0

Suppose I have the following bi-vector of $Cl_{3,1}(\mathbb{R})$:

$$ \begin{align} \mathbf{b}&=(E_x+IB_x)\gamma_0\gamma_1+(E_y+I B_y)\gamma_0\gamma_2+(E_z+IB_z)\gamma_0\gamma_2 \end{align} $$

where $I=\gamma_0\gamma_1\gamma_2\gamma_3$, and where I define its complex conjugate as:

$$ \begin{align} \mathbf{b}^*&=(E_x-IB_x)\gamma_0\gamma_1+(E_y-I B_y)\gamma_0\gamma_2+(E_z-IB_z)\gamma_0\gamma_2 \end{align} $$

What is the invariance group of $f$ such that $f(\mathbf{b})=f(T\mathbf{b})$, and where $T$ is a linear transformation.

$$ f(\mathbf{b})=\mathbf{b}^*\mathbf{b}\implies f(T\mathbf{b})=(T\mathbf{b})^*(T\mathbf{b}) $$

For reference:

$$ \mathbf{b}^*\mathbf{b}=E_x^2+E_y^2+E_z^2+B_x^2+B_y^2+B_z^2+\det \pmatrix{\gamma_0\gamma_1 & \gamma_0\gamma_2&\gamma_0\gamma_3\\E_x&E_y&E_z\\B_x&B_y&B_z} $$

$T$ must maintain $E_x^2+E_y^2+E_z^2+B_x^2+B_y^2+B_z^2$, as $U(3)$ would, but must also make the cross product between the real part of $\mathbf{b}$ and it's imaginary part also invariant. In this sense, $\mathbf{b}^*\mathbf{b}$ is reminiscent of complex numbers but with the added structure of the determinant.

Anon21
  • 2,699
  • Note that $b^*=-\gamma_0b\gamma_0$ (assuming $\gamma_0!^2=+1$), so maybe the group would have to leave the vector $\gamma_0$ invariant. – mr_e_man Feb 17 '20 at 02:27
  • @mr_e_man Another interesting thing is that the determinant has the form: $\gamma_0\gamma_1(E_y B_z-E_z B_z)+...$. Now, each term $(E_y B_z-E_z B_z)$ is the determinant of a $2 \times 2$ matrix. Since $E_y, B_z,E_z,B_y$ are real, then the determinant produces 3 copies of $SO(2)$. But then these $SO(2)$ are also related orthogonally by the vector elements $\gamma_0\gamma_1,...$, so something like $O(3,SO(2))$ if such a thing is possible. – Anon21 Feb 17 '20 at 17:20
  • @mr_e_man From this question https://math.stackexchange.com/questions/129227/cross-product-in-complex-vector-spaces, it appears that the cross-product is SO(3)... so I have a term that is U(3) and a term that is SO(3). Does that yield SU(3)? – Anon21 Feb 17 '20 at 20:44

0 Answers0