# Euler vector field is a $C^{\infty}$ vector field.

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, $$\chi=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}+z\frac{\partial}{\partial z}.$$