How can I show $\mathbf{e}_0\mathbf{e}_1\mathbf{e}_2\mathbf{e}_3=\sqrt{|g|}\gamma_0\gamma_1\gamma_2\gamma_3$

Question

Let me try to work it out with $Cl_2(\mathbb{R}$) with basis elements $\hat{x}, \hat{y}$ such that $\hat{x}^2=1, \hat{y}^2=1, \hat{x}\hat{y}+\hat{y}\hat{x}=0$. I define a non-orthonomal basis as follows:

$$ \mathbf{e}_1=a\hat{x}+b\hat{y}\\ \mathbf{e}_2=c\hat{x}+d\hat{y} $$

Then I am trying to show that $\mathbf{e}_1\wedge \mathbf{e}_2=\sqrt{|g|}\hat{x}\wedge\hat{y}$ where $\sqrt{|g|}$ is the square root of the determinant of the metric tensor associated with $\{\mathbf{e}_1,\mathbf{e}_2\}$.

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$$ \begin{align} \mathbf{e}_1 \wedge \mathbf{e}_2=(a\hat{x}+b\hat{y})\wedge(c\hat{x}+d\hat{y}) \end{align} $$

The wedge product is distributive, therefore I can write :

$$ \begin{align} (a\hat{x}+b\hat{y})\wedge(c\hat{x}+d\hat{y})&= ac\hat{x}\wedge\hat{x}+ad\hat{x}\wedge\hat{y}+bc\hat{y}\wedge\hat{x}+bd\hat{y}\wedge\hat{y}\\ &=(ad-bc)\hat{x}\wedge\hat{y} \end{align} $$

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Edit:

For reference, the metric tensor is:

$$ \mathbf{v}=x\mathbf{e}_1+y\mathbf{e}_2\\ \mathbf{v}^2=(x\mathbf{e}_1+y\mathbf{e}_2)(x\mathbf{e}_1+y\mathbf{e}_2)\\ \mathbf{v}^2=x\mathbf{e}_1x\mathbf{e}_1+x\mathbf{e}_1y\mathbf{e}_2+y\mathbf{e}_2x\mathbf{e}_1+y\mathbf{e}_2 y\mathbf{e}_2\\ \mathbf{v}^2=x^2\mathbf{e}_1\mathbf{e}_1+xy(\mathbf{e}_1\mathbf{e}_2+\mathbf{e}_2\mathbf{e}_1)+y^2\mathbf{e}_2\mathbf{e}_2 $$

Then finding the scalar values for $\{ \mathbf{e}_1\mathbf{e}_1,\mathbf{e}_1\mathbf{e}_2+\mathbf{e}_2\mathbf{e}_1, \mathbf{e}_2\mathbf{e}_2 \}$:

1. $\mathbf{e}_1\mathbf{e}_1$:

$$ \mathbf{e}_1\mathbf{e}_1=(a\hat{x}+b\hat{y})(a\hat{x}+b\hat{y})\\ =a^2+b^2 $$

2. $\mathbf{e}_2\mathbf{e}_2$:

$$ \mathbf{e}_2\mathbf{e}_2=(c\hat{x}+d\hat{y})(c\hat{x}+d\hat{y})\\ =c^2+d^2 $$

3. $\mathbf{e}_1\mathbf{e}_2+\mathbf{e}_2\mathbf{e}_1$:

$$ \mathbf{e}_1\mathbf{e}_2+\mathbf{e}_2\mathbf{e}_1 = (a\hat{x}+b\hat{y})(c\hat{x}+d\hat{y})+(c\hat{x}+d\hat{y})(a\hat{x}+b\hat{y})\\ = (a\hat{x}c\hat{x}+a\hat{x}d\hat{y}+b\hat{y}c\hat{x}+b\hat{y}d\hat{y})+(c\hat{x}a\hat{x}+c\hat{x}b\hat{y}+d\hat{y}a\hat{x}+d\hat{y}b\hat{y})\\ =ac+a\hat{x}d\hat{y}+b\hat{y}c\hat{x}+bd+ca+c\hat{x}b\hat{y}+d\hat{y}a\hat{x}+db\\ =2ac+2bd $$

consequently, the metric tensor is :

$$ g=\pmatrix{a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2} $$

Let us now find $\sqrt{|g|}$:

$$ \sqrt{|g|}=\sqrt{\det \pmatrix{a^2+b^2 & ac+bd \\ ac+bd & c^2+d^2}}\\ =\sqrt{(a^2+b^2)(c^2+d^2) - (ac+bd)^2}\\ =\sqrt{a^2c^2+a^2d^2+b^2c^2+b^2d^2-(ac)^2-acbd-bdac-(bd)^2}\\ =\sqrt{(ad)^2+(bc)^2-2acbd}\\ =ad-bc $$

*mind blown.

Answer 1

Assume that $\mathbf{e}_i = \sum_k a_i^k \gamma_k.$ Then, $$\begin{align} \mathbf{e}_0 \wedge \mathbf{e}_1 \wedge \mathbf{e}_2 \wedge \mathbf{e}_3 &= (\sum_i a_0^i\gamma_i) \wedge (\sum_j a_1^j\gamma_j) \wedge (\sum_k a_2^k\gamma_k) \wedge (\sum_l a_3^l\gamma_l) \\ &= \sum_i \sum_j \sum_k \sum_l a_0^i a_1^j a_2^k a_3^l \, \gamma_i \wedge \gamma_j \wedge \gamma_k \wedge \gamma_l \\ &= \sum_i \sum_j \sum_k \sum_l a_0^i a_1^j a_2^k a_3^l \, \epsilon_{ijkl} \, \gamma_0 \wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_3 \\ &= (\sum_i \sum_j \sum_k \sum_l a_0^i a_1^j a_2^k a_3^l \epsilon_{ijkl}) \, \gamma_0 \wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_3 \\ &= \det(a) \, \gamma_0 \wedge \gamma_1 \wedge \gamma_2 \wedge \gamma_3 . \end{align}$$

Now, $g = a^t a,$ so $\det g = \det(a^t a) = \det(a^t) \det(a) = \det(a)^2,$ so $\det(a) = \sqrt{\det(g)}.$