Recall that if the momentum of scattering amplitudes is taken to be complex and from little group scaling that the 3-particle interaction for massless particles of any spin is given as \begin{equation} A_3(1^{h_1}2^{h_2}3^{h_3}) = c\langle 12\rangle^{h_3-h_2-h_1}\langle 13\rangle^{h_2-h_1-h_3}\langle 23\rangle^{h_1-h_2-h_3}. \end{equation} The constant $c$ is also constrained based on the number of mass dimensions the amplitude must have which is $[A_n]=4-n$ for $n$-point amplitude in $d=4$. Lets do an example I am sure those familiar with the literature will recognize right away, that of the 4-particle scattering between 3-spin 2 particles namely of helicity $h_1 = h_2 = 2$ and $h_3 = -2$ (3 gravitons). Since $\sum_i^3h_i >0$ we need to change all of the angle brackets to square brackets in the equations above, but nonetheless we get (where I switch from $A$ to $\mathcal{M}$ to keep with the literature's notation) \begin{equation} \mathcal{M}(1^+2^+3^-) = a\frac{[12]^6}{[13]^2[23]^2}. \end{equation} Based on units, since $[\mathcal{M}] = 1$ then the units of $a$ needs to be $[a] = (\text{mass})^{-1}$. Since these are spin-2 particles and the only consistent spin-2 theory has a coupling of the gravitational constant given as $\kappa = m_{\text{planck}}^{-1}$. For it to fit to GR, we can replace $a\rightarrow \kappa/2$. Based on the power of $\kappa$ this 3-particle interaction corresponds to a single insertion of Riemann at the level of the action.
However, now consider the 3-particle interaction with $h_1 = h_2 = 2$ but $h_3 = 0$. Then the 3-particle amplitude is \begin{equation} \mathcal{M}_3(1^+2^+3^0) = c[12]^4. \end{equation} Based on units, we know that the units of $c$ must be $[c] = (\text{mass})^{-3}$. At the level of the EFT, we know that this amplitude corresponds to an operator of the form $\phi(\text{Riem})^2$ or $\beta\phi R_{\mu\nu\kappa\gamma}R^{\mu\nu\kappa\gamma}$ explicitly where $\beta$ is the coupling coefficient.
However, here is my question. Notice that seemingly, I can write down another operator that could correspond to this amplitude namely $\tilde{\beta}\frac{1}{2}\phi R^{\mu\nu\alpha\beta}\epsilon_{\alpha\beta}{}^{\gamma\delta}R_{\gamma\delta\mu\nu}$. The first operator I wrote down was parity even, but this one is parity odd. So, how can I tell the difference? Or, how can I build the EFT to account for both? For example, must I instead write the 3-particle interaction as \begin{equation} \mathcal{M}_3(1^+2^+3^0) = (\beta + \tilde{\beta})[12]^4. \end{equation} Or is this a matter of one operator is constrained to only have a specific channel structure, for example instead of $[12]$ it must be $[23]$ since it is a parity violating operator, but the particles are all bosonic, so shouldn't the amplitude have a bose symmetry? As you can tell, I am very confused on this particular point.
