Vector surface integral - pseudovector? I'm teaching myself differential forms in relation to vector calculus. I understand that for a vector field $$\mathbf{v}\left(x,y,z\right)=f_{1}\hat{\mathbf{e}}_{x}+f_{2}\hat{\mathbf{e}}_{y}+f_{3}\hat{\mathbf{e}}_{z}$$there is a corresponding 1-form$$\omega_{1}=f_{1}\,dx+f_{2}\,dy+f_{3}\,dz$$and 2-form$$\omega_{2}=f_{1}\,dy\wedge dz+f_{2}\,dz\wedge dx+f_{3}dx\wedge dy.$$ My question was prompted by reading Frankel, _The Geometry of Physics_ , where (p99) he gives the surface integral of a 2-form $\beta^{2}$, ![enter image description here]( and then goes on to say: “Suppose that one insists on writing this in terms of the vector, or rather the pseudovector $\mathbf{B}$, associated to $\beta^{2}$.” He seems to be saying that in the context of finding the surface integral of a 2-form, the corresponding vector is always a pseudovector. Why is that? And, in general, does the 2-form $\omega_{2}$ always correspond to a pseudovector?

The definition of a pseudovector is that it changes sign when you change the orientation of the ambient space. You start with the vector field $\mathbf v$ and it corresponds — given the duality coming from the metric — to the $1$-form $\omega_1$. Where does $\omega_2$ come from? Well, $\omega_2 = \star\omega_1$, and the Hodge star operator changes sign here when you change the orientation on $\Bbb R^3$. Thus, the recipe for $\mathbf B$ depends on the orientation choice.