Eq. (5.3.20) Weinberg Volume 1, p. 209 Weinberg claims that it is obvious that the $\sigma = 0$ component of $u^\mu$ at zero spatial momentum points in the 3-direction. This is supposed to follow from (5.3.6). Unfortunately I am not seeing it.  I thought that the $J=1, m = 0$ component is conventionally aligned in the 2-direction?
Any help would be appreciated.
(5.3.20)
$u^\mu(0,0) = (2m)^{-1/2} (0,0,1,0)$
(5.3.6)
$\sum_{\bar{\sigma}} u^\mu(0,\bar{\sigma}) \mathbf{J}^{(j)}_{\bar{\sigma}\sigma} = {\mathbf{\mathcal{J}}^\mu}_\nu u^\nu (0,\sigma)$
(5.3.8)
$(\mathcal{J}_k)^0_0 = (\mathcal{J}_k)^0_i = (\mathcal{J}_k)^i_0 = 0$
(5.3.9)
$(\mathcal{J}_k)^i_j = - i \epsilon_{ijk}$
 A: By the most universal rules and conventions for the angular momentum in quantum mechanics, $J=1$, $m=0$ is always associated with the third direction i.e. with the $z$-axis. It's because $m$ (and, in this Weinberg notation, I suppose that $\sigma$ as well) conventionally denotes the eigenvalue of $J_z$, the generator of rotations around the third axis.
The matrices ${\mathbf J}^{(1)}$ and ${\mathcal J}_3$ are intrinsically the same, representing the angular momentum generators for the spin $j=1$. In the case of ${\mathcal J}$, it's by definition because this is the matrix in the "preferred" vector representation; in the case of ${\mathbf J}$, it was depicted that we talk about the same spin-one representation by the subscript $(1)$.
The last thing one needs to know is what are the eigenvalues of $J_z$ in the three-dimensional space generated by $e_x,e_y,e_z$. Well, the eigenvalues are $\sigma=-1$ and $\sigma=0$ and $\sigma=+1$ and the corresponding eigenstates are proportional to $e_x+i e_y$, $e_z$, $e_x-i e_y$. In particular, $e_z$ itself corresponds to the vanishing eigenvalue $\sigma=0$ of $J_z$. It's because if you rotate the vector in the $z$-axis around the same $z$-axis, it doesn't change at all. So the infinitesimal generator remembering the change is zero.
Alternatively, you may use your last equation implying that
$$({\mathcal J}_z)^z_j = 0$$
because the epsilon-symbol is antisymmetric. It means that the matrix elements of $J_z$ in the column corresponding to $e_z$ (and similarly in the row corresponding to $e_z$) are equal to zero.
A: 
The issue becomes obvious once you write (5.3.6) in its z-component and the matrix representation of Jz explicilty (a diagonal matrix with -1,0,+1 on the diagonal).
A: I just wanted to add that the confusion originates because Weinberg is using an inconsistent notation $D(\cdot)$ to simultaneously represent two different things: 1. the unitary representation of the Little group rotation matrices $W(\Lambda,p) \in SO(3)$ and 2. the non-unitary representation of Lorentz transformations $\Lambda \in SO(1,3)$.
This inconsistency can be resolved by using a different notation e.g. $\mathcal{D}(\cdot)$ to denote the non-unitary representation of $\Lambda$. After modifying Weinberg's notation, we have
$$\forall \Lambda \in SO(1,3)\quad : \quad D\big(W(\Lambda,p)\big)^\dagger = D\big(W(\Lambda,p)\big)^{-1} \quad \quad \quad \textrm{see Eq. (2.5.13)}$$
but at the same time we know $\exists \Lambda \in SO(1,3)$ such that
$$
\mathcal{D}(\Lambda)^\dagger \neq \mathcal{D}(\Lambda)^{-1}$$
In the special case where $\Lambda=\mathcal{R}$
we learn from p. 69 that $W(\mathcal{R},p)=\mathcal{R}$, but it is important to note that the representation matrices are different
$$ \mathcal{D}(\mathcal{R}) \neq D(\mathcal{R}) $$ which was not evident from Weinberg's notation.
