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On an oriented pseudo-Riemannian vector space $({\mathbb{R}}^n,\eta ,\mathrm{vol})$, the Hodge star operator is the unique linear mapsatisfyingfor all $p$-forms $\alpha ,\beta$. If $s$ is the number of negative metric directions, thenon $p$-forms under this convention. For the two-forms in the question,whereasThe induced inner product is symmetric, so these expressions are negatives of one another. Hence a self-dual differential form and an anti-self-dual differential form are orthogonal and

Write $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The metric is conformal to the standard Euclidean metric, and the Hodge star on middle-degree differential forms is conformally invariant. Taking $e^{ij}=d{x}^i\wedge d{x}^j$, the three real forms arebecause ${\omega }_1+i{\omega }_2=dw\wedge dz$. For the orientation specified by
$dw\wedge dz\wedge d\bar{w}\wedge d\bar{z}$, one hasThe corresponding relations for the complementary basis forms immediately giveThus these forms give a real basis of the self-dual two-forms.

Let $D_{\alpha }={\partial }_{\alpha }+A_{\alpha }$ be the gauge covariant derivative. In these complex coordinates the Anti-self-dual Yang-Mills equations areIntroduce the spectral parameter $\lambda$ and the linear operatorsTheir commutator isTherefore the Lax pair for the anti-self-dual Yang-Mills equationsis compatible for every $\lambda$ exactly when the ASDYM equations hold.

In particular, $F_{wz}=0$ says that the connection restricted to each $(w,z)$ surface is a flat connection. On a simply connected coordinate patch, the compatible equationshave an invertible solution $g$. Applying the associated gauge transformation setsThis conclusion is local; global topology can obstruct a single such gauge over the whole space.

Because ${}^{\star }{}^{\star }=1$, the exterior algebra of two-forms has the orthogonal decompositionwhere ${}^{\star }H=H$ for a self-dual differential form and ${}^{\star }G=-G$ for an anti-self-dual differential form. The Hodge star is self-adjoint, soIt follows that

For an $SU(n)$ connection, use the positive normand define the Second Chern number byOrthogonality givesTherefore the Euclidean Yang-Mills actionobeys the Yang-Mills instanton Bogomolny boundEquality holds precisely when $F_{-}=0$ or $F_{+}=0$, according to the sign of $k$; these are the self-dual and anti-self-dual Yang-Mills instantons. Other trace and orientation conventions may reverse $k$ but leave the absolute-value bound unchanged.

Put $\alpha =\delta \eta$. The hypothesis says that the exact two-form $\beta =d\alpha$ is anti-self-dual. Since $M$ is compact without boundary, Stokes theorem givesHence $d\alpha =0$ by the exact anti-self-dual form on a compact four-manifold argument. Since $\delta$ is the formal $L^2$ adjoint of $d$,Therefore $\boxed{\delta \eta =0}$.

On $p$-forms in dimension $n$, the defining identity for the Hodge star operator givesFor $n=4$ and $p=2$, therefore, ${\ast }^2=1$. For every $\alpha \in {\Omega }^2(N)$ defineThen $\ast {\alpha }_{+}={\alpha }_{+}$, $\ast {\alpha }_{-}=-{\alpha }_{-}$, and $\alpha ={\alpha }_{+}+{\alpha }_{-}$. The two eigenspaces of the involution $\ast$ have zero intersection, which proves uniqueness. They are respectively the spaces of self-dual and anti-self-dual two-forms.

Now suppose $N$ is compact and let $\beta$ be an exact three-form, say $\beta =d\theta$. Apply the Hodge decomposition theorem to the two-form $\theta$:Set $a=\delta \psi$. Then $da=d\theta =\beta$ and $\delta a={\delta }^2\psi =0$. For a two-form in dimension four, $\delta =-\ast d\ast$, so $d\ast a=0$. The self-dual formsatisfiesThus every exact three-form is the exterior derivative of a self-dual two-form.

On a complex two-dimensional Kähler manifold, a primitive $(1,1)$ form satisfies $\ast \alpha =-\alpha$, making it an anti-self-dual differential form. For a nonzero real primitive harmonic form on a compact surface, $\int \alpha \wedge \alpha =-\Vert \alpha {\Vert }_{L^2}^2<0$.