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?

  • 4
    $\begingroup$ 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". $\endgroup$ – Chris Brooks Mar 8 '13 at 1:55
  • $\begingroup$ 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.) $\endgroup$ – Michael Hardy Mar 8 '13 at 2:05
  • $\begingroup$ Might even and odd permutations have something to do with it? $\endgroup$ – Michael Hardy Mar 8 '13 at 2:07
  • $\begingroup$ What happens in vector spaces over $\mathbb C$ instead of $\mathbb R$? $\endgroup$ – Michael Hardy Mar 8 '13 at 2:08
  • 1
    $\begingroup$ You can replace "positive" with "square." $\endgroup$ – Qiaochu Yuan Mar 8 '13 at 2:18

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.