$F$ transforms like a tensor $\Rightarrow B$ transforms like a pseudo vector

Question

Notation: In the following $*$ is the hodge operator from $\Lambda^1(\mathbb R^{1\times 3})\cong \mathbb R^{1\times 3}$ to $\Lambda^2(\mathbb R^{1\times 3})\cong A\subset\mathbb R^{3\times 3}$ (or its inverse), where $A$ is the subspace of antisymmetric matrices.

I would like to check whether the assumption that the electromagentic tensor $F$ transforms like a tensor implies that the magnetic field $B$ transforms like a pseudo vector. $\newcommand{\hodge}{{*}}$ The electromagnetic tensor w.r.t. to a chart $\phi$ is given by an antisymmetric matrix \begin{equation} F_\phi=\left[ \begin{array}{c|c} 0 & -E^t \\ \hline E & \mathscr B \end{array} \right]\in\mathbb R^{4\times 4} \end{equation} with $\mathscr B=*(B^t)\in\mathbb R^{3\times 3}$ and we assume that $F_\psi=M^tF_\phi M$ where $M\in\mathbb R^{4\times 4}$ is the matrix identified wtih $D(\phi\circ\psi^{-1})\in L(\mathbb R^4,\mathbb R^4)$. For the case $$M=\left[ \begin{array}{c|c} 1 & 0 \\ \hline 0 & O^t \end{array} \right]$$ with $O\in\mathbb R^{3\times 3}$ orthogonal this reduces to \begin{equation} F'=\left[\begin{array}{c|c} 0 & -E^tO^t \\ \hline OE & O\mathscr BO^t \end{array} \right] \end{equation} We immediately see that $E$ transforms like a vector, but it is not obvious that $B$ transforms like a pseudo vector. Thus, we want to prove the following implication: \begin{equation} \mathscr B'=O\mathscr BO^t\Rightarrow B'=(\det O)OB \end{equation} This implication is an immediate consequence of the following equation: \begin{equation} *(O\mathscr BO^t)=(\det O)B^tO^t \end{equation} Can someone give me a hint regarding how to prove the last equation?

Answer 1

You use rather a lot of complicated notation to express the simple fact that $$ B_i = \frac 12 \epsilon_{ijk}F^{jk}. $$ Now $F$ is a tensor, the Levi-Civita symbol is a pseudo-tensor (changes sign under orthogonal transformations $O$ that have ${\rm det} O=-1$), so their contracted product $B$ inherits the sign-change properties, making it a psudovector.

Answer 2

As @mikestone suggested, we can complete the proof by using the fact that the Levi-Civita-symbol transforms like a pseudo-tensor, i.e. $\epsilon_{klj}\det O=\epsilon_{pqr}O^p{}_kO^q{}_lO^r{}_j$: \begin{align} &\forall i:\det O(B^tO^t)_i=\det OB_jO^i{}_j=\det O\epsilon_{klj}\mathscr{B}_{kl}O^i{}_j&\\ &=\epsilon_{pqr}O^p{}_kO^q{}_lO^r{}_j\mathscr{B}_{kl}O^i{}_j=\underbrace{O^i{}_jO^r{}_j}_{=\delta_{ir}}\epsilon_{pqr}O^p{}_kO^q{}_l\mathscr{B}_{kl}&\\ &=\epsilon_{pqi}O^p{}_k\mathscr{B}_{kl}O^q{}_l=\epsilon_{pqi}(O\mathscr{B}O)^p{}_q=*(O\mathscr{B}O^t)_i& \end{align}