# The derivation of the irreducible representations of the Lorentz group

I took the way of classification of Lorentz group representations from Sexl, Urbantke, Relativity, groups and particles (Germ. ed. 1975). But I don't understand it as I outline in the following:

In the real Lie-algebra of the Lorentz group the following the commutation relations are valid: $[ M_\mu, M_\nu] = \epsilon_{\mu\nu\lambda} M_\lambda$, $[ N_\mu, N_\nu] =-\epsilon_{\mu\nu\lambda} M_\lambda$ and $[ N_\mu, N_\nu] = \epsilon_{\mu\nu\lambda} N_\lambda$.

An infinitesimal Lorentz group element is given by $Id +\vec{\alpha}\bf{M}+\vec{v}\bf{N}$, $\vec{\alpha}$ and $\vec{v}$ (rotation vector and velocity vector, velocity of light c=1) are real parameters. If a different basis is chosen $M^{\pm}_\mu = \frac{1}{2}(M_\mu\pm iN_\mu)$ the commutation relations go over in the following relations:

$[ M^\pm_\mu, M^\pm_\nu] = \epsilon_{\mu\nu\lambda} M^\pm_\lambda$, and $[M^+_\mu, M^-_\nu]=0$.

This is the Lie-algebra of the direct sum $so(3)\oplus so(3)$ (or $su(2) \oplus su(2)$ if you like) or and therefore all irreducible representations of the Lorentz group group are given by the irreducible representations of $SO(3) \times SO(3)$. The problem I have is that in order to make this argument it is necessary to pass through complex representations of the Lorentz group since the decomposition of the Lie-algebra of the Lorentz group only works with complex numbers. An infinitesimal Lorentz group element is now given by $Id + (\vec{\alpha} - i\vec{v})\bf{M}^+ (\vec{\alpha} +i\vec{v})\bf{M}^-$ but now the parameters are complex. Doing it this way it would not change the reducibility of representations according to Sexl, Urbantke.

Actually I have a counter example: Look at the following Lorentz group example. $$\left(\begin{array}{c} E'_1 \\E'_2 \\E'_3 \\B'_1 \\B'_2 \\B'_3 \end{array}\right)=\left(\begin{array}{cccccc} 1 & 0 & 0& 0 & 0&0\\ 0 & \gamma & 0 & 0 & 0& -\gamma v\\ 0 & 0& \gamma & 0 &\gamma v & 0\\ \gamma & 0 & 0& 1& 0 & 0\\0 & 0& \gamma v& 0 & \gamma & 0 \\ 0 & -\gamma v &0 & 0& 0& \gamma \end{array}\right)=\left(\begin{array}{c} E_1 \\E_2 \\E_3 \\B_1 \\B_2 \\B_3 \end{array}\right)$$

This is a real irreducible representation of the Lorentz group. However, if now complex representations are considered, the basis can be changed to a complex basis and within this basis the representation is reducible and decomposes into 2 irreducible complex representations:

$$\left(\begin{array}{c} E'_1 + iB'_1\\E'_2+iB'_2 \\E'_3+iB'_3 \\E'_1-iB'_1 \\E'_2 -iB'_2\\E'_3 -iB'_3\end{array}\right)=\left(\begin{array}{cccccc} 1 & 0 & 0 & 0&0 & 0\\ 0 &\gamma & i\gamma v& 0 & 0& 0\\ 0 &-i\gamma v & \gamma & 0 &0& 0\\ 0& 0& 0& 1 & 0 & 0\\ 0 & 0& 0& 0& \gamma & -i\gamma v \\ 0 &0& 0& 0 & i\gamma v& \gamma \end{array}\right)\left(\begin{array}{c} E_1 + iB_1\\E_2+iB_2 \\E_3+iB_3 \\E_1-iB_1 \\E_2 -iB_2\\E_3 -iB_3\end{array}\right)$$

With real parameters the representation is irreducible whereas using complex numbers it is reducible. That means, using real or complex numbers makes a difference in representation theory.

Therefore I cannot understand why Lorentz group representations can be (so easily) classified according to the given argument.

I learnt in the meantime using complex numbers in Lie group theory is rather convenient, but in physics almost all Lie groups are $\bf{real}$ groups and the $\bf{real}$ representations have to be classified and understood. I hope somebody here has a deeper understanding than I have and can explain it to me.

OK, you seem to have stumbled on the Weyl unitarian trick which validates complexification maneuvers so I won't dwell on it here. The non-unitary reducible 6dim rep you started with is this group's adjoint. But then, complexifying, you end up with the 3dim self-dual forms, together adding up to the equivalent curvature form rep.

Suppose you complex-conjugate your 2nd expression. That interchanges the 1,2,3 block with the 4,5,6 block, in both vectors and matrices. The reduced 6x6 matrix can be transformed to the equivalent complex conjugated one by the similarity symmetric matrix that has 3d identity matrices in the 2-off diagonal 3x3 sub-blocks and 0s in the 2 3x3 diagonal blocks. (it clearly squares to the identity).

Comparing with the WP article, you see that you are combining two 3dim irreps to a 6dim one...

Edit to comport with @ACuriousMind's valid point below. Indeed. You may see that complexification hardly matters if you consider the complex similarity transformation converting your 1st expression to the 2nd one, vectors and matrices, $$S=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1\!\!1 & i 1\!\!1 \\ 1\!\!1 & -i1\!\!1 \end{array}\right)= (S^T) ^{-1},$$ where $1\!\!1$ is the 3x3 identity matrix.

• The Lorentz group is $\mathrm{SO}(1,3)$, not $\mathrm{SU}(2)\oplus\mathrm{SU}(2)$. The two are different, because the former is non-compact, while the latter is compact. The correct statement is that the complexification of $\mathfrak{o}(1,3)$ is $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$, cf. this answer by Qmechanic. Since representations of $\mathfrak{sl}(2,\mathbb{C})$ are equivalent to representations of $\mathfrak{su}(2)$, this error has no consequence for the representation theory and is thus often overlooked. – ACuriousMind Feb 23 '16 at 0:00
• Thank you that finally somebody dedicated some effort on an answer. First of all, what is "WP article"? Second, how can a representation and its c.conjugate be related by a similarity transformation? For me both are different as well as a 2-Weylspinor $\xi_A$ transforms differently from a dotted 2-Weylspinor $\psi_\dot{A}$. – Frederic Thomas Feb 24 '16 at 17:25
• I apologize I did not specify what appeared obvious: The WikiPedia article is the third link, detailing all of these representations. We are not talking about Weyl spinors (doublets), but instead, about vectors, so then the (1,0) and (0,1) reps. These two are linked precisely by the similarity tfmation which interchanges the upper 3 with the lower three components of the 6-vector. – Cosmas Zachos Feb 24 '16 at 17:43
• To wit, $\left(\begin{array}{cc} 0 & 1\!\!1 \\ 1\!\!1 & 0 \end{array}\right)$, where $1\!\!1$ is the 3x3 identity matrix. – Cosmas Zachos Feb 24 '16 at 17:48
• Sorry, but your matrix only reorders the components of the vector. I don't think that a representation like (0, 1/2) or (0, 1) or similar is equivalent its c.conjugate. The latter would need if $E'+iB'= A(E+iB)$ and $E'-iB'=A^{\ast}(E-iB)$, then the similarity transformation fulfills $A^{\ast}=SAS^{-1}$. At the moment I don't see that this S exists. – Frederic Thomas Feb 24 '16 at 18:30