain an example we should start with a quadratic cone. If we take the cone with equation xy = z2, then the flock with coordinate functions f(t) = t, g(t) = 3t3 and h(t) = 3t2 gives the FTW flock as can be seen by calculating the herd space functions at the points of the carrier and observing that they are all permutations. (Notice that this cone does not contain the base point (0,0,1,0), so h(t) is not a permutation.)

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