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 !

  • $\begingroup$ $G$ is a group element; $J_i$ and $K_i$ are the generators, which are indeed daggered when going from $\rho_-$ to $\rho_+$. $\endgroup$ – Demosthene Dec 9 '17 at 2:23
  • $\begingroup$ I see what you mean, but I find that in general, if we take the dagger of a Lie Algebra representation, we do not necessarily obtain a representation since the daggered representation (if $\rho_\dagger(g) = \rho(g)^\dagger$ with $\rho$ a representation) would have $\rho_\dagger([x,y]) = -[\rho_\dagger(x),\rho_\dagger(y)]$. Thus I don't see how your comment can apply $\endgroup$ – Frotaur Dec 9 '17 at 2:43
  • $\begingroup$ The physicists notation, mathematically speaking, is taking a basis in the lie algebra and then exponentiating to get an element of the associated group (more precisely, in the connected component of the identity). $\endgroup$ – Mozibur Ullah Dec 9 '17 at 10:03
  • $\begingroup$ We can certainly talk about representations of the Lie algebra alone. In that case, you do have to take the complex conjugate of one to get the other. $\endgroup$ – Amara Dec 11 '17 at 4:22

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.


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