Let $S\subset M$ be an embedded submanifold. A vector field $X \in \mathfrak{X}(M)$ is said to be tangent to $S$ if $X_p \in T_pS$ for all $p \in S$. Show that for $Y \in \mathfrak{X}(M)$ tangent to $S$, there is a unique smooth vector field $X \in \mathfrak{X}(S)$ such that $X$ is $\iota$-related to $Y$.

I have seen that since each $Y_p \in \iota_{*p}(T_pS) \equiv T_pS$, we can assign tangent vectors $X_p \in T_pS$ pointwise, with $Y_p = \iota_{*p}X_p$. I'm having trouble showing that the resultant vector field $X$ is indeed smooth and unique.

One criterion is checking whether $Xf \in C^\infty(S)$ for all $f \in C^\infty(S)$. I am aware of the characterization $$ T_pS = \{X \in T_pM: Xf = 0 \;\text{ whenever } f \in C^\infty(M)\text{ and } f\big|_S = 0\}, $$ but I don't quite see if this helps.


1 Answer 1


Let $f\in C^\infty (S)$. We need only to show that $Xf \in C^\infty(S)$. Since smoothness is a local property, it suffices to check it locally around each $p\in S$.

Let $p\in S$. Let $U$ be an open neighborhood of $p$ in $M$, so that $f$ extends to a smooth function $F$ on $U$ (indeed one can choose $U$ so that $M\subset U$, see here). Since $Y$ is smooth, $YF$ is smooth. But since

$$ YF|_{S\cap U} = Xf|_{S\cap U}$$

(by definition of the action of vector fields as directional derivative), it implies that $Xf|_{U\cap S}$ is smooth. since this holds for all $p\in S$, $Xf$ is smooth for all smooth $f$ on $S$. Hence $X$ is a smooth vector field.


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