Notation: gradient as vector field

Question

Consider the tangent space $T_p\mathbb{R}^n$, and suppose $\{\big(\frac{\partial }{\partial x^i}\big)_p\}$ is a basis. So my textbook says that the gradient of a function $f$, $f\in C^\infty(U)$, $U\subseteq\mathbb{R}^n$ with $U$ open, is defined to be: $$\text{grad}(f):=\sum_{i=1}^n \frac{\partial f}{\partial x_i}$$ but I am failing to see why it would not be $$:=\sum_{i=1}^n \frac{\partial f}{\partial x^i}\frac{\partial }{\partial x^i}$$ so that evaluated at $p\in U$, we get the gradient vector at $p$. In other words, how is $\frac{\partial f}{\partial x^i}$ a vector field? Thanks

Answer 1

I agree with your interpretation. Given the conventions I am familiar with, $$ \nabla f=\sum_{i=1}^n \frac{\partial f}{\partial x^i}\cdot\frac{\partial}{\partial x^i}.$$ In this way, $\nabla f$ lives in the tangent bundle. The expression $$ \sum_{i=1}^n \frac{\partial f}{\partial x^i}\bigg|_p\in \mathbb{R}$$ which is probably not what we want our gradient to be.