$C$, $P$, $T$ symmetry of $O(3)$ non-linear sigma model Consider the $O(3)$ nonlinear sigma model with topological theta term in 1+1 D: $$\mathcal{L}=|d\textbf{n}|^{2}+\frac{i\theta}{8\pi}\textbf{n}\cdot(d\textbf{n}\times d\textbf{n}).$$
The time reversal symmetry is $$T: \textbf{n}(x,t)\to -\textbf{n}(x,-t),$$
the parity (reflection) symmetry (actually, it is $CP$ combination) is $$P^{*}: \textbf{n}(x,t)\to -\textbf{n}(-x,t),$$
and the charge conjugation symmetry is $$C: \textbf{n}(x,t)\to -\textbf{n}(x,t).$$
It is said that the model above has time reversal and parity symmetry for any $\theta$, while it is charge conjugation symmetric only when $\theta=0,\pi$. (references: https://arxiv.org/abs/1802.02153).

But it seems that
1).under time reversal symmetry $i\to -i$,then $$|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t))\to|d\textbf{n}(x,-t)|^{2}+\frac{-i\theta}{8\pi}(-)\textbf{n}(x,-t)\cdot(d\textbf{n}(x,-t)\times d\textbf{n}(x,-t)),$$
shortly,$\mathcal{L}[\textbf{n}(x,t)]\to\mathcal{L}[\textbf{n}(x,-t)]$, Is it time reversal symmetric? If yes, it is clearly that it has time reversal symmetry for any $\theta$.
2). for parity symmetry, $$|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t))\to|d\textbf{n}(-x,t)|^{2}+\frac{i\theta}{8\pi}(-)\textbf{n}(-x,t)\cdot(d\textbf{n}(-x,t)\times d\textbf{n}(-x,t)),$$ Is the model has parity symmetry? The $\theta$ term now has a minus sign. Why it is said that it has parity symmetry for any $\theta$?
3). for charge conjugation, since $i\to -i$,then $$|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t))\to|d\textbf{n}(x,t)|^{2}+\frac{-i\theta}{8\pi}(-)\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t)),$$
shortly,$\mathcal{L}[\textbf{n}(x,t)]\to\mathcal{L}[\textbf{n}(x,t)]$, it seems that the model has charge conjugation symmetry for any $\theta$, but why it is only a symmetry for $\theta=0,\pi$.
 A: If the T map changes $\theta$ to $-\theta$, and recalling that $\theta$ is an angle --- meaning that  $\theta$ is the same   as  $\theta+2\pi$ --- only $\theta=0$ and $\theta=\pi$ tranform into themselves.
The $\theta$ parameter is an angle because
$$
n=\frac 1{4\pi} \int \left\{\frac 12 \epsilon^{ijk} n^i dn^j dn^k\right\}
$$
is the integer-valued winding number (Brouwer degree) of the map ${\bf n}:S^2 \to S^2$, and the theta term is in an exponent where $e^{2\pi i n}=1$.
A: I make some progress on this question:
1).under time reversal symmetry $i\to -i$,then $$|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t))\to|d\textbf{n}(x,-t)|^{2}+\frac{-i\theta}{8\pi}(-)\textbf{n}(x,-t)\cdot(d\textbf{n}(x,-t)\times d\textbf{n}(x,-t)),$$
shortly,$\mathcal{L}[\textbf{n}(x,t)]\to\mathcal{L}[\textbf{n}(x,-t)]$.
Then we need the action $S=\int dxdt\mathcal{L}[\textbf{n}(x,-t)]=\int dxdt\mathcal{L}[\textbf{n}(x,t)]$. Since $\int_{-a}^a f(-t)dt=-\int_{a}^{-a}f(\tau)d\tau$, so if we take the integral on the imaginary time from $-\beta/2$ to $\beta/2$, then we find $\int_{-\beta/2}^{\beta/2} dt\mathcal{L}[\textbf{n}(x,-t)]=\int_{-\beta/2}^{\beta/2} dxdt\mathcal{L}[\textbf{n}(x,t)]$. So the system has time reversal symmetry for any $\theta$.
3). The charge conjugation should not act on the c-number, so there is no $i\to -I$. Then $$|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t))\to|d\textbf{n}(x,t)|^{2}+\frac{i\theta}{8\pi}(-)\textbf{n}(x,t)\cdot(d\textbf{n}(x,t)\times d\textbf{n}(x,t)).$$
Since the partition function is periodic under $\theta\to \theta+2\pi$, so the system has charge conjugation when $\theta=0,\pi$.
