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I have the following question:

If $(x^1,\ldots,x^n)$ are the standard coordinates on $R^n$ and $(y^1,\ldots,y^n)$ are other coordinates on $R^n$, how can we show that $\sum x^i\frac{\partial}{\partial x^i} =\sum y^i\frac{\partial}{\partial y^i}$ ?

Thank you!

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Welcome to MSE =) – Patrick Da Silva Nov 20 '11 at 20:34

In the title of your question you mention the Euler vector field, so I suspect that your question was about the following fact:

for any vector bundle $\pi:E\to M$, the Euler vector field $X$ on $E$, i.e. the infinitesimal generator of the homotheties of $E$, has the expression $X=u\frac{\partial}{\partial u}$ w.r.t. standard coordinates $(x,u)$.

If I have understood your question then below there is my answer.

If $\pi:E\to M$ is a vector bundle then its Euler vector field $X$ is defined as the vector field on $E$ which is the infinitesimal generator of the action $\Phi$ of $\mathbb{R}$ on $E$ given by the fiberwise scalar multiplication.
$$\Phi:(t,e)\in\mathbb{R}\times E\to\Phi_t(e):=\exp(t).e\in E.$$ $$X_e:=\left.\frac{d}{dt}\right|_{t=0}\ \Phi_t(e),\ \forall e\in E.$$

Let $x:U\to\mathbb{R}^n$ be coordinates on $M$ and $(x,u):\pi^{-1}(U)\to\mathbb{R}^n\times\mathbb{R}^n$ the associate standard coordinates on $E$. In such coordinates we find the following explicit expressions: $$\Phi(t,(x,u))=(x,\exp(t).u),$$ $$X_{(x,u)}=\left.\frac{d}{dt}\right|_{t=0}\ (x,\exp(t).u)=\left.u\frac{\partial}{\partial u}\right|_{(x,u)}.$$

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$y$'s have to be linear functions of $x$'s, otherwise the identity will not hold. If they are, notice that $(\sum x^i \partial_{x^i})f=f$ for any linear function $f$, in particular $(\sum x^i \partial_{x^i})y^k=y^k$, which gives the components of the vector field in the $y$-coordinates.

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