arXiv · 2204.12628
$\mathbb{S}^6$ (or any of $\mathbb{S}^2 \times \mathbb{S}^4$, $\mathbb{S}^2\times\mathbb{S}^6$, or $\mathbb{S}^6\times \mathbb{S}^6$, respectively) is not diffeomorphic to a complex manifold
Abstract
We identify all metrics on a closed $n$-manifold with their Nash isometric embeddings into a standard sphere of large, but fixed dimension, and use the Palais' isotopic extension theorem to identify their deformations with the isotopic deformations of their embeddings, the deformations of metrics in a conformal class identified with their corresponding isotopic conformal deformations. If $n\geq 3$, we characterize metrics of constant scalar curvature in terms of properties of extrinsic quantities of their associated embeddings, and prove that any metric on the manifold of constant positive scalar curvature, which can be minimally embedded into this background sphere, is a Yamabe metric in its conformal class. We then use Simons' gap theorem to study the extrinsic quantities of almost complex Hermitian deformations, by Yamabe metrics, of the standard minimal almost complex isometric embeddings of $\mb{S}^6$, $\mb{S}^2 \times \mb{S}^4$, $\mb{S}^2\times\mb{S}^6$, and $\mb{S}^6\times \mb{S}^6$, respectively, and prove that none of these manifolds carry integrable almost complex structures. :
Explore related subjects
Keep this discovery
Santiago R Simanca. 2022-04-26. $\mathbb{S}^6$ (or any of $\mathbb{S}^2 \times \mathbb{S}^4$, $\mathbb{S}^2\times\mathbb{S}^6$, or $\mathbb{S}^6\times \mathbb{S}^6$, respectively) is not diffeomorphic to a complex manifold. https://arxiv.org/abs/2204.12628
Cite the original work for its findings. Save a collection to share your selection of sources.