$\newcommand{\mc}{\mathcal} \newcommand{\set}[1]{\{#1\}} \DeclareMathOperator{\pr}{pr} \newcommand{\at}{\big|} \DeclareMathOperator{\GL}{GL}$

Let $\pi:E\to N$ be a smooth vector bundle over a smooth manifold $N$ with typical fibre a $k$-dimensional vector space $\mc V$ and $f:M\to N$ be a smooth map between smooth manifolds.

Is it true that the total space of the pullback bundle $\Pi:f^*E\to M$ is an embedded submanifold of $M\times E$?

(For reference, I have described the pullback construction I below).

I think the above is true. It is clear that $f^*E$ has the subspace topology induced from $M\times E$. So we only need to check that the inclusion map is a smooth immersion.

When passing to charts, this is equivalent to checking that the map $(m, v)\mapsto (m, f(m), v):f^{-1}(U)\times \mc V\to f^{-1}(U)\times U\times \mc V$ is an immersion, where $U$ is an open set in $N$ over which $E$ trivializes. But this is indeed an immersion.

If the assertion is correct, then to check if a map $g:Q\to f^*E$ is smooth, we would just need to check if $\pr_1\circ g$ and $\pr_2\circ g$ are smooth (and this was the motivation for the question).


Pullback Bundle Construction:

Let $\pi:E\to N$ be a smooth vector bundle over a smooth manifold $N$ with typical fibre a $k$-dimensional vector space $\mc V$ and $f:M\to N$ be a smooth map between smooth manifolds.

Define the set $f^*E=M\times_N E=\set{(p, x)\in M\times E:\ f(p)=\pi(x)}$. Define $\Pi:f^*E\to M$ as the restriction of $\pr_1:M\times E\to M$ to $M\times_N E$. We will write $(f^*E)_p$ to denote $\Pi^{-1}(p)$ for each $p\in M$.

Now for each smooth local trivialization $\phi:\pi^{-1}(U)\to U\times \mc V$ of $E$ over $U$, define a map $\Phi:\Pi^{-1}(f^{-1}(U))\to f^{-1}(U)\times \mc V$ as $\Phi(p, v)=(p, \pr_2\circ \phi(v))$ and note that $\Phi$ is a bijection whose restriction to $\Pi^{-1}(p)$ is a linear isomorphism from $(f^*E)_p$ to $\set{p}\times \mc V$ for each $p\in f^{-1}(U)$. For each $q\in U$, write $\phi_q$ to mean $\pr_2\circ \phi\at_{E_q}$ and note that $\phi_q:E_q\to \mc V$ is a linear isomorphism.

Let $\phi:\pi^{-1}(U)\to U\times \mc V$ and $\psi:\pi^{-1}(V)\to V\times \mc V$ be two overlapping smooth local trivializations of $U$ and $V$ over $V$ and $\Phi:\Pi^{-1}(f^{-1}(U))\to f^{-1}(U)\times \mc V$ and $\Phi:\Pi^{-1}(f^{-1}(V))\to f^{-1}(V)\times \mc V$ be the corresponding bijections. If $\tau:U\times V\to \GL(\mc V)$ is the transition function associated with $\psi\circ \phi^{-1}$, then note that $\Psi\circ \Phi^{-1}:f^{-1}(U\cap V)\times \mc V\to f^{-1}(U\cap V)\times \mc V$ is given by

\begin{equation*} \Psi\circ \Phi^{-1}(p, v)=(p, \tau_{f(p)}v) \end{equation*} for all $(p, v)\in (U\cap V)\times \mc V$. Defining $\Gamma:f^{-1}(U\cap V)\to \GL(\mc V)$ as $\Gamma=\tau\circ f$ we see that %equation \begin{equation*} \Psi\circ\Phi^{-1}(p, v) = (p, \Gamma_p(v)) \end{equation*} for all $(p, v)\in f^{-1}(U\cap V)\times \mc V$. It is clear that $\Gamma$ is smooth. Therefore there is a unique topology and smooth structure such that $\Pi:f^*E\to M$ is smooth vector bundle with typical fibre $\mc V$ whose smooth local trivializations are the maps $\Phi:f^{-1}(U)\to U\times \mc V$ as defined above.


1 Answer 1


$\newcommand{\M}{M}$ $\newcommand{\N}{N}$ $\newcommand{\brk}[1]{\left(#1\right)}$ $\newcommand{\be}{\beta}$ $\newcommand{\al}{\alpha}$ $\newcommand{\til}{\tilde}$

The pullback bundle is indeed an embedded submanifold of the product $M \times E$. The essential notion here is transversality. (Together with the fact that bundle projection is a submersion).

The full story with all the details is a bit long; I haven't seen it done in any textbook, I found all the steps here and there, and built my own picture of the things.


$(1)$ Let $\M,\N$ be a smooth manifolds. Suppose $F:\N \to \M$ is a smooth map, $S \subseteq \M$ is an embedded submanifold. We say $F$ is transverse to $S$ if $\forall x \in F^{-1}\brk{S} \, , \, T_{F\brk{x}}\M= T_{F(x)}S + dF_x(T_x\N)$.

$(2)$ Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. We say $F,F'$ are transverse to each other if $\forall x \in \N, x' \in \N' $ such that $F(x)=F'(x')$ , $T_{F(x)}\M=dF_x(T_x\N) + dF'_{x'}(T_{x'}\N')$.

Note: If either one of $F,F'$ is a submersion, then they are automatically transverse.

Note: Some of the proofs are at the end of the answer (So it will be possible to skim through the general scheme without all the details at first)

Lemma (1): $\M,\N$ be a smooth manifolds, $S \subseteq M$ is an embedded submanifold. Let $F:\N \to \M$ be transverse to $S$. Then $F^{-1}(S)$ is an embedded submanifold of $\N$ whose codimension is equal to the codimension of $S$ in $\M$.

proof: See Theorem 6.30 , in Lee. (pg 144).

Lemma (2): (This is exercise 13 in chapter 6, Lee)

Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. Then $F,F'$ are transverse to each other if and only if the map $F \times F' : \N \times \N' \to \M \times \M$ is transverse to the diagonal $\Delta_\M = \{(x,x)|x \in \M \}$

Lemma (3): Let $\M$ be a smooth manifold. Then $\Delta_\M = \{(x,x)|x \in \M \}$ is an embedded (smooth) submanifold of $\M \times \M$.

Lemma (4):

Let $\M$ be a manifold. Let $\Delta_\M$ be the diagonal manifold of $\M$. (see Lemma 3 ). Then $T_{\brk{x,x}}\Delta_\M = \text{diag}\brk{T_x\M \times T_x\M}=\{(v,v)| v \in T_xM \}$. (i.e the tangent space of the diagonal is the diagonal of the tangent space).

proof: Since any tangent vector can be realized as a derivative of a path, the tangent space to a manifold is identical to the set of derivatives of paths. Since $\Delta_\M$ is an embedded submanifold, a path $\be:I \to \Delta_\M $ is smooth if and only if it is smooth when considered as a path into the product $\M \times \M$ if and only if each of its components is smooth. So $\be(t)=\brk{\al\brk{t},\al\brk{t}}$ , where $\al : I \to \M$, so $\dot \be (0) \overset{(*)}= \brk{\dot \al (0),\dot \al (0)}$ , hence its clear the tangent space to the diagonal is exactly the diagonal of the tangent space. (Where in (*) we used the canonical isomorphism between $T_{(x,x')}\brk{\M \times \M'} = T_x\M \oplus T_{x'}\M'$ via the differentials of the projections onto the different components).

corollary (1):

Let $\M,\N,\N'$ be smooth manifolds, $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. (In short we write $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$). Assume $F,F'$ are transverse to each other. Then the fiber product of this diagram, which is defined as $\{\brk{x,x'} \in \brk{\N, \N'}|F(x)=F'(x') \}$ is an embedded smooth submanifold of the product $\N \times \N'$.

proof of corollary (1):

The fibered product $\N \times_{\M} \N'$ is the inverse image $(F \times F')^{-1}\brk{\Delta_{\M}}$. By Lemma 3, $\Delta_\M$ is a submanifold of $\M \times \M$. Now combine Lemma 2 and Lemma 1.

corollary (2):

Let $\M,\N,\N'$ be smooth manifolds, $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$. If either one of $F,F'$ is a submersion, then the fiber product $\N \times_\M \N' = \{\brk{x,x'} \in \brk{\N, \N}|F(x)=F'(x') \}$ is an embedded submanifold of the product $\N \times \N'$.

proof of corollary (2): If one of $F,F'$ is a submersion, then these two maps are automatically transverse to each other. Now use corollary (1).

In particular we get the following proposition:

Let $\pi: E \to B$ be a vector bundle, $f:B' \to B$. The pullack bundle $f^*\brk{E}$ is an embedded submanifold of the product $B' \times E$. (This is becuse the bundle projection $\pi$ is always a submersion).

proof of Lemma (2):

First, we need a sublemma: Sub-lemma:

Let $V$ be a vector space, $V_1,V_2 \subseteq V$ are subspaces. Let $\text{diag}(V \times V) = \{(v,v)|v \in V \} $. Then $V \oplus V = \text{diag}(V \times V) + \brk{V_1 \oplus V_2} \iff V = V_1 + V_2$

proof of the sublemma: $\Rightarrow :$ Let $v \in V$. Then $(v,0) \in V \oplus V$, hence by our assumption $\exists \til v \in V, v_1 \in V_1 , v_2 \in V_2$ such that $(v,0) = (\til v ,\til v) + (v_1,v_2)=(\til v + v_1, \til v + v_2) \Rightarrow \til v = -v_2, v = \til v + v_1 = v_1 - v_2 \in V_1 + V_2 $.

$\Leftarrow :$ Note that both sides of the left equation are subspaces. Hence, from symmetry it's enough to show that $\forall v \in V \, , \, (v,0) \in \text{diag}(V \times V) + \brk{V_1 \oplus V_2}$. The assumption $V =V_1 + V_2 \Rightarrow \exists v_i \in V_i$ such that $v = v_1 -v_2$. Define $\til v = -v_2$, so we get $(v,0)=(v_1-v_2,\til v +v_2)=(v_1 + \til v, v_2 + \til v) = (\til v,\til v) +(v_1,v_2)$.

Now to the actual proof of Lemma (2):

By definition (1), $F \times F'$ is transverse to the diagonal if

\begin{split} &\forall (x,x') \in (F \times F')^{-1}\brk{\Delta_\M} \, , \, T_{(F \times F')\brk{x,x'}}\brk{\M \times \M}= T_{(F \times F')(x,x')}\Delta_\M + d(F \times F')_{(x,x')}(T_{(x,x')}\brk{\N \times \N'}) \iff \\ & T_{\brk{F(x),F'(x')}}\brk{\M \times \M}= T_{\brk{(F(x),F'(x')}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\ & T_{\brk{F(x),F(x)}}\brk{\M \times \M}= T_{\brk{(F(x),F(x)}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\ &T_{\brk{F\brk{x}}}\M \oplus T_{\brk{F\brk{x}}}\M \overset{Lemma 4}= \text{diag}\brk{T_{F(x)} \M \times T_{F(x)}\M} + \brk{ dF_x \brk{T_x \N} \oplus dF'_{x'} \brk{T_{x'}\N'}} \overset{Sub-lemma} \iff \\ &T_{F(x)}\M =dF_x \brk{T_x \N} + dF'_{x'} \brk{T_{x'}\N'} \end{split}

Since the last row is the definition of transverse maps, we finished.

proof of Lemma (3): The diagonal is the graph of the smooth function $Id_\M$, and graphs of smooth functions are always embedded submanifolds of the product of the domain and the codomain. (See prop 5.4, Lee).

  • $\begingroup$ If anything is unclear don't hesitate to ask for clarification. $\endgroup$ Jan 21, 2016 at 12:14
  • $\begingroup$ Very informative. Thank you so much. I din't know this much about transversality. $\endgroup$ Jan 22, 2016 at 22:38
  • $\begingroup$ No problem. I am glad I could share this and save you some of the trouble I had to go through. (Although the process itself can be educating...). By the way, I am still feeling I do not grasp intuitivley what transversality is exactly... (even though I suceeded to work with it on the formal level). By the way, if you eventually write some nice anlaysis of the case of Hom-bundle (Or tensor product) , let me know. (I would like to see it). $\endgroup$ Jan 22, 2016 at 23:19
  • $\begingroup$ Sure no problem. $\endgroup$ Jan 23, 2016 at 4:09
  • $\begingroup$ I recently wrote something about transversality. dropbox.com/s/a3dc4pkhupoihel/Transversality.pdf?dl=0 I am planning to write such articles on different topics, just for my own edification. I am in the process of writing a small one on curvature, and one on Teichmuller spaces. Tell me if you would be interested in reading any of it. Also, if you have any such write ups which you may want to share, please let me know. $\endgroup$ Dec 17, 2017 at 11:45

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