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I have a calculation involving a gradient and a parametrization, but I haven't been able to find out the relation between them. Let me explain.

Let $f:X↦R$ be a smooth function and $\mathrm{grad}f\in \mathcal T X$ its gradient. $X$ is a Riemannian manifold, and therefore $\mathrm{grad}f$ has a norm. While my calculation requires this norm, the only expression that I have is through a parametrizion $\alpha:R^n \mapsto X$.

This parametrization, yields the vector $\mathrm{grad}f \circ \alpha \in \mathcal T R^n$, whose norm I can compute. However, I haven't been able to find how these two norms are related.

I hope someone can point me in the right direction. Thank you and have a nice day.

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You should pull back the Riemannian metric $g$ on $X$ under the map $\alpha$. This defines a Riemannian metric $\alpha^*g$ on $\mathbb R^n$. The norm of $\nabla (f\circ \alpha)$ in this metric gives the answer to your question.

I'll illustrate this by an example. Let $X$ be the unit sphere in $\mathbb R^3$ with the metric induced from $\mathbb R^3$. The map $\alpha$ is $$\alpha(\theta,\phi)=(\cos\theta \sin \phi, \sin\theta\sin\phi, \cos\phi)\tag1$$ What does it mean to pull back the metric under $\alpha$? It means that the length of vector $p\frac{\partial}{\partial \theta}+q\frac{\partial}{\partial \phi}$ from the tangent space to $\mathbb R^2$ at a point $(\theta,\phi)$ is the length of $$\left(p\frac{\partial}{\partial \theta}+q\frac{\partial}{\partial \phi}\right)\alpha = (-\sin\theta \sin\phi, \cos\theta\sin\phi,0)\,p+ (\cos\theta \cos \phi, \sin\theta\cos\phi, -\sin\phi)\,q $$ in the metric of $X$ (which, for me, comes from $\mathbb R^3$). Using the Pythagorean theorem, I calculate $$\bigg\|p\frac{\partial}{\partial \theta} + q\frac{\partial}{\partial \phi}\bigg\|_{\alpha^*g}^2 = (\sin^2\phi)\, p^2 +q^2 \tag2$$ Now suppose I want to calculate the norm of gradient of the function $f(x,y,z)=xyz$. Working through $\alpha$, I find $$f\circ\alpha(\theta,\phi)=\cos\theta\sin\theta \sin^2\phi\cos\phi $$ and then calculate the gradient of $f\circ \alpha$ from its partial derivatives $$\nabla(f\circ \alpha) = (\cos 2\theta \sin^2\phi\cos\phi)\frac{\partial}{\partial \theta} + \cos\theta\sin\theta (2\sin\phi\cos^2\phi-\sin^3\phi)\frac{\partial}{\partial \phi} $$ According to (2), the norm of the gradient is the square root of $$ \cos^2 2\theta \sin^6\phi\cos^2\phi + \cos^2\theta\sin^2\theta (2\sin\phi\cos^2\phi-\sin^3\phi)^2 $$ Ugh, but this how it works in general.

If your parametrization $\alpha$ is conformal, the computations simplify because the pullback of $g$ under $\alpha$ is a multiple of the standard metric on $\mathbb R^n$. However, most $n$-manifolds with $n>2$ do not admit conformal parametrization.

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  • $\begingroup$ This makes sense. I wasn't sure if the push forward of the gradient in $R^n$ was the gradient -or its multiple- in $X$. It makes sense when the map's conformal, and then the scaling factor would be the Jacobian. Thanks. $\endgroup$ – Diego May 2 '13 at 18:05

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