# LOG#097. Group theory(XVII).

The case of Poincaré symmetry

There is a important symmetry group in (relativistic, quantum) Physics. This is the Poincaré group! What is the Poincaré group definition? There are some different equivalent definitions:

i) The Poincaré group is the isometry group leaving invariant the Minkovski space-time. It includes Lorentz boosts around the 3 planes (X,T) (Y,T) (Z,T) and the rotations around the 3 planes (X,Y) (Y,Z) and (Z,X), but it also includes traslations along any of the 4 coordinates (X,Y,Z,T). Moreover, the Poincaré group in 4D is a 10 dimensional group. In the case of a ND Poincaré group, it has $N(N-1)/2+N$ parameters/dimensions, i.e., the ND Poincaré group is $N(N+1)/2$ dimensional.

ii) The Poincaré group formed when you add traslations to the full Lorentz group. It is sometimes called the inhomogenous Lorentz group and it can be denoted by ISO(3,1). Generally speaking, we will generally have $ISO(d,1)$, a D-dimensional ($D=d+1$) Poincaré group.

The Poincaré group includes as subgroups, the proper Lorentz transformations such as parity symmetry and some other less common symmtries. Note that the time reversal is NOT a proper Lorentz transformation since the determinant is equal to minus one.

Then, the Poincaré group includes: rotations, traslations in space and time, proper Lorentz transformations (boosts). The combined group of rotations, traslations and proper Lorentz transformations of inertial reference frames (those moving with constant relative velocity) IS the Poincaré group. If you give up the traslations in space and time of this list, you get the (proper) Lorentz group.

The full Poincaré group is a NON-COMPACT Lie group with 10 “dimensions”/parameters in 4D spacetime and $N(N+1)/2$ in the ND case.  Note that the boost parameters are “imaginary angles” so some parameters are complex numbers, though. The traslation subgroup of the Poincaré group is an abelian group forming a normal subgroup of the Poincaré group while the Lorentz grou is only a mere subgroup (it is not a normal subgroup of the Poincaré group). The Poincaré group is said, due to these facts, to be a “semidirect” product of traslations in space and time with the group of Lorentz transformations.

The case of Galilean symmetry

We can go back in time to understand some stuff we have already studied with respect to groups. There is a well known example of group in Classical (non-relativistic) Physics.

The Galilean group is the set or family of non-relativistic continuous space-time (yes, there IS space-time in classical physics!) transformations in 3D with an absolute time. This group has some interesting subgroups: 3D rotations, spatial traslations, temporal traslations and proper Galilean transformations ( transformations leaving invariant inertial frames in 3D space with absolute time). Thereforem the number of parameters of the Galilean group is 3+3+1+3=10 parameters. So the Galileo group is 10 dimensional and every parameter is real (unlike Lorentz transformations where there are 3 imaginary rotation angles).

The general Galilean group can be written as follows:

$G\begin{cases} \mathbf{r}\longrightarrow \mathbf{r}'=R\mathbf{r}+\mathbf{x_0}+\mathbf{V}t\\ t\longrightarrow t'=t+t_0\end{cases}$

Any element of the Galileo group can be written as a family of transformations $G=G(R,\mathbf{x_0},\mathbf{v},t_0)$. The parameters are:

i) $R$, an orthogonal (real) matrix with size $3\times 3$. It satisfies $RR^T=R^TR=I$, a real version of the more general unitary matrix $UU^+=U^+U=I$.

ii) $\mathbf{x_0}$ is a 3 component vector, with real entries. It is a 3D traslation.

iii) $\mathbf{V}$ is a 3 component vector, with real entries. It gives a 3D non-relativistic (or galilean) boost for inertial observers.

iv) $t_0$ is a real constant associated to a traslation in time (temporal traslation).

Therefore, we have 10 continuous parameters in general: 3 angles (rotations) defining the matrix $R$, 3 real numbers (traslations $\mathbf{x_0}$), 3 real numbers (galilean boosts denoted by $\mathbf{V}$) and a real number (traslation in time). You can generalize the Galilean group to ND. You would get  $N(N-1)/2+N+N+1$ parameters, i.e, you would obtain a $N(N+3)/2+1$ dimensional group. Note that the total number of parameters of the Poincaré group and the Galilean group is different in general, the fact that in 3D the dimension of the Galilean group matches the dimension of the 4D Poincaré group is a mere “accident”.

The Galilean group is completely determined by its “composition rule” or “multiplication operation”. Suppose that:

$G_3(R_3,\mathbf{z_0},\mathbf{V}_3,t_z)=G_2\cdot G_1$

with

$G_1(R_1,\mathbf{x_0},\mathbf{V}_1,t_x)$

and

$G_2(R_2,\mathbf{y_0},\mathbf{V}_2,t_y)$

Then, $G_3$ gives the composition of two different Galilean transformations $G_1, G_2$ into a new one. The composition rule is provided by the following equations:

$R_3=R_2R_1$

$\mathbf{z_0}=\mathbf{y_0}+R_2\mathbf{x_0}+\mathbf{V}_2 t_x$

$\mathbf{V}_3=\mathbf{V}_2+R_2\mathbf{V}_1$

$t_z=t_x+t_y$

Why is all this important? According to the Wigner theorem, for every continuous space-time transformation $g\in G$ should exist a unitary operator $U(g)$ acting on the space of states and observables.

We have seen that every element in uniparametric groups can be expressed as the exponential of certain hermitian generator. The Galilean group or the Poincaré group depends on 10 parameters (sometimes called the dimension of the group but you should NOT confuse them with the space-time dimension where they are defined). Remarkably, one can see that the Galilean transformations also act on “spacetime” but where the time is “universal” (the same for every inertial observer). Then, we can define

$iK_\alpha=\dfrac{\partial G}{\partial \alpha}\bigg| _{\alpha=0}$

These generators, for every parameter $\alpha$, will be bound to dynamical observables such as: linear momentum, angular momentum, energy and many others. A general group transformation for a 10-parametric (sometimes said 10 dimensional) group can be written as follows:

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10}=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}}$

We can apply the Baker-Campbell-Hausdorff (BCH) theorem or simply expand every exponential in order to get

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=\prod_{k=1}^{10}e^{iK_{\alpha_k}\alpha_k}=\exp \sum_{k=1}^{10}\omega_k (\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}}$

$\displaystyle{G(\alpha_1,\ldots,\alpha_{10})=I+i\sum_{k=1}^{10}\omega_k(\alpha_1,\ldots,\alpha_{10})K_{\alpha_k}+\ldots}$

The Lie algebra will be given by

$\displaystyle{\left[K_i,K_j\right]=i\sum_{k}c_{ij}^kK_k}$

and where the structure constants will encode the complete group multiplication rules. In the case of the Poincaré group Lie algebra, we can write the commutators as follows:

$\left[X_\mu,X_\nu\right]=\left[P_\mu,P_\nu\right]=0$

$\left[M_{\mu\nu},P_\alpha\right]=\eta_{\mu\alpha}P_\nu-\eta_{\nu\alpha}P_\mu$

$\left[M_{\mu\nu},M_{\alpha\beta}\right]=\eta_{\mu\alpha}M_{\nu\beta}-\eta_{\mu\beta}M_{\nu\alpha}-\eta_{\nu\alpha}M_{\mu\beta}+\eta_{\nu\beta}M_{\mu\alpha}$

Here, we have that:

i) $P$ are the generators of the traslation group in spacetime. Note that as they commute with theirselves, the traslation group is an abelian subgroup of the Lorentz group. The noncommutative geometry (Snyder was a pioneer in that idea) is based on the idea that $P$ and more generally even the coordinates $X$ are promoted to noncommutative operators/variables/numbers, so their own commutator would not vanish like the Poincaré case.

ii) $M$ are the generators of the Lorent group in spacetime.

If we study the Galilean group, there are some interesting commutation relationships fo the corresponding generators (rotations and traslations). There are 6 “interesting” operators:

$K_{i}\equiv \overrightarrow{J}$ if $i=1,2,3$

$K_{i}\equiv \overrightarrow{P}$ if $i=,4,5,6$

These equations provide

$\left[P_\alpha,P_\beta\right]=0$

$\left[J_\alpha,J_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma J_\gamma$

$\left[J_\alpha,P_\beta\right]=i\varepsilon_{\alpha\beta}^\gamma P_\gamma$

$\forall\alpha,\beta=1,2,3$

The case of the traslation group

In Quantum Mechanics, traslations are defined in the space of states in the following sense:

$\vert\vec{r}\rangle\longrightarrow\vert\vec{r}'\rangle =\exp\left(-i\vec{x_0}\cdot \vec{p}\right)\vert \vec{r}\rangle=\vert\vec{r}+\vec{x_0}\rangle$

Let us define two linear operators, $R$ and $R'$ associated, respectively, to initial position and shifted position. Then the transformation defining the traslation over the states are defined by:

$R\longrightarrow R'=\exp\left(-i\vec{x_0}\cdot\vec{p}\right)R\exp \left(i\vec{x_0}\cdot \vec{p}\right)$

where

$R_i\vert\vec{r}'\rangle=\vec{r}_i\vert\vec{r}'\rangle$

Furthermore, we also have

$\left[\vec{x_0}\cdot \vec{p},\vec{y_0}\cdot R\right]=-i\vec{x_0}\cdot\vec{y_0}$

$\left[R_\alpha,p_\beta\right]=i\delta_{\alpha\beta}I$

The case of the rotation group

What about the rotation group? We must remember what a rotation means in the space $\mathbb{R}^n$. A rotation is a transformation group

$\displaystyle{X'=RX\longrightarrow \parallel X'\parallel^2=\parallel X\parallel^2 =\sum_{i=i}^n (x'_i)^2=\sum_{i=1}^n x_i^2}$

The matrix associated with this transformation belongs to the orthogonal group with unit determinant, i.e., it is an element of $SO(N)$. In the case of 3D space, it would be $SO(3)$. Moreover, the ND rotation matrix satisfy:

$\displaystyle{I=X^TX=XX^T\leftrightarrow \sum_{i=1}^N R_{ik}R_{ij}=R_{ik}R_{ij}=\delta_{kj}}$

The rotation matrices in 3D depends on 3 angles, and they are generally called the Euler angles in some texts. $R(\theta_1,\theta_2,\theta_3)=R(\theta)$. Therefore, the associated generators are defined by

$iM_j\equiv\dfrac{\partial R}{\partial \theta_j}\bigg|_{\theta_j=0}$

Any other rotation matric can be decomposed into a producto of 3 uniparametric rotations, rotation along certain 2d planes. Therefore,

$R(\theta_1,\theta_2,\theta_3)=R_1(\theta_1)R_2(\theta_2)R_3(\theta_3)$

where the elementary rotations are defined by

Rotation around the YZ plane: $R_1(\theta_1)=\begin{pmatrix} 1 & 0 & 0\\ 0 & \cos\theta_1 & -\sin\theta_1\\ 0 & \sin\theta_1 & \cos\theta_1\end{pmatrix}$

Rotation around the XZ plane: $R_2(\theta_2)=\begin{pmatrix} \cos\theta_2 & 0 & \sin\theta_2\\ 0 & 1 & 0\\ -\sin\theta_2 & 0 & \cos\theta_2\end{pmatrix}$

Rotation around the XY plane: $R_3(\theta_3)=\begin{pmatrix} \cos\theta_3 & -\sin\theta_3 & 0\\ \sin\theta_3 & \cos\theta_3 & 0\\ 0 & 0 & 1\end{pmatrix}$

Using the above matrices, we can find an explicit representation for every group generator (3D rotation):

$M_1=-i\begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{pmatrix}$

$M_2=-i\begin{pmatrix}0 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 0\end{pmatrix}$

$M_3=-i\begin{pmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{pmatrix}$

and we also have

$\left[M_j,M_k\right]=i\varepsilon^{m}_{jk}M_{m}$

where the $\varepsilon^m_{jk}=\varepsilon_{mjk}$ is the completely antisymmetry Levi-Civita symbol/tensor with 3 indices. There is a “for all practical purposes” formula that represents a rotation with respect to some axis in certain direction $\vec{n}$. We can make an infinitesimal rotation with angle $d\theta$, due to the fact that rotation are continuous transformations, it commutes with itself and it is unitary, so that:

$R(d\theta)\vec{r}=\vec{r}+d\theta(\vec{n}\times\vec{r}+\mathcal{O}(d\theta^2)=\vec{r}-id\theta M_\alpha\vec{r}+\mathcal{O}(d\theta^2)$

In the space of physical states, with $\vec{k}=\theta\vec{n}$ some arbitrary vector

$\vec{r}'=R\vec{r}\longrightarrow\vert\vec{r}'\rangle=\vert R\vec{r}\rangle=U(R)\vert\vec{r}\rangle=e^{-i\vec{k}\cdot\vec{J}}\vert\vec{r}\rangle=e^{-i(k_xJ_x+k_yJ_y+k_zJ_z)}\vert \vec{r}\rangle$

Here, the operators $J=(J_x,J_y,J_z)$ are the infinitesimal generators in the space of physical states. The next goal is to relate these generators with position operators $Q$ through commutation rules. Let us begin with

$Q\longrightarrow Q'=e^{-i\vec{k}\cdot{J}}Qe^{i\vec{k}\cdot\vec{J}}$

$Q'\vert\vec{r}'\rangle =\vec{r}\vert\vec{r}'\rangle$

Using this last result, we can calculate for any 2 vectors $\vec{k},\vec{n}$:

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

or equivalent, in component form,

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

These commutators complement the above commutation rules, and thus, we have in general

$\left[\vec{k}\cdot\vec{J},\vec{n}\cdot\vec{Q}\right]=i(\vec{k}\times\vec{n})\cdot\vec{Q}$

$\left[J_j,J_k\right]=i\varepsilon_{jkm}J_m$

$\left[J_j,Q_k\right]=i\varepsilon_{jkm}Q_m$

In summary: a triplet of rotation operators generates “a vector” somehow.

The case of spinning particles

In fact, these features provide two different cases in the case of a single particle:

i) Particles with no “internal structure” or “scalars”/spinless particles. A good example could it be the Higgs boson.

ii) Particles with “internal” degrees of freedom/structure/particles with spin.

In the case of a particle without spin in 3D we can define the angular momentum operator as we did in classical physics ($L=r\times p$), in such a way that

$J=Q\times P$

Note that the “cross product” or “vector product” in 3D is generally defined if $C=A\times B$ as

$C=A\times B=\begin{vmatrix}i & j & k\\ A_x & A_y & A_z\\ B_x & B_y & B_z\end{vmatrix}$

or by components, using the maginal word XYZZY, we also have

$C_x=A_yB_z-A_zB_y$

$C_y=A_zB_x-A_xB_z$

$C_z=A_xB_y-A_yB_x$

Remember that the usual “dot” or “scalar” product is $A\cdot B=A_xB_x+A_yB_y+A_zB_z$

Therefore, the above operator $J$ defined in terms of the cross product satisfies the Lie algebra of $SO(3)$.

By the other hand, in the case of a spinning particle/particle with spin/internal structure/degrees of freedom, the internal degrees of freedom must be represented by some other operator, independently from $Q,P$. In particular, it must also commute with both operators. Then, by definition, for a particle with spin, the angular momentum will be a sum with two contributions: one contribution due to the “usual” angular momentum (orbital part) and an additional “internal” contribution (spin part). That is, mathematically speaking, we should have a decomposition

$J=Q\times P+S$

with $\left[Q,S\right]=\left[P,S\right]=0$

If $S$, the spin operator, satisfies the above commutation rules (in fact, the same relations than the usual angular momentum), we must impose

$\left[S_j,S_k\right]=i\varepsilon_{jkm}S_m$

The case of Parity P/Spatial inversions

This special transformation naturally arises in some applications. From the pure geometrical viewpoint, this transformation is very simple:

$\vec{r}'=-\vec{r}$

In coordinates and 3D, the spatial inversion or parity is represented by a simple matrix equals to minus the identity matrix

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

This operator correspods, according to the theory we have been studying, to some operator P (please, don’t confuse P with momentum) that satisfies

$PqP^{-1}=-q$

$PpP^{-1}=-p$

and where $q, p$ are the usual position and momentum operators. Then, the operator

$L=q\times p$ is invariant by parity/spatial inversion P, and thus, this feature can be extended to any angular momentum operator like spin S or angular momentum J. That is,

$PJP^{-1}=J$ and $PSP^{-1}=S$

The Wigner’s theorem implies that corresponding to the operator P, a discrete transformation, must exist some unitary or antiunitary operator. In fact, it shows that P is indeed unitary

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}$

If P were antiunitary we should get

$P\left[Q_i,P_j\right]P^{-1}=\left[Q_i,P_j\right]=P(i\hbar\delta_{ij})P^{-1}=-i\hbar\delta_{ij}$

Then, the parity operator P is unitary and $P^{-1}=P$. In fact, this can be easily proved from its own definition.

If we apply two succesive parity transformations we leave the state invariant, so $P^2=I$. We say that the parity operator is idempotent.  The check is quite straightforward

$\vert\Psi\rangle\longrightarrow\vert\Psi'\rangle=PP\vert\Psi\rangle\longrightarrow\vert\Psi\rangle=e^{i\omega}\vert\Psi\rangle$

Therefore, from this viewpoint, there are (in general) only 2 different ways to satisfy this as we have $PP=e^{i\omega}I$:

i) $e^{i\omega}=+1$. The phase is equal to $0$ modulus $2\pi$. We have hermitian operators

$P=P^{-1}=P^+$

Then, the effect on wavefunctions is that $\Psi (P^{-1}(\vec{r}))=\Psi (-\vec{r})$. That is the case of usual particles.

ii) The case $e^{i\omega}=-1$. The phase is equal to $\pi$ modulus $2\pi$. This is the case of an important class of particles. In fact, Steven Weinberg has showed that $P^2=(-1)^F$ where F is the fermion number operator in the SM. The fermionic number operator is defined to be the sum $F=L+B$ where L is now the leptonic number and B is the baryonic number. Moreover, for all particles in the Standard Model and since lepton number and baryon number are charges Q of continuous symmetries $e^{iQ}$  it is possible to redefine the parity operator so that $P^2=I$. However, if there exist Majorana neutrinos, which experimentalists today believe is quite possible or at least it is not forbidden by any experiment, their fermion number would be equal to one because they are neutrinos while their baryon and lepton numbers are zero because they are Majorana fermions, and so $(-1)^F$ would not be embedded in a continuous symmetry group. Thus Majorana neutrinos would have parity equal to $\pm i$. Beautiful and odd, isnt’t it? In fact, if some people are both worried or excited about having Majorana neutrinos is also due to the weird properties a Majorana neutrino would have under parity!

The strange case of time reversal T

In Quantum Mechanics, temporal inversions or more generally the time reversal is defined as the operator that inverts the “flow or direction” of time. We have

$T: t\longrightarrow t'=-t$ $\vec{r}'(-t)=\vec{r}(t)$

And it implies that $\vec{p}(-t)=-\vec{p}(t)$. Therefore, the time reversal operator $T$ satisfies

$TQT^{-1}=Q$

$TPT^{-1}=-P$

In summary: T is by definition the “inversion of time” so it also inverts the linear momentum while it leaves invariant the position operator.

Thus, we also have the following transformation of angular momentum under time reversal:

$TJT^{-1}=-J$

$TST^{-2}=-S$

Time reversal can not be a unitary operator, and it shows that the time reversal T is indeed an antiunitary operator. The check is quite easy:

$T\left[Q,P\right]T^{-1}=\left[TQT^{-1},TPT^{-1}\right]=-\left[Q,P\right]=Ti\hbar T^{-1}$

This equation matches the original definiton if and only if (IFF)

$TiT^{-1}=-i \leftrightarrow TT^{-1}=-1$

Time reversal is as consequence of this fact an antiunitary operator.

# 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#086. Group theory(VI).

We are going to be more explicit and to work out some simple examples/exercises about elementary finite and infinite groups in this post.

Example 1. Let us define the finite group of three elements as $(G,\circ)$ where $G=\left\{ I, M, M^2\right\}$, and such as the element $I_n$ and $M\in \mathcal{M}(n\times n)$ (matrices/arrays with n rows and columns, square matrices) with $M^3=I$ and where the matrix multiplication rule acts as group operation $\circ$. Then, this is an abelian group and its Cayley table can be written as follows:

 $\circ$ $I$ $M$ $M^2$ $I$ $I$ $M$ $M^2$ $M$ $M$ $M^2$ $I$ $M^2$ $M^2$ $I$ $M$

Example 2. Let $G$ be the following set of $2\times 2$ matrices:

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

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

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

$d=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$

If we choose the matrix multiplication as group multiplication, then the group $(G,\circ)$ is a group. The same set, with the operation “addition of matrices” $(G,+)$ is NOT a group. This is left as an exercise for you.

Example 3. Let $G$ denote the set of $2\times 2$ matrices with determinant equal to one (the unit) and complex “entries”. A generic element of this set is:

$X=\begin{pmatrix} a & b\\ c & d\end{pmatrix}$

It satisfies $\det X=+1$, where the determinant is defined to be (of course, a little of linear algebra is assumed to be known) $\det X= ad-bc=1$.

This group, with the ordinary matrix multiplication as group operation, is in fact a well defined group, since:

i) $\det (AB)=\det (A)\det B$

ii) $I$ is the identity matrix $\mathbb{I}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$

iii) $\forall A\in G\exists A^{-1}\in G$ since $\det A=1\Rightarrow \det A^{-1}=1$.

iv) The associative law holds for matrix multiplication and thus $(G,\circ)$ is a group (of matrices).

Example 4. The set $(\mathbb{Z},\circ)$, with $\circ$ defined by $n\circ m=n+m+1$ is a group. It can be easily proved checking that the neutral element is $e=-1$ and the inverse is $n^{-1}=-n-2$.

Example 5. The sets $(\mathbb{Z},+)$, $(\mathbb{Q},+)$, $(\mathbb{R},+)$, $(\mathbb{C},+)$ are abelian groups.

Example 6.  The set of every $2\times 2$ matrix with real or complex numbers with the operation + defined by

$\begin{pmatrix}a & b\\ c & d\end{pmatrix}+\begin{pmatrix}a' & b'\\ c' & d'\end{pmatrix}=\begin{pmatrix}a+a' & b+b'\\ c+c' & d+d'\end{pmatrix}$

is a group.

Example 7.  The “special” rotation group in two dimensions, commonly referred as $SO(2)$, is given by the matrices of the following class:

$R(\theta)=\begin{pmatrix}\cos \theta & \sin \theta\\ -\sin\theta & \cos \theta\end{pmatrix}$

This group is in fact abelian (something that it is NOT generally true with 3D or higher dimensional “rotation groups”). It can be easily proved that the composition of 2 different matrices of the above type is another matrix of the same type, and that the inverse or the unit element exist for every continuous and differentiable function f of two arguments. The neutral element is in fact $R(0)=I$ and $R(\theta)^{-1}=R(-\theta)$.

Example 8.   The 3D  “special” rotation group $SO(3)$ is defined by the set of the following 3 matrices (related to the celebrated Euler angles):

$R(\theta_1,\theta_2,\theta_3)=R(\theta_1)R(\theta_2)R(\theta_3)$

$R(\theta_1,\theta_2,\theta_3)=\begin{pmatrix}1 & 0 & 0\\ 0 & \cos\theta_1 & -\sin\theta_1\\ 0 & \sin\theta_1 & \cos\theta_1\end{pmatrix}\begin{pmatrix}\cos\theta_2 & 0 & -\cos\theta_2\\ 0 & 1 & 0\\ \sin\theta_2 & 0 & \cos\theta_2\end{pmatrix}\begin{pmatrix}\cos\theta_3 & -\sin\theta_3 & 0\\ \sin\theta_3 & \cos\theta_3 & 0 \\ 0 & 0 & 1\end{pmatrix}$

or equivalently R has the form

$\begin{pmatrix}\cos\theta_2\cos\theta_3 & -\cos\theta_2\sin\theta_3 & -\sin\theta_2\\ -\sin\theta_1\sin\theta_2\cos\theta_3+\cos\theta_1\sin\theta_3 & \sin\theta_1\sin\theta_2\sin\theta_3+\cos\theta_1\cos\theta_3 & -\sin\theta_1\cos\theta_2\\ \cos\theta_1\sin\theta_2\cos\theta_3+\sin\theta_1\sin\theta_3 &-\cos\theta_1\sin\theta_2\sin\theta_3+\sin\theta_1\cos\theta_3 & \cos\theta_1\cos\theta_2\end{pmatrix}$

This matrix is also important in neutrino oscillations and quark mixing. However, there the notation is a little bit different (beyond the fact that it has also extra “complex phases”). To simplify the notation and writing above, we can write $\sin\theta_i=s_i$ and $\cos\theta_j=c_j$ whenever $i,j=1,2,3$ (even the notation is useful with “extra dimensions” or higher dimensional rotation groups), so the above 3D matrix is rewritten as follows:

$R(1,2,3)=\begin{pmatrix}c_2c_3 & -c_2s_3 & -s_2\\ -s_1s_2c_3+c_1s_3 & s_1s_2s_3+c_1c_3 & -s_1c_2\\ c_1s_2c_3+s_1s_3 & -c_1s_2s_3+s_1c_3 & c_1c_2\end{pmatrix}$

In this case, the inverse element and the composition function are not “so easiy” as in the 2D case but they can be computed if you are patient and careful enough in a straightforward way. In particular, the composition rule for 3D rotations is NOT symmetrical in its arguments, and it shows that the group is non-abelian (although it has abelian subgroups, of course).

Example 9. The Lorentz group in 1D space and 2D spacetime is defined by the set of matrices ( with matrix multiplication and units $c=1$, i.e., natural):

$L(v)=\left\{\dfrac{1}{\sqrt{1-v^2}}\begin{pmatrix} 1 & -v\\ -v & 1\end{pmatrix}/ v\in (-1,1)\right\}$

This group, as it has OPEN intervals, and they are not closed, is “non-compact”. Non-compactness is an ackward property sometimes in physical/matheamtical applications, but it is important to know that feature. Compact groups are groups with parameters belonging to closed and bound/finite intervals (i.e., the interval limits can not be “infinite”). The Lorentz group in 2D spacetime IS abelian, and $L(0)=e=I$. However, it has a modified “cool” composition rule (the relativistic “addition” of velocities) given by the function:

$v_3=\dfrac{v_1+v_2}{1+v_1v_2}$

Moreover, the elements can be parametrized by a new parameter $\theta$, sometimes called “boost”, and defined as:

$\exp(\theta)=\dfrac{1}{\sqrt{1-v^2}}$

In terms of the boost parameter, Lorentz 2D spacetime transformations can be rewritten as follows:

$L(\theta)=\dfrac{1}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix}=\begin{pmatrix}\cosh\theta & -\sinh\theta\\ -\sinh\theta & \cosh\theta\end{pmatrix}$

The composition rule with this parametrization is in fact very simple and it is related to hyperbolic trigonometry.

Example 10. Groups with less than 6 elements. We can eneumerate the finite groups with the help of the Lagrange’s theorem and/or building explicitly every possible multiplication table for a given number of elements. The results are given by the following list:

i) Using the Lagrange theorem, as 1,2,3 and 5 are prime, we do know that there is only one group of each class, the respective cyclic group with 1, 2, 3 and 5 elements.

ii) There exists at least one group of 4 elements: the cyclic group with 4 elements. In fact, there are two different groups with 4 elements:

ii.1) The so-called Klein’s group.

ii.2) The proper cyclic group with 4 elements.

Remark: We can enumerate easily every element in a cyclic group with n-elements. Using complex analysis we get that the n-elements in a cyclic group can be written with the following formula

$g_k=\exp \left(\dfrac{2\pi i k}{n}\right)$ $\forall k=1,2,\ldots,n$

Example 11. Congruences (I). Given $m\in \mathbb{Z}$ and numbers $x,y$ in that set, we ay that $x$ is congruent with $y$ modulus $m$, and we denote it by $x\equiv y(m)$ or even by $x=y(m)$ (sometimes it is writtem $\mbox{mod}(m)$ as well), if and only if:

$\exists k\in \mathbb{Z}/x-y=km$

This is an equivalence class/relation and we can define classes $\left[x\right]=\left\{y\in \mathbb{Z}/x\equiv y(m)\right\}$ and the set of classes

$\mathbb{Z}_m=\left\{\left[x\right],\left[y\right],\ldots\right\}$

Example 12. Congruences (II). We define an addition and a product acting on the congruence equivalente classes:

$(+):\left[a\right],\left[b\right]\in \mathbb{Z}_m\Rightarrow\left[a\right]+\left[b\right]=\left[a+b\right]$

$(\cdot):\left[a\right],\left[b\right]\in \mathbb{Z}_m\Rightarrow\left[a\right]\cdot \left[b\right]=\left[a \cdot b\right]$

With these operations it is possible to define in some cases groups, in the following way:

i) The group $(\mathbb{Z}_m,+)$. It is an abelian group $\forall m\in \mathbb{Z}$.

ii) The group $(\mathbb{Z}_m,\cdot)$ is a group if and only if (iff) $m$ is a prime number.

Example 13. Congruences (III). Write down the Cayley tables for $\mathbb{Z}_m$ if $m=1,2,3$.

Example 14. Permutation groups, sometimes denoted by $S_n$. This group is the (sub)group of the symmetrical group $S_n$ formed by the permutations of elements of a set of n-elements. Note that the group of all permutations of a set is in fact the definition of  the symmetric group. Then, the term permutation group is usually restricted to mean a subgroup of the symmetric group. The permutation group is the group formed by those elements that are permutations of a given set.

Example 15. The translation group in space. $T_{\vec a}T_{\vec b}=T_{\vec a+\vec b}$

In every vector space, it forms an (abelian) group.

Example 16. Translation group in spacetime. The same as before, but in spacetime. Translations and Lorentz groups in space-time together form the so-called Poincare group.

Example 17. Fractional linear transformations over the real numbers, the complex numbers or some other “beautiful and nice” class of numbers:

$A(x)=\dfrac{Ax+B}{Cx+D}$

where A,B, C and D are generally real, complex numbers or “similar numbers” with $AD-BC\neq 0$.

Example 18. Important matrix groups. Let A be the set of square matrices $n\times n$ over certain “field” $\mathbb{K}=\mathbb{R},\mathbb{C}$  (field said in the mathematical sense, not in the physical sense of “field”, be aware). The following sets with the usual matrix multiplication form a group with a continuous number of parameters (they are continuous infinite groups):

i) $G=GL(n,\mathbb{K})$. The general linear group. It is a continuous group with $n^2$ parameters. Generally, it is understood that the matrix is non-singlular in order to have a well-defined inverse element. Then, $\det M\neq 0\in G$.

ii) $G=SL(n,\mathbb{K})$. The special linear group. It has $n^2-1$ parameters over the field. They are the subgroup of $GL(n,\mathbb{K})$ with determinant equal to one.

iii) $G=U(n,\mathbb{C}$. The unitary group. It is formed by complex matrices that verify the property $MM^+=M^+M=I$, where $+$ denotes “adjoint” and transposition (the so-called hermitian conjugate by physicists). It has $n^2$ complex parameters, or $2n^2$ real parameters if you count in terms of real numbers.

iv) $G=SU(n)$. The subgroup of the unitary group formed with unitary matrices whose value is equal to the unit. $\det M=1$ in this group. Its number of parameters is given by $n^2-1$. The number of real parameters doubles it to be $2(n^2-1)$.

v) $G=O(n)$. The group of orthogonal matrices over the real numbers in euclidean space. It has $n(n-1)/2$ real parameters(generally “angles”). Any orthogonal matrix satisfies the property that $AA^T=A^TA=I$. The $SO(n)$ group is the subgroup of $O(n)$ formed by orthogonal matrices of unit determinant. The special orthogonal group has the same number of real parameters than the orthogonal group.

vi) $G=Sp(2n)$. The symplectic group. Sp(2n, K) is given by the set of 2n×2n matrices A (with entries in K) that satisfy

$\Omega A+A^T\Omega$ with

$\Omega=\begin{pmatrix} 0 & I_n\\ -I_n & 0\end{pmatrix}$

This version of the symplectic group is sometimes non-compact (and it is important in classical mechanics). There is another “symplectic” group, the group $Sp(n)$. It is the subgroup of quaternionic matrices GL(n,H), invertible quaternionic matrices (I am not going to explain the quaternions here this time). It is compact and formed by any matrix nxn over the quaternions leaving the hermitian form $x\cdot y=\overline{X}Y$. Then, the compact symplectic group can be related to the unitary group with some careful analysis on the number of parameters of the hermitian form.

May the group theory be with you!