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Added the missing word 'complex' and a clarifying phrase
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Robert Bryant
Robert Bryant
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For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.

For your second question: The answer is 'no' in general. If this can be done, then the tangent bundle of $M$ can be written as the direct sum of $n$ oriented $2$-plane bundles (i.e., complex line bundles), and it is easy to give examples of manifolds for which such a decomposition cannot hold. For example, just take $\mathbb{CP}^2$ with its standard symplectic structure. Its tangent bundle is not the sum of two complex line bundles of this kind for reasons having to do with its Chern and Euler classes.

For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.

For your second question: The answer is 'no' in general. If this can be done, then the tangent bundle of $M$ can be written as the direct sum of $n$ oriented $2$-plane bundles, and it is easy to give examples of manifolds for which such a decomposition cannot hold. For example, just take $\mathbb{CP}^2$ with its standard symplectic structure. Its tangent bundle is not the sum of two line bundles of this kind for reasons having to do with its Chern and Euler classes.

For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.

For your second question: The answer is 'no' in general. If this can be done, then the tangent bundle of $M$ can be written as the direct sum of $n$ oriented $2$-plane bundles (i.e., complex line bundles), and it is easy to give examples of manifolds for which such a decomposition cannot hold. For example, just take $\mathbb{CP}^2$ with its standard symplectic structure. Its tangent bundle is not the sum of two complex line bundles of this kind for reasons having to do with its Chern and Euler classes.

Source Link
Robert Bryant
Robert Bryant
  • 108.4k
  • 8
  • 342
  • 453

For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.

For your second question: The answer is 'no' in general. If this can be done, then the tangent bundle of $M$ can be written as the direct sum of $n$ oriented $2$-plane bundles, and it is easy to give examples of manifolds for which such a decomposition cannot hold. For example, just take $\mathbb{CP}^2$ with its standard symplectic structure. Its tangent bundle is not the sum of two line bundles of this kind for reasons having to do with its Chern and Euler classes.