# Canonical Isomorphism Between $\Omega^2(\mathbb{R^3})$ and $\mathbb{R^3}$?

Let $\Omega^2(\mathbb{R}^3)$ represent the collection of differential 2-forms on $\mathbb{R}^3$. For this space we take as an (ordered) basis $\{dx \wedge dy, dx \wedge dz, dy \wedge dz\}$.

First question: Is there a principle, other than convention, that this is the "canonical" or "usual" basis of $\Omega^2(\mathbb{R}^3)$ in terms of the way it is ordered?

Next, let $\{e_1, e_2, e_3 \}$ denote the usual ordered Euclidean basis for $\mathbb{R}^3$. Since the dimension of $\Omega^2(\mathbb{R}^3)$ is $3$, it is isomorphic to $\mathbb{R}^3$ and an isomorphism is determined by assigning $dx$ to $e_1$, $dy$ to $e_2$ and $dz$ to $e_3$ and extending by linearity.

So, here's my primary question: It seems to me that some of these choices are really arbitrary. For instance, we could also construct an isomorphism by assigning $dx$ to $e_2$, $dy$ to $e_1$ and $dz$ to $e_3$ instead. Is there an overriding principle that can be invoked to construct a canonical isomorphism between $\Omega^2(\mathbb{R}^3)$ and $\mathbb{R}^3$? Note also that a similar question applies to an isomorphism between $\Omega^1(\mathbb{R}^3)$ and $\mathbb{R}^3$ since these spaces are also isomorphic.

Update: Based on feedback in the comments, I will try to clarify my question. I am considering differential forms and the alternating spaces in question are over the ring of smooth, real-valued functions. Similarly, by the space $\mathbb{R}^3$ I mean the set of all linear combinations of $fe_1 + ge_2 + he_3$ where $f, g, h$ are smooth real-valued functions. I think though that if we were just considering the algebra of alternating forms, essentially the same questions would apply, yes?

• You've got $dy \wedge dz$ in the basis twice. – joriki Aug 14 '11 at 22:16
• @3Sphere: Look up "inner product" and "Hodge star" on Wikipedia. – Ryan Budney Aug 14 '11 at 22:19
• You have a problem with dimensions here. Yes, $\Omega^2(\mathbb{R}^3)$ has (sort of) dimension $3$ over $C^{\infty}(\mathbb{R}^3)$ but over $\mathbb{R}$ this space is evidently infinite dimensional. – t.b. Aug 14 '11 at 22:20
• @3Sphere, after your edit, the problem is that your definition of $\mathbb{R}^3$ is nonstandard. In any case, follow Ryan Budney's advice to look up inner product and Hodge star would give you the answer you are looking for. – Soarer Aug 15 '11 at 0:12
• @3Sphere: that is not what anyone I know means by $\mathbb{R}^3$. – Qiaochu Yuan Aug 15 '11 at 0:16

Let me answer some questions you could be asking, which are more basic than the questions you're actually asking. Let $V$ be a $3$-dimensional real vector space. The exterior product gives a canonical pairing
$$V \times \Lambda^2(V) \to \Lambda^3(V).$$
Now, $\Lambda^3(V)$ is $1$-dimensional, but it doesn't come with a canonical isomorphism to $\mathbb{R}$. Such an isomorphism is equivalent to the data of a nonzero vector $v \in \Lambda^3(V)$, which defines a volume form on $V$. Given a volume form, $V$ is canonically dual to $\Lambda^2(V)$.
One way to get a volume form is to equip $V$ with an inner product. Then the wedge product of three unit vectors in $V$ gives one of two possible elements of $\Lambda^3(V)$, which gives two possible volume forms on $V$. Fixing one of these volume forms is equivalent to fixing an orientation on $V$ (a choice of which ordered orthonormal bases of $V$ are "right-handed").
Hence if $V$ is a $3$-dimensional real oriented inner product space, $V$ is canonically dual to $\Lambda^2(V)$; said another way, $V^{\ast}$ is canonically isomorphic to $\Lambda^2(V)$. But since $V$ is an inner product space, $V^{\ast}$ is canonically isomorphic to $V$. Hence $V$ is canonically isomorphic to $\Lambda^2(V)$.
The isomorphism can be written down explicitly as follows: if $e_1, e_2, e_3$ is an oriented orthonormal basis of $V$, then the corresponding basis of $\Lambda^2(V)$ is $e_2 \wedge e_3, e_3 \wedge e_1, e_1 \wedge e_2$. (Concretely, "canonical" here means that the above identification is covariant under the action of $\text{Aut}(V) \cong \text{SO}(3)$.)