# Push-forward vector space calculation

Let $v$ be the vector field $x_1\dfrac{\partial}{\partial x_2}-x_2\dfrac{\partial}{\partial x_1}$, and let $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ be the diffeomorphism $(x_1,x_2)\rightarrow (x_2,x_1)$. What is $f_*v$, the push-forward of $v$ by $f$?

So, at a point $p=(x_1,x_2)$, the vector $v(p)$ is $(-x_2,x_1)$. By definition, $f_*v$ is the vector field $w$ such that $v$ and $w$ are $f$-related. This means that $Df_pv(p)=w(f(p))$ for all $p$. So $Df_{(x_1,x_2)}(-x_2,x_1)=w((x_2,x_1))$. How can I calculate this $w$?

[By $Df_p$ I mean the derivative of $f$ at point $p$]

• Do you know how to compute $Df$? If so then your last equation is a recipe for $w$. – Anthony Carapetis Nov 4 '13 at 6:32
• Let's see... since $f((x_1,x_2))=(x_2,x_1)$, we should have $Df$ be the $2\times 2$ matrix $[0 1; 1 0]$. So $Df_{(x_1,x_2)}(-x_2,x_1)=(x_1,-x_2)$. Is that right? – Mika H. Nov 4 '13 at 6:40
• looks good to me. – Anthony Carapetis Nov 4 '13 at 6:51