Relation of Wigner $d$-matrix $d^l_{m',m} = d^l_{-m,-m'}$ How do you derive the symmetry relation of the Wigner $d$-matrix, i.e.,
$$
d^l_{m',m} = d^l_{-m,-m'}
$$
I know how Wikipedia proves this using the fact that $(Y_l^m)^* = (-1)^m Y_l^{-m}$ (basically using the property of time-reversal). However, if we think of the Wigner $d$-matrix as the matrix corresponding to the irreducible representation of $SO(3)$ with dimension $2l+1$, then the fact should be true without using time-reversal. 
More specifically, let $\Pi:SO(3) \rightarrow GL(V)$ denote the irrep of $SO(3)$ on $(2l+1)$-dim vector space $V$. Let $|l,m\rangle$ denote the usual orthonormal basis of $V$. Then $d_{m',m}^l=\langle l,m'|\Pi(R_2(\beta))|l,m\rangle$ where $R_2(\beta)$ denotes the rotation about $y$-axis of angle $\beta$. In such a general representation (not necessarilly coordinate so that $|l,m\rangle =Y_l^m$), the conjugate $|l,m\rangle^*$ is not well-defined. In that case, how would we prove the symmetry relation?
 A: I think I found a nice little proof for this question.
My proof consists of 2 parts. One is from this question in which I showed that for integer $l$, we have
$$
\exp {(-i\pi J_2)} |l,m\rangle = (-1)^{l-m} |l,-m\rangle
$$
The 2nd part is that we need to realize that $\langle l,m'|\exp{(-i\beta L_2)}|l,m\rangle$ is real. This follows from the fact that $2 i L_2 = L^+- L^-$ where $L^+,L^-$ are the ladder operators.
Then
\begin{align}
d_{-m,-m'}^l(\beta) &=\langle l,-m|\exp {(-i\beta L_2)}|l,-m'\rangle\\
&= (-1)^{m'-m} \langle l,m|\exp(i\pi L_2) \exp (-i\beta L_2) \exp(-i\pi L_2) |l,m'\rangle \\
&= (-1)^{m'-m}\langle l,m|\exp (-i\beta L_2)  |l,m'\rangle\\
&= (-1)^{m'-m}\langle l,m'|\exp (i\beta L_2)  |l,m\rangle\\
&= (-1)^{m'-m} d^l_{m',m}(-\beta)
\end{align}
Also notice that $\exp (-i\pi L_3) L_2 \exp (i\pi L_3) =-L_2$. Therefore,
\begin{align}
(-1)^{m'-m} d^l_{m',m}(-\beta) &= (-1)^{m'-m}\langle l,m'|\exp (i\beta L_2)  |l,m\rangle\\
&=\langle l,m'|\exp(-i\pi L_3) \exp(i\beta L_2)\exp(i\pi L_3) |l,m\rangle\\
&= d^l_{m',m} (\beta)
\end{align}
