I do understand why we are using the double cover, but why exactly do we make the transition to complex Lorentz transformations? Where and why are they needed?

To be precise:
The double cover of the Lorentz group is given by $SU(2)$. Where and why do we make the transition from $SU(2)$ to $SL(2,C)$? ($SL(2,C)$ is the complexification of $SU(2)$ as far as i understand)

Complexification means that we allow the vector space spanned by the generators of the group to be complex, i.e. we allow complex linear combinations of the generators. $SU(2)$ are all unitary 2x2 matrices with unit determinant, whereas $SL(2,C)$ are all complex matrices with unit determinant. As far as i understand this has something to do with allowing complex Lorentz transformations and therefore extending the Lorentz group to the complex Lorentz group. The double cover of the complex Lorentz group is then $SL(2,C)$, whereas the double cover of the (real) Lorentz group is given by $SU(2)$?!

If someone could clarify this i would be very thankful.

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    $\begingroup$ It is false that "The double cover of the Lorentz group is given by $SU(2)$". $SU(2)$ is the universal (double) covering of $SO(3)$. The double covering of the $SO(1,3)\uparrow$ is $SL(2,C)$ viewed as a real Lie group. $\endgroup$ – Valter Moretti Jan 13 '14 at 19:19
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    $\begingroup$ The point is that if you consider the complex Lorentz group, its complex Lie algebra is (complex) isomorphic to $su(2)\oplus su(2)$, that is the Lie algebra of a compact real Lie group and its representations are easily classified (essentially in view of Peter-Weyl theorem). Among these representations one finds also those of the real Lorentz group. $\endgroup$ – Valter Moretti Jan 13 '14 at 19:41
  • $\begingroup$ Thanks four you quick reply. I now do understand that the Lie algebra of the complex Lorentz group is isomorph to $su(2)\oplus su(2)$. We get the Lie Algebra of the (real?) Lorentz group $SO(1,3)\uparrow$ from the well known transformation matrices of special relativity and it can be seen that this Lie algebra is the same as the Lie algebra of $SL(2,C)$. What exactly is the relation between $SU(2)$ and $SL(2,C)$? I would guess, because $su(2)\oplus su(2)$ and $sl(2,C)$ are so close connected (what is the correct term for this connection?) there is some kind of link. $\endgroup$ – jak Jan 13 '14 at 20:35
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    $\begingroup$ Related: physics.stackexchange.com/q/28505/2451 , physics.stackexchange.com/q/141354/2451 and links therein. $\endgroup$ – Qmechanic Jan 13 '14 at 22:45
  • $\begingroup$ also related: physics.stackexchange.com/q/47339 $\endgroup$ – user1504 Jan 14 '14 at 0:16

As V. Moretti correctly pointed out $\mathfrak{su(2)}$ is the Lie Algebra of $SU(2)$ and also of $SO(3)$. You can easily check that the dimension of $SU(2)$ is $3$ while the dimension of $SO(1,3)$ is $6$ (three boosts and three rotations). Therefore $SU(2)$ can not cover $SO(1,3)$. It's "too little" to do so!

The point is that when we deal with non simply connected groups, we should always look for representations of the universal covering group. If we don't do that, we are going to miss some physical interesting irredicible representations. For example if you look at the representations of $SO(3)$, you'll find out that the particles that fit into multiplets (vectors) on which these irreducible representations work will always have integer spin. But we know fermions exist, therefore we look for representations of its cover: $SU(2)$.

The way one deals with the representation of $SO(1,3)$ is the following. First one sees that it is not simply connected. You can get convinced of this that noticing that $SO(1,3)$ has as a subgroup $SO(3)$ which is notoriously not simply connected itself. Therefore we look for the universal covering of $SO(1,3)$ which turn out to be $SL(2,\mathbb{C})$. This group has as a Lie Algebra $\mathfrak{sl(2)}$ (and it is a real algebra: you only allow real combination of the generators) and the problem is then to classify the irreducible reps of this real algebra.

This is hard to do, so we use a trick. We consider another algebra, which is $\mathfrak{sl(2)}_{\mathbb{C}}=\mathfrak{sl(2)}\otimes \mathbb{C}$, namely the complexification of $\mathfrak{sl(2)}$. This, as a complex Lie algebra is indeed isomorphic to $\mathfrak{su(2)}_{\mathbb{C}}\oplus \mathfrak{su(2)}_\mathbb{C}$, as you can show playing a little with the generators. The advantage is that its fairly easy to find all irrepses of $\mathfrak{su(2)}_\mathbb{C}$. However, seeing $\mathfrak{sl(2)}_{\mathbb{C}}$ as a real algebra you can understand that this is not the algebra of $SL(2,\mathbb{C})$ since its real dimension is double the one of $SL(2,\mathbb{C})$. (Of course it is, you complexified!)

Having found all the irrepses of $\mathfrak{su(2)}_{\mathbb{C}}\oplus \mathfrak{su(2)}_\mathbb{C}$ we have to "trow away" some, in order to keep only the ones of the real algebra we started from. This is called taking the real section of a complexified algebra.

Note I have never talked about complexification of groups (don't even know if the concept exists but guess it does). I only talked about complexification of algebras.

Also note that different Lie groups, with different Lie algebras may indeed have isomorphic complexified algebras. For example you can show that $SO(3)$ and $SO(2,1)$ have both a complexified algebra $\mathfrak{su(2)}_{\mathbb{C}}$.

Hope this makes a little more sense. If it's unclear to you, please ask.

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  • $\begingroup$ I have a little question left: If it is true that: $(\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq \mathfrak{su(2)}_{\mathbb{C}} \oplus \mathfrak{su(2)}_{\mathbb{C}}$ and $\mathfrak{su(2)}_{\mathbb{C}} \simeq \mathfrak{sl}(2,\mathbb{C})$ how is this related to $\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq \mathfrak{sl}(2,\mathbb{C})$ Surely one of the statements must be wrong, but i can't figure out which. $\endgroup$ – jak Jan 14 '14 at 21:48
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    $\begingroup$ What do you mean with $\mathfrak{sl}(2,\mathbb{C})$? Is it the Lie algebra of $SL(2,\mathbb{C})$, namely $sl(2)$ or is it its complexification, $\mathfrak{sl}(2)_{\mathbb{C}}$? $\endgroup$ – Federico Carta Jan 14 '14 at 22:07
  • $\begingroup$ I mean the Lie algebra of $SL(2,\mathbb{C})$ $\endgroup$ – jak Jan 15 '14 at 7:25
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    $\begingroup$ After your comment and having again a look at Wikipedia i would now guess that the third statement above is wrong and it must be $\mathfrak{so}(1,3)_+^{\uparrow})_\mathbb{C} \simeq \mathfrak{sl}(2)$ and in the Wikipedia Artikel $\mathfrak{sl}(2,\mathbb{C})$ means the complexification of the Lie algebra of $SL(2,\mathbb{C})$. Nevertheless the notation on the right-hand side at Wikipedia still confuses me. $\endgroup$ – jak Jan 15 '14 at 7:41
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    $\begingroup$ Suggestion to the answer (v2): Replace the disconnected group $SO(1,3)$ with the connected group $SO^+(1,3)$ in various places. $\endgroup$ – Qmechanic Feb 13 '19 at 15:45

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