How can I find the third spectrum of lines in $SU(2)$ gauge theory? In the article https://arxiv.org/abs/1305.0318, they take a gauge theory based on the algebra $su(2)$ as a first example of how to determine the allowed line operators. Once different lines can be constructed depending if we chose $SU(2)$ or $SO(3)$ as our gauge group, which agree in their algebra, but have different topologies.
Following the discussion of https://www.mit.edu/~elake/index.html in his math-diary, we know that for $SU(2)$ only $\sigma_z$ is in the Cartan sub-algebra and any weight is a linear combination of integers of the eigenvalues of $\sigma_z$, that is, $\pm 1$. And from the algebra $[\sigma_z,\sigma_\pm]=\pm 2\sigma_\pm$ we can identify the basis for the root lattice as $\pm 2$. Therefore, the weight and root lattices for $SU(2)$ are
\begin{align}
\Lambda_w(SU(2))=\mathbb{Z},
\\ 
\Lambda_r(SU(2))=2\mathbb{Z}.
\end{align}
Which gives the first spectrum of lines in figure 1.
Now, here is the problem, following the same logic, I'm finding that in $SO(3)$, we only have one possibility of spectrum of lines, the one in which
\begin{align}
\Lambda_w(SO(3))=2\mathbb{Z},
\\ 
\Lambda_r(SO(3))=\mathbb{Z}.
\end{align}
That is, the second spectrum in figure 1.
My logic is going like that. For $SO(3)$ we have the eigenvalues of $J_z$ as ours weights, and since we now have integers and half-integers spins only $2j_z$ respect the Dirac quantization, implying that $\Lambda_w(SO(3))=2\mathbb{Z}$. On the other hand, the algebra gives $[J_z,J_\pm]=\pm J_\pm$, that is, the roots are $\mathbb{Z}$.
But how can I find the third spectrum ? Which is defined by
\begin{align}
(\lambda_e,\lambda_m)\in \mathbb{Z}\times\mathbb{Z},
\end{align}
such that $\lambda_e+\lambda_m\in 2\mathbb{Z}$.

 A: The third spectrum comes from the Witten effect.
The point is that what you mean by $\lambda_\mathrm{e}$ depends on the theta angle of the theory. Namely, if you normalise your theta term as
$$S_\theta = \mathrm{i}\,\theta\ \underset{\mathrm{SO}(3)\ \text{instanton number}}{\underbrace{\frac{1}{2}\int_M\mathrm{tr}\left(\frac{F}{2\pi}\wedge \frac{F}{2\pi}\right)}},$$
then, if $M$ is a spin manifold (by the context of the linked paper you are looking at $\mathbb{R}^4$, which is obviously spin), the action is invariant under shifts $\theta\mapsto\theta+2\pi$,  but the spectrum of line operators is only invariant under $\theta\mapsto\theta+4\pi$. This means that, shifting $\theta$ by $2\pi$ is a half-period shift, so the electric charge of the shifted theory becomes
$$\lambda_\text{e}^{\theta+2\pi} = \lambda_\text{e}^{\theta}+\lambda_\text{m}^{\theta}, \qquad
\text{(but still}\  \lambda_\text{m}^{\theta+2\pi}=\lambda_\text{m}^\theta\text{)}$$
which is precisely the Witten effect.
With this in mind, you still have
$$\left(\lambda_\text{e}^{\theta+2\pi},\lambda_\text{m}^{\theta+2\pi}\right)\in\Lambda_\text{w}(\mathrm{SO}(3))\times\Lambda_\text{r}(\mathrm{SO}(3))=2\mathbb{Z}\times\mathbb{Z}.$$
It is only if you wish to express $\lambda_\text{e}^{\theta+2\pi}$ in terms of the unshifted charges, that you'll see the third spectrum
$$\left(\lambda_\text{e}^\theta,\lambda_\text{m}^\theta\right)\in\mathbb{Z}\times\mathbb{Z}, \quad \text{with}\quad \lambda_\text{e}^\theta+\lambda_\text{m}^\theta\in 2\mathbb{Z}.$$
Of course, if you shift again by $2\pi$, $\lambda_\text{e}^{\theta+4\pi} = \lambda_\text{e}^{(\theta+2\pi)+2\pi} = \lambda_\text{e}^\theta+2\lambda_\text{m}^\theta =\lambda_\text{e}^\theta \ \text{mod}\ 2\mathbb{Z},$ as it ought, so you do not get a fourth, unexpected spectrum.
