I know from Weinberg and Schwartz's book on Quantum Theory of Fields that $SL(2, \mathbb{C})$ double-cover $SO^{+}(1,3)$.

However, moving to the Lie algebra, based on the following wiki:


we have $so(1,3)_{\Bbb C} \cong sl(2,\Bbb C)_{\Bbb C} \cong sl(2,\Bbb C) \oplus sl(2,\Bbb C)$

where $_{\Bbb C}$ indicates the complexification.

Now, if we move from the Lie algebra to the group, by exponentiating, the last term above gives $SL(2,\Bbb C) \times SL(2,\Bbb C)$

So my questions are:

  1. What is the Lie group related to $so(1,3)_{\Bbb C}$?
  2. In $so(1,3)_{\Bbb C} \cong sl(2,\Bbb C)_{\Bbb C} \cong sl(2,\Bbb C) \oplus sl(2,\Bbb C)$, the algebras are viewed as real or complex algebraS?
  3. On most of the books in QFT from the following brackets for $so(1,3)$:

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

where $J^+=1/2(J_i+iK_i), \, J^-=1/2(J_i-iK_i)$

they deduce $so(1,3) \cong su(2) \oplus su(2)$

Shouldn't they deduce $so(1,3)_{\Bbb C} \cong su(2)_{\Bbb C} \oplus su(2)_{\Bbb C}$?

  • 3
    $\begingroup$ The physics textbooks are just being sloppy. Possible duplicates: physics.stackexchange.com/q/28505/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 21, 2021 at 9:42
  • 1
    $\begingroup$ @Qmechanic just read the post you mentioned, what is $SO(1,3;\Bbb C)$? I know complexification of vector space or algebra cause that structure have a field, not sure what is the complexification of a group $\endgroup$
    – Andrea
    Oct 21, 2021 at 12:13
  • 1
    $\begingroup$ $SO(1,3;\Bbb C)$ is the group of complex Lorentz matrices with unit determinant. $\endgroup$
    – Qmechanic
    Oct 21, 2021 at 13:25

1 Answer 1


The group $\mathrm{SL}(2,\mathbb{C})$ is isomorphic to the double cover of ${\rm SO}(1,3)$. In that case the relation between the two can be expressed as $${\rm SO}(1,3)\simeq {\rm SL}(2,\mathbb{C})/\mathbb{Z}_2\tag{1}.$$

The other kind of relation you mention is just a bit more subtle and is often stated without much care, leading to confusion. Starting form ${\rm SO}(1,3)$ build its Lie algebra ${\frak so}(1,3)$. This is a real Lie algebra and as with any vector space over $\mathbb{R}$ it can be complexified. Constructing the complexification ${\frak so}_\mathbb{C}(1,3)$ we can show that $${\frak so}_\mathbb{C}(1,3)\simeq \mathfrak{su}_\mathbb{C}(2)\oplus \mathfrak{su}_\mathbb{C}(2)\tag{2}.$$

In truth there is nothing complicated here. The generators $J_{\mu\nu}$ of ${\mathfrak{so}}(1,3)$ can be split into generators of rotations $\mathbf{J} = (J^{23},J^{31},J^{12})$ and of boosts $\mathbf{K} = (J^{10},J^{20},J^{30})$. Since $\mathfrak{so}(1,3)$ is a real Lie algebra you can only form combinations of such generators with real coefficients. Complexification is just a fancy terminology to say that we are now allowing to form linear combinations with complex coefficients. These linear combinations of the generators with complex coefficients do not live on $\mathfrak{so}(1,3)$ but rather on $\mathfrak{so}_\mathbb{C}(1,3)$.

The reason complexification is essential is that it allows us to form $$\mathscr{A}_i = \frac{1}{2}(J_i+iK_i),\quad \mathscr{B}_i=\frac{1}{2}(J_i-iK_i)\tag{3},$$

the advantage of this being that the Lorentz algebra of the generators $J_i, K_i$ is equivalent to the $\mathscr{A}_i$ and $\mathscr{B}_i$ separately satisfying the $\mathfrak{su}_\mathbb{C}(2)$ algebra. This is what (2) is all about.

So complexified $\mathfrak{so}(1,3)$ is isomorphic to a direct sum of two complexified $\mathfrak{su}(2)$. That is the proper relation. It is useful in the end because it allows you to build representations of $\mathfrak{so}_\mathbb{C}(1,3)$ out of those of $\mathfrak{su}_\mathbb{C}(2)$ which are known from the theory of angular momentum.

Finally I should mention that sometimes people write ${\rm SO}(1,3)\simeq {\rm SU}(2)\times{\rm SU}(2)$. This is incorrect and what they actually have in mind is (2) above.

  • $\begingroup$ Thanks! Is then correct to say that the lie group of the complexified algebra $so(1,3)_{\Bbb C}$ is isomorphic to $SL(2, \mathbb{C}) \times SL(2, \mathbb{C})$, not to $SO^+(1,3)$ ? $\endgroup$
    – Andrea
    Oct 21, 2021 at 13:51
  • $\begingroup$ You're welcome. As I said as far as I know the relation between complexified algebras (2) does not exponentiate. Moreover even if there were a sense in which that relation exponentiated I'm quite sure you would not find two copies of ${\rm SL}(2,\mathbb{C})$. The most important point, though, is that the relation (2) do not need to exponentiate. All we need is the relation at the level of the algebras. $\endgroup$
    – Gold
    Oct 21, 2021 at 13:57

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