[Math] Euler vector field and homogeneous function

differential-geometry

In $\mathbb{R}^n$ consider the Euler vector field $V$ $$V=x^1\partial_1+\dots+x^n\partial_n$$
Let $f:\mathbb{R}^n \setminus \{0\} \rightarrow \mathbb{R}$ be a $c-homogeneous$ smooth function, i.e. $$\exists\ c \in \mathbb{R}: f(\lambda x)=\lambda^cf(x), \quad \forall \lambda>0, \forall x\ \in \mathbb{R}^n \setminus \{0\}$$
Show that $V(f)=cf$.

This is where I've come so far:

Some basics: you can skip to the main point if you want

  1. Vector fields are sections of the tangent bundle. In general, being $M$ a smooth manifold and $TM$ the tangent bundle, we have (fixed a chart) $$X=X^i\partial_i: M \rightarrow TM, \quad X(P)=X_p=X^i(p) \partial_i|_p \in T_pM$$
    1. Per definiton of tangent space, a vector sends a germ in $\mathbb{R}$ $$X_p:C^\infty_p \rightarrow \mathbb{R}, \quad f_p \mapsto X_p(f_p) \in \mathbb{R}$$
    2. The space of vector field is isomorphic to the space of derivations on smooth functions $C^\infty(M)$, i.e. any vector field $X$ acts on a smooth function as follows: $$X: C^\infty(M) \rightarrow C^\infty(M), \quad f \mapsto X(f) $$ $$X(f):p \mapsto X(f)(p):=X_p(f_P) \in \mathbb{R} $$
    3. Given $f \in C^\infty(M)$, being $df_p:T_pM \rightarrow T_{f(p)} \mathbb{R} \cong \mathbb{R}$, we have $$ df_p(X_p)=X_p(f_p) \in \mathbb{R}$$

So 3. and 4. give $$df(X)=X(f) \in C^\infty(M)$$

The main point

Then in order to prove that $V(f)=cf$ we need to show that $df(V)=cf$. Since $V=x^i \partial_i$ and $df=\frac{\partial f}{\partial x^j}dx^j$ (being a 1-form), and $dx^j(\partial_i)=\delta^j_i,$ we have $$df(V): \mathbb{R}^n \setminus \{0\} \rightarrow \mathbb{R}, \quad x \mapsto \sum_i \frac{\partial f}{\partial x^i}(x)\ x^i \equiv \sum_i f^{(i)}(x) \ x^i $$ Now $$f(\lambda x) = \lambda^c f(x) \Rightarrow \lambda f^{(i)}(\lambda x) = \lambda^c f^{(i)}(x) \Rightarrow f^{(i)}(x)= \lambda^{1-c}f^{(i)}(\lambda x) $$ and pluggin this into the expression of $df(V)$ gives $$ df(V)(x)=\sum_i \lambda^{1-c} \ f^{(i)}(\lambda x) \ x^i = \sum_i \lambda^{-c} \ f^{(i)}(\lambda x) \ \lambda x^i = \lambda^{-c} \ df(V)(\lambda x) $$ $$ df(V)(\lambda x)=\lambda^c df(V)(x)$$ Looks like $df(V)$ is $c-homogeneous$ too! Does this make any sense? I don't see a way to bring that $c$ down from the exponent; using the fact that $f$ is homogeneous we get (setting $g\equiv df(V)$) $$g(x)=\frac{g(\lambda x)}{f(\lambda x)}f(x)$$ but $\frac{g(\lambda x)}{f(\lambda x)}$ does not look like $c$, as it should…

Best Answer

For brevity $df(V)\equiv V(f) \equiv Vf$. I stay with $Vf(\lambda x)= \lambda^c Vf(x)$ as derived from the post. Then $$ \frac{d}{d \lambda} \ f(\lambda x) = \sum_i f^{(i)}(\lambda x) \ x^i = \frac{d}{d \lambda} \ \lambda^c \ f(x) = c \lambda^{c-1}\ \ f(x) $$ and using the expression of $Vf(x)$ from the question, multiplying the second and fourth terms by $\lambda$ $$Vf(\lambda x)=c\lambda^c \ f(x) $$ Now using the expression of $Vf(\lambda x$ from the first row $$ \lambda^c \ Vf(x) = \lambda^c \ cf(x) \quad \blacksquare $$

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