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I am currently stuck on Problem 8-18 (b) in Lee's Introduction to Smooth Manifolds.

We have a smooth submersion $F: M \to N$ and a smooth vector field $Y$ on $N$. The problem in part (b) asks to show that if $\dim M \neq \dim N$, then $Y$ has a lift, but the lift is not unique. I am struggling to show the existence of a lift globally.

If we assume that $\dim M = \dim N$ instead (this is part (a) of the problem) then it is clear that $dF_p: T_pM \to T_pN$ is an isomorphism for all $p \in M$ so we can define the lift of $Y$ by $X_p = dF_p^{-1}\left(Y_{F(p)}\right)$ for all $p$ in $M$. By the rank theorem, there are smooth charts $(U, \varphi)$ for $M$ containing $p$ and $(V, \psi)$ for $N$ such that $F(U) \subseteq V$ and $\psi \circ F \circ \phi^{-1} = \text{id}$. Then working in the coordinate basis $\frac{\partial}{\partial x^i}$ associated to $(U, \varphi)$ and $\frac{\partial}{\partial y^i}$ associated to $(V, \psi)$ and using the Jacobian of $dF_p$ which is just $I_n$ in this case, we see that $X^i = Y^i \circ F$ so the components of $X$ are smooth, and so $X$ is a smooth lift of $Y$.

This argument fails when $\dim M \gt \dim N$ because there is no canonical way to choose a lift of $Y_{F(p)}$. We now have more than one choice for picking an element in $dF_p^{-1}\left(Y_{F(p)}\right)$ since $dF_p$ is not injective. And there is no clear way to make that arbitrary choice behave smoothly over all $p \in M$.

One idea I had was that by the local section theorem, for each $p \in M$, we can choose a smooth local section $\sigma_p: V_p \to M$ of $F$ defined on a neighborhood $V_p$ of $F(p)$ in $N$ whose image contains $p$, and define $X_p = d\left(\sigma_p\right)_{F(p)}\left(Y_{F(p)}\right)$. However it is not clear that this defines a smooth vector field because if we want to apply the gluing lemma to stitch together the smooth vector fields $d\sigma_p \circ Y$ defined on $V_p$, we need to know that $d\sigma_p \circ Y$ agrees with $d\sigma_{p'} \circ Y$ on their overlaps $V_p \cap V_{p'}$.

Another approach is to work in coordinates as in part (a) and try to define $X$ explicitly by $X = \sum\limits_{i=1}^n (Y^i \circ F) \frac{\partial}{\partial x^i}$ on each chart. But again it is not clear how to show that definitions agree on overlaps between different charts. For example, you could have a situation where $X_{p'} = \sum\limits_{i=1}^n (Y^i \circ F) \frac{\partial}{\partial x^i}\bigg\vert_{p'} + \sum\limits_{i=n+1}^mc_i\frac{\partial}{\partial x^i}\bigg\vert_{p'}$ for some $p'$ arbitrarily close to $p$, where the nonzero terms $c_i$ for $i \gt n$ arise from the change of coordinates when writing $\frac{\partial}{\partial {x'}^i}\bigg\vert_{p'}$ in terms of $\frac{\partial}{\partial x^i}\bigg\vert_{p'}$.

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    $\begingroup$ There's a hint that might be useful to you in the paragraph just below Exercise 2.24 on page 44 of my book. $\endgroup$
    – Jack Lee
    Commented Dec 21, 2022 at 22:28
  • $\begingroup$ @JackLee Ah I think I see it now, I got too focused on the first approach to constructing smooth functions and forgot about partitions of unity. So it looks like we can construct smooth vector fields $X^{(p)} = \sum_{i=1}^n (Y^i \circ F)\frac{\partial}{\partial x^i}$ on charts containing $p$, and then glue them together as $X = \sum_{p \in M} \psi_p X^{(p)}$ where $\left\{\psi_p\right\}$ is a smooth partition of unity subordinate to the coordinate cover of $M$. Each $X^{(p)}$ is $F$-related to $Y$ by construction, and so $X$ also is by linearity. $\endgroup$
    – Tob Ernack
    Commented Dec 22, 2022 at 5:00

1 Answer 1

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Here is my solution. Suppose $m=\dim M \neq \dim N=n$ and $Y \in \mathfrak{X}(N)$. For any p $p \in M$ let $(U_p,x^i)$ be a chart centered at $p$ and $(V_{F(p)},y^j)$ centered at $F(p)$ so the representation of $F: M \to N$ is $$ \hat{F}(x^1,\dots,x^m) = (x^1,\dots,x^n). $$ If there is a smooth vector field $X$ that is $F$-related to $Y$, then $\forall x \in U$ we have $$ Y_{F(x)} = Y^j(F(x)) \partial_{y^j}\big|_{F(x)} = dF_x(X_x) = X^i(x) \frac{\partial F^j}{\partial x^i}(x) \partial_{y^j} \big|_{F(x)} = X^i(x) \, \delta^j_i \partial_{y^j}\big|_{F(x)}. $$ So the first $n$ components of $X$ at $(U_p,x^i)$ must satisfy $X^i = Y^i \circ F|_{U_p}$. The rest of the components of $X$ can be chosen arbitrarily. So define local vector $X_p : U_p \to TM$ $F$-related to $Y$ as $X_p = X_p^i \partial/\partial x^i$ with $X_p^i = Y^i \circ F|_{U_p}$ for $i=1,\dots,n$. This construction can be done for every point in $M$, so by partition of unity we can blend them to get a global vector field $X$, that we hope still $F$-related to $Y$.

Suppose $(\psi_p)_{p \in M}$ is a partition of unity on $\{U_p\}_{p \in M}$, define a smooth vector field $X =\sum_p \psi_p X_p$, where $\psi_p X_p$ interpreted as local extension of $\psi_p|_{U_p} X_p$ to $M$ that zero outside $\text{supp }\psi_p \subseteq U_p$. This vector field $F$-related to $Y$ by the following computation \begin{align*} dF_x(X|_x) &= dF_x \Big( \sum_{p} (\psi_p X_p)(x) \Big) \\ &= dF_x\Big( \sum_{i=1}^N \psi_{p_i}(x) X_{p_i}|_x \Big) \\ &= \sum_{i=1}^N \psi_{p_i}(x)\, dF_x(X_{p_i}|_x) \\ &= \sum_{i=1}^N \psi_{p_i}(x)\, Y_{F(x)} \\ &= Y_{F(x)}. \end{align*} So $X$ is a lift of $Y$. It is not unique because the construction depends on the choice of partition of unity local components $X_p = X^i_p \partial/\partial x^i$.

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