Finding the minimum degree of a representation of an algebra I'm reading this paper, and in it the authors make the following claim (Eq. 27 in the paper):
They first show that a certain subspace of $\mathbb{C}^N$ is invariant under a set of unitary operators $\{Z_i, Y_i\}$, $i=1,...,\alpha$, satisfying


*

*$[Z_i,Z_j]=0$

*$[Y_i,Y_j]=0$

*$[Y_i,Z_j]=0\quad$ provided $i\neq j$

*$Y_iZ_i=e^{2\pi i p/q}Z_iY_i\qquad$  ($p/q$ is a fraction in lowest form)


They go on to say the algebra generated by the $\{Z_i,Y_i\}$ can be represented in this subspace; in other words, by restricting the $Z_i,Y_i$ to act only on this subspace, we can get a lower-dimensional representation of the algebra. That makes sense. They then say that this representation must have degree at least $q^{\alpha}$, where $q$ is the denominator in the fourth bullet point. I don't understand how they reach this conclusion.
How can we conclude from the relations between the generators that we cannot have a smaller complex representation than degree $q^\alpha$? Is there a method in general to find the minimal degree of a representation from some presentation of an algebra?
 A: Just listing some observations. Not entirely happy with this yet and not a full solution - I haven't studied the paper, and I need to leave the construction of the representation for later. At this point I will show that the dimension of the representation must be a multiple of $q^\alpha$. If we can then exhibit a representation of minimum dimension $q^\alpha$ we are done.
The point is that corollaries of the given relations force the dimensions of the representations to go up in a manner that is somewhat analogous to the way the commutator relations involving the ladder operators of a simple Lie algebra.
Let's look at a non-trivial representation of this algebra in the case $\alpha=1$.
Assume that $Z_1$ has a non-zero eigenvalue $\lambda$. Let $v$ be an eigenvector
belonging to this eigenvalue. Applying the fourth relation we get 
$$
\lambda Y_1 v= Y_1Z_1v=\zeta Z_1Y_1v, 
$$
where $\zeta=e^{2\pi i p/q}$ is the prescribed root of unity. Rewriting this in the form
$$Z_1(Y_1v)=\zeta^{-1}\lambda (Y_1v)$$
shows that $Y_1v$ is an eigenvector of $Z_1$ belonging to the eigenvalue $\zeta^{-1}\lambda$.
Rinse. Repeat. We see that for all integers $\ell$ the vector $Y_1^\ell v$ is an eigenvector of $Z_1$ belonging to the eigenvalue $\zeta^{-\ell}\lambda$. As $\zeta$ has order $q$ this forces $Z_1$ to have (at least) $q$ distinct eigenvalues. Recall that eigenvectors belonging to different eigenvalues are necessarily linearly independent.
Furthermore (denoting by $V_{\lambda,1}$ the eigenspace of $Z_1$ belonging to eigenvalue $\lambda$):


*

*By the above argument the action of $Y_1$ maps $V_{\lambda,1}$ injectively into $V_{\zeta^{-1}\lambda,1}$. Injectivity follows from the fact that $Y_1$ is unitary.

*After running through the full circle of eigenvalues we see that we must actually have (assuming a finite dimensional representation) 
$$\dim V_{\zeta^\ell\lambda,1}=\dim V_{\lambda,1}.$$

*In particular the dimension of the entire representation is a multiple of $q$ (need to check that $q$ is possible).



Let's add more generators. When $\alpha=2$ the commutator relations imply that both $Z_2$ and $Y_2$ keep the eigenspaces of $Z_1$ stable. In other words, the eigenspaces $V_{\lambda,1}$ carry a representation of the algebra generated by $Z_2$ and $Y_2$. Repeating the above argument once more shows that the dimensions of all the spaces 
$V_{\zeta^\ell\lambda,1}$ are multiples of $q$. Consequently, in view of the last bullet, the dimension of a representation of the algebra generated by $\{Z_1,Z_2,Y_1,Y_2\}$ is an integer multiple of $q^2$. That integer multiplier being the dimension of a common eigenspace $V_{\lambda,1,\mu,2}$ of $Z_1$ and $Z_2$.
It is hopefully clear that when we add $Z_3$ and $Y_3$, those subspaces $V_{\lambda,1,\mu,2}$ become subrepresentations, and we can keep going.
