# Basic question about related vector fields and pushforwards

Problem: Let $F \in C^{\infty}(M, N)$ be a diffeomorphism, $X,Y$ vector fields on the manifolds $M,N$ respectively. Then $X$ and $Y$ are $F$-related (i.e. $T_pF(X_p) = Y_{F(p)}$) if and only if $Y = F_{*}X$. I know $F_*v(f) = v(f \circ F)$ for $v \in T_p M$, $f \in C^{\infty}(N)$.

Attempt: I think my confusion with this basic exercise lies in my weak understanding of the pushforward operator on vector fields. In detail, I know that if $Y$ is $F$-related to $X$, then for $f \in C^{\infty}(N)$, $Y_{F(p)}(f) = T_pF(X_p)(f) = (F_*X_p)(f)$, but why does this equate $(F_*X)_{F(p)}(f)$? For the converse direction, similar question arises, how do I bring the $F(p)$ outside into the $(F_*X)$? Any help is appreciated!

• How do you define the vector field $F_*X$? – positrón0802 Mar 22 '18 at 1:42
• We did not define this in class. We only defined pushforward of tangent vectors... – Longti Mar 22 '18 at 1:44
• How will you prove $Y=F_*X$ if you don't have the definition of $F_*X?$ Maybe a property of $F_*X$ they gave you? I say it because the definition of $F_*X$ I have solves the problem almost trivially. – positrón0802 Mar 22 '18 at 1:49
• Could you tell me your definition of the $F_*X$? Thanks. – Longti Mar 22 '18 at 1:55

## 1 Answer

We say that $X$ and $Y$ are $F$-related if $T_pF(X_p)=Y_{F(p)}$ for all $p\in M.$

Definition 1. We can define the pushforward $F_*X$ by the formula $$(F_*X)_q=T_{F^{-1}(q)}F(X_{F^{-1}(q)})$$ for all $q\in N.$ With definition your problem is just the definition, because if $q=F(p)$ then $$(F_*X)_{F(p)}=T_{F^{-1}(F(p))}F(X_{F^{-1}(F(p))})=T_pF(X_p)=Y_{F(p)}.$$

Definition 2. Define $F_*X$ by the property that $$X(f\circ F)= (F_*X(f))\circ F$$ for all $f\in C^\infty(N).$

Hence, given $p\in M$ and $f\in C^\infty(N)$ we have $$X(f\circ F)(p)=X_p(f\circ F)=T_pF(X_p)(f)$$ and $$(F_*X(f))\circ F(p)=(F_*(X)(f))(F(p))=(F_*X)_{F(p)}(f),$$ so that $Y_{F(p)}:=T_pF(X_p)=(F_*X)_{F(p)}$ for all $p\in M$ and hence $Y=F_*X.$