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There have been a lot of questions in the past about the subtleties of the Lorentz algebra. In particular the usage of the real Lie algebra $\mathfrak{so}(1,3;\Bbb R)$ and it's complexification which is more convenient in the context of representation theory. I have searched through a lot of questions on the site and I would like to just outline a few things I've picked up and ask that if there is anything incorrect in the following, could someone please point it out.


The following Lie algebra isomorphisms are useful:

$$\mathfrak{so}(1,3;\Bbb R)_{\Bbb C}\cong\mathfrak{so}(1,3;\Bbb C)\cong \mathfrak{sl}(2;\Bbb C)\oplus \mathfrak{sl}(2;\Bbb C)\cong \mathfrak{su}(2)_{\Bbb C}\oplus \mathfrak{su}(2)_{\Bbb C}\tag{1}.$$

In particular the first with the last, as the representation theory of $\mathfrak{su}(2)_{\Bbb C}$ is well understood, and hence the classifications of the representations of the (complexified) Lorentz algebra can be obtained.


Next I note two forms that I have seen of the Lie bracket relations for the Lorentz algebra:

  1. $[J_i,J_j]=i\epsilon_{ijk}J_k, \quad [K_i,K_j]=-i\epsilon_{ijk}J_k,\quad [J_i,K_j]=i\epsilon_{ijk}K_k \tag{2}$

  2. $[J_i,J_j]=\epsilon_{ijk}J_k, \quad [K_i,K_j]=-\epsilon_{ijk}J_k,\quad [J_i,K_j]=\epsilon_{ijk}K_k \tag{3}$

Found here on page 3 and here on page 22 respectively. These are obviously very similar and only differ by a factor of $i$. My question is (since the subtlety of whether we are working with the Lorentz algebra or the complexified Lorentz algebra is almost never said in physics texts), is it sufficient in passing from the real Lorentz algebra to the complexified Lorentz algebra, to simply insert a factor of $i$ into the Lie bracket relations, as above? Or is there something more subtle being done here?

The text in which the first set of relations is found, still defines the following quantities:

$$J_j^{\pm}:=\frac{1}{2}(J_j\pm iK_j) \tag{4},$$

which is surprising to me since it appears we are already in the complexified Lorentz algebra, given the commutation relations they give in $(2)$. In other words, why is it necessary to take a complex linear combination of $J$ and $K$ here just as we would if we were going from the real algebra to the complex. I understand that in terms of these new $J_j^\pm$ generators we find the isomorphism $\mathfrak{so}(1,3)_{\Bbb C}\cong \mathfrak{su}(2)_{\Bbb C}\oplus \mathfrak{su}(2)_{\Bbb C}$ and things follow from there.

Are the $J_j$ and $K_j$ above perhaps not the same as in $(2)$?

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The $J_j$ and $K_j$ are the physically natural Lorentz generators: $J_j$ are the three spacial rotations; $K_j$ are the three Lorentz boosts.

However these variables are not the natural generators for displaying the reducibility of the Lorentz algebra into two irreducible $\mathfrak{su}(2)$ components. This is because $[J_i,K_j] \neq 0$ as your equations show.

With $J^+_j$ and $J^-_j$ the reducibility becomes explicit (at the expense of changing to non-physical generators which are complex linear combinations of spacial rotations and Lorentz boosts).

It can be shown that

$$[J^+_i,J^+_j] = i\epsilon_{ijk}J^+_k$$ $$[J^-_i,J^-_j] = i\epsilon_{ijk}J^-_k$$ $$[J^+_i,J^-_j] = 0$$

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  • $\begingroup$ I understand how to proceed once we've complexified to study the representations, my question is more about the two Lie brackets I gave and why the authors of $(2)$ take what appear to be generators of the complexified Lorentz algebra and construct complex linear combinations of them. $\endgroup$
    – Charlie
    May 23 at 16:44
  • $\begingroup$ So once you have moved to the complexification you are free to define a new set of generators $J'_j = iJ_j$ etc. This will have the effect of introducing an extra i in the commutation relations. You are then further free to take complex linear combinations of these, which is useful for the reason I have given. I would need to think more whether the first of these steps is strictly necessary or just convenient. $\endgroup$
    – isometry
    May 23 at 17:02
  • $\begingroup$ Ah I see what you mean, I thought about it for a minute and that makes complete sense. This solves my problem so I'll go ahead and accept it, if you want to add what you've written to the answer it might be useful for anyone who stumbles on this in the future but it doesn't matter too much to me. Thanks! $\endgroup$
    – Charlie
    May 23 at 17:07
  • $\begingroup$ If the generators are taken to be hermitian - which is typical in the physics literature - then the factor of $i$ is strictly necessary in the commutation relations for the real- i.e. un-complexified - Lie algebra spanned by the generators in question. $\endgroup$
    – J. Murray
    May 23 at 17:21
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The "physics convention" when working with orthogonal/unitary Lie groups and their associated Lie algebras is to take the elements $A$ of the algebra to be Hermitian, such that elements $g$ of the group are obtained by exponentiation with a factor of $i$ inserted, $g = \exp[iA]$. In contrast, the "mathematics convention" is to take the elements of the algebra to be anti-Hermitian, so group elements are obtained by bare exponentiation $g=\exp[A]$.

There is a subtle problem with the Physics convention, however. The product operation $\star:\mathfrak{g}\times\mathfrak g \rightarrow \mathfrak g$ on the algebra is ordinarily assumed to be the commutator:

$$A\star B := [A,B]$$ However, the commutator of two Hermitian operators is not Hermitian (unless it vanishes)! Therefore, if we are to have a well-defined product operation on our algebra, we need to modify the product operation, which we do via $$A\star B := \frac{[A,B]}{i}$$

As a concrete example, in the mathematics convention the standard basis for $\mathfrak{su}(2)$ is given by $\tau_j:= i \sigma_j$, which obey $\tau_j\star \tau_k := [\tau_j,\tau_k]=2\epsilon_{jk\ell} \tau_\ell$. In the physics convention, the standard basis is given by $\sigma_j$, which obey $$\sigma_j\star \sigma_k := \frac{[\sigma_j,\sigma_k]}{i}= 2\epsilon_{jk\ell} \sigma_\ell$$ $$\implies [\sigma_j,\sigma_k] = 2i\epsilon_{jk\ell}\sigma_\ell$$

So the presence (or absence) of a factor of $i$ in the commutation relations for the group's infinitesimal generators reflects a choice of whether the generators should be Hermitian or anti-Hermitian. This is true even for a real Lie algebra, despite the factors of $i$ involved.

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  • $\begingroup$ Interesting, I was aware of the physics/maths convention regarding the factor of $i$ but had never thought about how the bracket operation itself had to change to accommodate this. So $(2)$ really is the Lie bracket of the non-complexified Lorentz algebra, just with the factor of $i$ accounted for. $\endgroup$
    – Charlie
    May 23 at 17:31
  • $\begingroup$ I've already accepted the other answer but this is definitely useful information to have alongside it, hope that's alright. $\endgroup$
    – Charlie
    May 23 at 17:32
  • $\begingroup$ @Charlie It's your prerogative to accept whichever answer you feel best addresses your particular question. I would simply emphasize that the factor of $i$ in the commutation relations has nothing to do with the complexification of the algebra, and that the commutation relations don't change at all when you complexify since the essential definition of the complexification is extending the underlying real vector space to the natural complex one while leaving the commutation relations alone. $\endgroup$
    – J. Murray
    May 23 at 17:39
  • $\begingroup$ Since your answer does actually render the discussion in the comments of the other answer incorrect, I'll accept this. This addresses a very longstanding point of confusion I've had with this topic so thanks for clearing that up. $\endgroup$
    – Charlie
    May 23 at 17:45

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