By showing that the complexified Lie algebra of the proper Lorentz group ${\rm SO(3,1)}$ is equivalent to two the direct sum of two complexified ${\rm SU(2)}$ Lie algebras $$\mathfrak{so}(3,1)_\mathbb{C}=\mathfrak{su}(2)_\mathbb{C}\oplus \mathfrak{su}(2)_\mathbb{C}.$$ With this at our disposal, it is possible to find all the irreducible representations of that algebra.

With this technology, I do not need to know or mention anything about the group ${\rm SL(2, C)}$, its Lie algebra or its representations. The knowledge of ${\rm SO(3,1)}$ and ${\rm SU(2)}$ seems to be sufficient, I think.

  • So why learn about ${\rm SL(2, C)}$ in the first place? It seems to me to be unnecessary. Please tell me if I am missing something on a conceptual level.
  • $\begingroup$ This is a terribly ill-defined question. Surely you can follow the usual steps, implicitly using su(2) technology, but without saying so explicitly. Would this count as "without thinking about su(2)"? What if I use the standard angular momentum operators, but call them A,B,C instead of $J_x,J_y,J_z$? Would that count? That being said, all of representation theory of simple groups is done using su(2), i.e., you identify distinguished su(2) subgroups and diagonalise those, cf. roots and weights. So the answer is no: you cannot do representation theory without thinking about su(2). $\endgroup$ Mar 14, 2020 at 17:51
  • $\begingroup$ Are you talking about the complexified Lie algebras? To reopen this post (v2) consider to provide precise statements, cf. physics.stackexchange.com/q/28505/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 14, 2020 at 18:02
  • $\begingroup$ I tried to restate my discomfort. I cannot fully understand the need to learn ${\rm SL(2, C)}$ separately. $\endgroup$ Mar 14, 2020 at 18:17
  • $\begingroup$ This only holds for finite dimensional dims. Unitary infinite dimensional reps of SO(3,1) are not equivalent to SU(2)xSU(2). $\endgroup$ Mar 14, 2020 at 18:17
  • 1
    $\begingroup$ $sl(2.\mathbb{C})$ is isomorphic to the complexification of $su(2)$, so it seems the question (v4) is a triviality. $\endgroup$
    – Qmechanic
    Mar 14, 2020 at 18:26

2 Answers 2


To put it simply, to know that the Lie group $SL(2,\mathbb{C})$ of $2\times 2$ complex matrices with unit determinant is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$ gives a simple and direct way to understand how the restricted Lorentz group can act on a Weyl spinor $\psi\in\mathbb{C}^2$. In contrast, the group action (on a Weyl spinor) is somewhat mysterious/less intuitive from the point of $4\times 4$ spacetime Lorentz transformations $\Lambda\in SO^+(1,3;\mathbb{R})$.

Similarly, the subgroup $SU(2)\subseteq SL(2,\mathbb{C})$ is (the double cover of) the 3D rotation group $SO(3)\subseteq SO(1,3;\mathbb{R})$.

There is a corresponding double-copy version for the complexified proper Lorentz group $SO(1,3;\mathbb{C})$. For more details, see this & this related Phys.SE posts.


As far as the Fierz identities are concerned you are right. Turns out that usually we also want to define complex conjugation. For instance we want to preserve CPT, so the field content should be closed under CPT action.

If $a=1,2$ is the index of the first $SU(2)$ and $\dot a=1,2$ the second $SU(2)$, for the case $SO(1,3)$ the complex conjugation switch the representations:

$$ (\chi^{a})^{*}=\bar\chi^{\dot a} $$

Which breaks $SU(2)_{\mathbb{C}}\times SU(2)_{\mathbb{C}}$ down to $SL(2,\mathbb{C})$.

Different signatures will impose different reality conditions to the Lorentz generators, reducing the $SU(2)_{\mathbb{C}}\times SU(2)_{\mathbb{C}}$ to a subgroup.

For signature $SO(4)$ we get

$$ (\chi^{a})*=\bar\chi_{a},\qquad (\chi^{\dot a})^{*}=\bar\chi_{\dot a} $$

which leads to the $SU(2)\times SU(2)$ subgroup, without the complexification.

For signature $SO(2,2)$ we get

$$ (\chi^{a})*=\bar\chi^{a},\qquad (\chi^{\dot a})^{*}=\bar\chi^{\dot a} $$

which leads to the $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$.

So if you are interested in imposing Majorana conditions on spinors you see that $SO(4)$ this is not possible since $\varepsilon^{ab}\chi_{b}=\chi^{a}$ implies that $\chi^{a}=0$. The minimal number of components in that case is two complex, or four real.

For the $SO(1,3)$ the Majorana condition fix the spinor $\chi^{\dot a}$ in terms of $\chi^{a}$, or vice-versa. The minimal number of components in that case is two complex or four real.

For $SO(2,2)$ it is possible to impose a Majorana condition in a single chiral spinor, reducing the number of components to a total of two real components.


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