Hi everyone,


I'm currently studying the construction of the $Pin(1,3)$ group and given the definition I'm using to find its elements I'm having some problems with the signs associated with $2\pi$ and $4\pi$ rotations.


Definitions:


Let $\left(L\_{\alpha}^{\beta}\right) \in O(1,3)$ and $\{ \gamma\_{\alpha}\}$ the generators of the real Clifford Algebra $Cl(1,3)$ which obey:


$\{\gamma\_{\alpha}, \gamma\_{\beta}\} = 2\eta\_{\alpha \beta}\mathbb{I}$


with the metric $\eta\_{\alpha \beta} = diag(1,-1,-1,-1)$.


Then: 


$ \Lambda\_L \in Pin(1,3) \subset Cl(1,3) \iff \Lambda\_L\gamma\_{\alpha} \Lambda\_L^{-1} = \gamma\_{\beta}L\_{\alpha}^{\beta}$
and
$\Lambda \Lambda^T = \pm \mathbb{I}$


Where T is a reversion.


With this formula it's possible to explicitly find the elements of Pin, but when I try to do it for a $2\pi$ and a $4\pi$ rotation I can't see how to get the results I'm supposed to:


$\Lambda\_{R(2\pi)} = -\mathbb{I} $


$\Lambda\_{R(4\pi)} = \mathbb{I} $


The problem is that $L\_{R(n\pi)} = \mathbb{I}$ for every $n \in \mathbb{Z}$, so the solution will always be either $\mathbb{I}$ or $-\mathbb{I}$.


Which criteria is used to get the right signs?


Thanks!



2Pi and 4Pi rotations in the Pin(1,3) group
I can't recall the computation you need from the top of my head, but I will try to clarify the geometric picture, hoping it will help. From your question and the notation you use I suspect you are a physicist; if I am wrong and I spend too much time explaining standard mathematics, please forgive me.


As you noticed, the condition you gave to verify that $\Lambda\_L \in Pin(1,3)$ does not distinguish $\Lambda\_L$ from $- \Lambda\_L$. Moreover 
$R(2 \pi)$ and $R(4 \pi)$ are the same element in $O(1,3)$: they both are the identity. What we need to distinguish $R(2 \pi)$ from $R(4 \pi)$ is some "dynamical" information: you should regard them not as elements in $O(1,3)$ but as paths in $O(1,3)$ starting and ending at the identity. The first one will be a path which makes a single turn,while the second one will make two turns. Then you should try to lift these paths to paths in $Pin(1,3)$ starting from the identity: if you do the computation correctly you'll see that the lift of the first path will end at $- \mathbb I$ and the lift of the second path will end at $\mathbb I$.


In mathematical terminology this means that $Pin(1,3)$ is a double cover of $O(1,3)$: that is, there is a surjective map $Pin(1,3) \to O(1,3)$ such that the preimage of an element $L \in O(1,3)$ is the set $\lbrace \Lambda\_L, - \Lambda\_L \rbrace$. Moreover a non-contractible closed loop in $O(1,3)$ starting and ending at $L$ will lift to a path in $Pin(1,3)$ staring at $\Lambda\_L$ and ending at $- \Lambda\_L$,


