Given the commutation relation of rotations $$[X_i, X_j] = \sum_{k=1}^3 \epsilon_{ijk} X_k$$ where $\epsilon_{ijk}$ is the Levi-Civita symbol in dimension 3 and $X_1,X_2,X_3$ are square matrices.
How can we prove that for all triplet of irreducible solution $(X_1,X_2,X_3)$, the matrices have each only one zero eigenvalue?
By irreducible I mean that there is no non-trivial projector $P$ such that $P X_1 P^t, P X_2 P^t, P X_3 P^t$ are solution. (For instance the zero matrices of any dimension are solutions but only the $1\times1$ matrices are irreducible.)
I am also curious to see how to prove other things like for instance that the solutions are dimension $2l+1$ for $l$ integer.
Progress:
- I observe numerically that the eigenvalues are $-li,\dots,-i,0,i,\dots, li$.
- I will go to the library tomorrow to check in Representations of the Rotation and Lorentz Groups and Their Applications
Does anyone has a book in which this is explain in a way a physicist can understand?
Found the answer in the book The Rotation and Lorentz Groups and their representations for physicists at page 157.
Here is a resume of the proof:
Define $p=i X_3$, $\alpha = i(X_1 + iX_2)$ and $\beta = i(X_1 - iX_2)$. One can show that $\alpha$ and $\beta$ are the raising and lowering operators of the eigenvectors of $p$. Basically if $p v = \lambda v$ then $p \alpha v = (\lambda + 1) \alpha v$ and $p \beta v = (\lambda - 1) \beta v$.
Then one can easily show that the irreps must be dimension $2j+1$ with $j$ half-integer.