Gradient in level-set coordinates.

Question

If f:$\mathbb{R}^n \to \mathbb{R}$ is a smooth function and $x_0$ is a regular point, we know there is a diffeomorphism $\phi:U \to \phi(U)\subset \mathbb{R}^n$, with $x_0 \in \phi(U)$ such that

$$f(\phi(x)) = x_n, \quad \text{for } x=(x_1,\ldots,x_n) \in U.$$

Using Riemannian Geometry notation, can we say that that the coordinate frame induced by the local chart $(U,\phi)$ satisfies $\partial_n = \text{grad} f$? It seems to me that this should be the case, since the composition with $\phi$ depends only on the last variable.

Answer 1

I assumed that you use Euclidean metric $g=(\delta_{ij})$ in $\mathbb{R}^n$, so gradient $\text{grad }f$ in standard coordinate $(x^1,\dots,x^n)$ is the vector field $$ \text{grad }f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} \frac{\partial}{\partial x^i}. $$ First of all if we denote $(\widetilde{x}^1,\dots,\widetilde{x}^n)$ as any coordinate system other than the standard coordinate, then the vector field $$ \sum_{i=1}^n \frac{\partial f}{\partial \widetilde{x}^i} \frac{\partial}{\partial \widetilde{x}^i}, $$ is (in general) not the same as the gradient $\text{grad }f$ above since $\text{grad }f$ should transform as $$ \text{grad }f = \sum_{i,j=1}^n \widetilde{g}^{ij} \frac{\partial f}{\partial \widetilde{x}^i} \frac{\partial }{\partial \widetilde{x}^j}. $$

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The short answer is that the claim is generally false because of above misinterpretation. The following example will illustrate this.

Consider the following smooth submersion $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y)=e^x$ and diffeomorphism $\phi : (0,\infty)\times \mathbb{R} \to \mathbb{R}^2$ defined by $\phi(u,v) = (x,y) =(\ln u,v)$. Then $$ (f \circ \phi) (u,v) = f(\ln u,v) =u, \quad \forall (u,v) \in (0,\infty) \times \mathbb{R}. $$ So this is the chart we're willing to work with. Now, it is clear that $$ \frac{\partial}{\partial u}\Big|_{(u,v)} = \bigg[\frac{\partial f}{\partial u} \frac{\partial}{\partial u} + \frac{\partial f}{\partial v} \frac{\partial}{\partial v}\bigg]_{(u,v)} \quad \text{ since } (f\circ \phi) (u,v) = u. $$ But on standard coordinates $(x,y)$, gradient vector field $$ \text{grad }f|_{(x,y)}= \bigg[\frac{\partial f}{\partial x} \frac{\partial}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial}{\partial y}\bigg]_{(x,y)} = e^x \frac{\partial}{\partial x} \bigg|_{(x,y)}, \quad (x,y) = \phi(u,v), $$ is clearly not equal to \begin{align*} \frac{\partial}{\partial u}\bigg|_{\phi(u,v)=(x,y)} &= d\phi\Big(\frac{\partial}{\partial u}\Big|_{(u,v)}\Big) = \bigg[\frac{\partial x}{\partial u} \frac{\partial}{\partial x} + \frac{\partial y}{\partial u} \frac{\partial}{\partial y}\bigg]_{\phi(u,v)} \\ &= \frac{1}{e^x} \frac{\partial}{\partial x}\Big|_{(x,y)}. \end{align*} So now we see a phenomenon that $\text{grad }f$ is not equal to $\partial_u$ on the original chart $\mathbb{R}^2$ but $$ \frac{\partial f}{\partial u} \frac{\partial}{\partial u} + \frac{\partial f}{\partial v} \frac{\partial}{\partial v} \equiv \frac{\partial}{\partial u} \quad \text{on }(0,\infty) \times \mathbb{R}. $$ This is happen because vector field $\frac{\partial f}{\partial u} \frac{\partial}{\partial u} + \frac{\partial f}{\partial v} \frac{\partial}{\partial v}$ is not actually $\text{grad }f $, because $\text{grad }f$ transformed in $(u,v)$ coordinate as $$ \text{grad }f = g^{uu} \frac{\partial f}{\partial u} \frac{\partial}{\partial u} + 2g^{uv} \frac{\partial f}{\partial u} \frac{\partial}{\partial v} + g^{vv} \frac{\partial f}{\partial v} \frac{\partial}{\partial v} = u^2 \frac{\partial}{\partial u}. $$