Area form of $M^2 \subseteq \Bbb R^4$: does $(-1)^{i+j-1}\det_{i'j'}(n,\nu) = dx^i \wedge dx^j$?

Question

This is a follow up question from this one. I'm asking separately since one way or another, that one is sort of answered. Now, we know that if $M^{n-1} \subseteq \Bbb R^n$, then the volume element $dM$ satisfies $$(-1)^{i-1}n_i\,dM = dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge \cdots dx^n,$$where $(n_1,\cdots, n_n)$ is normal to $M$. For $M^2 \subseteq \Bbb R^4$ I got the expression $$dM = \sum_{1\leq i < j \leq 4} (-1)^{i+j-1}\begin{vmatrix} n_{i'} & n_{j'} \\ \nu_{i'} & \nu_{j'}\end{vmatrix}\,dx^i\wedge dx^j,$$where $i'<j'$ are the remaining indices (e.g., if $(i,j) = (1,2)$ then $(i',j') = (2,4)$, etc). I wanted a relation like the one I put above for hypersurfaces and I'm itching to say $$(-1)^{i+j-1}\begin{vmatrix} n_{i'} & n_{j'} \\ \nu_{i'} & \nu_{j'}\end{vmatrix}\,dM = dx^i\wedge dx^j, \quad \forall\,1\leq i < j \leq 4,$$where $n$ and $\nu$ are normal to $M$, following notation from my previous question. But I'm not sure that this is true, and I'm having trouble attacking this. I tried to mimic Spivak's proof of theorem $5.6$ in page $128$ of his Calculus on Manifolds, but I know it's hopeless to try and define a cross product for only two vectors in $\Bbb R^4$. Help?

Answer 1

HINT: Generalized Laplace expansion by cofactors. The way I like to think of it is this. Reduce to the case of an oriented orthonormal basis $v_1,\dots,v_n$ for $\Bbb R^n$ (in your case, some will be tangent to $M$ and the rest normal to $M$). Consider the Hodge star operator on both forms and multivectors. Then it boils down to $$(dx^i\wedge dx^j)(v_1\wedge v_2) = \big({\star}(dx^i\wedge dx^j)\big)\big({\star}(v_1\wedge v_2)\big) = \big({\star}(dx^i\wedge dx^j)\big)\big(v_3\wedge\dots\wedge v_n\big).$$