I want to know which of the affine geometries $AG(n,q)$ are self-dual matroids?
I have proved before that uniform matroids $U_{n,m}$ that are self duals are those who satisfies that $m = 2n.$ I am guessing that affine geometries that are uniform matroids are the only ones who will be self dual, but I think no affine geometry is a uniform matroid (I do not know how to prove this though), therefore there are no self - dual affine geometries.
Any educated proof of all these guesses or a correction of the wrong ones will be greatly appreciated.
First, recall that $r(M)+r(M^*) = E(M)$. If $M$ is self-dual, we thus have $2r(M) = E(M)$. In our case, this implies $2r = q^{r-1}$, since $AG(r-1, q)$ has rank $r$ and $q^{r-1}$ elements. We are interested in integer solutions to this equation, so $q=2k$ for some $k\in \mathbb{N}$. This only has solutions $r=2, q=4$ and $r=4, q=2$. Thus, the only potential candidates for self-dual affine geometries are $AG(1, 4)$ and $AG(3, 2)$.
Now, $AG(1,4) \cong U_{2,4}$, which is identically self-dual, meaning that $AG(1,4)$ is equal to its dual, not just isomorphic to it.
As for $AG(3,2)$, it is also identically self-dual, which can be shown by direct computation of the dual (there is likely a more clever way, but I cannot see it as of now). However, it is not isomorphic to a uniform matroid, since $AG(3,2)$ has non-spanning circuits, and all circuits in a uniform matroid are spanning.