# Lorentz group $SO(3,1)$ and $SU(2)\times SU(2)$

One way to classify Lorentz representations is to consider the Lie algebra isomorphic of Lorentz group to $SU(2)\times SU(2)$. So that we can classify it by two integers $(j_1,j_2)$. In this way I can think $j_1$ is a generator of the Lorentz group. My question is：

We can have $\exp(-ibJ_1)$, where $b$ is a parameter. Is this an element of the Lorentz group? Since $j_1$ is a generator of the Lorentz group.

First of all, the Lorentz-group and SU(2) (respectively $SU(2)\times SU(2)$) are groups represented by matrices. Therefore $j_1$ cannot be a generator, because, as you say $j_1$ is an integer. For being an generator, however, it must be matrix with some rather specific properties.
Secondly, looking at your $\exp(-ibJ_1)$, as the $J_1$ is uppercase, it suggests that it could be indeed a matrix. If not, i.e. you mean $\exp(-i b j_1)$ by it, it is no case a element of the Lorentz-group. But even if $J_1$ is meant to a matrix, it could be hardly be an element of the Lorentz-group. Actually the parametrisation of the Lorentz-group is $\exp(i(\vec{\alpha}-i \vec{u})\cdot\bf{J})\otimes\exp(i (\vec{\alpha}+i \vec{u})\cdot\bf{K})$ where $\bf{J}=(J_1,J_2,J_3)$ are generators of SU(2) with highest weight $j_1$ and $\bf{K}=(K_1,K_2,K_3)$ generators of SU(2) of highest weight $j_2$. The $\otimes$ is a kind of tensor product.
Example: $j_1=1/2, j_2=1/2$: $\bf{J}$ and $\bf{K}$ are the SU(2) generators of the same highest weight 1/2 yield the 4-vector representation.
Example: $j_1=1/2, j_2=0$: In this case $\exp(i (\vec{\alpha}+i \vec{u})\cdot\bf{K})=id$. The matrices $\exp(i(\vec{\alpha}-i \vec{u})\cdot\bf{J})$ are representations of the group SL(2,C). Weyl-Spinors transform according to this representation. There are "normal" spinors and "dotted" spinors either if $j_1=1/2, j_2=0$ or $j_1=0, j_2=1/2$. In this particular case $\bf{J}$ are the famous Pauli-matrices $(\sigma_1, \sigma_2, \sigma_3)$.