Is the length squared of the gradient ($g(\nabla f_p,\nabla f_p)$) independent of the coordinates?

Question

Put on $\mathbb{R}^2$ the standard riemannian metric $g=dx^2+dy^2$ and consider a differentiable function $f:\mathbb{R}^2\rightarrow \mathbb{R}$. I'll write $\nabla f_p$ for the gradient of $f$ computed in $p\in \mathbb{R}^2$. I want to understand if $g(\nabla f_p,\nabla f_p)$ is independent of the coordinates (the writing $g(\nabla f_p,\nabla f_p)$ seems independent of the coordinates. I'll explain myself better:

$g(\nabla f_p,\nabla f_p)$ is clearly equal to $(\partial_xf_p)^2+(\partial_y f_p)^2$. Now, consider $V$, $W$ vector fields which are orthonormal in every point of a neightborhood $U$ of $0\in \mathbb{R}^2$ with respect to $g$. Is it true $(\partial_xf_p)^2+(\partial_y f_p)^2=(Vf_p)^2+(Wf_p)^2$?

Intuitively this seems true, but I don't know how to do the right calculations:

suppose $V=v_x\partial_x+v_y\partial_y$ and $W=w_x\partial_x+w_y\partial_y$, with $g(V,V)=v_x^2+v_y^2=1$,$g(W,W)=w_x^2+w_y^2=1$ and $g(V,W)=v_xw_x+v_yw_y=0$.

Then $(Vf_p)^2+(Wf_p)^2=(\partial_xf_p)^2(v_x^2+w_x^2)+(\partial_yf_p)^2(v_y^2+w_y^2)+2\partial_xf\partial_yf(v_xv_y+w_xw_y)$ which doesn't seem equal to $(\partial_xf)^2+(\partial_yf)^2$... What am I doing wrong?

Answer 1

What you're forgetting is that if the columns of a matrix are orthonormal, then so are the rows! So there's nothing wrong with your calculation. :)

Answer 2

By definition, the gradient of a smooth function $f$ defined in a Riemannian manifold $(M,g)$ is the vector field that is metrically equivalent to the differential ${\rm d}f$, that is, it is the vector field ${\rm grad}\,f$ such that $${\rm d}f(X) = g({\rm grad}\,f,X)$$for all vector field $X$. This clearly do not depend on coordinates. The notation $\nabla f$ for the gradient is not too good in this case, since if $\nabla$ is a linear connection in $M$, we have $\nabla_Xf = X(f) = {\rm d}f(X)$, so $\nabla f = {\rm d}f$ instead (in $\Bbb R^n$, we're sort of identifying ${\rm grad}\,f$ and ${\rm d}f$).

In coordinates $x^1,\cdots,x^n$, if $g_{ij} = g\big(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\big)$, we have $${\rm grad}\,f = \sum_{i,j=1}^n g^{ij}\frac{\partial f}{\partial x^j}\frac{\partial}{\partial x^i},$$where $(g^{ij})$ is the inverse matrix of $(g_{ij})$, which exists since $g$ is non-degenerate. If $M = \Bbb R^n$ and we work with cartesian coordinates so that $g_{ij} = \delta_{ij}$, we get the expression from early calculus courses.

To summarize: $g({\rm grad}\,f,{\rm grad}\,f)$ is independent of coordinates. You are not doing anything wrong.