second-class constraints of the original 2$D$ chiral Schwinger model into the first-class system so that we could have the “classical" gauge and “quantum" (anti-)BRST symmetries in the theory (see, e.g. \[19\] for details). The other symbols $(\bar C)C$ and $B(x)$ have already been explained earlier. The Lagrangian density (5) can be re-expressed as $$\begin{aligned}
\mathcal{L}_{B}^{(a)}\,&=&\,{\cal B} E\,-\,\frac{{\cal B}^2}{2}
+\,\frac{1}{2}\,\partial_{\mu}\phi\,\partial^{\mu}\phi\,
+\,\frac{a}{2}\,A_{\mu}\,A^{\mu}\\ \nonumber
&+&\,(\eta^{\mu\nu}\,-\,\varepsilon^{\mu\nu})\,\partial_{\mu}\phi\,A_{\nu}\,+\,
\sigma \,\Bigl[(a-1)\,(\partial \cdot A)\,+\,\varepsilon^{\mu\nu}\,\partial_{\mu}\,A_{\nu}\Bigr] \\ \nonumber
&+&\frac{(a-1)}{2}\,\partial_{\mu}\,\sigma\,\partial^{\mu}\,\sigma\,+\,B\,(\partial \cdot A)\,+\,\frac{B^2}{2}
\,+\,i\,\partial_{\mu}\,\bar C\,\partial^{\mu}\,C,\end{aligned}$$ which is endowed with the following (anti-)co-BRST symmetries $$\begin{aligned}
&&s_{ad}\,A_{\mu}\,=\,-\varepsilon_{\mu\nu}\,\partial^{\nu}\,C, \quad s_{ad}C\,=\,0, 
\quad s_{ad}\bar C\,=\,-i{\cal B}, \quad s_{ad}{\cal B }\,=\,0,\quad s_{ad}\phi\,=\,- C, \\ \nonumber
&&s_{ad}E\,=\,\square C,\quad s_{ad}(\partial \cdot A)\,=\,0, s_{ad}B\,=\,0, \quad
s_{ad}\sigma\,=\,-\frac{C}{a-1} \cong C \; (1 + a),\\ \nonumber
&&s_{d}A_{\mu}\,=\,-\varepsilon_{\mu\nu}\partial^{\nu} {\bar C}, \quad s_{d}\bar C\,=\,0,
\quad s_{d}C\,=\,i{\cal B}, \quad s_{d}{\cal B}\,=\,0, \quad s_{d}\phi\,=\,-\bar C, \\ \nonumber
&& s_{d}E\,=\,\square \bar C, \quad s_{d}(\partial \cdot A)\,=\,0,\quad s_{d}B\,=\,0, \quad
s_{d}\sigma\,=\,-\frac{\bar C}{(a-1)} \, \cong \bar C \; (1 + a),\end{aligned}$$ where we have introduced an auxiliary field ${\cal B}(x)$ to linearize the kinetic term ($-\,\frac{1}{4}\,F_{\mu\nu}F^{\mu\nu} = \frac{1}{2}\,E^{2} \equiv {\cal B}\,E
- \frac{1}{2}\,{\cal B}^2$) of our modified 2$D$ anomalous Abelian 1-form gauge theory. The symmetry invariance can be explicitly checked, by using the above transformations, where the action integral $S=\int d^{2}x\,{\cal L}_{B}^{(a)}$ 