Relation of Wigner $d$-matrix $d^l_{m',m} = d^l_{-m,-m'}$ I asked this in physics stackexchange, but I migrated over here.
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?
EDIT. Thanks for the answers. I forgot to mention over here that I had already given an answer in physics stackexchange.
 A: The operator for rotation by $\theta$ about the $y$-axis is $e^{-i\theta l_y}$:
$$ d^l_{m',m}(\theta) = \langle l,m'|e^{-i\theta l_y}|l,m\rangle.$$ Write the exponential operator as a Taylor series: $$e^{-i\theta l_y} = \sum_n^\infty {(-i\theta)^n\over n!} (l_y)^n. $$ Now, using $l_y = {1\over2i}(l_+-l_-)$ it is easy to show that $\langle l,m'|l_y|l,m\rangle=\langle l,-m|l_y|l,-m'\rangle$. This identity is in fact true for all $(l_y)^n$, where $n=0,1,2,3,\ldots$. For example, for the square of $l_y$ we have
$$\langle l,m'|(l_y)^2|l,m\rangle=\sum_{m''}\langle l,m'|l_y|l,m''\rangle\langle l,m''|l_y|l,m\rangle=\sum_{m''}\langle l,-m''|l_y|l,-m'\rangle\langle l,-m|l_y|l,-m''\rangle=\sum_{m''}\langle l,-m|l_y|l,-m''\rangle\langle l,-m''|l_y|l,-m'\rangle=\langle l,-m|(l_y)^2|l,-m'\rangle.$$ Repeating this derivation for $(l_y)^3=(l_y)^2l_y$, for $(l_y)^4=(l_y)^3l_y$, and so on, shows this property to be true for all $(l_y)^n$, and therefore, by the Taylor series, also for $e^{-i\theta l_y}$. So we have
$$\langle l,m'|e^{-i\theta l_y}|l,m\rangle = \langle l,-m|e^{-i\theta l_y}|l,-m'\rangle,$$ or $d^l_{m',m}(\theta)=d^l_{-m,-m'}(\theta)$.
Reference: Properties of the Wigner $d$-matrices by Howard E. Haber.
