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#088. Group theory(VIII).

Schur’s lemmas are some elementary but very useful results in group theory/representation theory. They can be also used in the theory of Lie algebras so we are going to review these results in this post (for completion).

FIRST SCHUR’S LEMMA. If $D_1$ and $D_2$ are 2 finite-dimensional irreducible representations of G, and if A is certain linear map (generally a matrix) from $D_1$ to $D_2$ such as it commutes with the action of the group (for any element $g\in G$), i.e.,

$AD_1(g)=D_2(g)A$

then at least one of the following consequences holds:

1) A is invertible, and then the representation are necessarily equivalent.

2) A=0.

SECOND SCHUR’S LEMMA. If $A$ is a complex matrix of order $n$ that commutes with every matrix from an irreducible representation $D(G)$

$AD(g)=D(g)A\forall g\in G$

Then, A must be a scalar matrix multiple of the identity matrix, i.e., $A=\lambda I$.

Schur’s lemmas and their corollaries are used to prove the so-called (Schur’s) orthogonality relations and to develop the basics of the representation theory of finite groups. Schur’s lemmas admits some generalizations to Lie groups and some other kind of structures, such as Lie algebras and other sets of operators or matrices. Therefore, it is important to understand Schur’s lemmas!

Consequences of these lemmas:

1) Irreducible representations of an abelian finite group are one dimensional.

2) Orthonormality of group representations. Let $G$ be a finite group, and $D_\mu, D_\nu$ two irreducible representations of G. Then,

$\displaystyle{\dfrac{N_{(\mu)}}{N_G}\sum_{g\in G}\left[D^{-1}_\mu (g')\right]_{ki}\left[D_\nu (g)\right]_{jl}=\delta_{\nu\mu}\delta_{jl}\delta_{kl}}$

3) Completeness of irreducible representations. Let $G$ be a finite group and $(D_\mu)_{\mu\in A}$ the set of every irreducible representation of G. Then,

$\displaystyle{\sum_{\mu \in A}\dfrac{N_{(\mu)}}{N_G}\mbox{Tr}\left[D_\mu (g)D^{-1}_\mu (g')\right]=\delta_{gg'}}$

4) The number of irreducible representations of a finite group is finite and they can be computed with the aid of the following formula:

$\displaystyle{\sum_{\mu\in A}(\mbox{dim}D_\mu)^2=\vert G\vert}$

We can give some elementary examples of irreducible representations of abelian (finite) groups.

1) For $G_2=\left\{e,a\right\}$, then $2=1^2+1^2$ and so there are two and only two irreducible representations, both one dimensional:

$D_1(e)=1$ $D_1(a)=1$

$D_2(e)=1$ $D_2(a)=-1$

2) For $G_3=\left\{e,a,a^2\right\}$, then $3=1^2+1^2+1^2$ and there are 3 irreducible representations:

$D_1(e)=1$ $D_1(a)=1$ $D_1(a^2)=1$

$D_2(e)=1$ $D_2(a)=\exp\left(\dfrac{2\pi i}{3}\right)$ $D_2(a^2)=\exp\left(\dfrac{4\pi i}{3}\right)$

$D_3(e)=1$ $D_3(a)=\exp\left(-\dfrac{2\pi i}{3}\right)$ $D_3(a^2)=\exp \left(-\dfrac{4\pi i}{3}\right)$

3) For $G_4$ we have two cases: $4=1^2+1^2+1^2+1^2$ and $2^2$. And then we have:

3.1) The case of the cyclic group $C_4=\left\{e,a,a^2,a^3\right\}$ with 4 unidimensional representations

$D_1(e)=1$ $D_1(a)=1$ $D_1(a^2)=1$ $D_1(a^3)=1$

$D_2(e)=1$ $D_2(a)=i$ $D_2(a^2)=-1$ $D_2(a^3)=-i$

$D_3(e)=1$ $D_3(a)=-1$ $D_3(a^2)=1$ $D_3(a^3)=-1$

$D_4(e)=1$ $D_4(a)=-i$ $D_4(a^2)=-1$ $D_4(a^3)=i$

3.2) The case of the Klein group $K_4=\left\{e,a,b,ab\right\}$ where $a^{-1}=a$,$b^{-1}=b$ and $(ab)^{-1}=ab$ with $ab=ba$. The representations of this group, in the unidimensional case, are given by:

$D_1(e)=1$ $D_1(a)=1$ $D_1(a^2)=1$ $D_1(a^3)=1$

$D_2(e)=1$ $D_2(a)=-1$ $D_2(a^2)=1$ $D_2(a^3)=-1$

$D_3(e)=1$ $D_3(a)=-1$ $D_3(a^2)=-1$ $D_3(a^3)=1$

$D_4(e)=1$ $D_4(a)=1$ $D_4(a^2)=-1$ $D_4(a^3)=-1$

Of course, you can build the matrix representation of the above group representations as a nice homework. :).

See you in another blog post!