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A paper I'm reading defines the following $\left(\frac{(k-1)k}{2}+1\right)\times n$ matrix:

\begin{align*} \Omega_f(x) := \begin{bmatrix} \omega_{1,2}(x)\\ \vdots \\ \omega_{i,j}(x) \\ \vdots \\ \omega_{k-1,k}(x) \\ x_1, \ldots, x_n \end{bmatrix} \end{align*}

where $f\colon \mathbb{R}^n\rightarrow\mathbb{R}^k$, $f = (P_1,\ldots,P_k)$ and $\omega_{i,j} = P_i(x)\nabla P_j(x) - P_j(x)\nabla P_i(x)$. It is then stated that $\operatorname{rank} \Omega_f(x) = k$ is equivalent to saying that $\frac{f}{\|f\|}$ is a submersion. I've tried to work out why this is the case, and all I've come up with is that $\frac{\omega_{i,j}(x)}{P_i^2(x) + P_j^2(x)} = \nabla\chi(x)$ if $\chi(x) = \arctan\left(\frac{P_j(x)}{P_i(x)}\right)$.

I assume it has something to do with some parametrization of the sphere by angles, but I can't see how to get from $\Omega_f$ to $J(\frac{f}{\|f\|})$ or back.

Any help is appreciated.

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  • $\begingroup$ I'm a bit confused: is $p=k$ and $m=n$? $\endgroup$ Nov 19, 2013 at 14:33
  • $\begingroup$ If I'm not wrong, the $ij$ entry of the Jacobian of $f/\|f\|$ is $$\frac{1}{\|f\|^3}\langle f,((\omega_{1i})_j,\ldots,(\omega_{ki})_j)\rangle,$$ where $(\omega_{li})_j$ is the $j$-th component of $\omega_{li}$. $\endgroup$ Nov 19, 2013 at 15:15
  • $\begingroup$ @A.Bellmunt Yes, sorry. Changed now. $\endgroup$
    – Greebo
    Nov 20, 2013 at 9:07

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