Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
2  
For the record, multiplication by $i$ does not preserve $TS^{2n-1}$, because this is an odd-rank bundle. In fact, the space of complex tangencies to $S^{2n-1}\subset \mathbb{C}^n$ is the standard contact structure on $S^{2n-1}$. –  Marco Golla Aug 27 '11 at 8:52

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.