Let $P$ be the parity operator of the Lorentz group,
$$P=\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}$$ the commutation relations of $so(3,1)$ be :
$$[M_i,M_j] = \epsilon_{ijk}M_k$$
$$[M_i,N_j] = \epsilon_{ijk}N_k$$
$$ [N_i,N_j] = -\epsilon_{ijk}M_k$$
and define : $L_i=1/2(M_i+iN_i)$ and $\overline{L_i}=1/2(M_i – iN_i)$.
From these, we have : $[P,M_i]=0$, $PN_i=-N_iP$.
Hence : $PL_i=\overline{L_i}P$
We have the more general representation of $so(3,1)$ being labelled by two indices $(j,j')$ because of the following isomorphism : $so(3,1)=so(3)\oplus \overline{so(3)}$. We can by exponentiation get the representations of $L_{+}^{\uparrow}$ (the proper Lorentz transformation preserving direction of time). I try to get from these representations those of $L^{\uparrow}$ (generated by just adding $P$ in $L_{+}^{\uparrow}$). Assuming we have an irreducible representation of $L^{\uparrow}$, it is also a representation of $L_{+}^{\uparrow}$ and we can generally write it as $T=\oplus_{j,j'} (j,j')$.
Until here I think it is clear. But then the textbook says that if we take $v$ a vector in a invariant subspace associated to the representation $(j,j')$ then $T(P)v$ transforms with $(j',j)$.
I imagine that we can start with :
$$T(L_i)T(P)v=T(P)T(\overline{L_i})v.$$
But I don't understand how from it we can show that it is a transformation with $(j',j)$.
Edit :
I think we can also see it this way (not as complete as the answer of @Stephen Blake, but it gives the idea) :
$$T_{(j,j')}(L_i)T_{(j,j')}(P)v=T_{(j,j')}(P)T_{(j,j')}(\overline{L_i})v$$
and because of the equivalence between the complex conjugate of these representations. We have : $$\overline{(j,j')}\sim (j',j)$$ where it is the complex conjugate representation of $M_i$ and $N_i$ which was taken (we have a real vector space for $so(3,1)$) and hence : $$\overline{T}_{(j,j')}(L_i)= T_{(j,j')}(\overline{L_i})\ast$$ (where $*$ denotes the complex conjugate matrix).
Eventually :
$$T_{(j,j')}(P)\left(\overline{T}_{(j,j')}(L_i)*\right)v = T_{(j,j')}(P)\left(T_{(j',j)}(L_i)*\right)v$$
Here, we see that it is $v$ that transforms by $(j',j)$, not $T(P)v$ but it seems that Stephen Blake comes to the same result. Can I make it better in this way ?
Best Answer
Parity flips the sign of a Lorentz boost $\eta$, $$ T(P)T(\eta)T(P^{-1})=T(-\eta) $$ and commutes with a spatial rotation R, $$ T(P)T(R)T(P^{-1})=T(R) \ . $$ Consider a Lorentz spinor $\psi^{A}\in V_{2}$ with indices A=1,2 and try to set up a 2x2 matrix $T(P)$ for parity. A Lorentz boost along the z-axis is, $$ T(\eta)=\begin{pmatrix}\ e^{-\eta/2}&0\\0&\ e^{\eta/2}\end{pmatrix} $$ and a rotation about z by angle $\phi$ is the unitary matrix, $$ T(R)=\begin{pmatrix}\ e^{-i\phi/2}&0\\0&\ e^{i\phi/2}\end{pmatrix} $$ It's easy to see that the conditions imposed upon $T(P)$ by the boost and rotation are impossible to satisfy so the matrix $[T(P)]^{A}_{B}$ does not exist. However, there are also dotted Lorentz spinors $\phi^{\dot{A}}\in \tilde{V}^{*}_{2}$ and so it's natural to try the parity matrix as $[T(P)]^{\dot{A}}_{B}$. A dotted spinor transforms as, $$ \phi'^{\dot{A}}=[T(g^{-\dagger})]^{\dot{A}}_{\dot{B}}\phi^{\dot{B}} $$ and taking parity as the diagonal matrix, $$ [T(P)]^{\dot{A}}_{B}=i\delta^{\dot{A}}_{B} $$ and, $$ [T(P)]^{A}_{\dot{B}}=i\delta^{A}_{\dot{B}} $$ so that $P^{2}=-1$ for Lorentz spinors makes the boosts and rotations work nicely.
A general (m,n) spinor is a symmetric tensor, $$ F^{A_{1}\ldots A_{m}\dot{B}_{1}\ldots \dot{B}_{n}} $$ and if this is transformed under parity by tensoring $[T(P)]^{A}_{\dot{B}}$ and $[T(P)]^{\dot{A}}_{B}$ the (m,n) spinor is changed into a (n,m) spinor.