What is the gradient of length of gradient of function $f$?

Question

Suppose $f\in C^3(M)$ and $\nabla f$ denotes the gradient of $f$ w.r.t Riemannian metric $g$. Then what is the equivalent expression of the following? $$\nabla \langle \nabla f, \nabla f\rangle=?$$

Background problem: By definition $\mathrm{Hess} f(X, Y) = \langle \nabla _X(\nabla f), Y\rangle$; $\triangle f= \mathrm{tr}(\mathrm{Hess} f)$. Now I want to calculate $\triangle|\nabla f|^2$. By above definitions we have: $$\triangle|\nabla f|^2=\mathrm{tr}(\mathrm{Hess} \langle\nabla f,\nabla f\rangle)=\langle \nabla _{X_i}(\nabla \langle \nabla f, \nabla f\rangle), X_i\rangle.$$

But I saw somewhere that $$\triangle|\nabla f|^2=\sum_iX_iX_i\langle\nabla f,\nabla f\rangle.$$

Why these two (my calculation and last one) are equal?

Answer 1

I'm using tensor notation.

$$ \nabla\langle\nabla f,\nabla f\rangle\equiv \nabla^a(\nabla_b f\:\nabla^bf)=(\nabla^a\nabla_bf)\nabla^bf+\nabla_bf(\nabla^a\nabla^bf)=2(\nabla^a\nabla^bf)\nabla_bf $$

$$ \nabla_XZ \equiv X^a\nabla_aZ^b $$

$$ \mathbf{Hess}\langle\nabla f,\nabla f\rangle(X,Y) \equiv X^a\{\nabla_a \nabla_c(\nabla_b f\:\nabla^b f)\}Y^c = X^a(\nabla_a \nabla_c h)Y^c $$

To compute trace of (2,0)-tensor, we have to rise one index and then contract them. To do this we need dual basis to $X_i$, and when $\langle X_i,X_j\rangle=\delta_{ij}$ we get it by lowering index of $(X_i)^a$.

So $\mathbf{Tr\:T}=\sum_i{T_a}^b(X_i)^a(X_i)_b=\sum_iT_{ab}(X_i)^a(X_i)^b$

Action of a vector field on a function is $Xf\equiv X^a\nabla_a f$, so

$$ X\:Y h\equiv X^c\nabla_c(Y^a\nabla_ah)=(X^c\nabla_cY^a)\nabla_ah+X^c Y^a \nabla_c \nabla_a h $$

If $X_i$ satisfy geodesic equation then $\nabla_{X_i}X_i\equiv (X_i)^c\nabla_c(X_i)^a=0$ and we are done, because

$$ \mathbf{Tr}(\mathbf{Hess}\:h)=\sum_i{X_i X_i}\:h $$

when $X_i$ are orthonormal, geodesic coordinate vector fields.

Answer 2

I think I got the answer. Thanks for posting the question I have faced the same problem in the last few days. Here is a generalized solution. If I made any error kindly let me know. We know that if $g$ is the Riemannian metric then $\langle\nabla f,\nabla h\rangle =g(\nabla f,\nabla h)=g_{pq}\nabla^p f\nabla^q h=\nabla_q f\nabla^q h$ (called lowering the index [similarly raising the index is another term]) thus

$$\begin{eqnarray} \nabla\langle\nabla f,\nabla h\rangle &\equiv& \nabla^a(\nabla_b f\nabla^b h)\\ &=& \nabla^a(\nabla_b f)\nabla^b h+\nabla^a(\nabla^b h)\nabla_b f\\ &=& \nabla^a(\nabla_b f)g^{kb}\nabla_k h+\nabla^a(\nabla^b h)\nabla_b f\\ &=& \nabla^a(g^{kb}\nabla_b f)\nabla_k h+\nabla^a(\nabla^b h)\nabla_b f\text{ (Since $\nabla g=0$)}\\ &=& \nabla^a(\nabla^k f)\nabla_k h+\nabla^a(\nabla^b h)\nabla_b f \end{eqnarray},$$ Changing the dummy index $k$ to $b$ we obtain. $$\begin{eqnarray}&=&\nabla^a(\nabla^b f)\nabla_b h+\nabla^a(\nabla^b h)\nabla_b f\\ &\equiv& \text{Hess }f\ \nabla h+\text{Hess }h\ \nabla f\end{eqnarray}$$ In particular for $f=h$ we have the answer by user $87091403130$ which is $\nabla \langle\nabla f,\nabla f\rangle=2\text{Hess }f\ \nabla f$

You've asked another question about calculating $\Delta |\nabla f|^2$, I think we can use the Bochner formula to calculate over Euclidean space, Riemannian manifold and other suitable spaces. One can try to see the proof of it for better understanding.