# Show that there is a $\pi_i$-related smooth vector field for each smooth vector field $X_i \in \Gamma(M_i,TM_i)$

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 $$p\in M$$, $$dF_p(X_p)=Y_{F(p)}$$.

Lemma:

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.

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 $$M \times \{q\}$$.

• 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/…. Jun 1, 2020 at 18:15
• 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 $(X \oplus Y)_{(p,q)} = \alpha^{-1}(X_p,Y_q)$. Jun 1, 2020 at 18:17