About the pullback of a $1$-form

Question

Suppose $M$, $N$ are smooth manifolds, $f: M\to N$ is a smooth map. Let $X$ be a vector field on $N$, which is function that assigns to each $n\in N$ a vector $X_n\in T_n N$, and let $\omega$ be the smooth $1$-form on $N$ dual to that vector field: $\omega(n)(X_{n})=1$ whenever $X_{n}\in T_{n} N$. Now $f^\ast\omega$ is a smooth $1$-form on $M$ defined by $(f^\ast\omega)(m)(Y_m)=\omega(f(m))(f_\ast Y_m)$, where $Y_m\in T_mM$. Since $f_\ast Y_m\in T_{f(m)} N$, it follows from definition that $\omega(f(m))(f_\ast Y_m)=1$ and so $f^\ast \omega \equiv 1$, which is a $0$-form. But how can this be possible if $f^\ast\omega$ is supposed to be a $1$-form?

Answer 1

My interpretation of your confusion is as follows:

• You state that $\omega_n(X_n)=1$ for all $n\in N$. This is for a particular (fixed) vector field $X$. Here, you state that this is true whenever $X_n\in T_nN$, i.e., $n\in N$.

• Later, you state that $\omega_{f(m)}(f_\ast Y_m)=1$, and it seems that your reasoning is that this is true because $f_\ast Y_m\in T_{f(m)}N$.

• The problem with this reasoning is that the condition above is not that $X_n\in T_nN$ (this is automatic from the fact that $X$ is a vector field). You would only be guaranteed to get $\omega_{f(m)}(f_\ast Y_m)=1$ if $f_\ast Y_m=X_{f(m)}$. There is no reason for this to be true.

What appears to be the case is that you're using the condition $X_n\in T_nN$ as the definition for $\omega_n(X_n)=1$, in other words, $\omega_n(X_n)=1$ whenever $X_n$ is a vector, so $\omega$ is the constant function, which is not true.