This short entry is a continuation of this one. I was listening to a very nice talk by Anton Galaev on his article arXiv:2502.07543. In it, he explained various ways to see curvature on sub-Riemannian manifolds $(M,D,g)$, and that sometimes it is useful to consider extended connections (operations $\Gamma(TM)\times \Gamma(D)\to \Gamma(D)$ rather than $\Gamma(D)\times \Gamma(D)\to \Gamma(D)$).

Definition

Let $\rho:A\to TM$ be an anchored bundle, $V\to M$ a vector bundle and $\tilde \nabla:\Gamma(TM)\times \Gamma(V)\to \Gamma(V)$ a connection. We call $\nabla=\rho^*\tilde\nabla$ the pullback connection. It is defined by $\nabla_a\sigma:=\tilde\nabla_{\rho(a)}\sigma$.

Extending connections

In general, not every connection can be realized as a pullback. For instance, for pullback connections $\nabla_a$ is zero when $a\in \ker(\rho)$; for general $A$-connections this might not be the case. Even when a connection vanishes on $\ker(\rho)$, it is not clear to me whether it can always be extended. In the case of constant rank anchors, everything works fine:

Lemma

Let $\rho: A\to TM$ be of constant rank and $\nabla$ an $A$-connection on $V$ which vanishes on $\ker(\rho)$. Then there exists a $TM$-connection $\tilde\nabla$ on $V$ such that $\nabla=\rho^*\tilde\nabla$.

The proof is based on two elementary facts:

  • If $\nabla$ is an $A$-connection on $V$ and $K\subset \ker(\rho)$ a subbundle on which $\nabla$ vanishes, then $\nabla$ induces a unique $A/K$-connection on $V$.
  • If $A$ and $A'$ are anchored bundles then $A\oplus A'$ is an anchored bundle and given $A$- and $A'$-connections on $V$, there is a natural $A\oplus A'$-connection on $V$ given by their sum.

Proofof the Lemma

Starting with $A,\nabla$ as in the Lemma, one has an anchored bundle $\bar A=A/\ker(\rho)$ and a corresponding connection $\bar \nabla$. This can be equivalently seen as a $\rho(A)$-connection on $V$, since $\bar A$ is isomorphic to the image of $\rho$. Picking a complement $D$ of $\rho(A)$ in $TM$ and a $D$-connection $\nabla^D$ on $V$, one obtains the desired $TM$-connection as $\bar\nabla + \nabla^D$ (using $TM\cong\rho(A)\oplus D$).

A particular case of the above is when $A$ is a subbundle of $TM$, i.e., $\rho$ is injective. Then the Lemma simply states that any $A$-connection can be extended to a full $TM$-connection.

The curvature of a pulled back connection

When $\nabla=\rho^*\tilde\nabla$ is an $A$-connection on $A$, which is a pullback of a $TM$-connection $\tilde \nabla$ on $A$, the curvature $R$ of $\nabla$ can be written as:

$$ R(a,b)c = [\nabla_a, \nabla_b]c - \tilde \nabla_{\rho(\llbracket a,b\rrbracket)}c = [\nabla_a, \nabla_b]c - \tilde \nabla_{[\rho (a),\rho(b)]}c - \tilde\nabla_{\tau(a,b)}c $$

The first two terms are simply the pullback via the anchor of the curvature $\tilde R^{\tilde \nabla}$ of $\tilde\nabla$, i.e.,

$$\rho^*\tilde R^{\tilde \nabla}(a,b)c = [\nabla_a, \nabla_b]c - \tilde \nabla_{[\rho (a),\rho(b)]}c.$$

This implies:

Lemma

The expression $\rho^*\tilde R^{\tilde \nabla}(a,b)c$ is tensorial in all three entries $a,b$ and $c$.

In the previous post, we managed to decompose the curvature into a tensorial and a torsion-based term using the fact that $A$ carries a Lie algebroid structure (hence allowing us to lift the torsion to $A$). Here, we instead use the auxiliary datum of a connection to perform the same type of separation, which now depends on the ambient connection $\tilde \nabla$:

$$ R(a,b)= \rho^*\tilde R^{\tilde \nabla}(a,b) -\tilde\nabla_{\tau(a,b)}. $$