# On nowhere zero self-dual 2-forms

Let $(X, g_x)$ be a smooth, oriented, Riemannian 4-diemnsional manifold. Let $\Lambda^2$ denote the bundle of 2-forms over $X$. Then the Hodge-star decomposes $\Lambda^2$ into the space of self-dual and anti-self-dual 2-forms $$\Lambda^2 = \Lambda^2_+ \oplus \Lambda^2_- .$$ Assume that there exists a non-vanishing section of $\Lambda^2_+$, say $\omega$. Then $\omega$ defines an almost-complex structure $I$ over $X$. Consider the complexification with respect to the almost-complex structure: $$\Lambda^2 \otimes \mathbb{C} = \Lambda^{2,0} \oplus \Lambda^{0,2} \oplus \Lambda^{1,1}.$$

Then $$\Lambda^2_+ = \Lambda^{2,0} \oplus \Lambda^{0,2} \oplus \mathbb{C}\cdot \omega~~\text{and}~~ \Lambda^2_- = \Lambda^{1,1}_{0}$$ where, $\Lambda^{1,1}_{0}$ is the space orthogonal to the space spanned by $\omega$ in $\Lambda^{1,1}$.

Let $\mathcal{Z}$ denote the twistor space of $X$ which is just the sphere bundle inside $\Lambda^2_+$. We can write $\omega$ as a real-valued function times a section of $\mathcal{Z}$ $$\omega = |\omega |\cdot \left(\frac{\omega}{|\omega|}\right):= |\omega|\cdot \omega', ~~ \omega' \in \Gamma(\mathcal{Z}).$$

My question is the following:

Consider $\nabla \omega$, where $\nabla$ is the Levi-Civita connection. In four dimensions $\nabla \omega \in \Lambda^{1,2} \oplus \Lambda^{2,1}$, the first component being the Nijenhuis tensor and the second one $d\omega$. However, $$d \omega = d|\omega| \wedge \omega' + |\omega|\cdot d\omega'.$$

The first term contains both $(2,1)$ and $(1,2)$-components, contradicting the fact that $d\omega \in \Lambda^{2,1}$. Am missing something here?

• Your assertion that $\nabla\omega$ can be interpreted as a section of $\Lambda^{1,2}\oplus \Lambda^{2,1}$ is in error. In fact, $\nabla\omega$ is a section of $\Lambda^2_+\otimes \Lambda^1$, a vector bundle of (real) rank $3\times 4 = 12$, whereas $\Lambda^{1,2}$ and $\Lambda^{2,1}$ are vector bundles of (real) rank $4$. You are missing another $4$-dimensional representation of $\mathrm{U}(2)$. By the way, as a representation of $\mathrm{SO}(4)$, $\Lambda^2_+\otimes \Lambda^1$ splits as the sum of two vector bundles, one of rank $4$ and the other of rank $8$. – Robert Bryant Jul 10 '17 at 12:16
• My apologies. Let me be more specific. The almost complex structure give the isomorphism $$T^{\ast}X \otimes_{\mathbb{R}} \overline{K} \cong T^{\ast}X \otimes_{\mathbb{C}} K \oplus T^{\ast}X \otimes_{\mathbb{C}} \overline{K}.$$ where $K$ is a complex line bundle. Moreover, the wedge product gives a complex, bi-linear map $$T^{\ast}X \times T^{\ast}X \longrightarrow \Lambda^2 T^{\ast}X = K.$$ using which, we can identify $TX \cong T^{\ast}X \otimes_{\mathbb{C}} \overline{K}$. These are the 2 components of $\nabla \omega$ which I mention in my question. – Varun Jul 10 '17 at 13:37
• You are confusing various decompositions of the vector bundles involved. For example, $\mathrm{d}\omega$ is a real 3-form (since $\omega$ is a real $2$-form), and, as such, $\mathrm{d}\omega$ does not lie in either $\Lambda^{2,1}$ or $\Lambda^{1,2}$, but 'diagonally' in the sum of the two, as $\Lambda^3(T^*)\otimes\mathbb{C} = \Lambda^{2,1} \oplus \Lambda^{1,2}$. (In fact, each of the terms in your 'formula' for $\mathrm{d}\omega$ are sections of $\Lambda^3(T^*)$.) – Robert Bryant Jul 10 '17 at 13:58