Pushfoward of smooth vector field is smooth

differential-geometrydifferential-topologygeneral-topologymanifoldsVector Fields

My books are Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).

Let $F : N \to M$ be a diffeomorphism of manifolds that have dimensions. Let $X$ be a smooth vector field on $N$. Then the pushforward $F_*X$ is a defined vector field on $M$ by Volume 1 Example 14.15

  1. Is $F_*X: M \to TM$ smooth? This is a step of a proof in another question. I think $F_*X$ is smooth because:

    • 1.1. Let $F_{*,p}: T_pN \to T_{F(p)}M$ be the differential of $F$ at $p$, defined in Volume 1 Section 8.2.

    • 1.2. Let $F_*: TN \to TM$ be the map given by $F_*(X_p) = F_{*,p}(X_p)$. I think $F_*$ is the same as what would be known as $\tilde{F}$ in Volume 1 Section 12.3.

    • 1.3. $F_*X: M \to TM$ is actually $F_*X: M \to N \to TN \to TM$, $F_*X = F_* \circ X \circ F^{-1}$

    • 1.4. $F_*$ is smooth because $F_*$ is a smooth embedding by this because $F$ is a smooth embedding (Hopefully the definitions there are the same as in Volume 1 Definition 11.11).

      • Note: We might say $F_*$ is smooth by some other route. I ask about the other routes here.

      • Note: I'm not sure Volume 1 Section 12.3 explicitly says $F_*$, also known as $\tilde{F}$, is smooth.

      • (I think this might be proved in Volume 2, but I actually stopped Volume 2 at Section 6, and I did not study Sections 1-6 in too much detail because I noticed Volume 2 is not really a prerequisite of Volume 3 and because I was hoping to learn more of vector bundles from Volume 3 before continuing Volume 2.)

    • 1.5 Therefore, $F_*X$ is smooth by (1.3), (1.4), smoothness of $F^{-1}$, smoothness of $X$ and Volume 1 Proposition 6.9.

  2. What can $F$ alternatively be if not a diffeomorphism for $F_*X$ to be smooth? Some guide questions:

    • 2.1. Must $F$ be injective (and smooth) for $F_*X$ to be defined in the first place? (Answer must be the opposite of the answer of 2.2, I think.)

    • 2.2. Can $F$ be a local diffeomorphism (defined in Volume 1 Section 6.7 and further described in Volume 1 Remark 8.12)?

      • 2.2.1 Can $F$ be a local diffeomorphism onto its image? (I guess that $F$ is a local diffeomorphism onto its image is defined as that $F$ with restricted range, $\tilde F: N \to F(N)$ is a local diffeomorphism. I actually don't know and haven't yet thought of the relationship between $F$ local diffeo and $F$ local diffeo onto image)

    • 2.3. Can $F$ be a smooth embedding (defined in Volume 1 Definition 11.11)?

      • I think yes because we would still have that $F_*X$ defined by $F$'s injectivity and that $F_*$ smooth by this. The problem might be the $F^{-1}$, but I think that's not too difficult to fix.

        • Update: Indeed not a difficult (maybe) fix. Just use $\tilde F ^{-1}$ for $\tilde F: N \to F(N)$. The fix is complete when we show $\tilde F ^{-1}$ is smooth. This may be by your definition of smooth embedding (not difficult) or by a property of your definition of smooth embedding (difficulty depends on your understanding of the proof of the property).

Best Answer

Here is the main issue: how can one define the pushforward of a vector field? Rather – when? Take two copies of the real line, parametrized as $M=\mathbb{R}\times \{0\}\sqcup\mathbb{R}\times \{1\}\subseteq \mathbb{R}^2$ and define a map $$ \pi:M\to \mathbb{R}$$ by $\pi(x,n)=x$. Then take the vector field defined by $X\in \mathfrak{X}(M)$ with $X_{(x,0)}=-1$ and $X_{(x,1)}=1$ for all $x\in \mathbb{R}$. Then, if we calculate $\pi_{*,(x,0)}X_{(x,0)}=-1\in T_x\mathbb{R}$ and $\pi_{*,(x,1)} X_{(x,1)}=1\in T_x\mathbb{R}$. The sensible way to (try to) define the pushforward vector field here is to set $Y=\pi_{*}X$ to be $Y_p=\pi_{*,q}(X_q)$ for some $q\in M$ with $\pi(q)=p$.

Unfortunately, this won't work because if for instance we choose $p=1$, then its preimages under $\pi$ are $(1,0)$ and $(1,1)$. If we choose $q=(1,0)$ we get $Y_p=-1$ and if we choose $q=0$ we get $Y_p=1$. So, the pushforward vector field is not well-defined in general.

The only way this could be defined is if the following criterion is met:

Let $F:M\to N$ denote a map of $\mathscr{C}^\infty$ manifolds. Then given $X\in \mathfrak{X}(M)$, there exists a vector field $F_*X\in \mathfrak{X}(F(M))$ defined as above if and only if for each $p\in N$, $F_{*,q}(X_{q})=F_{*,q'}(X_{q'})$ for all $q,q'\in F^{-1}(p)$.

An immediate Corollary is that when each $F^{-1}(p)$ contains only one point, the pushforward vector field is defined. So, if we have a smooth embedding $F:M\to N$ then the pushforward vector field is defined. It suffices to have a smooth injective map, too. An immersion might not work in general, however.

Now, for the first question a criterion (found in Tu's Introduction to Manifolds) says that a vector field $X\in \mathfrak{X}(M)$ is $\mathscr{C}^\infty$ if and only if for any $f\in \mathscr{C}^\infty(M)$ the map $p\mapsto X_pf$ is a smooth function. For convenience, assume $F$ is aan embedding. Now, for $g\in \mathscr{C}^\infty(N)$, fix $p=F(q)\in N$. Then $(F_*X)_p(g)=X_q(g\circ F)$ where we note that $g\circ F\in \mathscr{C}^\infty(M)$. So, the association $p\mapsto (F_*X)_p(g)$ is given by the composition, $p\mapsto F^{-1}(p)=q\mapsto X_q(g\circ F)$. By $X\in \mathfrak{X}(M)$ smooth, the second map is smooth. By $F$ admitting a smooth inverse defined on $F(M)$, the first map is smooth. Hence, the pushforward of a smooth vector field (by an embedding) is again smooth.

Related Question