How to show the invariant nature of some value by the group theory representations? Let's have Dirac spinor $\Psi (x)$. It transforms as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$ representation of the Lorentz group: 
$$
\Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot {a}}\end{pmatrix}, \quad \Psi {'} = \hat {S}\Psi .
$$ 
Let's have spinor $\bar {\Psi} (x)$, which transforms also as $\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right)$, but as cospinor: 
$$
\bar {\Psi} = \begin{pmatrix} \kappa^{a} & \psi_{\dot {a}}\end{pmatrix}, \quad \bar {\Psi}{'} = \bar {\Psi} \hat {S}^{-1}.
$$ 
How to show formally that 
$$
\bar {\Psi}\Psi = inv?
$$
I mean that if $\Psi \bar {\Psi}$ refers to the direct product (correct it please, if I have done the mistake) 
$$
\left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right]\otimes \left[\left( \frac{1}{2}, 0 \right) \oplus \left( 0, \frac{1}{2} \right) \right],
$$
what group operation corresponds to $\bar {\Psi} \Psi$?
This question is strongly connected with this one.
 A: You need to work out the tensor product and will find a direct sum of different contributions
\begin{multline} [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] =\\ \big((1/2, 0) \otimes (1/2, 0)\big) \oplus \big((1/2, 0) \otimes (0, 1/2) \big)\oplus \quad \\\big((0, 1/2) \otimes (1/2, 0)\big) \oplus \big((0, 1/2) \otimes (0, 1/2)\big) =
\\
(0, 0) \oplus (1, 0) \oplus (1/2, 1/2) \oplus (1/2, 1/2) \oplus (0, 1) \oplus (0, 0)\end{multline}
The states now can be classified:


*

*$(0, 0)$ is a scalar or pseudoscalar, i.e. the $\bar \psi \psi$ you are looking for as well as $\bar \psi \gamma_5 \psi$

*$(1/2, 1/2)$ is the vector / pseudovector component $\bar \psi \gamma^\mu \psi$ or $\bar \psi \gamma^\mu \gamma_5 \psi$

*(1, 0) and (0, 1) are the (anti)-self dual parts of the tensor $\bar \psi \sigma^{\mu \nu } \psi$


All these transform well-definedly under Lorenty boosts. The $(0, 0)$ part tells you that this rep will transform neither under the left-chirality nor the right-chirality $sl(2)$ that you classify the reps by.
Edit: Let me add that the distribution law I used above to get from the first to the second line is one of reasons we speak of a "direct sum" vs. "direct product".
A: If we assume that 
$$\Psi {'} = \hat {S}\Psi$$
and 
$${\bar{\Psi}}{'} = \bar {\Psi} \hat {S}^{-1},$$
it follows that the product of the two transforms as 
$$(\bar{\Psi}\Psi)'={\bar{\Psi}}{'}\Psi {'}=\bar {\Psi} \hat {S}^{-1}\hat {S}\Psi=\bar{\Psi}\Psi,$$
which is a consequence of 
$$\hat {S}^{-1}\hat {S}=\mathbb{1}.$$
A: short answer if $ \hat {S}^{-1} S = \mathbb{I}$
I can give you a general example of  $\psi^\dagger\psi$ not being invariant.
because for  Dirac spinor $\psi$  whe have the following transformation rules
 $$\psi(x) \rightarrow S[\Lambda] \psi(\Lambda^{-1}x)=S[\Lambda] \psi(x^\prime) \\
\psi^\dagger(x) \rightarrow \psi^\dagger(\Lambda^{-1}x) S[\Lambda]^\dagger $$
So $\psi^\dagger\psi \rightarrow \psi^\dagger(\Lambda^{-1} x)S[\Lambda]^\dagger S[\Lambda] \psi(\Lambda^{-1} x) $ is invarieant if and only if $S[\Lambda]^\dagger S[\Lambda] = \mathbb{I}$
however for the case where $S[\Lambda]$ are formed by the Clifford algebra it can be shown this is not they case. I do not have the capability to show you that they dirac adjoint does satisfy this condition.   
