How are $SL(2, \mathbb{C})$ and $SL(2, \mathbb{C}) \times SL(2, \mathbb{C})$ related to the Lorentz Group?

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:

https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group

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 $$so(1,3)_{\Bbb C} \cong su(2)_{\Bbb C} \oplus su(2)_{\Bbb C}$$?

• The physics textbooks are just being sloppy. Possible duplicates: physics.stackexchange.com/q/28505/2451 and links therein. Commented Oct 21, 2021 at 9:42
• @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 Commented Oct 21, 2021 at 12:13
• $SO(1,3;\Bbb C)$ is the group of complex Lorentz matrices with unit determinant. Commented Oct 21, 2021 at 13:25

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.

• 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 $SO^+(1,3)$ ? Commented Oct 21, 2021 at 13:51
• 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.
– Gold
Commented Oct 21, 2021 at 13:57