Obtaining Euler-Lagrange equation from action with constraint - Witten's topological sigma model 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.
 A: Here is one method, perhaps not the shortest, but at least it is consistent and hopefully transparent. 


*

*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.) 

*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.)

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

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

*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.)
A: 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 $\delta_{ij}$.
