Let $A$ be a finite dimensional vector field over $\mathbb{R}$, we defined the Euler vector field $\chi$ such that $\chi(f)(a) = \frac{d}{d\lambda}\mid_{\lambda=1} f(\lambda a) $ for $a\in A$ and $f \in C^{\infty}(A)$.

I am asked to show that if $f \in C^{\infty}(A)$ then $\chi(f) \in C^\infty(A)$.

I am also asked to calculate the Euler vector field on $\mathbb{R}^3$ in terms of the coordinate vector fields $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$

I know that a vector field is basically just a function from $C^\infty(A) \to C^\infty(A)$ that satisfies Leibniz rule. I don't know how to show that $\chi(f) \in C^{\infty}(A)$.

For the second part I have no idea what it means to calculate the Euler vector field on $\mathbb{R}^3$ in terms of the coordinate vector fields $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}$, I don't get how $\frac{\partial}{\partial x}$ is a vector field.

I am completely lost so any help or reference would be appreciated.


1) Let us first check that if $f\in C^{\infty}(A)$, then $\chi(f)\in C^{\infty}(A)$. To do this, we have to prove that $\chi(f)$ is a smooth function, when expressed in a coordinate chart. For the vector space $A$, any choice of basis $\{a_{1},\ldots,a_{n}\}$ for $A$ induces global coordinates $(x_{1},\ldots,x_{n})$ on $A$. The coordinates of a vector $a$ are just its coefficients with respect to this basis: $$ a=x_{1}v_{1}+\cdots+x_{n}a_{n}. $$ In these coordinates, we now compute: \begin{align} \chi(f)(x_{1},\ldots,x_{n})&=\left.\frac{d}{d\lambda}\right|_{\lambda=1}f(\lambda x_{1},\ldots,\lambda x_{n})\\ &=\frac{\partial f}{\partial x_{1}}(x_{1},\ldots,x_{n})x_{1}+\cdots+\frac{\partial f}{\partial x_{n}}(x_{1},\ldots,x_{n})x_{n},\tag{1}\label{1} \end{align} using the chain rule. The expression $(1)$ is clearly smooth in $x_{1},\ldots,x_{n}$ so that $\chi(f)\in C^{\infty}(A)$.

2) As you say, a vector field on $\mathbb{R}^{3}$ is nothing else but an operator on $C^{\infty}(\mathbb{R}^{3})$ that satisfies the Leibniz rule. Examples of such operators are the partial derivatives \begin{align} &\frac{\partial}{\partial x}:C^{\infty}(\mathbb{R}^{3})\rightarrow C^{\infty}(\mathbb{R}^{3}):f\mapsto\frac{\partial f}{\partial x},\\ &\frac{\partial}{\partial y}:C^{\infty}(\mathbb{R}^{3})\rightarrow C^{\infty}(\mathbb{R}^{3}):f\mapsto\frac{\partial f}{\partial y},\\ &\frac{\partial}{\partial z}:C^{\infty}(\mathbb{R}^{3})\rightarrow C^{\infty}(\mathbb{R}^{3}):f\mapsto\frac{\partial f}{\partial z}. \end{align} These span all vector fields on $\mathbb{R}^{3}$. In particular, the Euler vector field $\chi$ on $\mathbb{R}^{3}$ can be written as a $C^{\infty}(\mathbb{R}^{3})$-linear combination of the partial derivatives. And from $(1)$, we already know what its coefficients are: for any $f\in C^{\infty}(\mathbb{R}^{3})$ we have $$ \chi(f)=\frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}y+\frac{\partial f}{\partial z}z=\left(x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}\right)(f). $$ Therefore, $$ \chi=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}. $$


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