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?

  • $\begingroup$ You've got $dy \wedge dz$ in the basis twice. $\endgroup$ – joriki Aug 14 '11 at 22:16
  • 2
    $\begingroup$ @3Sphere: Look up "inner product" and "Hodge star" on Wikipedia. $\endgroup$ – Ryan Budney Aug 14 '11 at 22:19
  • 2
    $\begingroup$ 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. $\endgroup$ – t.b. Aug 14 '11 at 22:20
  • 2
    $\begingroup$ @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. $\endgroup$ – Soarer Aug 15 '11 at 0:12
  • 2
    $\begingroup$ @3Sphere: that is not what anyone I know means by $\mathbb{R}^3$. $\endgroup$ – 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 $\text{Aut}(V) \cong \text{SO}(3)$.)


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.