Assume $M_1, \dots,M_k$ are smooth manifolds and define $M:=M_1\times \dots \times M_k$. Denote the projections on the $i$-th factor with $\pi_i: M \rightarrow M_i$. I want to show that for each smooth vector field $X_i \in \Gamma(M_i,TM_i)$ there is a $\pi_i$-related smooth vector field $Y\in \Gamma(M,TM)$.

Since I don't know any theorems about the existence for related vector fields my approach was to prove the existence by constructing one. I know that if $Y$ is a smooth vector field over $M$ then for all smooth function $f\in C^\infty(M)$, $fY:M\rightarrow TM$, defined by $$(fY)_p=f(p)Y_p$$ is a smooth vector field as well.

From the Lemma below I know that for each real-valued smooth function $g$ on an open subset of $M_i$, we have $$Y(g\circ \pi_i)=(Xg)\circ \pi_i.$$

Well, that's basically how far I am right. I have read the chapter about this topic in Introduction to smooth manifolds by John M. Lee but I am still lacking intuition for this situation. If anyone could lead me in the right direction I would appreciate it.

Definition of $F$-related vector fields:

Suppose $F: M\rightarrow N$ is a smooth, where $M,N$ are smooth manifolds. Smooth vector fields $X\in \Gamma(M,TM)$ and $Y\in \Gamma(N,TN)$ are called $\mathbf{F}$-related, if for each $p\in M$, $dF_p(X_p)=Y_{F(p)}$.


Assume $X,Y$ and $F$ are as specified in the definition above. $X$ and $Y$ are $F$-related if and only if for every smooth real-valued function $f$ on an open subset $U\subseteq N$ we have $X(f\circ F)=(Yf)\circ F$. This Lemma follows basically by inserting in the definitions.


1 Answer 1


Suppose $X \in \mathfrak{X}(M), Y \in \mathfrak{X}(N)$ we can define a vector field $X \oplus Y : M \times N \to T(M \times N)$ on product manifold $M \times N$ as $$ (X \oplus Y)_{(p,q)} = (X_p,Y_q) $$ under natural identification of $T_{(p,q)}(M \times N)$ with $T_p M \oplus T_qN$ (by isomorphism $\alpha : T_{(p,q)}(M \times N) \to T_pM \oplus T_qN$ defined as $\alpha (v) = (d\pi_M(v), d\pi_N(v))$, one can show that it is a smooth vector field on the product manifold.

So, wlog, given $X \in \mathfrak{X}(M_1)$ it can be checked that for any $X_j \in \mathfrak{X}(M_j)$ for $j=2,\dots,k$, the resulting product $X \oplus X_2 \oplus \cdots \oplus X_k$ is $\pi_1$-related to $X$ by the way the product vector field defined. So vector field on product manifold that $\pi_1$-related to $X$ is not unique. Of course we can choose $X \oplus \mathbf{0}\, \oplus \cdots\oplus \mathbf{0}$ for convenient.

Since you read Lee's, i want to point out that the construction of product vector field above is in fact an exercise in Lee's Introduction to Smooth Manifold (See Problem 8-17 and more general setting in Problem 8-18). However vector fields on the product manifold that $\pi_1$-related to a vector field $X \in \mathfrak{X}(M_1)$ is not necessarily in form of product vector field.

After read this post, i've come to conclusion that

  • $\mathfrak{X}(M \times N) \supsetneq \mathfrak{X}(M) \oplus \mathfrak{X}(N)$ (as shown in that answer),

  • Any vector vector field $V$ in product manifold $M \times N$ is in form of $V= X \oplus Y$ for some $X \in \mathfrak{X}(M)$ and $Y \in \mathfrak{X}(N)$ if and only if $V$ and $X$ are $\pi_M$-related and $V$ and $Y$ are $\pi_N$-related.

In more general setting, we know that for any smooth surjective submersion $F : M \to N$ and $X \in \mathfrak{X}(M)$, the pushforward $F_{*}(X)$ is a well-defined smooth vector field on $N$ that is $F$-related to $X$ is and only if $dF_p(X_p) = dF_q(X_q)$ whenever $p$ and $q$ are in the same fiber. So by applying this to the map $\pi_M : M \times N \to M$ and $\pi_N : M \times N \to N$, we have the following criteria :

Any vector vector field $V \in \mathfrak{X}(M \times N)$ is also in $\mathfrak{X}(M) \oplus \mathfrak{X}(N)$ if and only if $d\pi_M(V_{(p,q)})$ constant on each fiber $\{p\} \times N$ and $d\pi_N(V_{(p,q)})$ is constant on each fiber $M \times \{q\}$.

  • $\begingroup$ Are you defining the vector field by $(X \oplus Y)_{(p,q)}=(X_p,Y_q)$ or with the image of $(X_p,Y_q)$ under the identification? Hence with a map defined in an answer for this question: math.stackexchange.com/questions/413766/…. $\endgroup$ Jun 1, 2020 at 18:15
  • 1
    $\begingroup$ Yes that is exactly what i meant. To be precise, If $\alpha : T_{(p,q)}(M \times N) \to T_pM \oplus T_qN$ is the identification $\alpha(v) = (d\pi_M(v), d\pi_N(v))$, then define $(X \oplus Y)_{(p,q)} = \alpha^{-1}(X_p,Y_q)$. $\endgroup$ Jun 1, 2020 at 18:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .