How to define the orientation of a vector space over an arbitrary field?

Question

I know the construction of the Hodge star operator in the context of (pseudo-)euclidean real vector spaces. Apart from the scalar product it involves a orientation of the vector space, which one has to choose (at least if one is not willing to deal with forms of odd parity).

While I was wondering how to extend this definition to vector spaces over arbitrary fields, I realised that I even do not know what an orientation of a vector space $V$ over a general field $k$ is. In case there is such a notion: How could one use the definition in order to define a Hodge star operator?) Or is there a better definition?

A slight extension of this question I am also interested in is the case of a finitely presented flat module over a local ring (which includes the case where the module is the germ of sections of a vector bundle over a locally ringed space at a point).

Answer 1

An orientation of the $n$-dimensional real vector space $V$ is an equivalence class of generators of the $1$-dimensional vector space $det(V)=\Lambda^n(V)$ under the relation $\omega\sim c\omega$, $c>0$.

A basis-free description of the usual Hodge star for a real vector space with positive inner product is exactly as Mariano has described it (up to some sign depending on $k$ and $n$, I believe). This description can be used to define a "Hodge star" for any finite-dimensional $V$ over any field, as soon as an inner product and a volume are given.

But in the usual (real, positive) case we also require the volume to be compatible with the inner product in the sense that $\langle\omega,\omega\rangle=1$, following a standard convention for extending an inner product on $V$ to exterior powers of $V$. A positive definite inner product on $V$ gives a positive definite inner product on $det(V)$, and there are exactly two possible compatible choices of $\omega$, one for each orientation of $V$.

Without this compatibility, the star that you get will not have the usual formal properties. The scalar $\langle\omega,\omega\rangle$ will enter into the formula for star composed with star.

For other fields, or even for indefinite inner products in the real case, there will not always be a compatible $\omega$, but if there is one then there will be exactly two: the question of existence of a compatible volume for a given inner product has to do with $k^{\star}/k^{\star 2}$, while the uniqueness has to do with the kernel of squaring $k^{\star}\to k^{\star}$. (In the real case these groups are easily confused, since they are isomorphic, and in fact isomorphic via the obvious map.)

I suppose that in the real but not definite case one settles for using a volume for which $|\langle\omega,\omega\rangle|=1$; such a volume exists and is unique up to sign.

Answer 2

Let $n=\dim V<\infty$. If you fix a non-degenerate bilinear form $V\times V\to k$, you get an isomorphism $\beta:V\to V^*$. Now fix a non-zero element $\omega\in\bigwedge^nV$. Multiplication in the exterior algebra gives you a non-degenerate pairing $ \bigwedge^kV\times\bigwedge^{n-k}V\to\bigwedge^nV$ which, using $\omega$ as a basis of the codomain, gives you a non-degenerate pairing $\bigwedge^kV\times\bigwedge^{n-k}V\to k$, and consequently an isomorphism $\bigwedge^kV\to(\bigwedge^{n-k}V)^*$. Finally, since $(\bigwedge^{n-k}V)^*$ is canonically the same as $\bigwedge^{n-k}(V^*)$, and $\beta$ induces an isomorphism $\bigwedge^{n-k}(V^*)\to \bigwedge^{n-k}(V)$, we end up with isomorphisms $\bigwedge^kV\to\bigwedge^{n-k}V$. These deserve to be called the Hodge star with respect to the pair $(\beta,\omega)$.

Answer 3

I would say the orientation of $V$ is a $(k^*)^2$-orbit on $\det (V) \setminus \{0\}$. It generalizes to rings. I don't think it gives Hodge-star but I may be wrong.

If you need Hodge-star, you don't monkey around, choose a basis and define Hodge-star using this basis. Now your "orientation" is an equivalence class of bases defining the same Hodge-star.