Is the isomorphism between $\Lambda^2(\mathbb{R}^n)$ and $\mathfrak{so}(n)$ typical? I recently came across an interesting explanation why the curl of a vector field represents infinitesimal rotations.
For simplicity lets work in $\mathbb{R}^n$ as opposed to some other $n$-dimensional manifold. In general the curl operation can be identified with the exterior derivative taking 1-forms to 2-forms. We can of course think of 1-forms as vector fields but 2-forms, i.e. elements of $\Lambda^2(\mathbb{R}^n)$ consists of antisymmetric bilinear forms on $\mathbb{R}^n$ which also comprise the lie algebra $\mathfrak{so}(n)$ of infinitesimal rotations. To make this more concrete we could fix a basis so both are identified with antisymmetric $n \times n$ matrices. This lets us view the curl of a vector field at a point as an infinitesimal rotation at that point.
This is a cute fact but is it any more than that? I know next to nothing about lie algebras so I have to wonder if this is indicative of anything general regarding lie algebras or differential forms? For instance, I never thought of $\mathfrak{so}(n)$ as being comprised of bilinear forms, so I'm wondering if this is a general fact about lie algebras. 
 A: I think that there is a mix of different issues in your question, and most of them refer to linear algebra rather than to differential forms or Lie algebras. First, on an inner product space $(V,\langle\ ,\ \rangle)$, you can identify $L(V,V)$ with the space of bilinear forms on $V$. This associates to $f:V\to V$ the form $b_f(v,w):=\langle f(v),w\rangle$. This allows you to define symmetric and skew-symmetric endomorphisms, and the skew symmetric ones are closed under the commutator, thus forming a Lie subalgebra of $\mathfrak{gl}(V)$. It turns out that being skew symmetric is the infinitesimal version of preserving the inner product, so this Lie subalgebra is $\mathfrak{o}(V)$ (which coincides with $\mathfrak{so}(V)$. This generalizes to Lie subalgebras of $\mathfrak{o}(V)$ (for example $\mathfrak{u}(V)$ or $\mathfrak{su}(V)$ for a complex vector space) which can be identified with spaces of skew-symmetric bilinear forms.
Indeed, you can play the same game with an even dimensional vector space endowed with a non-degenerate, skew symmetric bilinear form $\omega$. The maps which are skew-symmetric with respect to $\omega$ are again closed under the commutator and form the symplectic Lie algebra $\mathfrak{sp}(V)$, but this time they are identified with the space $S^2V^*$ of symmetric bilinear forms on $V$. Again Lie subalgebras of $\mathfrak{sp}(V)$ can be viewed as space of symmetric bilinear forms. 
Finally, when talking about the curl of a vector field, you use the fact that on a vector space of dimension $n$ with a fixed volume form (which in particular is provided by an inner product and by a non-degenerate skew symmetric bilinear form), you can identify $V$ with $\Lambda^{n-1}V^*$ by inserting a vector into the volume form. So in $\mathbb{R^3}$, this identifies $2$-forms with vectors.  
All these linear algebra facts have their counterparts on manifolds by applying them to each tangent space (and observing that the result depends smoothly on the base point): On a Riemannian manifold, you can interpret a two-form as defining a familiy of infinitesimal rotations in the tangent spaces. On a symplectic manifold, infinitesimal symplectic transformations of the tangent spaces are described by symmetric 2-tensor fields. Finally for an $n$-manifold with a distinguished volume form, you can identify vector fields with $(n-1)$-forms
