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For finite-dimensional $\mathbb R$-vector spaces, we define an orientation to be an equivalence class of ordered bases, where $B_1 \sim B_2$ iff the change of basis matrix $A$ taking $B_{1}$ to $B_{2}$ has positive determinant. Then there are two equivalence classes, one which we call, "positive" and the other we call, "negative".

I wanted to know if any work has been done to extend this to finite-dimensional vector spaces over finite fields. My idea was to do everything as above, but replace the condition $\det A>0$ with $\det A$ is a quadratic residue. This should split the bases into two equivalence classes just as above.

I played around with these and noticed some interesting things. For example let $q=p^{f}$ and $\mathbb F_{q}$ be an $\mathbb F_{p}$ vector space. Unlike the case for $\mathbb R$, if $q \not\equiv 3 \pmod 4$ then switching any two vectors in your basis doesn't change the equivalence class, because the determinant of the corresponding change of basis matrix is $-1$, which is a quadratic residue in this case.

Has any work been done on this, or are there any other definitions or related concepts that might be of interest?

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Nice question. Here is a random thought that might inspire someone: The fact that there are two orientations of an $\mathbf{R}$-vector space comes from the fact that $\operatorname{GL}_n\mathbb{R}$ has exactly two connected components. Connected doesn't make sense if your field is finite, but maybe there is something about $\operatorname{GL}_n\mathbb{F}_q$ which might be related to "orientations". –  Daenerys Naharis Mar 8 '13 at 1:55
    
You don't need to write $B_1$~$B_2$. You can write $B_1\sim B_2$, entirely within $\TeX$, so that it follows standard conventions of size and spacing and harmony between fonts, etc. (I changed it above.) –  Michael Hardy Mar 8 '13 at 2:05
    
Might even and odd permutations have something to do with it? –  Michael Hardy Mar 8 '13 at 2:07
    
What happens in vector spaces over $\mathbb C$ instead of $\mathbb R$? –  Michael Hardy Mar 8 '13 at 2:08
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You can replace "positive" with "square." –  Qiaochu Yuan Mar 8 '13 at 2:18

1 Answer 1

up vote 4 down vote accepted

The concept which generalizes to all fields is not an orientation but a "volume form," by which I mean a nonzero element of the top exterior power $\Lambda^n(V)$ of an $n$-dimensional vector space over a field $k$. When $k = \mathbb{R}$, the space of volume forms has two connected components (indeed it can be noncanonically identified with $k^{\times} = \mathbb{R}^{\times}$), and a choice of such a connected component gives an orientation. More generally, if $G$ is any topological group, looking at connected components gives a natural homomorphism $G \to \pi_0(G)$, so one can think about a choice of orientation as a choice of element in the image of the natural map $$\Lambda^n(V)^{\times} \to \pi_0 \left( \Lambda^n(V)^{\times} \right)$$

Over an arbitrary field you can just pick any quotient group of $\Lambda^n(V)^{\times}$ and consider the corresponding choice of "generalized orientation," e.g. above I suggested quotienting by the subgroup of squares. A choice of volume form will then always naturally give rise to a "generalized orientation."

But it's unclear whether this is of any use.

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