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One can show that the Lie algebra of $SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$. And the representations of $SU(2)\times SU(2)$ exhausts all possible representations of $SO(3,1)$.

$\bullet$ Why does one consider, then, the irreducible representations of $SL(2,\mathbb{C})$?

$\bullet$ At the level of Lie algebra what is the relation between (i) $SL(2,\mathbb{C})$ and $SO(3,1)$ , (ii) $SU(2)\times SU(2)$ and $SL(2,\mathbb{C})$?

$\bullet$ I have found that, at the group level, $SO(3,1)$ is isomorphic not to $SU(2)\times SU(2)$ but to $SU(2)\times SU(2)/\mathbb{Z}_2$. Is this statement correct?

SRS
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  • This question (v2) is essentially a duplicate of http://physics.stackexchange.com/q/28505/2451 , http://physics.stackexchange.com/q/108212/2451 and links therein. – Qmechanic Oct 19 '16 at 13:51
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    At the level of the (complexified) Lie algebra, so(3,1) = su(2) $\oplus$ su(2) = sl(2,$\mathbb{C}$). At the level of the group, SO(3,1)$^+$ = SL(2, $\mathbb{C})/\mathbb{Z}_2$, where the $+$ indicates that we consider only time-orientation-preserving Lorentz transformations. Note that your third bullet point cannot be correct, since the Lorentz group is not compact whilst $SU(2) \times SU(2)$ is compact. Only when we complexify do we see a correspondence between the Lorentz group and SU(2) $\times$ SU(2). Note that SO(4) = SU(2) $\times$ SU(2)/$\mathbb{Z}_2$. – gj255 Oct 19 '16 at 13:59
  • @gj255 Why does one write $so(3,1)=su(2)\oplus su(2)$ instead of $so(3,1)=su(2)\otimes su(2)$? Can you please explain what is meant by comlexified Lie algebra of $SO(3,1)$? I'm not familiar with such a term. I understand the rest of your answer. – SRS Oct 19 '16 at 14:08
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    I use $\oplus$ because Lie algebras are vector spaces, and the natural way of combining two vector spaces is with a direct sum. For groups, the natural product is $\times$, the direct product. One can show that the Lie algebra of a direct product of groups is the direct sum of the corresponding Lie algebras. – gj255 Oct 19 '16 at 14:11
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    To complexify a Lie algebra is to allow oneself to take complex linear combinations of elements of the Lie algebra, rather than just real linear combinations. When trying to establish a correspondence between the Lorentz group and $SU(2) \times SU(2)$, one typically looks at the combinations $J_i \pm i K_i$, where $J_i$ and $K_i$ span the (real) Lie algebra of the Lorentz group. It is only in terms of these complex linear combinations that we find the su(2) $\oplus$ su(2) commutation relations. – gj255 Oct 19 '16 at 14:13
  • @gj255 -Thanks. It was really helpful. Does this complexification of generators lead to any conceptual problem? If not, why one need to separately study the representations of $SL(2,\mathbb{C})$? – SRS Oct 19 '16 at 14:29
  • I'm not sure I can give a good answer to this question, but as far as I'm aware, whether one looks for representations of SL(2,$\mathbb{C}$) or of SU(2) $\times$ SU(2) is a matter of choice and convenience. In some cases it might be more natural and straightforward to look at one group over the other. – gj255 Oct 19 '16 at 14:39
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    @gj255 So when finding representations of $SU(2) \times SU(2)$, one does really find representations of the complexified proper ortho. Lorentz group (and not of the real p.o.L.g.), is this correct? So the representations (0,0), (0,1/2) etc. which one encounters in qft are representation of $SO^+(1,3;\mathbb{C})$, right? ... What is now the relation between $SO^+(1,3;\mathbb{C})$ and $SO^+(1,3;\mathbb{R})$, and is this difference relevant in QFT, where one primarily wants to work with $SO^+(1,3;\mathbb{R})$ ? – Breaking Mad Apr 06 '17 at 14:08

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