\section{Introduction}
One of the ways to search for New Physics is
precision measurement of the electric dipole moment $d_l$ of a charged lepton ($l=e,\mu,\tau$). The  value of  $d_l$ predicted by the Standard Model (SM) is too small for experimental measurement. Therefore, the observation of electric dipole moment or its manifestation would directly demonstrate the existence of New Physics.

The manifestation of the lepton electric dipole moment can be sought in the process of $l\bar l$ pair production in $e^+e^-$ annihilation. The general form of  $\gamma l \bar l$ vertex can be represented as
\begin{align}\label{genvert}
	&\Gamma^\mu=-ie\Bigg\{ F_1^l(k^2)\gamma^\mu +\dfrac{\sigma^{\mu\nu}k_\nu}{2m_l} \Big [iF_2^l(k^2) +F_3^l(k^2)\gamma_5 \Big ] +\Big( \gamma^\mu - \dfrac{2k^\mu \, m_l}{k^2}\Big)\gamma_5 F_4^l(k^2) \Bigg\}\,,
\end{align}
where $m_l$ is the lepton mass, $e<0$ is the electron charge, $k$ is the 4-momentum of the photon, $F_1^l(k^2)$ is the Dirac form factor, $F_2^l(k ^2)$ is the Pauli form factor, $F_3^l(k^2)$ is the electric dipole form factor, $F_4^l(k^2)$ is the anapole form factor, $\sigma^{\mu\nu} =i/2[\gamma^\mu,\gamma^\nu]$, $\gamma_5=-i\gamma^0\gamma^1\gamma^2\gamma^3$. In the limit $k^2 \to 0$ these form factors are
\begin{align*}
	& F_1^l(0)=1\, ,\quad F_2^l(0)=\mu'_l\,\dfrac{2m_l}{e}\, ,\quad F_3^l(0)=d_l\,\dfrac{2m_l}{e}\,,\quad
	F_4^l(0)=0\,,
\end{align*}
where $\mu'_l$ is the anomalous magnetic moment. It follows from \eqref{genvert} that a violation of $\mbox{P}$ and $\mbox{T}$ parities leads to the appearance of $F_3^l(k^2)$, while $F_4^l(k^2)$ is related to  violation of $\mbox{P}$-parity alone. Assuming that the $\mbox{CPT}$ theorem holds, violation of $\mbox{T}$-parity is equivalent to violation of $\mbox{CP}$-parity. Thus, $d_l$ occurs due to $\mbox{CP}$ violations.

For all charged leptons, predictions of  $\mu'_l$  in  SM \cite{Aoyama:2019ryr,Volkov:2019phy,Aoyama:2020ynm,Eidelman:2007sb}  can be experimentally verified \cite{Fan:2022eto,Muong-2:2021ojo,DELPHI:2003nah}. For $d_l$ the situation is essentially different. An estimate of $d_l$ in  SM  gives $d_l=(m_l/\mbox{MeV})\times 1.6\times 10^{-40}\,e\,\mbox{cm} $ \cite{Booth:1993af,Mahanta:1996er}. Hence $F_3^e(0) < F_3^\mu(0) < F_3^\tau(0)\approx 5 \times10^{-23}\ll 1$. The sensitivity of modern experiments does not allow one to  measure  $F_3^\tau(0)$ with an accuracy of $10^{-23}$. Therefore, extracting a non-zero value of $F_3^\tau(0)$ from the experiment would be a discovery of New Physics.  In Refs.~ \cite{Barr:1988mc,Grifols:1990ha,Escribano:1993pq,Escribano:1996wp,Taylor:1997tn,OPAL:1998dsa,L3:1998gov,DELPHI:2003nah,L3:2004ful,Blinov:2008mu,Grozin:2008nw} upper limits were set to $|F_3^\tau(k^2)|$, and in Refs.~\cite{ARGUS:2000riz,Belle:2002nla,Belle:2021ybo} upper limits were set to $\mbox{Re} \, F_3^\tau(k^2)$ and $\mbox{Im}\, F_3^\tau(k^2)$ separately.

In our work,  the processes $e^+e^-\rightarrow \tau^+\pi^-\nu_\tau$, 
 $e^+e^-\rightarrow \pi^+\tau^-\bar\nu_\tau$, $e^+e^-\rightarrow \tau^+\rho^-\nu_\tau$,  and $e^+e^-\rightarrow \rho^+\tau^-\bar\nu_\tau$ 
are studied for longitudinally polarized electron beams, and  the   process  $e^+e^-\rightarrow \pi^+\pi^-\nu_\tau\bar\nu_\tau$ is discussed for unpolarized electrons. The asymmetric with respect to $\mbox{CP} $ transformation parts of the corresponding cross sections  are obtained for $e^+e^-$ invariant masses  $\sqrt{s}\ll m_Z$.  To derive these results, it is sufficiently to consider   $\gamma\tau^+\tau^-$ vertex  in the form
\begin{align}\label{tauvert}
	&\Gamma^\mu=-ie\Bigg [ \gamma^\mu +\dfrac{\sigma^{\mu\nu}k_\nu}{2M} F_3^\tau(k^2)\gamma_5\Bigg ]\,,
\end{align}
where $M$ is the $\tau$ lepton mass and $k^2=s$.

Measurement of $\mbox{CP}$ odd parts of the cross sections  can diminish the upper limits of both $\mbox{Re}\,F_3^\tau(s)$ and $\mbox{Im}\,F_3^\tau(s)$. Moreover,  polarization of beams simplifies the measurement of $\mbox{Re}\,F_3^\tau(s)$. At present, there are no corresponding experiments with    longitudinally polarized electrons. However, such  experiments are planned to be carried out at the Super-Charm-Tau factory (SCTF) \cite{TauFactory}. This  collider, having high luminosity $\sim 10^{35}\,\mbox{cm}^{-2}\,\mbox{s}^{-1}$ and $\sqrt{s}$ from  
$3$ to $5-7\,\mbox{GeV}$,  will become an intense source of  $\tau$ leptons.

\section{$e^+e^-\to \tau^+\tau^-$}
To study the effect of polarization, let us consider a longitudinally polarized electron beam and an unpolarized positron beam. Since $\sqrt{s}\ll m_Z$, we neglect the contribution of the Z boson. Then the cross section $d\sigma_0$ of the process $e^+e^-\to \tau^+\tau^-$ in the center-of-mass frame is
\begin{align}
	&d\sigma_0=\dfrac{\beta\alpha^2}{4s}\,\left |\phi_\tau^\dagger\,H\,\chi_\tau\right|^2\,d\Omega_{\bm q}\,,\nonumber\\
	&H=\bm\sigma\cdot\bm e_\lambda-\dfrac{(\bm\sigma\cdot\bm q)(\bm e_\lambda\cdot\bm q)}{E(E+M)}+i\dfrac{b}{M}\,(\bm e_\lambda\cdot\bm q)\,,\nonumber\\
	&b=F_3^\tau(s)\,,\quad s=4E^2\,,\quad \beta=q/E\,,\quad \bm e_\lambda=\frac{1}{\sqrt{2}}(\bm e_x+i\lambda\bm e_y)\,.
\end{align}
Here $\alpha$ is the fine structure constant charge, $E$ is the electron energy, $\bm q$ is the momentum of the $\tau^-$ lepton, the vector $\bm e_z$ is directed along the electron momentum, $\lambda$ is the electron helicity, $\phi_\tau$ and $\chi_\tau$ are two-component spinors entering, respectively, into the positive-frequency and negative-frequency solutions of the Dirac equation for the $\tau$ lepton,
\begin{align}\label{wfin}
	&U_{\bf q}=\sqrt{\dfrac{E_q+M}{2E_q}}
	\begin{pmatrix}
	\phi_\tau\\
	\dfrac{\bm\sigma\cdot\bm q}{E_q+M}\,\phi_\tau\end{pmatrix}\, ,\quad 
V_{\bf -q}=\sqrt{\dfrac{E_q+M}{2E_q}}
\begin{pmatrix}
\dfrac{-\bm\sigma\cdot\bm q}{E_q+M}\,\chi_\tau\\
\chi_\tau\end{pmatrix}\,,
	\end{align}
where $E_q=\sqrt{\bm q^2+M^2}$.
The cross section $d\sigma_0$ summed over polarizations of $\tau^+$ is
\begin{align}
	&d\sigma_0=\dfrac{\beta\alpha^2}{4s}\,\left[ 1-\dfrac{ q_\perp^2}{2E^2}+\bm\zeta\cdot\bm Z\right]\,d\Omega_{\bm q}\,,\nonumber\\
	&\bm Z=\mbox{Im}\,b\,\dfrac{q_\perp^2\,\bm q}{ME(E+M)}-\mbox{Im}\,b\,\dfrac{\bm q_\perp}{M}-\,\mbox{Re}\,b\,\dfrac{[\bm q_\perp\times\bm \Lambda]}{M}\nonumber\\
	&+\dfrac{M}{E}\bm \Lambda+\dfrac{(\bm q\cdot\bm \Lambda)\,\bm q}{E(E+M)}\,,
\end{align}
where $\bm q_\perp=\bm q- \bm \Lambda (\bm q \cdot \bm \Lambda)$, $\bm \zeta$ is the spin of $\tau^-$, $\bm\Lambda =\lambda\bm e_z$,  terms quadratic in $b$ are omitted. It is seen that the  linear in $b$ terms contribute only to the $\bm \zeta$-dependent part of the cross section. Measurement of $\tau$ decay products is a way to measure polarization, which in turn makes it possible to measure $b$. The total cross section summed over the $\tau^-$ polarization is
 \begin{align}\label{sig0}
 	&\sigma_0=\dfrac{\pi\beta\alpha^2}{3E^2}\,\left( 1+\dfrac{M^2}{2E^2}\right)\,.
 \end{align}
 

\section{$e^+e^-\to \tau^+\pi^-\nu_\tau$ , $e^+e^-\to \tau^-\pi^+{\bar\nu}_\tau$}
Consider the cross section of the process $e^+e^-\to \tau^+\tau^-$ followed by the decay  $\tau^-\to \pi^-\nu_\tau$. Taking into account the smallness of the $\tau$  lepton  width,  $\Gamma_\tau\approx 2.27\,\mbox{meV}$ \cite{Workman:2022ynf},  we can make the substitutions
\begin{align}
&\dfrac{1}{{\hat q}-M}\rightarrow  \,\dfrac{2E_q}{{\cal E}^2-E_q^2+i\Gamma_\tau M}\,\sum_{\mu}U_{\bm q,\mu}\bar U_{\bm q,\mu} \,,\nonumber\\
&\dfrac{4E_q^2}{({\cal E}^2-E_q^2)^2+\Gamma_\tau^2 M^2}\rightarrow \dfrac{2\pi E_q}{M\Gamma_\tau}\,\delta({\cal E}-E_q)\,,
\end{align}
where ${\cal E}=q^0$. After that, the cross section of the process $e^+e^-\to \tau^+\pi^-\nu_\tau$ can be represented as
\begin{align}
&d\sigma_\pi^{(-)}(\bm k)=B_\pi\dfrac{\beta\alpha^2(E+M)\,d\Omega_{\bm q}\,d\bm k}{4\pi sM^2\,\omega_k}\,R^{(-)}\,\delta(E-\omega_k-|\bm q-\bm k|)\,,\nonumber\\
&R^{(-)}=\left |\phi^+_\nu\left[1+\dfrac{\bm\sigma\cdot\bm q}{E+M}\right]\,H\,\chi_\tau\right|^2\,,
\end{align}
where $B_\pi\approx 10.8\%$ \cite{Workman:2022ynf} is the branching ratio of $\tau\to \pi\nu$  decay, $\omega_k=\sqrt{\bm k^2+m_\pi^2 }$ and $\bm k$ are the pion energy and momentum, respectively, $\phi_\nu$ is a two-component spinor in the Dirac  spinor $U_{\bm Q}$ for neutrino, $\bm Q=\bm q-\bm k$ and $(\bm \sigma\cdot\bm Q)\phi_\nu=-Q\phi_\nu$.
We also neglected the pion mass $m_\pi$ compared to $M$ and took into account that the  matrix element of the decay $\tau\to \pi\nu$ is proportional to $\bar U_{\bm Q}\,U_{\bm q }$ \cite{Okun:1982ap}. After summing over polarizations $\chi_\tau$, we find
\begin{align}
&R^{(-)}=A_0-\bm n_{\bm Q}\bm A\,,\quad \bm n_{\bm Q}=\bm Q/Q\,,\nonumber\\
&A_0=\dfrac{2E}{E+M}\left[ 1-\dfrac{ q_\perp^2}{2E^2}+\dfrac{(\bm q\cdot\bm \Lambda)}{E}- \mbox{Im}\,b\,\dfrac{q_\perp^2}{E^2}\right]\,,\nonumber\\
&\bm A=\dfrac{2}{(E+M)}\Bigg\{\dfrac{M^2}{E}\bm \Lambda+\left[1+\dfrac{(\bm q\cdot\bm \Lambda)}{E}-\dfrac{ q_\perp^2}{2E^2}\right]\,\bm q\nonumber\\
&-\mbox{Im}\,b\,\bm q_\perp-\mbox{Re}\,b\,[\bm q\times\bm \Lambda]\Bigg\}\,.
\end{align}
Integrating over the angles of the vector $\bm q$, we obtain
\begin{align}\label{sm}
	&d\sigma_\pi^{(-)}(\bm k)=B_\pi\dfrac{\alpha^2}{4E^3}\,\Big[\xi_0+\xi_1\cos\theta+\xi_2(3\cos^2\theta-1)\Big]\,d\omega_k\,d\Omega_{\bm k}\,,\nonumber\\
	&\xi_0=\dfrac{1}{3}\left(1+\dfrac{M^2}{2E^2}\right)+\dfrac{\mbox{Im}\,b}{3}\left(1-\dfrac{2\omega_k}{E}\right)\,,\nonumber\\
	&\xi_1=\dfrac{M^2}{4E\omega_k}-\dfrac{1}{2}\left(1-\dfrac{2\omega_k}{E}\right)\,,\nonumber\\
	&\xi_2=\dfrac{M^2(M^2-4E\omega_k)}{32E^2\omega_k^2}+\dfrac{1}{12}\left(1+\dfrac{M^2}{2E^2}\right)+\dfrac{\mbox{Im}\,b}{3}\left(1-\dfrac{\omega_k}{2E}-\dfrac{3M^2}{8E\omega_k}\right)\,.
\end{align}
Here $\cos\theta=\bm k\cdot\bm\Lambda/k=\lambda k_z/k$, the available pion energy range is determined by the relation $|2\omega_k-E|\leq q$. It is seen that the only term, which contains the helicity $\lambda$, is the term proportional to $\xi_1$. However, this term is independent of $b$. Moreover, the cross section $d\sigma_\pi^{(-)}(\bm k)$ in \eqref{sm} contains only the imaginary part of $b$, which does not require electron polarization to be measured.
For the cross section $d\sigma_\pi^{(+)}(\bm k)$ of the process $e^+e^-\to \tau^-\pi^+{\bar\nu}_\tau$, we similarly find
\begin{align}\label{sp}
	&d\sigma_\pi^{(+)}(\bm k)=B_\pi\dfrac{\alpha^2}{4E^3}\,\Big[\xi'_0+\xi'_1\cos\theta+\xi'_2(3\cos^2\theta-1)\Big]\,d\omega_k\,d\Omega_{\bm k}\,,\nonumber\\
	&\xi'_0=\dfrac{1}{3}\left(1+\dfrac{M^2}{2E^2}\right)-\dfrac{\mbox{Im}\,b}{3}\left(1-\dfrac{2\omega_k}{E}\right)\,,\nonumber\\
	&\xi'_1=-\xi_1=-\dfrac{M^2}{4E\omega_k}+\dfrac{1}{2}\left(1-\dfrac{2\omega_k}{E}\right)\,,\nonumber\\
	&\xi'_2=\dfrac{M^2(M^2-4E\omega_k)}{32E^2\omega_k^2}+\dfrac{1}{12}\left(1+\dfrac{M^2}{2E^2}\right)-\dfrac{\mbox{Im}\,b}{3}\left(1-\dfrac{\omega_k}{2E}-\dfrac{3M^2}{8E\omega_k}\right)\,.
\end{align}
We define the differential asymmetry $dA_\pi$ as follows
\begin{align}\label{api}
	&dA_\pi=\dfrac{d\sigma_\pi^{(-)}(\bm k)-d\sigma_\pi^{(+)}(-\bm k)}{2\sigma_0}\,,
\end{align}
where $\sigma_0$ is defined in \eqref{sig0}.
The asymmetry is proportional to $\mbox{Im}\,b$, so its measurement is important from the experimental point of view. We have
\begin{align}\label{da}
	&dA_\pi=\dfrac{B_\pi{\mbox{Im}\,b}\,d\omega_k\,d\Omega_{\bm k}}{4\pi q(1+M^2/2E^2)}\,\left[1-\dfrac{2\omega_k}{E}+(3\cos^2\theta-1)\left(1-\dfrac{\omega_k}{2E}-\dfrac{3M^2}{8E\omega_k}\right)\right]\,.
\end{align}
After integration over angles, the asymmetry is
\begin{align}\label{da1}
	&dA_\pi=\dfrac{{B_\pi\mbox{Im}\,b}\,d\omega_k}{ q(1+M^2/2E^2)}\,\left(1-\dfrac{2\omega_k}{E}\right)\,.
\end{align}
As it should be, after integration over the pion energy, the asymmetry vanishes. Therefore, we define the total asymmetry $A_\pi$  as $dA_\pi$ \eqref{da1} integrated over $\omega_k$ from $(E-q)/2$ to $E/2$ (half of the allowed energy range),
\begin{align}\label{apiE}
	&A_\pi=\dfrac{B_\pi\,\mbox{Im}\,b}{ 2(2+M^2/E^2)}\,\sqrt{1-\dfrac{M^2}{E^2}}\,.
\end{align}


\section{$e^+e^-\to \tau^+\rho^-\nu_\tau$ , $e^+e^-\to \tau^-\rho^+{\bar\nu}_\tau$}
To measure $\mbox{Re}\,b$, consider the decay of one $\tau$ lepton into $\rho$ meson with the momentum
$\bm p$, energy $\varepsilon_p=\sqrt{\bm p^2+m_\rho^2}$ and 4-polarization vector $f=(f_0,\,\bm f)$, where $m_\rho $ is the mass of $\rho$ meson. We define the differential asymmetry $dA_\rho$ as
\begin{align}\label{arho}
	&dA_\rho=\dfrac{d\sigma_\rho^{(-)}(\bm p,\,\bm f)-d\sigma_\rho^{(+)}(-\bm p,\,-\bm f)}{2\sigma_0}\,,
\end{align}
where $d\sigma_\rho^{(-)}(\bm p,\,\bm f)$ is the cross section of the process $e^+e^-\to \tau^+\rho^-\nu_\tau $ and $d\sigma_\rho^{(+)}(-\bm p,\,-\bm f)$ is the cross section of the process $e^+e^-\to \tau^-\rho^+{ \bar\nu}_\tau$. Using the  matrix element of decay $\tau\to \rho\nu$ \cite{Okun:1982ap},  we obtain as a result of straightforward calculations
\begin{align}\label{darhopol}
	&dA_\rho=\dfrac{3B_\rho\,d\varepsilon_p\,d\Omega_{\bm p}\,\,[C_1\,\mbox{Re}\,b+C_2\,\mbox{Im}\,b]}{2\pi pM^2(1+M^2/2E^2)(2+M^2/m_\rho^2)(1-m^2_\rho/M^2)^2}\,,\nonumber\\
	&C_1=\left[\dfrac{q}{p}\varepsilon_p\,P_2(x_0)-E P_1(x_0)\right]\,([\bm\Lambda\times\bm f]\cdot\bm p)\,f_0\,,\nonumber\\ &C_2=\dfrac{qp}{3}\left[\left(2+\dfrac{\varepsilon_p}{E}\right)f_0^2+\left(1-\dfrac{\varepsilon_p}{E}\right)\bm f^2-(\bm\Lambda\cdot\bm f)^2\right]\nonumber\\
	&+P_1(x_0)\,\Big[\dfrac{1}{2}[p^2-(\bm\Lambda \cdot\bm p)^2](\bm f^2-f_0^2) \nonumber\\
	&+Ef_0[(\bm\Lambda\cdot\bm p)(\bm\Lambda\cdot\bm f)-(\bm p\cdot\bm f)]
	+\dfrac{2q^2}{5E}f_0[(\bm\Lambda\cdot\bm p)(\bm\Lambda\cdot\bm f)-2(\bm p\cdot\bm f)]\Big]\nonumber\\
	&+\dfrac{q}{p}\,P_2(x_0)\,\Bigg\{\left(\dfrac{\varepsilon_p}{2E}(\bm f^2-f_0^2)-f_0^2\right)\left[(\bm\Lambda\cdot\bm p)^2-\dfrac{p^2}{3}\right]+\left[(\bm f\cdot\bm p)^2-\dfrac{p^2\bm f^2}{3}\right]\nonumber\\
	&-(\bm f\cdot\bm\Lambda)\left[(\bm\Lambda\cdot\bm p)(\bm f\cdot\bm p)-\dfrac{p^2}{3}(\bm f\cdot\bm\Lambda)\right]\Bigg\}\nonumber\\
	&+P_3(x_0)\,\dfrac{q^2}{Ep^2}f_0\,\Big\{(\bm f\cdot\bm p)(\bm\Lambda\cdot\bm p)^2-\dfrac{p^2}{5}[(\bm f\cdot\bm p)+2(\bm f\cdot\bm\Lambda)(\bm p\cdot\bm\Lambda)]\Big\}\,,\nonumber\\
	&x_0=\dfrac{2E\varepsilon_p-M^2-m_\rho^2}{2qp}\,,
\end{align}
where $B_\rho\approx25.5\%$ \cite{Workman:2022ynf} is the branching ratio of $\tau\to\rho\nu$ decay and $P_n(x)$ are Legendre polynomials. It is seen that for a polarized electron beam and  a polarized $\rho$ meson the contribution of  $\mbox{Re}\,b$ does not vanish. Note that
\begin{align}
	& dA_\rho\Big|_{\lambda=+1}-dA_\rho\Big|_{\lambda=-1}=\dfrac{3B_\rho\,\mbox{Re}\,b\,C_1\,d\varepsilon_p\,d\Omega_{\bm p}}{\pi pM^2(1+M^2/2E^2)(2+M^2/m_\rho^2)(1-m^2_\rho/M^2)^2}\,.
\end{align}
Then we perform summation over polarizations of $\rho$ meson, using the formula
$$\sum_{pol} f^\mu f^\nu= -g^{\mu\nu}+\dfrac{p^\mu p^\nu}{m_\rho^2}\,,$$
and obtain 
\begin{align}\label{darho1}
		&\sum_{pol}C_1=0\,,\quad \sum_{pol}C_2=C_{21}\,F+C_{22}\,,\nonumber\\ 
		&C_{21}=-\dfrac{qp}{3}+P_1(x_0)\,\left(-\dfrac{3}{2}m_\rho^2+\dfrac{7}{5}E\varepsilon_p
		-\dfrac{2\varepsilon_p}{5E}M^2\right)\nonumber\\
	&+\dfrac{q}{p}P_2(x_0)\left(\dfrac{3\varepsilon_p}{2E}m_\rho^2-\dfrac{5}{3}\varepsilon_p^2
	+\dfrac{2}{3}m_\rho^2\right)+P_3(x_0)\,\dfrac{3\varepsilon_pq^2}{5E}\,,\nonumber\\
&C_{22}=\dfrac{p(M^2-2m_\rho^2)[m_\rho^2 E+M^2(E-2\varepsilon_p)]}{6m_\rho^2E\,q}\,,\nonumber\\
&F=\dfrac{1}{3m_\rho^2}\left[3(\bm\Lambda\cdot\bm p)^2-	p^2\right]\,.
\end{align}
In this formula, the contribution $\propto\mbox{Re}\,b$ is absent even for the case of polarized electrons. After integration over the angles of the vector $\bm p$, the term  $\propto F$ vanishes, and  the asymmetry reads
\begin{align}\label{darhoeps1}
	&dA_\rho=\dfrac{6B_\rho\,d\varepsilon_p\,C_{22}\,\mbox{Im}\,b}{ pM^2(1+M^2/2E^2)(2+M^2/m_\rho^2)(1-m^2_\rho/M^2)^2}\,.
	\end{align}
The allowed region of the energy $\varepsilon_p$ is given by the relation
$$\left|2\varepsilon_p-\left(1+\dfrac{m_\rho^2}{M^2}\right)E\right|\leq   q\left(1-\dfrac{m_\rho^2}{M^2}\right)\,.      $$
After integration over $\varepsilon_p$ in Eq.~\eqref{darhoeps1}, the asymmetry vanishes. Therefore, we define the total asymmetry  $A_\rho$ as a result of integration of Eq.~ \eqref{darhoeps1}  over $\varepsilon_p$ in the range 
$$(1+m_\rho^2/M^2)E-(1-m_\rho^2/M^2)q<2\varepsilon_p< (1+m_\rho^2/M^2)E\,.$$
One has
\begin{align}\label{arhoE}
	&A_\rho=\dfrac{B_\rho\,\mbox{Im}\,b}{ 2(2+M^2/E^2)}\,\sqrt{1-\dfrac{M^2}{E^2}}\Bigg(\dfrac{M^2-2m_\rho^2}{M^2+2m_\rho^2}\Bigg)\,.
\end{align}
Note that
 \begin{align}\label{arhoE1}
 	&A_\rho=\dfrac{B_\rho}{ B_\pi}\Bigg(\dfrac{M^2-2m_\rho^2}{M^2+2m_\rho^2}\Bigg)\,A_\pi\,.
 \end{align}


To determine  the polarization of  $\rho$ meson, it is possible to measure the main decay channel of  $\rho$ meson with momentum $\bm p$ into two pions with momenta $\bm k_1$ and $\bm k_2$. The corresponding asymmetry can be obtained from Eq.~
\eqref{darhopol} with obvious substitution
\begin{align}\label{darhopipi}
	&\frac{d\bm p}{2\varepsilon_p(2\pi)^3}\,f^\mu\,f^\nu\rightarrow  \,\dfrac{f^2_{\rho\pi\pi}\,d\bm k_1\,d\bm k_2\,(k_1-k_2)^\mu(k_1-k_2)^\nu}{ 4\omega_{1}\omega_{2}(2\pi)^6 [(s_1-m_\rho^2)^2+\Gamma_\rho^2m_\rho^2]}\,.
\end{align}
Here $s_1=(k_1+k_2)^2$, $\Gamma_\rho=149.1\,\mbox{MeV}$ \cite{Workman:2022ynf} is the $\rho$ meson width, $\omega_1=|\bm k_1|$, $\omega_2=|\bm k_2|$, the constant $f^2_{\rho\pi\pi}$ is
$$ f^2_{\rho\pi\pi}=\dfrac{48\pi\Gamma_\rho}{m_\rho\,(1-4m_\pi^2/m_\rho^2)^{3/ 2}}\,.$$

\section{$e^+e^-\to \tau^+\tau^-\to \pi^+\pi^-\nu_\tau{\bar\nu}_\tau$}
If each $\tau$ lepton decays into a pion and a neutrino (antineutrino), both the imaginary and real parts of $b$ can be measured even in the case of an unpolarized electron beam. It is this case that we consider in this section.  We define asymmetry as
\begin{align}\label{apipi}
	&dA_{\pi\pi}=\dfrac{d\sigma_{\pi\pi}(\bm k_1,\,\bm k_2)-d\sigma_{\pi\pi}(-\bm k_2,\,-\bm k_1)}{2\sigma_0}\,,
\end{align}
where $d\sigma_{\pi\pi}(\bm k_1,\,\bm k_2)$ is the cross section of the process $e^+e^-\to \pi^+\pi^-\nu_\tau{\bar\nu}_\tau$, $\bm k_1$ is the momentum of $\pi^-$ and $\bm k_2$ is the momentum  of $\pi^+$. For $dA_{\pi\pi}$ we obtain
\begin{align}\label{dapipi}
	&dA_{\pi\pi}=\dfrac{3B^2_{\pi}\,d\bm k_1\,d\bm k_2}{ (2\pi)^3M^4(1+M^2/2E^2)\omega_{1}\omega_{2}(E-\omega_{1})(E-\omega_{2})}\nonumber\\
	&\times\,\int\,d\Omega_{\bm q}\,\delta(E-\omega_{1}-|\bm q-\bm k_1|)\,\,\delta(E-\omega_{2}-|\bm q+\bm k_2|)\nonumber\\
	&\times\Bigg<\mbox{Re}\,b\,\Bigg\{(\bm q\cdot\bm\Lambda)\,\Big([\bm q\times\bm\Lambda]\cdot(\omega_{2}\bm k_1+\omega_{1}\bm k_2)\Big)\nonumber\\
	&-\dfrac{\Big(M^2+(\bm q\cdot\bm\Lambda)^2\Big)}{E}\,\Big(\bm q\cdot[\bm k_1\times\bm k_2]\Big)
	+E(\bm q\cdot\bm\Lambda)\big(\bm \Lambda\cdot[\bm k_1\times\bm k_2]\Big)\Bigg\}\nonumber\\
	&+
	\mbox{Im}\,b\,\dfrac{M^2}{2}\Bigg\{\dfrac{(\omega_{2}-\omega_{1})}{E}[M^2+(\bm q\cdot\bm\Lambda)^2]+(\bm q\cdot\bm\Lambda)\big(\bm \Lambda,\bm k_1+\bm k_2\Big)\Bigg\}\Bigg>\,.
\end{align}
Integrating over the angles of  vector $\bm q$, we find
\begin{align}\label{dapipi1}
	&dA_{\pi\pi}=\dfrac{3B^2_{\pi}\,d\bm k_1\,d\bm k_2}{ (2\pi)^3M^4(1+M^2/2E^2)\omega^2_{1}\omega^2_{2}\,\sqrt{(1-x^2)\,{ q}^2(q^2-{ P}^2)}}\nonumber\\
	&\times\Bigg<\mbox{Re}\,b\,\Big(\bm\Lambda\cdot[\bm k_1\times\bm k_2]\Big)\Bigg[2(q^2-P^2)(\bm N_2\cdot\bm\Lambda)\nonumber\\
	&+\dfrac{(\bm P\cdot\bm\Lambda)}{E}\Big(M^2+2P^2-q^2-\dfrac{Ea_2}{1-x}\Big)\Bigg]+
	\mbox{Im}\,b\,\dfrac{M^2}{2}\Bigg[\dfrac{(\omega_{2}-\omega_{1})}{E}
	\Big(M^2\nonumber\\
	&+(\bm N_3\cdot\bm\Lambda)^2(q^2-{ P}^2)+(\bm P\cdot\bm\Lambda)^2\Big)+(\bm P\cdot\bm\Lambda)\big(\bm \Lambda,\bm k_1+\bm k_2\Big)\Bigg]\Bigg>\,,
\end{align}
 Here the following notation is introduced:
 \begin{align}\label{not}
 	&\bm N_1=\frac{\bm n_1+\bm n_2}{2(1+x)}\,,\quad  \bm N_2=\frac{\bm n_1-\bm n_2}{2(1-x)}\,,\quad \bm N_3=\frac{[\bm n_2\times\bm n_1]}{\sqrt{1-x^2}}\,,\nonumber\\
 	&\bm P=a_1\,\bm N_1+a_2\,\bm N_2\,,\quad a_1=\dfrac{M^2(\omega_{1}-\omega_{2})}{2\omega_{1}\omega_{2}}\,,\quad a_2=2E-\dfrac{M^2(\omega_{1}+\omega_{2})}{2\omega_{1}\omega_{2}}\,,\nonumber\\
 	&x=(\bm n_1\cdot\bm n_2)\,,\quad \bm n_1=\dfrac{\bm k_1}{\omega_{1}}\,,\quad \bm n_2=\dfrac{\bm k_2}{\omega_{2}}\,.
 \end{align}
 It is also interesting to consider the asymmetry that remains after the integration over the angles of  vectors $\bm k_1$ and $\bm k_2$.
The corresponding result reads
 \begin{align}\label{dapipiEE}
 	&dA_{\pi\pi}=\dfrac{2B^2_{\pi}\,\mbox{Im}\,b\,(\omega_2-\omega_1)d\omega_1\,d\omega_2}{ Eq^2(1+M^2/2E^2)}\,.
 \end{align}
Here only the imaginary part of $b$ contributes. Integrating over $\omega_2$ in the region $|2\omega_2-E|<q$, we obtain the result
\begin{align}\label{dapipiE}
	&dA_{\pi\pi}=\dfrac{B^2_{\pi}\,\mbox{Im}\,b\,(1-2\omega_1/E)d\omega_1}{ q(1+M^2/2E^2)}\,,
\end{align}
which is consistent with \eqref{da1}. Integrating \eqref{dapipiE} over $\omega_2$ in the range $(E-q)/2<\omega_2 <  E/2$, we get
\begin{align}\label{apipiE}
	&A_{\pi\pi}=\dfrac{B_\pi^2\,\mbox{Im}\,b}{ 2(2+M^2/E^2)}\,\sqrt{1-\dfrac{M^2}{E^2}}\,.
\end{align}
 Naturally, $A_{\pi\pi}={B_\pi}A_\pi$.

\section{Discussion of the results}
In our work we have obtained the asymmetries which contain both $\mbox{Re}b$ and $\mbox{Im}\,b$, as well as asymmetries proportional only to $\mbox{Re}\,b$ or only to $\mbox{Im}\,b$. To measure the asymmetry, which is proportional only to $\mbox{Re}\,b$, it is necessary to have longitudinal polarization of electrons.

\begin{figure}[h]
	\centering
	\includegraphics[width=0.7\linewidth]{"dfig"}
	\caption[AsimFig]{Total asymmetry $A$ in units of $\mbox{Im}\,b$. Solid curve: $A_\pi$; dotted curve: $A_\rho$; dashed curve: $A_{\pi\pi}$.}
	\label{fig:plot-for-article}
\end{figure}
Asymmetries \eqref{api},\eqref{arho}, and \eqref{apipi} are the odd quantities  with respect to $\mbox{CP}$ transformations. Indeed, as a result of this transformation
\begin{align*}
	&d\sigma_\pi^{(-)}(\bm k)\rightarrow \,d\sigma_\pi^{(+)}(-\bm k)\,,\quad d\sigma_\pi^{(+)}(-\bm k)\rightarrow\, d\sigma_\pi^{(-)}(\bm k)\,;\\
	&d\sigma_\rho^{(-)}(\bm p,\,\bm f)\rightarrow\, d\sigma_\rho^{(+)}(-\bm p,\,-\bm f)\,,\quad d\sigma_\rho^{(+)}(-\bm p,\,-\bm f)\rightarrow\, d\sigma_\rho^{(-)}(\bm p,\,\bm f)\,;\\
	&d\sigma_{\pi\pi}(\bm k_1,\,\bm k_2)\rightarrow\, d\sigma_{\pi\pi}(-\bm k_2,\,-\bm k_1)\,,\quad d\sigma_{\pi\pi}(-\bm k_2,\,-\bm k_1)\rightarrow\, d\sigma_{\pi\pi}(\bm k_1,\,\bm k_2)\,.
\end{align*}
The term $\propto\gamma^\mu$ in \eqref{tauvert} is $\mbox{CP}$-even while the term 
 $\propto\,b$  is  $\mbox{CP}$-odd. Therefore, asymmetries appear due to interference between $\mbox{CP}$-odd and  $\mbox{CP}$-even terms and are linear in $b$.   
The energy dependence of total asymmetries is shown in Fig. \ref{fig:plot-for-article}. The solid curve corresponds to $A_\pi$, the dotted curve to $A_\rho$, and the dashed curve to $A_{\pi\pi}$. All curves have the same energy dependence and differ only in scale. Indeed, it follows from Eqs.~ \eqref{apiE}, \eqref{arhoE}, \eqref{apipiE}  that
$$A_\rho=1.06\,A_{\pi}\,,\quad A_{\pi\pi}=0.108\,A_{\pi}\,.$$

\section{Conclusion} 
In conclusion, we have considered the processes $e^+e^-\rightarrow \tau^+\pi^-\nu_\tau$, $e^+e^-\rightarrow \pi^+\tau^-\bar\nu_ \tau$, $e^+e^-\rightarrow \tau^+\rho^-\nu_\tau$, and $e^+e^-\rightarrow \rho^+\tau^-\bar\nu_\tau $ with longitudinally polarized electrons, as well as the process $e^+e^-\rightarrow \pi^+\pi^-\nu_\tau\bar\nu_\tau$ with unpolarized electrons for the invariant masses of the initial electron and positron $\sqrt {s}\ll m_Z$.  We have  calculated the $\mbox{CP}$-odd  asymmetries $\propto\mbox{Re}\,b$ and $\propto\mbox{Im}\,b$, as well as the asymmetries proportional only to $\mbox{ Re}\,b$ or only to $\mbox{Im}\,b$. Measuring these quantities can improve the upper limits for $\mbox{Re}\, b$ and $\mbox{Im}\, b$. It is shown that to measure $\mbox{Im}\,b$, polarization is not needed, and to measure $\mbox{Re}\,b$, the polarization is not necessary, but simplifies the measurement.


\section*{Acknowledgements}
We are grateful to D.A. Epifanov and V.S. Vorobiev for valuable discussions.


