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I'm wondering if the notion of an orientable/non-orientable manifold has any reasonable extension that allows for a similar classification of finite geometries.

For example, the real projective plane is non-orientable. Does this somehow mean that a finite projective plane is "non-orientable" too?

I would also be interested in a definition that applies over a narrower domain, e.g. a notion of orientability for finite ordered geometries or something like that.

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    $\begingroup$ There is a classical notion of oriented matroid, also known as chirotope, so I guess you could define an "oriented projective plane" as a projective plane structure together with an oriented matroid whose underlying matroid is that of alignment in the projective plane. $\endgroup$
    – Gro-Tsen
    Commented Mar 22, 2017 at 12:39
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    $\begingroup$ I don't understand the second paragraph. In particular, the complex projective plane is orientable. Is there some reason to regard the real field as closer to finite fields than the complex field? $\endgroup$
    – j.c.
    Commented Mar 22, 2017 at 13:26
  • $\begingroup$ Oriented matroids are used in MacPherson's theory of combinatorial differential manifolds, but I don't think that this is the kind of thing GMB is asking for. $\endgroup$
    – Timothy Chow
    Commented Mar 22, 2017 at 15:21

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To do that you would need a notion of a non-orientable and an orientable linear transformation, i.e., essentially a notion of a "positive" and "negative" determinant, where "positive" determinants would form a subgroup not containing the element $-1$. This works for the field $\mathbb{Z}/p\mathbb{Z}$ if and only if the prime $p$ satisfies $p\equiv 3 (mod\; 4)$.

Thus denoting by $G\subset (\mathbb{Z}/p\mathbb{Z})^\ast$ the index-2 subgroup consisting of the quadratic residues modulo $p$, one obtains an "orientable double cover" $M=(F^3\setminus\{0\})/G$, where $F=\mathbb{Z}/p\mathbb{Z}$. Quotienting $M$ by the ("orientation-reversing") antipodal map, you get the projective plane over $F$.

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    $\begingroup$ Why does '$-1$' have to be 'negative' (other than psychology)? Couldn't one view negative as 'not the square class of $1$' (as it seems that you do), and divide up that way? I guess that, as someone who doesn't study orientability, I mean to ask whether the additive, or only the multiplicative, structure of the determinants matters. If the latter, then there's nothing particularly special about $-1$. $\endgroup$
    – LSpice
    Commented Mar 22, 2017 at 16:19

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