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.


2 Answers 2


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.)

  • $\begingroup$ I don't understand why the equality should hold modulo eoms? Aren't we deriving an equation of motion? $\endgroup$
    – Mtheorist
    Jul 30, 2017 at 13:58
  • $\begingroup$ So what you meant is that $\approx$ is used because equation (2.15) is an equation of motion, am I right? $\endgroup$
    – Mtheorist
    Jul 30, 2017 at 17:15
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jul 30, 2017 at 17:21
  • $\begingroup$ Qmechanic, something is not clear to me. At the classical level, your derivation works perfectly. But what about path integrals, which are Witten's main concern (eqn (2.15) leads to the path integrals' localization to the moduli space of holomorphic maps). By the change of variables (B), there should be a corresponding change in the path integral measure, captured by a Jacobian $j$. To implement your derivation, det $j$ should be constant, or at least $\tilde{H}$-independent (otherwise we won't have a Gaussian integral over $\tilde{H}$), but I am unable to show it. Could you elucidate this? $\endgroup$
    – Mtheorist
    Oct 8, 2017 at 16:57

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 $\delta_{ij}$.

  • $\begingroup$ Wouldn't a different decomposition (instead of that given in your second equation) lead to a different Euler-Lagrange equation? E.g., instead of $H^{\alpha i}=\frac{1}{2}(H^{\alpha i}+H^{\alpha i})$ one could use $H^{\alpha i}=\frac{1}{3}(H^{\alpha i})+\frac{2}{3}(H^{\alpha i})$. Then, replacing (2.5) into the coefficient of $\frac{2}{3}$ will give us an expression different from the second equation in your answer. I think this will lead to a different Euler-Lagrange equation. $\endgroup$
    – Mtheorist
    Jul 28, 2017 at 17:47
  • $\begingroup$ Also, the Latin indices are target space indices, the Greek indices are worldsheet indices. $\endgroup$
    – Mtheorist
    Jul 28, 2017 at 17:58

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