Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 22 / 1

An almost complex structure is a smooth real vector bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular, $M$ has even real dimension $2n$. Extend $J$ complex-linearly to $T_{\mathbb{C}}M=TM{\otimes }_{\mathbb{R}}\mathbb{C}$. Its minimal polynomial divides $(u-i)(u+i)$, whose roots are distinct, so the type decomposition of the complexified tangent bundle isThe smooth projections are $(I-iJ)/2$ and $(I+iJ)/2$. Complex conjugation exchanges the two rank-$n$ eigenbundles. A $(1,0)$-covector annihilates $T^{0,1}M$ and a $(0,1)$-covector annihilates $T^{1,0}M$. Thus the differential forms of type (p, q) are the smooth sections ofHere the tensor product is identified with its image under the wedge product of differential forms. For a real tangent vector $v$, the vectors $v-iJv$ and $v+iJv$ exhibit the two types directly.

We now construct the almost complex structure from a decomposable volume form. Locally write $\Omega ={\theta }_1\wedge \cdots \wedge {\theta }_n$. The nonvanishing of $\Omega \wedge \overline{\Omega }$ says that the $2n$ covectors ${\theta }_1,\dots ,{\theta }_n,{\overline{\theta }}_1,\dots ,{\overline{\theta }}_n$ form a complex coframe. In particular, the span of the ${\theta }_j$ has the intrinsic descriptionIndeed, expansion in that coframe shows that every coefficient of a conjugate factor must vanish if $\alpha \wedge {\Omega }_x=0$. Therefore this span is independent of the chosen local factorization. The local coframes show that $W$ is a smooth rank-$n$ vector subbundle, andDefine $J_{\Omega }^{\ast }$ to be multiplication by $i$ on $W$ and by $-i$ on $\overline{W}$. It commutes with complex conjugation, so is the complexification of a real smooth bundle endomorphism, whose dual defines $J_{\Omega }$ on $TM$. It has square $-I$, and its $(1,0)$-covectors are exactly $W$. Conversely, any such almost complex structure must have this same action on the entire coframe. Thus $J_{\Omega }$ exists globally and is unique, and every factor in every allowed decomposition is a $(1,0)$-form.

Finally suppose $d\Omega =0$. Since ${\theta }_j\wedge \Omega =0$, the exterior derivative product rule givesDecompose $d{\theta }_j$ into types $(2,0)$, $(1,1)$ and $(0,2)$ using $J_{\Omega }$. The first two wedge products with the $(n,0)$-form $\Omega$ vanish by dimension. On $(0,2)$-forms, wedging with $\Omega$ is injective: the monomials $\Omega \wedge {\overline{\theta }}_a\wedge {\overline{\theta }}_b$ are linearly independent. For $n=1$ the $(0,2)$ space is zero, so the statement still holds. Hence every $d{\theta }_j$ has no $(0,2)$ component. For an arbitrary $(1,0)$-form $\alpha ={\sum }_j{a}_j{\theta }_j$, the extra terms $d{a}_j\wedge {\theta }_j$ likewise have only types $(2,0)$ and $(1,1)$. The bracket and differential-form criteria for integrability proved below now giveLocal decomposability has been used essentially; the nonvanishing top-degree product alone does not provide the required coframe.

These are local criteria: multiplying by a smooth bump function extends any local test field or form to a global one without changing it near a chosen point. Assume the Lie bracket of any two sections of $T^{1,0}M$ again lies in $T^{1,0}M$. Because the almost complex structure is real, conjugating this condition shows that the Lie bracket of any two sections $U,V$ of $T^{0,1}M$ lies in $T^{0,1}M$.

For a $(1,0)$-form $\alpha$, the exterior derivative formula isThe first two terms vanish because $\alpha$ annihilates $T^{0,1}M$, and the last vanishes by bracket closure. The $(0,2)$ component is exactly the restriction of $d\alpha$ to two arguments in $T^{0,1}M$. Thus it is zero. Since the only types of a two-form are $(2,0)$, $(1,1)$ and $(0,2)$, the bracket condition implies the stated differential-form condition. This argument uses neither holomorphic coordinates nor an integrability theorem as an unproved intermediate step.

Conversely, suppose the exterior derivative of every $(1,0)$-form has no $(0,2)$ part. For sections $U,V$ of $T^{0,1}M$, the same exterior derivative formula givesThe common annihilator of all $(1,0)$-covectors is $T^{0,1}M$. Local $(1,0)$-coframes therefore show that $[U,V]$ belongs to $T^{0,1}M$. Conjugating gives bracket closure of $T^{1,0}M$. Thus the two criteria are equivalent. These are the bracket and differential-form criteria for integrability, and establish integrability in the sense used here.