# Is electrodynamics associated with $O(3)$?

Let $$\mathbf{q}$$ be a complex vector of three elements defined as:

$$\mathbf{q}:=\pmatrix{ E_x + iB_x\\ E_y + i B_y\\ E_z +i B_z }$$

I define the function $$f(\mathbf{q})$$:

\begin{align} f(\mathbf{q})&=\mathbf{q}^T\mathbf{q}=\pmatrix{ E_x + iB_x& E_y + i B_y& E_z +i B_z }\pmatrix{ E_x + iB_x\\ E_y + i B_y\\ E_z +i B_z }\\ &=E_x^2+E_y^2+E_z^2-B_x^2-B_y^2-B_z^2+2i(E_xB_x+E_yB_y+E_zB_z)\\ &=||\mathbf{E}||^2-||\mathbf{B}||^2 +2i\mathbf{E}\cdot\mathbf{B} \end{align}

where $$\mathbf{E}:=(E_x, E_y,E_z)$$ and $$\mathbf{B}:=(B_x,B_y,B_z)$$.

The equation produces the Lorentz invariant of electromagnetism.

What is the invariance group of $$f(\mathbf{q})\to f(O\mathbf{q})$$ under a linear transformation $$O$$?

\begin{align} f(O\mathbf{q})&=(O\mathbf{q})^T(O\mathbf{q})\\ &=\mathbf{q}^TO^TO\mathbf{q}\\ &\implies O^TO=I \end{align}

Consequently, since $$Dim (\mathbf{q})$$ is 3, we have $$O(3)$$.

I am a bit baffled as to why am I getting $$O(3)$$ here? I was expecting anything else; for instance $$SO(3,1)$$ or even $$U(1)$$, as the usual group associated with electromagnetism in the literature. Why are the Lorentz invariants of electromagnetism not Lorentz invariant but $$O(3)$$ invariant - where is the mistake?

• Fundamentally it's because $\mathfrak{so}(3, 1)$ (when complexified, since you're using complex variables) is just two copies of $\mathfrak{su}(2)$, which is the Lie algebra of $SO(3)$. This trick only works in $3+1$ dimensions. – knzhou Feb 8 at 0:25
• @knzhou For $\mathfrak{so}(3,1)$, I need my vector $\mathbf{q}$ to have 4 components. Right now I have three: $E_x+iB_x,E_y+iB_y, E_z+iB_z$. What would the 4th component be? – Alexandre H. Tremblay Feb 8 at 0:34
• This is the Riemann-Silberstein vector from 1907. – G. Smith Feb 8 at 0:36
• @AlexandreH.Tremblay The point is that the EM field strength is a 6-dimensional representation of $\mathfrak{so}(3, 1)$ (not every representation of the Lorentz group has to be 4-dimensional), but this is in turn a combination of two 3-dimensional representations of $\mathfrak{su}(2)$. This works only because of the connection between $\mathfrak{so}(d-1, 1)$ and $\mathfrak{su}(2)$ when $d = 4$. – knzhou Feb 8 at 0:41
• OP note that your group is not $O(3,R)$ but $O(3,C)$ this matters. @knzhou if you would write up an answer explaining the subtleties regarding complexification and algebra/group distinctions I think that might clarify this? – jacob1729 Feb 8 at 0:57