I'm trying to understand the Lie algebra of the Lorentz group and am almost there, but am stuck at the final hurdle! It's easy to prove that

$$\frak so(1,3)^\uparrow_{\mathbb{C}}=sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$$

by considering generators. Indeed $\frak so(1,3)^\uparrow$ has generators $J_i$ for rotations and $K_i$ for boosts. The complexification has basis

$$L_i^{\pm}=J_i\pm iK_i$$

and it's not hard to show [$L_i^{\pm}, L_j^{\pm}]=\epsilon_{ijk}L^\pm_k$ and $[L_i^+,L_j^-]=0$ yielding two commuting copies of the complexification of $\frak su(2)$ which is $\frak sl(2,\mathbb{C})$. Is this correct?

Now my notes say that a generic representation of $\frak so(1,3)^\uparrow_{\mathbb{C}}$ is the tensor product of the spin-$j_1$ representation of $\frak sl(2,\mathbb{C})$ and the spin-$j_2$ conjugate representation of $\frak sl(2,\mathbb{C})$. Where does this conjugate business come from? I can't make head or tail of it!

Note: I know that this makes physical sense, since then the $(0,\frac 12)$ representation yields right handed spinors and the $(\frac 12,0)$ representation gives left handed spinors. But where does it come from mathematically?!

Many thanks in advance for your help!


For anyone who is interested - I've worked out the solution myself. It turns out it is just sloppy wording.

The generic representation of the Lorentz algebra is the tensor product of two spin representations of $\mathfrak{sl}(2,\mathbb{C})$, labelled $(j_1,j_2)$. Now we can see that the $(j_1,j_2)$ representation is conjugate to the $(j_2,j_1)$ representation, by plugging in the definitions of $J,K$ in terms of $L$ and seeing what happens.

This means that one can regard the $(0,j)$ representation as the conjugate of the $(j,0)$ representation. Now identifying the $(j,0)$ representation with the spin-$j$ representation of $\mathfrak{sl}(2,\mathbb{C})$ as a complex Lie algebra, the nomenclature makes sense.

It's a pretty circular way of looking at things though, and I certainly won't be using this terminology in any of my own work!

  • 1
    $\begingroup$ How do we see that the (j1,j2) irrep is conjugate to the (j2,j1) irrep? Could you please elaborate? $\endgroup$ – label Feb 19 '18 at 5:41
  • $\begingroup$ @label: That is indeed the question. Compare my answer. $\endgroup$ – Torsten Schoeneberg Jul 19 '19 at 22:04

Much is said in the other answer, but I want to amend a crucial missing point, whose importance is expressed in my lengthy recent answer to Isomorphic Lie algebras and their Representations (particularly its "Final Footnote").

Indeed the complexification of the Lorentz algebra is isomorphic to two copies of $\mathfrak{sl}_2(\mathbb C)$,

$$\mathfrak{so}(1,3)_\mathbb C \simeq \mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C).$$

Now complex irreps of $\mathfrak{sl}_2(\mathbb C)$ are indexed, via highest weight theory, by non-negative integers $j \in \mathbb Z_{\ge0}$ (math notation) or half-integers $j \in \frac12 \mathbb Z_{\ge 0}$ (physics notation). Irreps of direct sums are tensor products of irreps of the factors, so the complex irreps of the above complexification are indeed indexed by pairs $(j_1,j_2)$ of non-negative integers (math) or half-integers (physics).

Now for each real form of the complex Lie algebra above, this means that its complex irreps are also given, via restriction, by those irreps indexed by $(j_1, j_2)$. However, to see how conjugation acts on these irreps, we must have a closer look at what real form we are looking at.

Namely, besides our Lorentz algebra

$\mathfrak{g}_0 = \mathfrak{so}(3,1)$

there are other real Lie algebras which have the above complexification, notably

$\mathfrak{g}_1 = \mathfrak{sl}_2(\mathbb R) \oplus \mathfrak{sl}_2(\mathbb R)$ and

$\mathfrak{g}_2 = \mathfrak{so}_4 \simeq \mathfrak{su}_2\oplus \mathfrak{su}_2$.

$\mathfrak{g}_1$ is a split and $\mathfrak{g}_2$ is a compact form. All the Satake-Tits diagrams of these forms have underlying Dynkin diagram of type $D_2 = A_1 \times A_1$, i.e. two vertices without an edge, where

for the quasi-split $\mathfrak{g}_0$ there is an arrow between the vertices, and both vertices are white,

for the split $\mathfrak{g}_1$ there are no arrows and both vertices are white,

for the compact $\mathfrak{g}_2$ there are no arrows and both vertices are black.

The general outline in https://math.stackexchange.com/a/3298058/96384 now says that on $\mathfrak{g}_0$, because of that arrow, complex conjugation switches the two basis roots in $D_2$, hence the two fundamental weights, hence it turns the irrep indexed by $(j_1, j_2)$ into the one indexed by $(j_2, j_1)$.

But both for $\mathfrak{g}_1$ and $\mathfrak{g_2}$, conjugation acts as identity on the roots, hence on the weights, hence just leaves $(j_1, j_2)$ as itself; which means for those, all irreps are actually equivalent to their own conjugate.

So the fact that for the Lorentz Lie algebra, conjugation acts as described, is subtle, and really (pun intended) has more to do with its actual real structure. As in the other answer, and in the first part of my first answer quoted at the beginning, one can of course also see that by "plugging in [actual matrices of $\mathfrak{so}(3,1)$] and seeing what happens".


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