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I am trying to solve exercise 1.9.2 from "An Introduction to Differential Geometry" by Vincent Minerbe (link: https://webusers.imj-prg.fr/uploads//vincent.minerbe/Geodiff/m2dg.pdf page 33). In short, we consider $\mathbb{S}^{2n+1}$ to be the unit sphere in $\mathbb{C}^{n+1}$, and we define a 1-form $\eta$ on it, given by $\eta_z(V)=\langle iz | V\rangle$ for each $z\in\mathbb{S}^{2n+1}$ (here $\langle\cdot | \cdot\rangle$ denotes the standard hermitian product on $\mathbb{C}^{n+1}$). I managed to prove that there is a unique 2-form $\Omega$ on $\mathbb{C}P^n\cong \mathbb{S}^{2n+1}/\mathbb{S}^1$ such that $\pi^*\Omega=d\eta$, where $\pi$ is the quotient map $\mathbb{S}^{2n+1}\rightarrow \mathbb{C}P^n$. The exercise then asks to prove that $\Omega$ is symplectic (closed and non-degenerate), and the hint is to show that $\eta\wedge (d\eta)^n$ is a volume form on $\mathbb{S}^{2n+1}$. I am stuck precisely here: I can't show that $\eta \wedge (d\eta)^n$ is always non-zero. The idea is to prove that for every point in $z\in\mathbb{S}^{2n+1}$ there exists a basis of the tangent space $(v_1, \dots, v_{2n+1})$ such that $\eta\wedge (d\eta)^n_z(v_1, \dots, v_{2n+1})\neq 0$. Choosing $v_1=iz$ and looking for the remaining vectors in the orthogonal (w.r.t. the hermitian metric of $\mathbb{C}^{n+1}$) makes the calculations a bit easier, but I still don't manage to compute an expression for $d\eta$, not even with these restrictions.

My question is: may I have a hint on how to prove that $\eta\wedge (d\eta)^n\neq 0$ without using an explicit formula for $d\eta$? If this tasks appears impossible, given an atlas for $\mathbb{S}^{2n+1}$ how can I find an expression for the hermitian product between two tangent vectors as a function of their coordinates?

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I have no interest in trying to do this in a coordinate-free manner. You should note that, using complex coordinates $(z_0,\dots,z_n)$ on $\Bbb C^{n+1}$, the $1$-form $\eta$ is given by $$\eta = \sqrt{-1}\sum z_j\,d\bar z_j,$$ since $\langle z,V\rangle = \sum z_j\overline V_j = \sum z_j\,d\bar z_j(V)$. Then $d\eta = \sqrt{-1}\sum dz_j\wedge d\bar z_j$, and the rest is easy.

(By the way, if you've never thought about it, since $z$ and $v$ are real orthogonal, the hermitian inner product has real part $0$ and so $i\langle z,V\rangle$ is necessarily real.)

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  • $\begingroup$ Thank you, I managed to conclude using your answer. $\endgroup$ – Qu0ll Oct 3 '19 at 14:55

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