Suppose that $f:M \to N$ is a smooth map of two manifolds and $E$ is a vector bundle over $N$. Given a connecton $\nabla$ on $E$ one can pull it back to $f^*N$ (which is a bundle over $M$) and obtain a connection $f^* \nabla$.

If $E=TN$ is a tangent bundle one can consider the Levi-Civita connection $\nabla$, with respect to the chosen metric $g$ on $N$. It is characterised uniquely as torsion free metric connection. Given a map $f:M \to N$ one can pullback the metric $g$ to $f^*g$. If it happens that $f^*g$ is again a metric (which is the case for example for immersions) one can consider the Levi Civita connection on $M$ with respect to $f^*g$. I wonder how these two constructions are related, in particular

Is there a way to pullback affine connections to get again affine connection (i.e. connections on tangent bundle)?

Additionally I would like to know:

What is the relation between Christoffels symbols for the connection and for its pullback?

  • $\begingroup$ How one can do this? I know how to pull connection via the map of the underlying manifolds, not by a map between bundles $\endgroup$
    – truebaran
    Commented Oct 4, 2017 at 23:56
  • $\begingroup$ Hmm, okay, maybe it's not so straightforward. $\endgroup$ Commented Oct 5, 2017 at 0:03

1 Answer 1


Is there a way to pullback affine connections to get again affine connection?

The standard pullback construction you describe gives a connection $f^* \nabla$ on the bundle $f^* TN$, so the question is how to naturally turn this in to a connection on $TM$ using the bundle map $Df : TM \to f^* TN.$ More abstractly, we have two bundles $E \to M, F \to M$ and a bundle map $\phi :E \to F$, and we want to know what conditions on $\phi$ are required in order to pull back a connection $\nabla$ on $F$ to a connection $D$ on $E$.

In order to get a feel for what this "pullback" should do, we can first assume that $\phi$ is an isomorphism (which corresponds in the original problem to $f$ being a local diffeomorphism). In this case we can use $\phi$ to identify $E$ and $F$, and so all structures on $F$ transfer naturally to $E$ (and vice versa). In particular we have the formula $$ D_X \xi = \phi^{-1} \nabla_X (\phi(\xi))$$ for a section $\xi \in \Gamma(E).$

Now we can try to relax our assumptions: what do we really need to assume about $\phi$ in order for this formula to define a connection on $E$? Well, immediately we see that for $\phi^{-1}$ to make sense we need $\phi$ to be injective. But something else can go wrong: we need $\nabla_X(\phi(\xi))$ to be in the image of $\phi$, but connections are surjective on fibers; so in order for this to make sense we need to assume that $\phi$ is surjective. Thus we can't gain anything if we want to stick strictly to this formula: we need $\phi$ to be an isomorphism.

To reconcile this with the induced connection in Riemannian geometry, remember that we use extra structure there: the induced connection is the tangential projection of the pullback connection. That is, for a Riemannian immersion $f : M \to N$ with corresponding injective $\phi = Df : TM \to f^*TN,$ we have $D_X Y = \pi (\nabla_X \phi(Y))$ where $\pi : f^*TN \to TM$ is the left-inverse of $\phi$ given by orthogonal projection on to $TM$. Without the Riemannian structure to single out the correct choice of $\pi$, we have no way of distinguishing between the many left-inverses of $\phi$ and thus no canonical induced connection.

What is the relation between Christoffels symbols for the connection and for its pullback?

If we fix local coordinates $x^i$ on $M$, $y^\alpha$ on $N$ and a local frame $E_A$ for a vector bundle $E \to N$, then the pullback bundle $f^* E$ has a local frame given by $E_A \circ f$, and the corresponding connection coefficients defined by $$^{f^* \nabla}\Gamma^B_{Ai}(E_B \circ f)=(f^* \nabla)_{\partial/\partial x^i}(E_A \circ f),\;\;\;^{\nabla}\Gamma^{B}_{A\alpha} E_B=\nabla_{\partial/\partial y^\alpha} E_A$$ are related simply by $$^{f^* \nabla}\Gamma^B_{Ai} = {}^\nabla \Gamma^B_{A \alpha} (df)^\alpha_i.$$

  • $\begingroup$ I believe one can pull a connection back in some cases where $\phi:E\to F$ is injective but not surjective. All one needs is that the connection $\nabla$ on $F$ respects $\phi$, in the sense that parallel transport with respect to $\nabla$ carries elements of $E$ to elements of $E$. $\endgroup$ Commented Oct 15, 2017 at 18:59
  • $\begingroup$ So, what is the conclusion here? I fail to see the top view picture of this answer... $\endgroup$
    – user537667
    Commented Aug 17, 2019 at 10:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .