I'm currently reading through Hughes and Piper's textbook Design Theory and am stuck on a section on projective geometries (Chapter 1, pages 17-19).
The authors begin with the following definitions and construction.
A few pages on, they present the following exercise:
It is in attempting this final exercise - in particular showing that $\mathscr{P}_{2,2}(K)$ is not a projective plane - I've come to suspect I've not properly understood the above construction. My understanding is that in the construction of $\mathscr{P}_{n,i}(K)$, the points are the 1-dimensional subspaces of $V$ and the lines are the 2-dimensional subspaces of $V$. However, if this is the case, then whether a given $\mathscr{P}_{n,i}(K)$ is a projective plane shouldn't depend on $i$, the $g$-dimension of the blocks of $\mathscr{P}_{n,i}(K)$. But this can't be the case if Exercise 1.28 makes sense, as its truth implies $\mathscr{P}_{2,1}(K)$ is a projective plane while $\mathscr{P}_{2,2}(K)$ is not.
As an alternative, I tried assuming that the blocks of $\mathscr{P}_{n,i}(K)$ are themselves the "lines", but this leads to some of the $\mathscr{P}_{n,i}(K)$'s not even being projective geometries, which appears not to make sense, given that projective geometries are the objects supposedly being constructed.
Any input would be appreciated.
Exercise 1.28 is indeed asking you to consider the blocks of $\mathscr{P}_{n,i}$ to be the "lines" of a geometry, and show that this incidence structure does not make a projective plane.
In the definitions given, a projective geometry only has points and lines; and $\mathscr{P}_{n,i}$ only has points and blocks (with blocks being the $i$-spaces of $\mathscr{P}(n, K)$). So this is the only interpretation that makes sense.