The following is a continuation of this question. The action of Witten's topological sigma model (defined on a worldsheet, $\Sigma$, with target space an almost complex manifold denoted $X$) takes the form $$ S=\int d^2\sigma\big(-\frac{1}{4}H^{\alpha i}H_{\alpha i}+H^{\alpha }_i\partial_{\alpha}u^i -i\rho^{\alpha}_{i}D_{\alpha}\chi^i-\frac{1}{8}\rho^{\alpha}_i\rho_{\alpha t}\chi^k\chi^l{R_{kl}}^{it}\big), $$ as shown in equation 2.14 of his paper (for simplicity we have assumed that $X$ is Kaehler, whereby $D_k{J^i}_j=0$). The fields $\rho^{\alpha i}$ and $H^{\alpha i}$ also obey the "self-duality" constraints $$ \rho^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_j\rho^{\beta j}, \tag{2.4} $$

$$ H^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}, \tag{2.5} $$ where $\varepsilon$ and $J$ are respectively the almost complex structures of $\Sigma$ and $X$.

As explained here, using the unconstrained fields $\tilde{H}^i_{\alpha}$, where $$ H^i_{\alpha}=\tilde{H}_{\alpha}^i+\varepsilon_{\alpha\beta}{J^{i}}_j\tilde{H}^{\beta j}, $$ the Euler-Lagrange equation for $H_{\alpha}^{ i}$ can be found to be $H^i_{\alpha}=\partial_{\alpha}u^i+\varepsilon_{\alpha\beta}{J^{i}}_j\partial^{\beta}u^j. $ Likewise, we have $ \rho^i_{\alpha}=\tilde{\rho}_{\alpha}^i+\varepsilon_{\alpha\beta}{J^{i}}_j\tilde{\rho}^{\beta j} $ in terms of unconstrained fermionic fields $\tilde{\rho}_{\alpha}^i$.

We can then obviously obtain an action in terms of unconstrained fields. My question is, can we completely express the supersymmetry transformations of the model in terms of the unconstrained fields, so that all the information of the model is in terms of unconstrained fields?

Attempted solution: Let us consider the SUSY transformation of $\rho_{\alpha}^i$ (equation 2.9). Using unconstrained fields we have $$ \delta (\tilde{\rho}_{\alpha}^i+\varepsilon_{\alpha\beta}{J^{i}}_j\tilde{\rho}^{\beta j})=\varepsilon(\tilde{H}_{\alpha}^i+\varepsilon_{\alpha\beta}{J^{i}}_j\tilde{H}^{\beta j})-i\varepsilon\Gamma^{i}_{jk}\chi^j(\tilde{\rho}_{\alpha}^k+\varepsilon_{\alpha\beta}{J^{k}}_l\tilde{\rho}^{\beta l}). $$ Setting $\varepsilon=1$, and using $D_k{J^i}_j=0$, we obtain $$ \delta \tilde{\rho}_{\alpha}^i+{\varepsilon_{\alpha}}^{\beta}\delta\tilde{\rho}_{\beta}^{ j}{J^{i}}_j=(\tilde{H}_{\alpha}^i-i\Gamma^{i}_{jk}\chi^j\tilde{\rho}_{\alpha}^k)+{\varepsilon_{\alpha}}^{\beta}(\tilde{H}_{\beta}^j-i\Gamma^{j}_{kl}\chi^k\tilde{\rho}_{\beta}^l){J^{i}}_j. $$ It is then tempting to conclude that $$ \delta \tilde{\rho}_{\alpha}^i=\tilde{H}_{\alpha}^i-i\Gamma^{i}_{jk}\chi^j\tilde{\rho}_{\alpha}^k, $$ but I am not sure if this is the unique solution.


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