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I have a question related to the definition of the pull-back connection, more specifically about its uniqueness or the canonical way to induce it.

The definition that one finds in general goes along the following lines: let $F:P\rightarrow B$ be map between differentiable manifolds, let $\pi:E\rightarrow B$ be a vector bundle and $\nabla$ a connection on $E$. Then the connection $F^*\nabla$ is uniquely determined by the following relation

$(F^*\nabla)_X(F^* s):=F^*(\nabla_{df(X)} s)$.

This should uniquely determine the connection right?

Let us start with the most trivial case, when $B$ is a point. Then $E$ is a vector space and the pull-back of $E$ is a trivial vector space over $P$. A connection on $E\rightarrow pt$ is an endomorphism of $E$ (which trivially satisfies the Leibniz relation). Could anybody explain how does that canonically induce a connection on $P\times E$, presumably the trivial connection $d$ if one starts with the zero endomorphism of $E$? Leibniz relation does not suffice. One could say let us make a convention here. But the more general question is how does one define $(F^*\nabla)_X$ when $X\in\ker{dF}$, in general?

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    $\begingroup$ What is $F^*s$? $\endgroup$
    – Deane Yang
    Commented Dec 13, 2010 at 16:06
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    $\begingroup$ I don't know how to define $F^*\nabla$ functorially (i.e., without using local co-ordinates and/or trivializations). I suggest first working this out using local co-ordinates and trivializations. $\endgroup$
    – Deane Yang
    Commented Dec 13, 2010 at 16:17
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    $\begingroup$ I take it back. Your equation is right. But I still recommend playing around with it using local co-ordinates and trivialization. It's a rather confusing equation (at least for me). $\endgroup$
    – Deane Yang
    Commented Dec 13, 2010 at 16:26
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    $\begingroup$ I don't understand the statement "a connection on $E \to pt$ is an endomorphism of $E$''. I would have thought it is a map $E \to E\otimes \Omega^1_{\pt},$ and since $\Omega^1_{\pt} = 0$, there is a unique connection, namely the zero map. This makes sense in terms of parallel transport: over a point there is nowhere to transport anything! $\endgroup$
    – Emerton
    Commented Dec 13, 2010 at 17:04
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    $\begingroup$ Over a pt, connections vanish (since vector fields & 1-forms are 0). In general "$dF(X)$" makes no sense as a vector field on $B$. View connections as additive maps from sections of $E$ to sections of $E \otimes \Omega^1_B$ over varying opens in $B$. Local sections of $F^{\ast}(E)$ are function-linear combinations of $F$-pullbacks of local sections of $E$, so the pullback rule (using pullback of 1-forms and of local sections) and Leibniz yield uniqueness. Construction with local coords gives local existence (& yields d when $B$ is pt and $E$ trivial), so by uniqueness get global existence. $\endgroup$
    – BCnrd
    Commented Dec 13, 2010 at 17:15

6 Answers 6

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Here's my summary of the situation:

1) First, observe that the space of local sections of the pullback bundle is generated by the space of sections of the original bundle composed with the map $F$. (This is better stated using sheaf language)

2) So, using the Leibniz rule, it suffices to define the pullback connection on a section obtained by composing a section of the original bundle with $F$.

3) The formula given above accomplishes this. It is worth noting that in this formula you should view $X$ as a single vector and not as a vector field.

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    $\begingroup$ As an aside, I stumbled onto this, because the use of a pullback connection is implicit when analyzing variations of geodesics and Jacobi fields on a Riemannian manifold. I've never seen this discussed explicitly in any textbook or paper, but it's actually quite necessary to make all the arguments rigorous. $\endgroup$
    – Deane Yang
    Commented Dec 13, 2010 at 19:06
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    $\begingroup$ Dear Deane: Aha, I didn't think of the interpretation with $X$ as a single vector (since the notation $\nabla_X$ is usually used with $X$ a vector field). OK, that then makes sense, but working that pointwise then requires a separate argument to verify the smoothness aspect (i.e., recognizing how to bypass the single-vector formulation after all). If we use the 1-form formulation then we work with local smooth sections throughout, so smoothness "comes along for the ride", and the single-vector interpretation can be inferred afterwards. Well, six of one, half dozen of the other. :) $\endgroup$
    – BCnrd
    Commented Dec 13, 2010 at 19:41
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    $\begingroup$ @AliTaghavi: I think that "generate" is read with coefficients in the ring of smooth functions over the domain, not just with coefficients being the reals. (May be simpler to reduce your question more and look first at the case where you have $f(x) = 0$. The pullback sections are only the constant sections.) $\endgroup$
    – Willie Wong
    Commented Sep 3, 2019 at 19:14
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    $\begingroup$ @AliTaghavi: not in general. On $\mathbb{R}^n$ the mapping $x \mapsto Ax$ where $A$ is any invertible $n\times n$ matrix is a diffeomorphism. Unless $A\in O(n)$ the mapping is in general not an isometry. But as an affine mapping the connection is preserved. This notion of affine mappings generalize to arbitrary Riemannian manifolds (in fact, to arbitrary manifolds with affine connections). In general, the affine group is bigger than the isometry group. But there are times when the two can be identified. ... $\endgroup$
    – Willie Wong
    Commented Sep 4, 2019 at 14:23
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    $\begingroup$ ... for example, a theorem of Yano (final theorem of the paper) states that on a compact orientable Riemannian manifold, any infinitesimal affine transformation is necessarily an infinitesimal isometry. (Or that restricted to the connected component of the identity, the affine group is the same as the isometry group; this restriction to the connected component is necessary, as you can take two spheres of unequal sizes as your manifold, and make your map swap their metrics.) $\endgroup$
    – Willie Wong
    Commented Sep 4, 2019 at 14:26
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Just to put what others have said into an explicit formula, note that any section of $F^*E$ is of the form

$\sum_j \varphi_j F^*s_j$,

for certain $\varphi_1, \ldots, \varphi_n \in C^{\infty}(P)$, and $s_1, \ldots, s_n \in \Gamma^{\infty}(E)$. Then in the notation of the question (with $X \in T_pP$ a single tangent vector), at $p \in P$ we have by the Leibniz rule and the defining relation of $F^*\nabla$,

$(F^*\nabla)_X \left(\sum_j \varphi_j F^*s_j\right) = \sum_j \left( d\varphi_j(X) F^*s_j + \varphi_j (F^*\nabla)_X F^*s_j \right) = \sum_j \left( d\varphi_j(X) F^*s_j + \varphi_j F^*(\nabla_{dF(X)}s_j) \right) $

(with everything evaluated at $p$ where appropriate). This gives back $F^*\nabla = d$ in your example where $B$ is a point and you take the zero endomorphism of $E$.

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    $\begingroup$ Can you prove your first claim? It sounds just false to me. For example, take $F : [0,1] \to B$ a curve with a self-intersection. $\endgroup$
    – seub
    Commented Sep 9, 2018 at 17:22
  • $\begingroup$ say $F(t_0)=F(t_1)$ is the self-intersection, then let $\varphi_j(t_j)=1$ and $\varphi_j(t_{1-j})=0$ and $\varphi_0+\varphi_1=1$. then in the supports of these coefficients the curve is injective. $\endgroup$
    – peter
    Commented Mar 4, 2019 at 16:01
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    $\begingroup$ @seub as another counter example $f:\mathbb{R}\to \mathbb{R}\quad f(x)=x^2$ pull back sections are just even functions. $\endgroup$
    – Ali Taghavi
    Commented Sep 3, 2019 at 18:21
  • $\begingroup$ If we define by this, then well-definedness problem comes out: for any section, it may admit different linear combinations. So how to prove the definition is independent of choice? $\endgroup$
    – Sherose
    Commented Nov 11, 2021 at 0:48
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Hi, I have seen the equation you gave as a definition many times. For example, I think it is also used in a corresponding Wikipedia article. Nevertheless, as you correctly pointed out, it does not give a reasonable/unique description. A better formulation/definition of the pullback connection can be found for example in Milnor's and Stasheff's book 'Characteristic classes' on p. 292, Lemma 3 and its proof (definition by universal property/commutative diagram; proof: computation in local coordinates; it's the precise version of what the equation you gave tries to capture). I hope this helps more or less.

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    $\begingroup$ The commutative diagram in Milnor and Stasheff is exactly the same equation, only written in a different language. As others pointed out, it does define the pull-back connection uniquely but you can not use it directly for arbitrary sections of the pull-back bundle. $\endgroup$
    – Sergei Ivanov
    Commented Dec 14, 2010 at 9:03
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See 19.12.6 (page 250) in this book.

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To Matt: The relation $(F^*\nabla)_X=0$ does not satisfy Leibniz relation. Meanwhile I found out the answer to my question (a friend clarified it for me).

It turns out that there is an isomorphism $\Gamma(P;\pi^*E)\simeq \Gamma(P;\mathbb{C})\otimes \Gamma(B;E)$ where the tensor product is over $\Gamma(B;\mathbb{C})$. It is obviously true when $E$ is a vector space. Now use Leibniz relation $\nabla_X(f\otimes s)=X(f)\otimes s+f\otimes\nabla_Xs$ to extend the connection from $\Gamma(B;E)$ to $\Gamma (P;\pi^*E)$. In the trivial case one gets indeed that $\nabla_X(f\otimes s)=X(f)\otimes s$.

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  • $\begingroup$ @Unknown: see my comment above. The equation you wrote down is only meaningful for pull-backs of sections $s$. If you multiply $F^*s$ by an arbitrary function $h$, it will no longer be the pull back of a section over $B$ if $h$ is not constant along $F^{-1}B$. In other words, you were trying to force Leibniz rule somewhere it has no business being. The given expression is, indeed, enough to specify the pull-back connection, but it is not so simple as plug-and-play. $\endgroup$
    – Willie Wong
    Commented Dec 13, 2010 at 17:15
  • $\begingroup$ Willie i didn't say it was easy:) In the background lurks the isomorphism $\Gamma(P;\pi^*E)\simeq \Gamma(P;\mathbb{C})\otimes \Gamma(B;E)$ which is by not obvious(at least not to me). But Leibniz relation does make sense and it does define the connection everywhere and it also explains the trivial connection on the trivial bundle. $\endgroup$
    – user11538
    Commented Dec 13, 2010 at 17:29
  • $\begingroup$ Dear unknown: the isom. you're using is false in complex-analytic and algebraic cases (unsure in $C^\infty$-case, but seems silly to use a defn that only works there when there's a simple procedure applicable in all cases). Think more locally (it's good for you!): use local existence and global uniqueness to get global existence. What you propose with "global" tensors works locally with no hard work; then use uniqueness to globalize. Your comment to Matt is unclear since your original statement of the pullback-relation with global vector fields makes no sense ($dF(X)$ makes no sense on $B$). $\endgroup$
    – BCnrd
    Commented Dec 13, 2010 at 18:26
  • $\begingroup$ Dear BCnrd, Thank you for your interesting comments but... 1) I was only interested in the differentiable case. I am not sure what you mean by a connection in the analytic or algebraic cases. (an analytic(algebraic) splitting of the tangent bundle?) so I am really satisfied with the isomorphism of smooth sections i was talking about. By the way, it's not mine. 2)I admit the comment to Matt lacked details, probably thinking that it was clear $X$ should be thought as a tangent vector and not a vector field (as I wrote); (retrospectively that was really not the main issue of the question); $\endgroup$
    – user11538
    Commented Dec 13, 2010 at 20:30
  • $\begingroup$ Dear unknown: later in life you'll want to use complex manifolds or complex algebraic varieties. (Connections remain all about making vector fields act as "directional deriv." operators on sections of the bundle, as in diff'ble case.) Then no bump functions, so cannot work entirely so "globally" as above. That's why I outlined an alternative to exploit your preferred calculations on a local level, coupled with global uniqueness (which can be proved by local calculations!) to infer global existence. It really is easier that way (e.g., no non-obvious isoms needed). You'll appreciate it later. $\endgroup$
    – BCnrd
    Commented Dec 14, 2010 at 0:01
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OP's example is not correct for a point already.

First thing, connection is a differential operator $\nabla: \Gamma(E) \otimes TX \rightarrow \Gamma (E)$ of first order with a symbol $\partial$.

This amounts to saying that it satisfies Leibniz rule in the following sense:

$\nabla_{v} (f \gamma) = f\nabla_v(\gamma) + \partial_v (f) \gamma$

For a point, there is no need to even check Leibniz rule, because $TX = 0$, so there is no place from where you could obtain an automorphism! Connection allows you to derive the sections along a vector field, and every vector field on a point is $0$.

Moreover, pullback of the connection (which is defined correctly by the formula in the post) is a trivial connection on a trivial bundle (and it is canonical, because all fibers are canonically identified).

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