13

Let $f:M\rightarrow N$ a diffeomorfism between differentiable manifolds. $X$ is a $C^{\infty}$ vector field over N. If $\omega \in \Omega^{k}(N)$. (i.e. $\omega$ is a $k$ - form), prove that $$f^{\ast}(i_X \;\omega)=i_{f^{\ast} X}f^{\ast}\omega $$ where $f^{\ast} $ denotes the pullback, and $i_X$ denotes interior derivative (or interior product).

My attempt is to use the fact with exterior derivative $d$; because I know that $d(f^{\ast} \omega)= f^{\ast}(d\omega)$.
I don't know if there is a relationship between interior and exterior derivative that helps me prove my proposition.

Kelvin Lois
  • 7,430
Moe
  • 196
  • 1
    I think you need to get into the details to understand. Take a $k$-tuple of vectors on $M$ at $p$ and let $X$ be a vector to $N$ at $f(p)$ then use the definition of the pull-back and interior product. – James S. Cook Nov 28 '17 at 02:27

1 Answers1

14

$(f^*\omega)_y(Y_1,..,Y_k)=\omega_{f(y)}(df_y(Y_1),...,df_y(Y_k))$

$i_X\omega_x(X_1,..,X_{k-1})=\omega_x(X,X_1,..,X_{k-1})$ we deduce that

$(f^*i_X\omega)_y(Y_1,..,Y_{k-1})=\omega_{f(y)}(X(f(y)),df_yX_1(y),..,df_yX_{k-1}(y))$

$(f^*X)(y)=df^{-1}_{f(y)}X(f(y))$, we deduce that $(i_{f^*X}f^*\omega)_y(Y_1,..Y_{k-1})=$

$(f^*\omega)_y(df_{f(y)}^{-1}(X(f(y)),Y_1,..,Y_{k-1}))$=

$\omega_{f(y)}(df_y(df_{f(y)}^{-1}(X(f(y))),df_yY_1,..,df_y.Y_{k-1})=$

$\omega_{f(y)}(X(f(y)),df_yX_1(y),..,df_yX_{k-1}(y))$.

Tsemo Aristide
  • 89,587