Put $\alpha=\delta\eta$. The hypothesis says that the exact two-form $\beta=d\alpha$ is <anti-self-dual differential form>[anti-self-dual]. Since $M$ is compact without boundary, <Stokes theorem> gives
$$
0=\int_Md(\alpha\wedge d\alpha)
=\int_M\beta\wedge\beta
=-\int_M\beta\wedge *\beta
=-\|\beta\|_{L^2}^2.
$$
Hence $d\alpha=0$ by the <exact anti-self-dual form on a compact four-manifold> argument. Since $\delta$ is the formal $L^2$ adjoint of $d$,
$$
\|\delta\eta\|_{L^2}^2
=\langle\delta\eta,\alpha\rangle_{L^2}
=\langle\eta,d\alpha\rangle_{L^2}=0.
$$
Therefore $\boxed{\delta\eta=0}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 115 / 4 / e / Solution

If an exact two-form $\beta=d\alpha$ on a compact oriented Riemannian four-manifold is anti-self-dual, then
$$
0=\int_Md(\alpha\wedge d\alpha)=\int_M\beta\wedge\beta=-\lVert\beta\rVert_{L^2}^2,
$$
so $\beta=0$.


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