Hi,
This is my question. Can we compute easily the differential of the following map ?
$$ f:(x,\xi^\star)\in TS^{2n-1} \mapsto \xi^\star(ix)\in \mathbb{R} $$
where $TS^{2n-1}$ is the cotangent bundle of the odd sphere $S^{2n-1}\subset \mathbb{C}^n$. Notice that $f$ is well defined because if $x\in S^{2n-1}\subset \mathbb{C}^n$ then $ix\in T_x S^{2n-1}$.
Thanks
The map is the restriction to $TS^{2n-1}$ of a quadratic map, so its differential is the restriction of the differential of that map. In coordinates $z^{\mu}$ on $\mathbb{C}^n$, we get coordinates $z^{\mu},\xi_{\mu}$ on $T^*\mathbb{C}^n$, and $z^{\mu},\xi^{\mu}$ on $T\mathbb{C}^n$. The function is $f=\left<\xi, \sqrt{-1} z\right>=-\sqrt{-1}\xi^{\mu} z^{\bar{\mu}} +\sqrt{-1} \xi^{\bar{\mu}} z^{\mu}$, so has differential $df=-\sqrt{-1}\left(\xi^{\mu} dz^{\bar{\mu}} + z^{\bar{\mu}} d\xi^{\mu}\right)+\sqrt{-1}\left(\xi^{\bar\mu} dz^{\mu} + z^{\mu} d\xi^{\bar\mu}\right)$. The equations on the sphere are $z^{\mu} z^{\bar{\mu}} = 1$ so the tangent bundle of the sphere is $z^{\mu} \xi^{\bar{\mu}} + \xi^{\mu} z^{\bar{\mu}} = 0$. Taking exterior derivative, you can simplify $df$ a little.
