# What physical meaning do the dimension of Wigner d-matrices have?

Wigner's D-matrices is defined as $D_{m'm}^j(\phi,\theta,\psi)=\langle jm'|R(\phi,\theta,\psi)|jm\rangle$; it produces a square matrix (indices $m$ and $m'$) of dimension $2j+1$. It is also the case that these matrices (for all positive $j$ multiple of $1/2$) are a representation of the rotation group $SU(2)$, which is a double cover of $SO(3)$.

Now, I can see this for the specific case $j=1$ (3x3 matrices representing rotations in 3-dimensional Euclidean space), but what does it mean for other values of $j$? Say for $j=3/2$, does that represent the subgroup of 3-dimensional rotations given some parameter, in a 4-dimensional ambient geometry?

Thanks.

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Wigner matrices are always unitary, so in fact they are rotations in a complex space, they preserve length.Therefore they are indeed a(n) (irreducible) representation of the rotation group SU(2). In particular the spaces with dimension $2j+1$ where $j$ is a positive integer are tensor representations, one can think of them as made up of the Euclidean 3 dimensional space by tensoring up. For instance if you have two 3-vectors $a$ , $b$ you may form the antisymmetric tensor $T_{ij}=(a_ib_j -a_j b_i)$ and this will transform again as a 3 vector (it only has 3 independent components $T_{12}, T_{23},T_{13}$ which you can conceive as the components of a new vector $c_i$). The subtle point is that when $j$ is half-integer (or should we say half-odd?) the Wigner matrices do NOT represent SO(3), but rather SU(2) (in physics terminology one says that they do represent SO(3) but are double valued representations). One can not think of the half integer representations as made up of vectors by using tensor products, one must introduce spinors whithin the tensor product to achieve this kind of representations, and so actually spinors are the building blocks of all representations of SU(2), which include those of SO(3) (the tensor ones)