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.
Answer
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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