# Property of the gradient in Differential Geometry

For $$f \colon \mathbb{R}^n \to \mathbb{R}$$ the gradient of $$f$$ is a vector field over $$\mathbb{R}^n$$ defined as $$\nabla f := \sum_{i=1}^n \frac{\partial f}{\partial x_i}\frac{\partial}{\partial x_i}.$$

How I can prove that $$df_p(v) = \langle v, \nabla_p f \rangle$$?

If $$h \colon \mathbb{R}^n \to \mathbb{R}$$ is another differential map, how to compute $$[\nabla f, \nabla g]h?$$

• What do you mean by your last question? Nov 5 '18 at 15:26
• Lie brackets of vector fields! :)
– LH8
Nov 5 '18 at 15:28
• So do you want to find $[\nabla f, \nabla g]h$ explicitly? Nov 5 '18 at 15:28
• Yes, that´s right
– LH8
Nov 5 '18 at 15:33
• See math.stackexchange.com/questions/370851/… for computation of Lie bracket of vector fields.
– edm
Nov 5 '18 at 15:40

Since $$v = \sum_{k=1}^n v^k\frac{\partial}{\partial x^k}$$ then by definition of differential of a function you get \begin{align} df_p(v) & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p}dx^i(v) = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k dx^i\left(\frac{\partial}{\partial x^k} \right) \\ & = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} \sum_{k=1}^n v^k \delta^i_k = \sum_{i=1}^n \frac{\partial f}{\partial x^i}_{p} v^i = \langle v, \nabla_p f \rangle, \end{align} where $$\langle \cdot, \cdot \rangle$$ is the Euclidean metric on $$\mathbb{R}^n$$ and $$\delta^i_k$$ the Kronecker delta.
A hint for the second part: if $$X = \sum_i X^i \partial_i$$ and $$Y = \sum_k Y^k \partial_k$$ are two vector fields, then if $$h$$ is a smooth function \begin{align} [X,Y]h& = X(Yh)-Y(Xh) \end{align} with $$Xh = \sum_i X^i \partial_i h$$, and similarly for $$Yh$$.