Lower bounds on the norm of the gradient of a dot product on an $n$-sphere I'm working with an integral that involves terms of the form $\nabla (x\cdot \omega)$, where $x, \omega \in \mathbb{R}^n$, and $\omega$ is a point on $\mathbb{S}^{n-1}$. That is to say $$ \omega = (\omega_1(\phi_0, \ldots, \phi_{n-2}), \ldots, \omega_n(\phi_0, \ldots, \phi_{n-2})),$$ where $\phi_0 \in [0,2\pi]$ and $\phi_1, \ldots,\phi_{n-2} \in [0,\pi]$ are the usual $n$-dimensional spherical coordinates (i.e. assuming $r = 1$ is fixed). So, for example, in $\mathbb{R}^3$ on $\mathbb{S}^2$ we have $\omega = (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$, whereas $\mathbf{x} = (x,y,z)$, meaning
$$\mathbf{x}\cdot \omega = x\cos\theta \sin\phi + y\sin\theta \sin\phi + z\cos\phi.$$
In particular, I have shown that $\nabla_{\phi}(x\cdot \omega) \cdot \nabla_{\phi} e^{(x\cdot \omega)} = \left| \nabla_{\phi}(x\cdot \omega) \right|^2 e^{(x\cdot \omega)}$ from standard multivariable calculus identities, which leads me to (hopefully) conclude the following identity:
$$e^{(x\cdot \omega)} = \left( \frac{\nabla_{\phi} (x\cdot \omega)}{\left| \nabla_{\phi}(x\cdot \omega) \right|^2}  \cdot \nabla_{\phi} \right) e^{(x\cdot \omega)}.$$
However in order to do that I would need to conclude that $\left| \nabla_{\phi}(x\cdot \omega) \right| > 0$, which is where I'm struggling.
I have that $x \neq 0$, and that it is constant with respect to the $\phi_k$ variables. Shouldn't I be able to relate the norm of the gradient to the norm of $x$ somehow? I feel like I'm getting lost in the weeds here, so any help is much appreciated.
 A: I'm not sure I understand entirely. 
In the 3-space case, we have  $\omega_1(\theta, \phi) = \cos \theta \sin \phi$, is that right? 
In other words, $\omega_i( \ldots )$ is the $i$th rectangular coordinate of the point whose spherical coordinates are the arguments. So the $\omega_i$ are known functions, and predetermined, right? In particular, you've written out the three of them for $S^2$. 
And your hope is that for any fixed vector $x$, and any location $\phi_0$, we have
$$
| \nabla_\phi (x \cdot \omega(\phi)|(\phi_0) > 0,
$$
is that right? 
If that's right, then you're having a hard time proving it because it's not true. :( 
Look at $x = (0,0,1)$, and $(\theta_0, \phi_0) = (0, 0)$, so that $x$ is a vector pointing to the north pole, and $\omega(\theta_0, \phi_0)$ happens to point in the same direction as $x$. As it happens, in this case $x \cdot \omega(\theta, \phi)$ is easy to compute -- it's just $\omega_3 (\theta, \phi) = \cos \phi$. The gradient of this with respect to $\theta, \phi$ is $(0, \sin \phi)$. Evaluated at $(\theta_0, \phi_0) = (0,0)$, this gives the vector $(0,0)$, whose norm is not, in fact, positive. 
