If $e_1e_2 e_3=\sqrt{|\det g|}\hat{x}_1 \hat{x}_2 \hat{x}_3$ then what is $e_1e_2+e_1e_3$?

Question

Let

$$ \hat{\mathbf{x}}_1^2=1\\ \hat{\mathbf{x}}_2^2=1\\ \hat{\mathbf{x}}_3^2=1\\ \hat{\mathbf{x}}_1\hat{\mathbf{x}}_2+\hat{\mathbf{x}}_2\hat{\mathbf{x}}_1=0\\ \hat{\mathbf{x}}_1\hat{\mathbf{x}}_3+\hat{\mathbf{x}}_3\hat{\mathbf{x}}_1=0\\ \hat{\mathbf{x}}_2\hat{\mathbf{x}}_3+\hat{\mathbf{x}}_3\hat{\mathbf{x}}_2=0 $$

Then $$ \mathbf{e}_1\wedge \mathbf{e}_2 \wedge \mathbf{e}_3=\sqrt{|\det g|}\hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_2\wedge \hat{\mathbf{x}}_3 $$

for reference see my previous question (How can I show $\mathbf{e}_0\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3=\sqrt{|g|}\gamma_0\gamma_1\gamma_2\gamma_3$)

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My question is what is the following expression in terms of $g$?

$$ (\mathbf{e}_1\wedge \mathbf{e}_2+\mathbf{e}_1\wedge \mathbf{e}_3)^2 $$

Let

$$ g=\pmatrix{ g_{11}&g_{12}&g_{13}\\ g_{21}&g_{22}&g_{23}\\ g_{31}&g_{32}&g_{33} } $$

This is what I got so far:

$$ \begin{align} (\mathbf{e}_1\wedge \mathbf{e}_2+\mathbf{e}_1\wedge \mathbf{e}_3)^2&=(\mathbf{e}_1\wedge \mathbf{e}_2+\mathbf{e}_1\wedge \mathbf{e}_3)(\mathbf{e}_1\wedge \mathbf{e}_2+\mathbf{e}_1\wedge \mathbf{e}_3)\\ &=(\mathbf{e}_1\wedge \mathbf{e}_2)^2+(\mathbf{e}_1\wedge \mathbf{e}_2)(\mathbf{e}_1\wedge \mathbf{e}_3)+(\mathbf{e}_1\wedge \mathbf{e}_3)(\mathbf{e}_1\wedge \mathbf{e}_2)+(\mathbf{e}_1\wedge \mathbf{e}_3)^2 \end{align} $$

Let us treat each term individually:

1. The part $(\mathbf{e}_1\wedge \mathbf{e}_2)^2$ is:

$$ (\mathbf{e}_1\wedge \mathbf{e}_2)^2=((g_{11}g_{22}-g_{12}g_{21})\hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_2)^2=-(g_{11}g_{22}-g_{12}g_{21})^2 $$

2. The part $(\mathbf{e}_1\wedge \mathbf{e}_3)^2$ is:

$$ (\mathbf{e}_1\wedge \mathbf{e}_3)^2=((g_{11}g_{33}-g_{13}g_{31})\hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_3)^2=-(g_{11}g_{33}-g_{13}g_{31})^2 $$

3. The part $(\mathbf{e}_1\wedge \mathbf{e}_2)(\mathbf{e}_1\wedge \mathbf{e}_3)$ is:

$$ \begin{align} (\mathbf{e}_1\wedge \mathbf{e}_2)(\mathbf{e}_1\wedge \mathbf{e}_3)&=((g_{11}g_{22}-g_{12}g_{21}) \hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_2)((g_{11}g_{33}-g_{13}g_{31}) \hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_3)\\ &=-(g_{11}g_{22}-g_{12}g_{21})(g_{11}g_{33}-g_{13}g_{31}) (\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_1)( \hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_3)\\ &=-(g_{11}g_{22}-g_{12}g_{21})(g_{11}g_{33}-g_{13}g_{31}) ( \hat{\mathbf{x}}_2\wedge \hat{\mathbf{x}}_3) \end{align} $$

4. The part $(\mathbf{e}_1\wedge \mathbf{e}_3)(\mathbf{e}_1\wedge \mathbf{e}_2)$ is:

$$ \begin{align} (\mathbf{e}_1\wedge \mathbf{e}_3)(\mathbf{e}_1\wedge \mathbf{e}_2)&=((g_{11}g_{33}-g_{13}g_{31}) \hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_3)((g_{11}g_{22}-g_{12}g_{21}) \hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_2)\\ &=-(g_{11}g_{33}-g_{13}g_{31})(g_{11}g_{22}-g_{12}g_{21}) (\hat{\mathbf{x}}_3 \wedge \hat{\mathbf{x}}_1)( \hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_2)\\ &=-(g_{11}g_{33}-g_{13}g_{31})(g_{11}g_{22}-g_{12}g_{21}) ( \hat{\mathbf{x}}_3\wedge \hat{\mathbf{x}}_2)\\ &=(g_{11}g_{33}-g_{13}g_{31})(g_{11}g_{22}-g_{12}g_{21}) ( \hat{\mathbf{x}}_2\wedge \hat{\mathbf{x}}_3) \end{align} $$

So the cross-terms cancel, and we get a sum of areas:

$$ (\mathbf{e}_1\wedge \mathbf{e}_2+\mathbf{e}_1\wedge \mathbf{e}_3)^2=-(g_{11}g_{22}-g_{12}g_{21})^2-(g_{11}g_{33}-g_{13}g_{31})^2 $$

What is the geometric interpretation of what I have done? Do the cross-terms really cancel for areas even in the midst of arbitrary curved space -- this result so so surprising to me.

Answer 1

I will post a tentative answer, rather than an edit because I think I may have missed the true generalization. The key to proofs of this type is to start with:

$$ \mathbf{e}_i=\sum_{k=1}^n a_i^k \hat{\mathbf{x}}_k $$

There are a few interesting cases:

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Suppose $\mathbf{u}=\mathbf{e}_1$ and $n=1$, then

$$ \mathbf{e}_1=a_1^1\hat{\mathbf{x}}_1\implies (\mathbf{e}_1)^2= (a_1^1\hat{\mathbf{x}}_1)^2=(a_1^1)^2 $$

Suppose $\mathbf{u}=\mathbf{e}_1$ and $n=2$, then

$$ \mathbf{e}_1=a_1^1\hat{\mathbf{x}}_1+a_1^2\hat{\mathbf{x}}_2\implies (\mathbf{e}_1)^2=(a_1^1\hat{\mathbf{x}}_1+a_1^2\hat{\mathbf{x}}_2)^2=(a_1^1)^2+(a_1^2)^2 $$

Suppose $\mathbf{u}=\mathbf{e}_1\wedge \mathbf{e}_2$ and $n=1$, then

$$ \mathbf{e}_1\wedge \mathbf{e}_2= (a_1^1 \hat{\mathbf{x}}_1) \wedge (a_2^1 \hat{\mathbf{x}}_1)=a_1^1 a_2^1 (\hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_1)=0 $$

Suppose $\mathbf{u}=\mathbf{e}_1\wedge \mathbf{e}_2 \wedge \dots \wedge \mathbf{e}_m$ and $n=m$, then

$$ \mathbf{e}_1\wedge \mathbf{e}_2 \wedge \dots \wedge \mathbf{e}_m = (\det a) \hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_2 \wedge \dots \wedge \hat{\mathbf{x}}_m $$

Now the case closer to our interest:

Suppose $\mathbf{e}_1\wedge \mathbf{e}_2$ and $n=3$

$$ \begin{align} \mathbf{e}_1\wedge \mathbf{e}_2 &= (a_1^1\hat{\mathbf{x}}_1+a_1^2\hat{\mathbf{x}}_2+a_1^3\hat{\mathbf{x}}_3) \wedge (a_2^1\hat{\mathbf{x}}_1+a_2^2\hat{\mathbf{x}}_2+a_2^3\hat{\mathbf{x}}_3) \\ &= a_1^1\hat{\mathbf{x}}_1\wedge a_2^1\hat{\mathbf{x}}_1+a_1^1\hat{\mathbf{x}}_1\wedge a_2^2\hat{\mathbf{x}}_2 + a_1^1\hat{\mathbf{x}}_1\wedge a_2^3\hat{\mathbf{x}}_3\\ &\quad+ a_1^2\hat{\mathbf{x}}_2 \wedge a_2^1\hat{\mathbf{x}}_1 + a_1^2\hat{\mathbf{x}}_2 \wedge a_2^2\hat{\mathbf{x}}_2 + a_1^2\hat{\mathbf{x}}_2 \wedge a_2^3\hat{\mathbf{x}}_3\\ &\quad+a_1^3\hat{\mathbf{x}}_3 \wedge a_2^1\hat{\mathbf{x}}_1 + a_1^3\hat{\mathbf{x}}_3 \wedge a_2^2\hat{\mathbf{x}}_2+ a_1^3\hat{\mathbf{x}}_3 \wedge a_2^3\hat{\mathbf{x}}_3\\ &= a_1^1\hat{\mathbf{x}}_1\wedge a_2^2\hat{\mathbf{x}}_2 + a_1^2\hat{\mathbf{x}}_2 \wedge a_2^1\hat{\mathbf{x}}_1 \\ &\quad +a_1^3\hat{\mathbf{x}}_3 \wedge a_2^1\hat{\mathbf{x}}_1 + a_1^1\hat{\mathbf{x}}_1\wedge a_2^3\hat{\mathbf{x}}_3\\ &\quad + a_1^2\hat{\mathbf{x}}_2 \wedge a_2^3\hat{\mathbf{x}}_3+ a_1^3\hat{\mathbf{x}}_3 \wedge a_2^2\hat{\mathbf{x}}_2\\ &= a_1^1 a_2^2 (\hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_2) + a_1^2a_2^1 (\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_1) \\ & \quad +a_1^3a_2^1 (\hat{\mathbf{x}}_3 \wedge \hat{\mathbf{x}}_1) + a_1^1 a_2^3(\hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_3)\\ & \quad + a_1^2a_2^3 (\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_3)+ a_1^3 a_2^2 (\hat{\mathbf{x}}_3 \wedge \hat{\mathbf{x}}_2)\\ &= a_1^1 a_2^2 (\hat{\mathbf{x}}_1\wedge \hat{\mathbf{x}}_2) - a_1^2a_2^1 (\hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_2) +a_1^3a_2^1 (\hat{\mathbf{x}}_3 \wedge \hat{\mathbf{x}}_1) - a_1^1 a_2^3(\hat{\mathbf{x}}_3\wedge \hat{\mathbf{x}}_1) + a_1^2a_2^3 (\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_3)- a_1^3 a_2^2 (\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_3)\\ &= (a_1^1 a_2^2 - a_1^2a_2^1 )(\hat{\mathbf{x}}_1 \wedge \hat{\mathbf{x}}_2) +(a_1^3a_2^1 - a_1^1 a_2^3)(\hat{\mathbf{x}}_3\wedge \hat{\mathbf{x}}_1) + (a_1^2a_2^3 - a_1^3 a_2^2 )(\hat{\mathbf{x}}_2 \wedge \hat{\mathbf{x}}_3) \end{align} $$