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I'm currently trying to read up on the Lorentz-group and it's representations. I've found a couple of posts here on stack-exchange that I find helpful and confusing at the same time, so I would be glad if someone could help me connect the dots. I will be referencing the following two posts: [1] and[2]. From here on I will call $SO(1,3; \mathbb{R})$ the (proper) Lorentz-group (if the distinction between proper and not-proper Lorentz group should be relevant for the representation-theory, please mention it!).


By studying [2] and [1], sooner or later we make the switch from $SO(1,3;\mathbb{R})$ to $SO(1,3;\mathbb{C})$. I'm having a hard time seeing why... During a class my professor wrote down the following: $$\mathfrak{so}(1,3;{\mathbb{C}})\cong \mathfrak{su}(2)_{\mathbb{C}} \oplus \mathfrak{su}(2)_{\mathbb{C}}\cong \mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C}),$$ which agrees with what is written in [3] as far as I understand. But agian, the question why we consider the complexification of the Lorentz group comes up. In [1] I only see the statement that $SO^+(1,3;\mathbb{R})$ is a subgroup of $SO(1,3;\mathbb{C})$, but why is this relevant for the representation theory of $SO(1,3;\mathbb{R})$ if $SO(1,3;\mathbb{C})$ is not simply connected and the universal cover of $SO(1,3;\mathbb{R})$?


I'm sorry if the question is badly structured or confusing, but I just can't put a better structured question together.

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  1. The passage from real to complex is harmless, as the complex linear representations of the complexified Lie algebra are in 1 to 1 correspondence with the real linear representations of the real form. (Given, that we are working with representations on a $\mathbb{C}$ vector space.)
  2. As you wrote, the complexified Lorentz algebra can be written as a sum of $\mathfrak{sl}_2(\mathbb{C})$, which has the significant advantage, that the representation theory of $\mathfrak{sl}_2(\mathbb{C})$ is completly understood. See, for instance, this article.
  3. Using the representations of these copies of $\mathfrak{sl}_2(\mathbb{C})$, we can label the representations of the complexified Lorentz algebra, and thus those of the Lorentz algebra (see 1.) by pairs $(i,j) \in \mathbb{N}/2 \times \mathbb{N}/2$, which helps when talking about particles 'living in certain representations'. More on these can be found in Knapp - Representation Theory of Semisimple Groups_an overview based on examples.
  4. Having classified all the representations of the Lie algebra, you get those of the Lie group, but have to take care of so called 'projective representations'. An overview can be found in this article. Ignoring technical difficulties, it doesn't matter as you would want to allow projective representations when talking about quantum physics anyways. An then, classifying all representations of the Lorentz algebra gives rise to a classification of all possible spin configurations (assuming that spin is described by the Lorentz group). A source for further reading might be D.J. Simms - Lie groups and quantum mechanics, though you should be familiar with basic representation theory if you want to give it a try (as a first read for representation theory, I would recommend B. Hall - Lie Groups, Lie Algebras, and Representations: An Elementary Introduction).
  5. The universal cover of the Lorentz group is $SL_2(\mathbb{C})$, viewed as a real Lie group (it is sometimes written as 'Spin$(1,3)$'). The classification of its Lie algebra does indeed classify its representations as it's a simply connected Lie group.
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  • $\begingroup$ Thank you for your answer! Can you suggest some further reading for point 1? I‘m having a bit trouble understanding why this should be true... $\endgroup$ – Sito Feb 18 at 8:02
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    $\begingroup$ Sure, this can be found in Hall's book, mentioned in point 4. Actually, I can sketch the proof: Given a real representation $\pi: \mathfrak{g} \to \mathfrak{gl}(V)$ on a complex vector space $V$, we can define $\pi_{\mathbb{C}}: \mathfrak{g}_{\mathbb{C}} \to \mathfrak{gl}(V)$ by setting $\pi_{\mathbb{C}}(X+iY) := \pi(X) + i \pi(Y)$ (note that this is possible as $V$ is a $\mathbb{C}$ vector space. Now, given a $\mathbb{C}$ linear representation of $\mathfrak{g}_{\mathbb{C}}$, it mus also be of the form (continued below...): $\endgroup$ – Creo Feb 18 at 12:36
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    $\begingroup$ (...continued from above) $\pi_{\mathbb{C}}(X+iY) = \pi_{\mathbb{C}}(X) +i \pi_{\mathbb{C}}(Y)$ (by $\mathbb{C}$ linearity). So by setting $\pi: \mathfrak{g} \to \mathfrak{gl}(V)$ to be $\pi(X) := \pi_{\mathbb{C}}(X)$, we get the sought representation. Its 'complexified' version (in the sense of the preceeding comment) does coincide with $\pi_{\mathbb{C}}$, giving the desired corresponendence. (my notation is taken from Hall's book). $\endgroup$ – Creo Feb 18 at 12:36
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    $\begingroup$ addendum: This might seem quite different from the 'physicists style' - they do coincide, which can be seen by writing, for instance, $\pi(X)v = Xv = a^iX_i v$ for generators $\{X_i\}$ of the Lie algebra $\mathfrak{g}$ $\endgroup$ – Creo Feb 18 at 12:43
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There are several reasons:

  1. The complexified version is semisimple, and hence can be decomposed by the classification of semisimple complex Lie groups. This is much simpler since the field of complex numbers is algebraically closed.

  2. From the complexified version one can always go back to the real version by considering in the Lie algebra of generators only the Hermitian elements (invariant under conjugation). Thus nothing is lost.

  3. The representation theory of a Lie group and its complexification is closely related. In the present case, this is of great use since the representation theory of the factors of the complexified group is particularly simple, and allows one to deduce the representation theory of the original group.

  4. In quantum physics, one wants to work with Hermitian generators rather than antihermitian ones as in mathematics. This already introduces complex coefficients.

  5. It is often convenient to look at linear combinations of generators with complex coefficients, which are naturally in the complexified Lie algebra. For example, in the Heisenberg algebra, one gets the creation and annihilation operators in this way.

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  • $\begingroup$ Thank you very much for your answer! A question to point 1, do you have a resource where it is shown/explained why $SO(1,3;\mathbb{C})$ is semisimple? And what exactly do you mean by "can be decomposed by the classification of semisimple complex Lie groups"? $\endgroup$ – Sito Feb 15 at 20:21
  • $\begingroup$ @Sito; See en.wikipedia.org/wiki/Simple_Lie_group $\endgroup$ – Arnold Neumaier Feb 15 at 21:34
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    $\begingroup$ @Sito: Showing semisimplicity is done simplest by actually exhibiting it as a direct product through giving generators of the two SL(2,C). $\endgroup$ – Arnold Neumaier Feb 17 at 17:09
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The reason why the complex Lorentz group ($O(1,3;\mathbb{C}$) or any of its subgroups) is needed in physics is furnished by Streater & Wightman in their famous book „PCT, Spin-Statistics, And All That” on page 13. I quote: „It is essential in the proof of the PCT theorem as we shall see”. Indeed, if you turn to page 142 and read that entire section on the PCT theorem, the complex Lorentz group and its local isomorphism to $ SL(2,\mathbb{C}) \otimes SL(2,\mathbb{C})$ are needed.

Your other question

In [1] I only see the statement that SO+(1,3;R) is a subgroup of SO(1,3;C), but why is this relevant for the representation theory of SO(1,3;R) if SO(1,3;C) is not simply connected and the universal cover of SO(1,3;R)?

has a simple answer: it is not. There are separate topics. The representation theory of $SO(1,3;\mathbb{R})$ can be very well tackled without using complex group elements.

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  • $\begingroup$ I'm having a hard time seeing what this has to do with the representation theory of the Lorentz group. I'm also not familiar with the PCT-Theorem (or QFT in general), so if there is a connection between the representation theory of the group and this theorem I would be really glad if you could elaborate a bit more (or point me to a possible reference). $\endgroup$ – Sito Feb 13 at 22:59
  • $\begingroup$ This is not an answer to the question. You can use the comment section to point ou references that might help the OP $\endgroup$ – Hugo V Feb 14 at 0:35
  • $\begingroup$ @HugoV. Which question? The OP has at least two questions. I gave an answer to why one needs $ SO(1,3;\mathbb{C})$ in physics, in QFT in particular. I will address the other question, too. $\endgroup$ – DanielC Feb 14 at 19:55
  • $\begingroup$ The quoted book by Buchbinder by our expert in group theory @Qmechanic has a very good chapter 1, but you need previous exposure to general group theory. Try the book by Wu Ki Tung from the 1980s. It is brilliant. $\endgroup$ – DanielC Feb 14 at 20:11
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The answer is really simple. The Weyl spinor representation as the representation of the electron field includes besides the Lorentz group with its Space-Time generators also an Internal degree of freedom: the phase.

A physical representation should have the form:

$$\mbox{Space-Time generators}~~ \times ~~\mbox{Internal generators}$$

These generators should act independently of each other. They must be order independent so they must commute. The complexification increases the group to include the phase of the field. The internal operator to vary the phase of the field is.

$$e^{-i\phi}\psi$$

The space-time generators are the rotation and boost generators so that you can go from one reference frame to any other.


Addendum: With respect to your other question in the comment first this:


From $\mathfrak{su}(2)_{\mathbb{C}} \oplus \mathfrak{su}(2)_{\mathbb{C}}$ to the relativistic physics of spin 1/2

Each of the two spinors $\xi_L$ and $\xi_R$ represents a precessing vector $S$, the total spin. See the image:

enter image description here

The precession axis (typically $S_z$ but it can be anyone) can have any arbitrary orientation in space, it can point into any direction.

The operator $e^{-i\phi}$ acting on $\xi$ makes the total spin $S$ precess around $S_z$. This corresponds with the phase of the spinor.

The operators $e^{-i\sigma^2*\phi}$ and $e^{\sigma^2* \phi}$ precess the total spin around $S_x$ and $S_y$ respectively. (with the complex conjugate $*$ acting to the right and $e^{G\phi} = \cos\phi + G\sin\phi$, where $\phi$ is the half angle).

Mathematically, the calculation of the three spin vectors $S_x$, $S_y$ and $S_z$ and their sum $S$, the total spin, goes back to Leonard Euler (1775) and Rodrigues. (The Euler Rodrigues formula) using the equivalent real valued Euler parameter representation long before the discovery of QM and mostly forgotten.

I recently found a new formula to calculate all spin vectors at once with a single matrix multiplication explained here or in more detail in sections 1.2, 1.3 and 1.4 of chapter 1 and the whole chapter 2 here.


We now have a physical image of what a spinor represents and we know how to calculate it.


The equivalent Lorentz rotation operators acting on a spinor $\xi$ rotate the spin vectors $S_x$, $S_y$ and $S_z$ and their sum $S$, the total spin, around the $x$, $y$ and $z$ axis respectively. The spin of the bispinor field $\psi$ is given by the sum $S_R^z+S_L^z = S^z$ representing the axial current.

The Lorentz boost operators as they act on the bispinor field $\psi$ can be understood as follows: Since $\xi_R$ and $\xi_L$ propagate at opposing directions (in the rest frame) we must subtract $S_R^z-S_L^z = \vec{j}$ to find the vector current density which is proportional to the momentum. This is how a Lorentz boost in a certain direction provides a momentum in a that direction.

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    $\begingroup$ Can you maybe explain how the vector space generated by the space-time generators and internal generators is related to $\mathfrak{sl}(2,\mathbb{C})$? And why should there be an "Internal generator", I'm having a hard time seeing what this has to with the Lorentz-group in general... $\endgroup$ – Sito Feb 15 at 20:24
  • $\begingroup$ The Weyl spinor representation is more then only a representation of the Lorentz-group. It represents a phase as well. $\endgroup$ – Hans de Vries Feb 15 at 20:35

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