# LOG#095. Group theory(XV).

The topic today in this group theory thread is “sixtors and representations of the Lorentz group”.

Consider the group of proper orthochronous Lorentz transformations $\mathcal{L}^\uparrow_{+}$ and the transformation law of the electromagnetic tensor $F_{\mu\nu}c^{\mu\nu}$. The components of this antisymmetric tensor can be transformed into a sixtor $F=E+iB$ or $F=(E,B)$ and we can easily write how the Lorentz group acts on this 6D vector ignoring the spacetime dependence of the field.

Under spatial rotations, $E,B$ transform separately in a well-known way giving you a reducible representation of the rotation subgroup in the Lorent orthochronous group. Remember that rotations are a subgroup of the Lorentz group, and it contains Lorentz boosts in additionto those rotations. In fact, $L_R=L_E\oplus L_B$ in the space of sixtors and they are thus a reducible representation, a direct sum group representation. That is, rotations leave invariant subspaces formed by $(E,0)$ and $(0,B)$ invariant. However, these two subspaces mix up under Lorentz boosts! We have written before how $E,B$ transform under general boosts but we can simplify it without loss of generality $E'=Q(E,B)$ and $B'=P(B,E)$ for some matrices $Q,P$. So it really seems that the representation is “irreducible” under the whole group. But it is NOT true! Irreducibility does not hold if we ALLOW for COMPLEX numbers as coefficients for the sixtors/bivectors (so, it is “tricky” and incredible but true: you change the numbers and the reducibility or irreducibility character does change. That is a beautiful connection betweeen number theory and geometry/group theory). It is easy to observe that  using the Riemann-Silberstein vector

$F_\pm=E+iB$

and allowing complex coefficients under Lorent transformations, such that

$\overline{F}_\pm =\gamma F_\pm -\dfrac{\gamma-1}{v^2}(F_\pm v)v\mp i\gamma v\times F_\pm$

i.e., it transforms totally SEPARATELY from each other ($F_\pm$) under rotations and the restricted Lorentz group. However, what we do have is that using complex coefficients (complexification) in the representation space, the sixtor decomposes into 2 complex conjugate 3 dimensional representaions. These are irreducible already, so for rotations alone $\overline{F}_\pm$ transformations are complex orthogonal since if you write

$\dfrac{\mathbf{v}}{\parallel \mathbf{v}\parallel}=\mathbf{n}$

with $\gamma =\cos\alpha$ and $i\gamma v=\sin\alpha$. Be aware: here $\alpha$ is an imaginary angle. Moreover, $\overline{F}_\pm$ transforms as follows from the following equation:

$\overline{x}=\dfrac{\alpha\cdot x}{\alpha^2}\alpha +\left( x-\dfrac{\alpha\cdot x}{\alpha^2}\alpha\right)\cos\alpha -\dfrac{\alpha}{\vert \alpha\vert}\times x\sin\alpha$

Remark: Rotations in 4D are given by a unitary 4-vector $\alpha$ such as $\vert \alpha\vert\leq \pi$ and the rotation matrix is given by the general formula

$\boxed{R^\mu_{\;\;\; \nu}=\dfrac{\alpha^\mu\alpha_\nu}{\alpha^2}+\left(\delta^\mu_{\;\;\; \nu}-\dfrac{\alpha^\mu\alpha_\nu}{\alpha^2}\right)\cos\alpha+\dfrac{\sin\alpha}{\alpha}\varepsilon^{\mu}_{\;\;\; \nu\lambda}\alpha^\lambda}$

or

$\boxed{R^\mu_{\;\;\; \nu}=\cos\alpha\delta^\mu_{\;\;\; \nu}+(1-\cos\alpha)\dfrac{\alpha^\mu\alpha_\nu}{\alpha^2}+\dfrac{\sin\alpha}{\alpha}\varepsilon^{\mu}_{\;\;\; \nu\lambda}\alpha^\lambda}$

If you look at this rotation matrix, and you assign $F_\pm\longrightarrow x$ with $n\longrightarrow \alpha/\vert\alpha\vert$, the above rotations are in fact the same transformations of the electric and magnetic parts of the sixtor! Thus the representation of the general orthochronous Lorentz group is secretly complex-orthogonal for electromagnetic fields (with complex coefficients)! We do know already that

$F_\pm^2=(E+iB)^2=(E^2- B^2)\pm 2E\cdot B$

are the electromagnetic main invariants. So, complex geometry is a powerful tool too in group theory! :). The real and the imaginary part of this invariant are also invariant. The matrices of 2 subrespresentations formed here belong to the complex orthogonal group $SO(3,\mathbb{C})$. This group is a 3 dimensional from the complex viewpoint but it is 6 dimensional from the real viewpoint. The orthochronous Lorentz group is mapped homomorphically to this group, and since this map has to be real and analytic over the group $SO(3,\mathbb{C})$ such that, as Lie groups, $\mathcal{L}^\uparrow_+\cong SO(3,\mathbb{C})$. We can also use the complex rotation group in 3D to see that the 2 subrepresentations must be inequivalent. Namely, pick one of them as the definition of the group representation. Then, it is complex analytic and its complex parameter provide any equivalent representation. Moreover, any other subrepresentation is complex conjugated and thus antiholomorphic (in the complex sense) in the complex parameters.

Generally, having a complex representation, i.e., a representation in a COMPLEX space or representation given by complex valued matrices, implies that we get a complex conjugated reprentation which can be equivalent to the original one OR NOT. BUT, if share with original representation the property of being reducible, irreducible or decomposable. Abstract linear algebra says that to any representation in complex vector spaces $V$ there is always a complex conjugate representation in the complex conjugate vector space $V^*$. Mathematically, one ca consider representations in vector spaces over various NUMBER FIELDS. When the number field is extended or changed, irreducibility MAY change into recubibility and vice versa. We have seen that the real sixtor representation of the restricted Lorentz group is irreducible BUT it becomes reducible IF it is complexified! However, its defining representation by real 4-vectors remains irreducible under complexification. In Physics, reducibility is usually referred to the field of complex numbers $\mathbb{C}$, since it is generally more beautiful (it is algebraically closed for instance) and complex numbers ARE the ground field of representation spaces. Why is this so? There are two main reasons:

1st. Mathematical simplicity. $\mathbb{C}$ is an algebraically closed filed and its representation theory is simpler than the one over the real numbers. Real representations are obtained by going backwards and “inverting” the complexification procedure. This process is sometimes called “getting the real forms” of the group from the complex representations.

2nd. Quantum Mechanics seems to prefer complex numbers (and Hilbert spaces) over real numbers or any other number field.

The importance of $F_\pm=E\pm iB$ is understood from the Maxwell equations as well. In vacuum, without sources or charges, the full Maxwell equations read

$\nabla\cdot F_+=0$ $i\partial_t F_+=\nabla\times F_+$

$\nabla\cdot F_-=0$ $-i\partial_t F_-=\nabla\times F_-$

These equations are Lorentz covariant and reducibility is essential there. It is important to note that

$F_+=E+iB$ $F_-=E-iB$

implies that we can choose ONLY one of the components of the sixtor, $F_+$ or $F_-$, or one single component of the sixtor is all that we need. If in the induction law there were a plus sign instead of a minus sign, then both representations could be used simultaneously! Furthermore, Lorentz covariance would be lost! Then, the Maxwell equations in vacuum should satisfy a Schrödinger like equation due to complex linear superposition principle. That is, if $F_+$ and $F'_+$ are solutions then a complex solution $f=c_+F_++c'_+F'_+$ with complex coefficients should also be a solution. This fact would imply invariance under the so-called duality transformation

$F_+\longrightarrow F_+e^{i\theta}$ $\theta \in \mathbb{R}$

However, it is not true due to the Nature of Maxwell equations and the (apparent) absence of isolated magnetic  charges and currents!

# LOG#093. Group theory(XIII).

The sixtor or 6D Riemann-Silberstein vector is a complex-valued quantity $\Psi=E+iB$ up to one multiplicative constant and it can be understood as a bivector field in Clifford algebras/geometric calculus/geometric algebra. But we are not going to go so far in this post. We remark only that a bivector field is something different to a normal vector or even, as we saw in the previous post, a bivector field can not be the same thing that a spinor field with $1/2$ spin. Moreover, the electric and magnetic parts of the sixtor transform as vectors under spatial rotations

$(x^0,\mathbf{x})\longrightarrow (x'^0,\mathbf{x'})=(x^0,R\mathbf{x})$

where $C'(x')=RC(x)$ and R being an orientation preserving rotation matrix in the spcial orthogonal group $SO(3,\mathbb{R})$. Remember that

$SO(3)=G=\left\{R\in M_{3\times 3}(\mathbb{R})/RR^T=R^TR=I_3,\det R=+1\right\}$

The group $SO(3,\mathbb{R})$ is bases on the preservation of the scalar (inner) product defined by certain quadratic form acting on 3D vectors:

$(x,y)=x\cdot y=X^TY=(RX,RY)=X^TR^TRY$

and $R^TR=RR^T=I_3$ so $\det R=+1$ for proper rotations. This excludes spatial reflections or parity transformations P, that in fact are important too. Parity transformations act differently to electric and magnetic fields and they have $\det P=-1$. Parity transformations belong to the group of “improper” rotations in 3D space.

However, the electromagnetic field components are NOT related to the spatial components of a 4D vector. That is not true. With respect to the proper Lorentz group:

$SO^+(1,3)=\left\{\Lambda \in M_{4\times 4}(\mathbb{R}/\Lambda^T\eta\Lambda=\eta,\Lambda^0_{\;\;\;0}\geq 1,\det \Lambda=+1\right\}$

and where the metric $\eta=\mbox{diag}(-1,1,1,1)$ is the Minkovski metric. In fact, the explicit representation of

$\Lambda\in SO(1,3)^+$ by its matrix elements

$\Lambda =\Lambda^\mu_{\;\;\; \nu}=\begin{pmatrix}\Lambda^{0}_{\;\;\; 0} & \Lambda^{0}_{\;\;\; 1} & \Lambda^{0}_{\;\;\; 2} & \Lambda^{0}_{\;\;\; 3}\\ \Lambda^{1}_{\;\;\; 0} & \Lambda^{1}_{\;\;\; 1} & \Lambda^{1}_{\;\;\; 2} & \Lambda^{1}_{\;\;\; 3}\\ \Lambda^{2}_{\;\;\; 0} & \Lambda^{2}_{\;\;\; 1} & \Lambda^{2}_{\;\;\; 2} & \Lambda^{2}_{\;\;\; 3}\\ \Lambda^{3}_{\;\;\; 0} & \Lambda^{3}_{\;\;\; 1} & \Lambda^3_{\;\;\; 2} & \Lambda^{3}_{\;\;\; 3}\end{pmatrix}$

Despite this fact, that the electromagnetic field sixtor $\Psi$ transforms as a vector under the COMPLEX special orthogonal group $SO(3,\mathbb{C})$ in 3D space

$SO(3,\mathbb{C})=\left\{Q\in M_{3\times 3}(\mathbb{C})/Q^TQ=QQ^T=I_3,\det Q=1\right\}$

This observation is related to the fact that the proper Lorentz group and the complex rotation group are isomorphic to each other as Lie groups, i.e. $SO(3,\mathbb{C})\cong SO(3,1;\mathbb{R})^+$. This analogy and mathematical result has some deeper consequences in the theory of the so-called Dirac, Weyl and Majorana spinors (quantum fields describing fermions with differnt number of independent “components”) in the massive case.

The puzzle, now,  is to understand why the mass term is forbidden in

$i\Gamma^\mu\partial_\mu \Psi=0$

as it is the case of the electromagnetic (classical) field. Moreover, in this problem, we will see that there is a relation between symmetries and operators of the Lie groups $SO(1,3;\mathbb{R}), SO(3,\mathbb{C}),SO(3,\mathbb{R})$ and the corresponding generators of their respective Lie algebras. Let’s  begin with pure boosts in some space-time plane:

$\Lambda =\begin{pmatrix}\gamma & -\gamma\beta & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$

and where we defined, as usual,

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$ and $\beta c=v$, with $\gamma^2-\gamma^2\beta^2=1$.

Then,

$\Lambda=I_4+\beta\begin{pmatrix}0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}=I_4+\beta L_1$

and where we have defined the boost generator as $L_1$ in the plane $(t,x^1)$, and the boost parameter will be given by the number

$\cosh (\xi_1)=\gamma_1$

Remark: Lorentz transformations/boosts corresponds to rotations with an “imaginary angle”.

Moreover, we also get

$e^{\xi_1L_1}=\begin{pmatrix}\cosh\xi_1 & -\sinh\xi_1 & 0 & 0\\ -\sinh\xi_1 & \cosh\xi_1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 &0 & 1\end{pmatrix}$

Equivalently, by “empathic mimicry” we can introduce the boost generators in the remaining 3 planes $L_2, L_3$ as follows:

$L_2=\begin{pmatrix}0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

$L_3=\begin{pmatrix}0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 0 & 0\end{pmatrix}$

In addition to these 3 Lorentz boosts, we can introduce and define another 3 generators related to the classicla rotation in 3D space. Their generators would be given by:

$S_1=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\end{pmatrix}$

$S_2=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\end{pmatrix}$

$S_3=\begin{pmatrix}0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$

Therefore, $S=(S_1,S_2,S_3)$ and $L=(L_1,L_2,L_3)$ span the proper Lorent Lie algebra $so(1,3)^+$ with generators $L_i, S_j$. These generators satisfy the commutators:

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

$\left[L_a,L_b\right]=+\varepsilon_{abc}S_c$

$\left[L_a,S_B\right]=-\varepsilon_{abc}L_c$

with $a,b,c=1,2,3$ and $\varepsilon_{abc}$ the totally antisymmetric tensor. This Levi-Civita symbol is also basic in the $SO(3)$ structure constants. Generally speaking, in the Physics realm, the generators are usually chosen to be hermitian and an additional imaginar $i$ factor should be included in the above calculations to get hermitian generators. If we focus on the group $SO(3)$ over the real numbers, i.e., the usual rotation group, the Lie algebra basis is given by $(S_a)_{bc}=\varepsilon_{abc}$, or equivalently by the matrices

$S_1=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}$ $S_2=\begin{pmatrix} 0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}$ $S_3=\begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$

and the commutation rules are

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

If the rotation matrix is approximated to:

$R=I_3+\delta R\leftrightarrow R^TR=RR^T=(I_3+\delta R)(I_3+\delta R)^T=I_3+\delta R+\delta R^T+\mathcal{O}(\delta^2)$

then we have that the rotation matrix is antisymmetric, since we should have

$\delta R+\delta R^T=0$ or $\delta R=-\delta R^T$

so the matrix generators are antisymmetric matrices in the $so(3)$ Lie algebra. That is, the generators are real and antisymmetric. The S-matrices span the full $so(3)$ real Lie algebra. In the same way, we could do the same for the complex group $SO(3,\mathbb{C})$ and we could obtain the Lie algebra $so(3)$ over the complex numbers with $\tilde{S_k}=iS_k$ added to the real $S_k$ of the $SO(3)$ real Lie group. This defines a “complexification” of the Lie group and it implies that:

$\left[S_a,S_b\right]=-\varepsilon_{abc}S_c$

$\left[\tilde{S}_a,\tilde{S}_b\right]=+\varepsilon_{abc}S_c$

$\left[\tilde{S}_a,S_b\right]=-\varepsilon_{abc}\tilde{S}_c$

Do you remember this algebra but with another different notation? Yes! This is the same Lie algebra we obtained in the case of the Lorentz group. Therefore, the Lie algebras of $SO(1,3)^+$ and $SO(3,\mathbb{C})$ are isomorphic if we make the identifications:

$S_a\leftrightarrow S_i$ The rotation part of $SO(1,3)^+$ is the rotation part of $SO(3,\mathbb{C})$

$L_a\leftrightarrow \tilde{S}_a$ The boost part of $SO(1,3)^+$ is the complex conjugated (rotation-like) part of $SO(3,\mathbb{C})$

Then for every matrix $\Lambda\in SO(1,3)^+$ we have

$\Lambda=e^{\xi_1L_1+\xi_2L_2+\xi_3L_3+\alpha_1S_1+\alpha_2S_2+\alpha_3S_3}$

and for every matrix $Q\in SO(3,\mathbb{C})$ we have

$Q=e^{\xi_1\tilde{S}_1+\xi_2\tilde{S}_2+\xi_3\tilde{S}_3+\alpha_1S_1+\alpha_2S_2+\alpha_3S_3}$

For instance, a bivector boost in some axis provides:

$e^{\xi_1\tilde{S}_1}=\begin{pmatrix}1 & 0 & 0\\ 0 & \cosh\xi_1 & i\cosh\xi_1\\ 0 & -i\sinh\xi_1 & \cosh\xi_1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0\\ 0 & \gamma_1 & i\gamma_1\beta_1\\ 0 & -i\gamma_1\beta_1 & \gamma_1\end{pmatrix}$

and where in the first matrix, it acts on $\Psi=\dfrac{1}{\sqrt{2}}(E+iB)=\overline{\Psi}_a$, the second matrix (as the first one) acts also on a complex sixtor, and where the rotation around the axis perpendicular to the rotation plane is defined by the matrix operator:

$e^{\alpha_1S_1}=\begin{pmatrix}1 & 0 & 0\\ 0 & \cos\alpha_1 & \sin\alpha_1\\ 0 & -\sin\alpha_1 & \cos\alpha_1\end{pmatrix}$

and this matrix would belong to the real orthogonal group $SO(3,\mathbb{R})$.

Note: Acting $\exp (\xi\tilde{S}_1)$ onto the sixtor $\Psi$ as a bivector field shows that it generates the correct Lorentz transformation of the full electromagnetic field! The check is quite straightforward, since

$\begin{pmatrix}E'_1+iB'_1\\ E'_2+iB'_2\\ E'_3+iB'_3\end{pmatrix}=\begin{pmatrix}1 & 0 & 0\\ 0 & \gamma_1 & i\gamma_1\beta_1\\ 0 & -i\gamma_1\beta_1 & \gamma_1\end{pmatrix}\begin{pmatrix}E_1+iB_1\\ E_2+iB_2\\ E_3+iB_3\end{pmatrix}$

From this complex matrix we easily read off the transformation of electric and magnetic fields:

$E'_1=E_1$

$E'_2=\gamma_1(E_2-\beta_1B_3)$

$E'_3=\gamma_1(E_3+\beta_1B_2)$

$B'_1=B_1$

$B'_2=\gamma_1(B_2+\beta_1E_3)$

$B'_3=\gamma_1(B_3+\beta_1E_2)$

Note the symmetry between electric and magnetic fields hidden in the sixtor/bivector approach!

For the general Lorentz transformation $x'^\mu=\Lambda^\mu_{\;\;\; \nu}x^\nu$ we have the invariants

$\Psi^T\Psi=\Psi^TQ^+Q\Psi=\dfrac{1}{2}(E^2-B^2)+iE\cdot B$