The integer eigenvalues for a generator of $SU(N)$ I am studying gauge theory by V.P. Nair's QFT textbook.
He explains in p. 455 that all components of all fields which are $\mathbb{Z}_N$ invariant will have integer eigenvalues for $Y=diag(1/N,1/N,\cdots,-1+1/N)$. Y is a generator of $SU(N)$. The fields in adjoint representation satisfies the invariant condition $\phi\to\phi^{\prime}=g\phi g^{-1}=\phi, \phi=\phi^at^a, g\in \mathbb{Z}_N \subset{SU(N)}$. I computed the eigenvalues of $Y$ $(N=2,3)$ in adjoint representation as an example but they were zero and fractional numbers.
I think the matrix Y must be multiplied by some constant number to make the eigenvalues integers...any ideas?
 A: For adjoint representation $Z_N$ invariance, you demonstrated that for adjoint generator action  $[Y,\phi]=\lambda \phi$,
$$\phi= e^{2\pi iY}\phi e^{-2\pi i Y}=\phi+ (2\pi i)[Y,\phi]+\frac{1}{2!}  (2\pi i)^2[Y,[Y,\phi]]+ ... =e^{2\pi i\lambda/N}\phi ,$$
so $\lambda$ must be an integer.
So for all tensor products of this adjoint $\phi$, the over-all composite $\lambda$ is also an integer.
For N=2, $Y= \sigma_3 /2$, and you know $$[\sigma_3, \vec \varphi \cdot \vec \sigma  ]/2= \lambda \vec \varphi \cdot \vec \sigma  $$
has eigenvectors $( \sigma_1\pm  i\sigma_2)$ for eigenvalues $\pm 1$, and of course $ \sigma_3$ for eigenvalue 0, all integers.
For N=3, in Gell-Mann matrix notation, $Y= \lambda_8/\sqrt 3$.  Here, you had best use the wonderful panoply of the weights in the Cartan-Weyl basis  (your Y normalization is already that of their  hypercharge $\hat Y$), and utilize its simplest commutation relations with U- and V-spin raising and lowering operators, 4 eigenvectors; hypercharge  commutes with isospin, so isospin is out of the picture, with 3 null eigenvalues right there! Done.
$$[\hat{Y},\hat{I}_3]=0, \\
 [\hat{Y},\hat{I}_\pm]=0, \\
 [\hat{Y},\hat{U}_\pm]=\pm \hat{U_\pm}, \\
 [\hat{Y},\hat{V}_\pm]=\pm \hat{V_\pm}.
$$

*

*Indeed,  all components of all fields which are $ℤ_N$-invariant will have integer eigenvalues  under commutation (adjoint action) by .

