Obtaining Euler-Lagrange equation from action with constraint - Witten's topological sigma model

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 $$ S=\int d^2\sigma\big(-\frac{1}{4}H^{\alpha i}H_{\alpha i}+H^{\alpha }_i\partial_{\alpha}u^i+\ldots\big), \tag{2.14} $$ as shown in equation 2.14 of his paper. The auxiliary fields $H^{\alpha i}$ also obey the "self-duality" constraint $$ H^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}, \tag{2.5} $$ where $\varepsilon$ and $J$ are respectively the almost complex structures of $\Sigma$ and $X$.

Now, the Euler-Lagrange equation for $H_{\alpha}^{ i}$ is given in equation 2.15 as $$ H^i_{\alpha}=\partial_{\alpha}u^i+\varepsilon_{\alpha\beta}{J^{i}}_j\partial^{\beta}u^j. \tag{2.15} $$ How does one show this? I have tried including the "self-duality" constraint in the action via Lagrange multipliers, but I have not been able to obtain the correct Euler-Lagrange equation in this way.

Answer 1

Here is one method, perhaps not the shortest, but at least it is consistent and hopefully transparent.

1. Before we begin let us introduce a hopefully obvious matrix notation $$ J^i{}_j\quad \longrightarrow \quad J $$ $$ \varepsilon^{\alpha}{}_{\beta}\quad \longrightarrow \quad \varepsilon $$ $$ u^i{}_{,\alpha} \quad \longrightarrow \quad u_{,} $$ $$ H^{\alpha}{}_i\quad \longrightarrow \quad H $$ $$ H^i{}_{\alpha}\quad \longrightarrow \quad H \tag{A}$$ etc, for notational simplicity. (The last two lines in eq. (A) may seem ambiguous, but in practice one can tell them apart from context.)

2. Write Witten's tensor field $$H ~:=~\frac{1}{2} ( \tilde{H} - \varepsilon \tilde{H} J ) \tag{B}$$ in terms of an un-constrained tensor field $\tilde{H}$ with same type of indices. (The perhaps surprising minus sign in eq. (B) has to do with the ordering of the matrices.)

3. It is easy to check that the definition (B) is manifestly self-dual $$ H ~=~ -\varepsilon H J, \tag{2.5}$$ by using $$J^2 ~=~ -{\bf 1}, \qquad \varepsilon^2 ~=~ -{\bf 1}. \tag{C}$$

4. The Lagrangian density becomes $$ {\cal L}~:=~{\rm tr}\left(-\frac{1}{4} H^2 + H u_{,}\right) +\ldots ~\stackrel{(B)}{=}~\frac{1}{2}{\rm tr}\left(-\frac{1}{4} \tilde{H}^2 + \frac{1}{4} \varepsilon \tilde{H} J \tilde{H} + ( \tilde{H} - \varepsilon \tilde{H} J ) u_{,}\right) +\ldots.\tag{2.14}$$

5. Vary the Lagrangian density (2.14) wrt. the un-constrained tensor field: $$ 0~\approx~ \delta {\cal L}~\stackrel{(2.14)}{=}~\frac{1}{2}{\rm tr}\left\{\left(- \frac{1}{2}(\tilde{H} - J \tilde{H}\varepsilon ) + (u_{,} - J u_{,}\varepsilon)\right)\delta \tilde{H}\right\}. \tag{D}$$ In other words, $$ H~\stackrel{(B)}{=}~\frac{1}{2} (\tilde{H} - J \tilde{H}\varepsilon )~\stackrel{(D)}{\approx}~u_{,} - J u_{,}\varepsilon \tag{2.15}$$ which is OP's sought-for equation. (Here the $\approx$ symbol means equality modulo eoms.)

Answer 2

Isn't the trick to take $$ H^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j},\tag{2.5} $$ and note $$ H^{\alpha i}=\frac{1}{2}\left(H^{\alpha i} + {\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}\right) $$ then plug this into the action to rewrite it as $$ S=\int d^2\sigma\left(-\frac{1}{4}H^{\alpha i}H_{\alpha i}+\frac{1}{2}\delta_{ik}\left(H^{\alpha i} + {\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}\right)\partial_{\alpha}u^k+\ldots\right), $$ where I am using the fact that contraction over Latin indices is done using the "Worldsheet metric" which is (by conformal invariance) equal to $\delta_{ij}$.