How to calculate the gradient of the function $\frac{1}{2}|\nabla f|^2$?

Question

Let $(M,g)$ be a Riemannian manifold. Denote the covariant differentiation by $\nabla$. Let $f$ be a smooth function on $M$. Denote the gradient of $f$ by $\nabla f$. I want to prove that $$\nabla\left(\frac{1}{2}|\nabla f|^2\right)=\nabla_{\nabla f}\nabla f.$$ My attempt: Take a vector field $X$ on $M$. Using the metric property of covariant differentiation, we have $$D_Xg\left(\nabla f,\nabla f\right)=2g(\nabla_{X}\nabla f, \nabla f).$$ Using the definition of gradient, we have $$g\left(\nabla\left(\frac{1}{2}|\nabla f|^2\right),X\right)=\frac{1}{2}D_X|\nabla f|^2.$$ Using the definition of the norm, $|\nabla f|^2=g\left(\nabla f,\nabla f\right)$, we get $$g\left(\nabla\left(\frac{1}{2}|\nabla f|^2\right),X\right)=g(\nabla_{X}\nabla f, \nabla f).$$ It seems that I need to prove that $$g(\nabla_{X}\nabla f, \nabla f)=g(\nabla_{\nabla f}\nabla f, X),$$ but I have no idea.

Answer 1

Note that in a local coordinates,

$$ \frac{1}{2} |\nabla f|^2 = \frac{1}{2} g^{ij} f_i f_j,$$ and for any function $h$, $$\nabla h = g^{ij} h_i \frac{\partial}{\partial x^j}$$ hence

$$\nabla \left( \frac{1}{2} |\nabla f|^2\right) = g^{ij} \left(\frac{1}{2} g^{kl} f_k f_l\right)_i \frac{\partial}{\partial x^j}.$$

If we calculate at the center of a normal coordinates, $g^{kl} = \delta_{kl}$ and $\partial _i g^{kl}=0$. Hence

$$\nabla \left( \frac{1}{2} |\nabla f|^2\right) = \sum_k f_k f_{ki} \frac{\partial}{\partial x^i}.$$

On the other hand,

$$\nabla_{\nabla f} \nabla f = f_i \nabla_{\frac{\partial}{\partial x^i}} \left( g^{jk} f_k \frac{\partial }{\partial x^j}\right) = \sum_k f_i f_{ki} \frac{\partial}{\partial x^i}$$ since $\Gamma_{ij}^k = 0$.

Answer 2

Consider a vector field $X$ defined on the manifold $(M,g)$. By the definition of the gradient, we have:

$$g(\nabla |\nabla f|^2,X) = X|\nabla f|^2 = Xg(\nabla f,\nabla f) = 2g(\nabla_{X} \nabla f,\nabla f)$$

Now by symmetry of the Hessian, that is $$\operatorname{Hess}f(X,Y) = g(\nabla_{X} \nabla f, Y) = g(\nabla_{Y} \nabla f, X) = \operatorname{Hess}f(Y,X).$$ Then we have that $$ g(\nabla |\nabla f|^2,X)=2g(\nabla_{X} \nabla f, \nabla f) = 2g(\nabla_{\nabla f} \nabla f, X). $$ Now, since $X$ is arbitrary this allows us to deduce that: $$ \nabla |\nabla f|^2=2\nabla_{\nabla f} \nabla f. $$