Why the Minkowski and Euclidean spinors need to be fermions? 
*

*Minkowski spinors are the spinor representations of the spin group $Spin(1,d)$ of spacetime rotational symmetry.


*Euclidean spinors are the spinor representations of the spin group $Spin(1+d)$  of spacetime rotational symmetry.
The group elements of the spin group are generated by some Clifford algebra with $\gamma_j$ obeying
$$
\{\gamma_j, \gamma_k\} = 2 g_{jk}
$$
where the $g_{jk}$ is the metric signature for the Minkowski or Euclidean signatures.
At this stage, we do not yet know whether the Minkowski and Euclidean spinors need to be fermions or bosons, or any other statistics.
But when we dicuss the Minkowski and Euclidean spinors, we regard them in QFT as Grassman variables
with fermion anticommutator relations $$
\{ \psi_a(x) ,\psi_b(y) \}=\delta_{ab}\delta(x-y),
$$ thus they are fermions --- obeying fermions statistics.

question: So, why the Lorentz and Euclidean spinors need to be fermions? Could they have some other statistics, like


*

*bosons?

*anyons in 2+1d?

 A: There are various levels of generality you can aim for when proving the spin statistics theorem but it's probably best to start with free fields coming from the Dirac action
\begin{equation}
S = \int i\bar{\psi} \gamma^\mu \partial_\mu \psi - m \bar{\psi} \psi \, dx.
\end{equation}
The canonical quantization procedure then tells us to define a conjugate momentum
\begin{equation}
\pi_\alpha = \frac{\delta L}{\delta \partial_0 \psi^\alpha} = i \psi^\dagger_\alpha.
\end{equation}
(If the spinors are Majorana there is a $\frac{1}{2}$ from Dirac brackets but this doesn't change the conclusion.) Now why do we have to impose $\left \{ \psi^\alpha(\textbf{x}), i\psi^\dagger_\beta(\textbf{y}) \right \} = i\delta^\alpha_\beta \delta(\textbf{x} - \textbf{y})$? Well let's try imposing $\left [ \psi^\alpha(\textbf{x}), i\psi^\dagger_\beta(\textbf{y}) \right ] = i\delta^\alpha_\beta \delta(\textbf{x} - \textbf{y})$ instead and seeing what this says about the creation an annihilation operators. We know the fundamental field needs to have one $a$ and one $b^\dagger$ to make the energy bounded from below.
\begin{equation}
\psi(x) = \int a(\textbf{p})^r u_r(\textbf{p}) e^{-ipx} + b^\dagger(\textbf{p})^r v_r(\textbf{p}) e^{ipx} \, \frac{d\textbf{p}}{(2\pi)^3}
\end{equation}
Along with this expression, we also need to know how to contract the plane wave solutions to the Dirac equation.
\begin{equation}
\bar{u}_r(\textbf{p}) u_s(\textbf{p}) = 2m \delta_{rs}, \;\; \bar{v}_r(\textbf{p}) v_s(\textbf{p}) = -2m \delta_{rs}
\end{equation}
The algebra then leads to
\begin{align}
[a(\textbf{p})^r, a^\dagger(\textbf{q})^s] = \delta^{rs} \delta(\textbf{p} - \textbf{q}), \;\; [b(\textbf{p})^r, b^\dagger(\textbf{q})^s] = -\delta^{rs} \delta(\textbf{p} - \textbf{q})
\end{align}
from which we see that $b^\dagger(\textbf{p})_r \left | 0 \right >$ has negative norm. This problem goes away when we use ant-commutators instead.
What this demonstrates is that the Pauli principle is obeyed by half-integer spin fields because of unitarity. As with all no-go theorems, the exceptions are interesting. In particle physics, ghosts are able to get around this because they are an artifact of fixing the gauge, not something physical. In statistical physics, many theories get around this too because they are intrinsically Euclidean. Therefore we don't mind at all if the Wick rotated versions of them turn out not to be unitary.
