Relation between $(0,1/2)$ and $(1/2,0)$ representations of $SL(2,\mathbb{C})$ I have trouble understand how the representations $(1/2,0)$ and $(0,1/2)$ of $SL(2,\mathbb{C})$ are related. Let's consider the physicist notation where the lie group elements of $SL(2,\mathbb{C})$ are written as $G = \exp(i(\theta^i J_i+\beta^iK_i)$ where $J_i$ and $K_i$ are the generators of rotations and boosts respectively.
Now, in the decomposition of $so(1,3)_{\mathbb{C}} = su(2)\oplus su(2)$, I consider the first $su(2)$ to be associated with the generators $J_i^{-} = \frac{J_i-iK_i}{2}$ and the second one with $J_i^{+} = \frac{J_i+iK_i}{2}$.
Now, consider the $(1/2,0)$ rep (call it $\rho_-$). Then, the representations of the generators can be expressed using the Pauli matrices, yielding $\rho_-(J_i) = \frac{1}{2}\sigma_i$, $\rho_-(K_i) = \frac{i}{2}\sigma_i$.
Conversely, in the $(0,1/2)$ rep (call it $\rho_+)$, we get $\rho_+(J_i) = \frac{1}{2}\sigma_i$m $\rho_+(K_i) =-\frac{i}{2}\sigma_i$.
Now, here is my problem : in my course I wrote "We see that the two generators are related by the conjuguate transpose operation, hence they are conjuguate representations of one another."
Now, I tried to verify that, but I don't get what I expect. Indeed, writing out, $G\in SL(2,\mathbb{C})$ :
$$\rho_-(G) = \exp(\frac{1}{2}(i\theta_i-\beta_i)\sigma_i) = \left[\exp(\frac{1}{2}(-i\theta_i+\beta_i)\sigma_i)\right]^{-1} = \left[\left[\exp(\frac{1}{2}(i\theta_i+\beta_i)\sigma_i)\right]^{\dagger}\right]^{-1} = \left[\left[\rho_+(G)\right]^{\dagger}\right]^{-1}$$
Which shows that they are clearly not the complex conjuguate of each other, but something much more contorted... The only possibility that comes to mind is if $\rho_+^{-1} = \rho_+^T$, but that does not seem to hold.
So either I wrote something wrong in my notes, or I am doing something wrong here, but I really cannot spot the mistake !
 A: Actually, in order to show that a representation is complex-conjugated to another one, you only have to show the equivalence:
Two representations $R$ and $R'$ are considered as the same if $R' =S^{-1}RS$ with $S$ some invertible matrix $S$.
Actually I change a bit the notation: the 3 components $\theta_i$ form the vector $\vec{\theta}$. Furthermore I replace the 3 components $\beta_i$ by the rapidity vector $\vec{u}$ (Only this one is an additive parameter in SR). We have:
$$
D^{(1/2,\,0)} = \exp\left[-\frac{1}{2}( \vec{u} +i\vec{\theta})\cdot\vec{\sigma}\right]=\exp\left[-\frac{i}{2}(\vec{\theta}-i\vec{u})\cdot\vec{\sigma}\right]
$$
$$
D^{(0,\,1/2)} = \exp\left[+\frac{1}{2}( \vec{u} -i\vec{\theta})\cdot\vec{\sigma}\right]= \exp\left[-\frac{i}{2}(\vec{\theta}+i\vec{u})\cdot\vec{\sigma}\right]
$$
Now we are going to use the  following properties of the  Pauli-matrices ($\sigma_2$ is the second Pauli-matrix):
$$
(\sigma_2) \vec{\sigma}(\sigma_2)^{-1}=-(\vec{\sigma})^{\ast}=-(\vec{\sigma})^T
$$
from which we get 
$$
(\sigma_2)^{-1}\vec{\sigma}(\sigma_2)= -(\vec{\sigma})^{\ast}.
$$
We now apply this to the representations:
$$
(\sigma_2) \left(\exp\left[-\frac{i}{2}(\vec{\theta}+i\vec{u})\cdot\vec{\sigma}\right]\right)(\sigma_2)^{-1}=\exp\left[\frac{i}{2}(\vec{\theta}+i\vec{u})\cdot(\vec{\sigma})^{\ast}\right]=
$$
$$
=\left(\exp\left[-\frac{i}{2}(\vec{\theta}-i\vec{u})\cdot(\vec{\sigma})\right]\right)^{\ast}.
$$ 
(Note that a change in the initial sign in the exponents does not change the result) 
We conclude that $D^{(0,\,1/2)}$ is equivalent to $(D^{(1/2,\,0)})^{\ast}$ which shows the required property. 
