# Derivative of norm of vector field

Let $M$ be a smooth riemannian manifold with metric $\langle \cdot, \cdot \rangle$ and Levi-Civita connection $\nabla$. For $X$ a vector field on $M$, define $f : M \to \mathbb{R}$ by $$f(p) = \langle X(p), X(p) \rangle.$$

I want to calculate the derivative of $f$. Is the following correct?

$$df(p) \cdot v = v(f) = v \langle X, X\rangle = 2 \langle \nabla_v X, X(p) \rangle, \qquad p \in M, \, v \in T_p M.$$

• yes, this is correct!! Commented Jul 30, 2016 at 19:48

Namely, fix a "Gauss" coordinate system centred at $p$ (the one constructed with the exponential map), so that $$g_{ij}(p)= \delta_{ij} \ \ \ \ \ \ \ \ \frac{\partial g_{ij}}{\partial x^\ell}(p)=0 \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \Gamma_{ij}^k(p)=0 \ .$$ Write $X = X^i \frac{\partial}{\partial x^i}$ (in Einstein notation) and observe that $$df = \frac{\partial f}{\partial x^\ell} dx^\ell = \frac{\partial }{\partial x^\ell} \big< X, X \big> dx^\ell = \frac{\partial }{\partial x^\ell} \Big< X^i\frac{\partial}{\partial x^i}, X^j\frac{\partial}{\partial x^j}\Big> dx^\ell = \frac{\partial }{\partial x^\ell} \Big( X^i X^j g_{ij} \Big) dx^\ell \ .$$
Thus, because of the Liebeniz rule $$df = \frac{\partial X^i }{\partial x^\ell} X^j g_{ij} dx^\ell +X^i\frac{\partial X^j }{\partial x^\ell} g_{ij} dx^\ell + X^i X^j \frac{\partial g_{ij}}{\partial x^\ell} dx^\ell \\ = 2\frac{\partial X^i }{\partial x^\ell} X^j g_{ij} dx^\ell + X^i X^j \frac{\partial g_{ij}}{\partial x^\ell} dx^\ell\ .$$ Evaluating at $p$ the expression simplifies to $$df(p) = 2\frac{\partial X^i }{\partial x^\ell}(p) X^j(p) \delta _{ij} dx^\ell = 2\frac{\partial X^i }{\partial x^\ell}(p) X^i(p) dx^\ell \ .$$ Now take $$v=v^i \frac{\partial}{\partial x^i}\Big|_p \in T_pM \ ,$$ and observe that (since $\Gamma_{ij}^k=0$ at $p$) $$df(p) \cdot v = 2\frac{\partial X^i }{\partial x^\ell}(p) X^i(p) v^\ell= 2 \cdot (\nabla_v X)^i \delta_{ij} X^j(p)= 2 \cdot \big<\nabla_v X , X\big>_p \ .$$ You can do something similar to prove the other equality.