How to calculate curl curl E using differential forms? We can calculate $\mathbf{curl}\,\mathbf{E}$ by $d(E^1dx_1+E^2dx_2+E^3dx_3)$. But how to calculate $\mathbf{curl}\,\mathbf{curl}\,\mathbf{E}$ using differential forms? I know the first $\mathbf{curl}\,\mathbf{E}$ needs to be contracted with a tensor, but which tensor is it here: the metric?
Moreover, how can we prove $\nabla\times(\mathbf{A}\times\mathbf{B})=\nabla^T(\mathbf{B}\mathbf{A}^T-\mathbf{A}\mathbf{B}^T)$ using differential forms?
 A: Yes, this is a metrical operation.  In geometric calculus, the calculus built using clifford algebra, all you're asking for is
$$\nabla \times (A \times B) = \epsilon^{-1} (\nabla \wedge [A \times B]) = \nabla \cdot [\epsilon^{-1}(A \times B)] = -\nabla \cdot (A \wedge B)$$
The dot here, like the dot in traditional vector algebra, is inherently metrical.
Converting this to differential forms parlance isn't difficult.
$$-\nabla \cdot (A \wedge B) = -\epsilon \epsilon [\nabla \cdot (A \wedge B)] = -\epsilon (\nabla \wedge (\epsilon [A \wedge B]) )= \star d (\star [A \wedge B])$$
Similarly, the curl of the curl of $E$ is
$$\nabla \times (\nabla \times E) = -\epsilon \nabla \wedge (-\epsilon \nabla \wedge E) = -\nabla \cdot (\nabla \wedge E) = \star[ d (\star dE)]$$
Remember, the Hodge star is metrical!

Edit: about that last identity.  It helps to use a common identity from vector algebra called the BAC-CAB rule.  In clifford algebra, it takes this form:
$$A \cdot (B \wedge C) = C(A \cdot B)- B(A \cdot C)$$
You only need this and the product rule to expand the double curl.  It comes out to
$$-\nabla \cdot (A \wedge B) = A (\nabla \cdot B) + (B \cdot \nabla) A - B(\nabla \cdot A) - (A \cdot \nabla) B$$
I'm not familiar with mixing vector calculus with explicitly matrix-like notation the way you have with transposes of vectors and whatnot, but I'm familiar enough with this identity that I suspect the form you posed can be written in this way.  A purely differential forms approach almost certainly would have to prove the BAC-CAB rule and proceed in this manner also.  The BAC-CAB rule proof is rather tedious even in clifford algebra (one of the few proofs that I think is much easier and cleaner in a basis) and isn't really that interesting.
