Symmetry group of the vector field $V=x \partial /\partial x + y \partial /\partial y$

Question

I was trying to solve an exercise in one of Arnold's book that asks for the symmetry group of the vector field $V=x \partial /\partial x + y \partial /\partial y$, that is the diffeomorphisms $g$ of $\mathbb{R}^2$ such that $g_*V = V$. I can see that the dilations and rotations are in this group and in general all the linear isormorphisms of $\mathbb{R}^2$. For the general solution I guess I have to look for the $g$ such that $Jg|_p V_p = V_{g(p)}$ but this is a PDE and I guess there should be another way to do it.

Answer 1

I guess that with the identification of $\mathbb{R}^2$ with its tangent space then $V$ is the identity map and then the solution are just the diffeormorphism that are equal to its Jacobian, therefore the solution must be $GL(2)$. To be more specific: we need to prove that $g_{*,p} V(p)= V(g(p))$, but $V(p)$ is the tangent vector at $p$ with coordinates equal to those of $p$ so we can write $V(p) = p \in T_p \mathbb{R}^2 \cong \mathbb{R}^2$ so the equation is then, $(Jg_p) (p) = g(p) \forall p \in \mathbb{R}^2$.

Answer 2

An equivalent condition for $g_\ast V=V$ is that $g$ commutes with the flow of $V$. You showed that the flow of $V$ is given by $$\theta_t:\mathbb{R}^2\to\mathbb{R}^2 \ , \ (x,y)\mapsto (xe^t,ye^t).$$Suppose that $g(x,y)=(g_1(x,y),g_2(x,y))$. We want that $g\circ\theta_t=\theta_t\circ g$. That is,

$g(\theta_t(x,y))=g(xe^t,ye^t)=(g_1(xe^t,ye^t),g_2(xe^t,ye^t))$ needs to equal

$\theta_t(g(x,y))=\theta_t(g_1(x,y),g_2(x,y))=(g_1(x,y)e^t,g_2(x,y)e^t)$.

Hopefully this helps. I believe this shows that $g$ has to be linear, where-as we also need $g$ to be a diffeomorphism.