Vtyjkyuk

Let $(M,g)$ be a $4$-dimensional smooth Riemannian manifold. I am trying to understand the following exterior algebra computation:

Let $x^1,x^2,x^3,x^4$ be local coordinates on $M$ such that the Riemannian volume form of $g$ is $mathrmdx^1wedgemathrmdx^2wedgemathrmdx^3wedgemathrmdx^4$.

Then there exist a local frame $alpha_1,alpha_2,alpha_3,alpha_4$ of $1$-forms such that the following identities hold:
$$
beginaligned
ast_g(mathrmdx^1wedgemathrmdx^2)&= alpha_3wedgealpha_4\
ast_g(mathrmdx^1wedgemathrmdx^3)&= alpha_4wedgealpha_2\
ast_g(mathrmdx^1wedgemathrmdx^4)&= alpha_2wedgealpha_3\
endalignedqquad
beginaligned
ast_g(mathrmdx^3wedgemathrmdx^4)&= alpha_1wedgealpha_2\
ast_g(mathrmdx^4wedgemathrmdx^2)&= alpha_1wedgealpha_3\
ast_g(mathrmdx^2wedgemathrmdx^3)&= alpha_1wedgealpha_4\
endaligned
tag1
$$
(The pattern is that the indices on both sides of every equation are complementary.)

In fact, the solution of these equations (which is unique up to replacing each $alpha_i$ by $-alpha_i$) is given by
$$
alpha_i = g_ij,mathrmdx^jqquadqquadtextwherequad
g = g_ij,mathrmdx^imathrmdx^j.
$$

Question: Why does the formula $alpha_i = g_ij,mathrmdx^j$ solve the equations?

Here is what I understood:

$mathrmdx^1,mathrmdx^2$ are $g$-orthogonal to $alpha_3,alpha_4$. Since $ast_g(mathrmdx^1wedgemathrmdx^2)$ is decomposable*, we can write
$$
ast_g(mathrmdx^1wedgemathrmdx^2)=beta_1 wedge beta_2$$
for some one-forms $beta_i$, which implies $mathrmdx^1,mathrmdx^2$ are $g$-orthogonal to $beta_1,beta_2$. Thus $textspanbeta_1,beta_2=textspanalpha_3,alpha_4$, so $beta_1 wedge beta_2=falpha_3 wedge alpha_4$ for some function $f$. We now need to prove that $f=1$, which is (up to a sign) equivalent to the statement
$$
| ast_g(mathrmdx^1wedgemathrmdx^2)|=|alpha_3 wedge alpha_4|.
$$
Since the Hodge dual operator is an isometry, this is equivalent to
$$
| mathrmdx^1wedgemathrmdx^2|=|alpha_3 wedge alpha_4|.
$$
This is where the assumption $det(g_ij)=1$ is supposed to enter. I tried to expand both sides in terms of the $g_ij$, but so far I don't see how the result follows.

Perhaps there is another easier way to see this.

For the interested, this computation came up in a question about the existence of "higher-order" harmonic coordinates.

Comment: Of course, everything here is "pointwise", i.e. this is really a result about $4$-dim inner product spaces. I have kept the manifold notation since it might be more familiar.

*The Hodge star operator preserve decomposability of elements.

You have not defined the Hodge dual precisely, so I am gonna assume that you use the "determinant" convention, meaning that $$ (omegawedgeeta)_i_1...i_kj_1...j_l=frac(k+l)!k!l!omega_[i_1...i_keta_j_1...j_l]. $$ In this convention $$ (staromega)_i_k+1...i_n=frac1k!epsilon_i_1...i_nomega^i_1...i_k $$ where indices are raised and lowered with $g^ij$ and $g_ij$ respectively and $$ epsilon_i_1...i_n=sqrtdet gpi_i_1...i_n $$ with $pi$ being the Levi Civita symbol.

If $omegainOmega^k(M)$, the relationship between the tensor and differential form representations is $$ omega=omega_i_1...i_kdx^i_1otimes...otimes dx^i_k=frac1k!omega_i_1...i_kdx^i_1wedge...wedge dx^i_k $$ where the summation is unrestricted in both cases and the array of components is completely antisymmetric.

How does one evaluate $star dx^i_1wedge...wedge dx^i_k$ then? Well, $$ dx^i_1wedge...wedge dx^i_k=delta^i_1_j_1...delta^i_k_j_kdx^j_1wedge...wedge dx^j_k=frack!k!delta^i_1_[j_1...delta^i_k_j_k]dx^j_1wedge...wedge dx^j_k, $$ so the tensor components are $$ k!delta^i_1_[j_1...delta^i_k_j_k]. $$

The Hodge dual is then $$ (star dx^i_1wedge...wedge dx^i_k)_j_k+1...j_n=delta^i_1_[j_1...delta^i_k_j_k]epsilon^j_1...j_n=g^i_1j_1...g^i_k j_kepsilon_j_1...j_n=epsilon^i_1...i_k _j_k+1...j_n, $$ hence $$ star dx^i_1wedge...wedge dx^i_k=frac1k! epsilon^i_1...i_k _j_k+1...j_n dx^j_k+1wedge...wedge dx^j_n=sum_<epsilon^i_1...i_k _j_k+1...j_n dx^j_k+1wedge...wedge dx^j_n,$$ where the sum with $<$ underneath is restricted summation (to lexicographically ordered indices).

Let us apply this in your case:

$$ star dx^iwedge dx^j=sum_k<lepsilon^ij _kldx^kwedge dx^l=sum_k<lepsilon^ijklalpha_kwedgealpha_l $$ where $alpha_k=g_ikdx^i$.

Now if we fix the indices and use $sqrtdet g=1$, we obtain that $$ star dx^1wedge dx^2=sum_k<lpi^12klalpha_kwedgealpha_l=pi^1234alpha_3wedgealpha_4=alpha_3wedgealpha_4, $$ and so on. In fact, one may obtain a bit more general result without having to fix the indices.

Since if any two indices agree, we'll get a vanishing expression, let us assume that $i$ and $j$ are different. We can then assume that $k,l$ are different from $i,j$, thus allowing two different values for $k,l$. If $k,l$ can take on two different values, then for $sum_k<lepsilon^ijklalpha_kwedgealpha_l$ there is only one term, we don't actually sum. If we then further assume that we order indices such that $i,j,k,l$ is an even permutation of $1,2,3,4$, we can also get rid of $pi$, since it will be identially 1. So we obtain $$ star dx^iwedge dx^j=alpha_kwedgealpha_l, $$ with $i,j,k,l$ all different and are positive permutations of $1,2,3,4$.