I was recently reading the very nice preprint arXiv:2608.07351, on how to construct a Lie algebroid over $\mathbb R^n$ whose underlying foliation is given by the vector fields vanishing up to a prescribed order $k$ and tried to translate the notion of left-symmetric algebroid used therein, in terms of connections. The following is a writeup of my thoughts.

All of the below is certainly well-known, I will try to add more references when I find them. In the sequel $M$ will be a smooth manifold and all objects are smooth unless otherwise stated.

Connections

Definition

An anchored bundle is a vector bundle $A\to M$ together with a vector bundle homomorphism $\rho: A\to TM$ over the identity of $M$, such that $\rho_*\Gamma(A)\subset \mathfrak X(M)=\Gamma(TM)$ is involutive (i.e. stable under the Lie bracket), where $\rho_*$ is the pushforward by $\rho$.

The anchor is exactly what we need to be able to define connections:

Definition

Let $A$ be an anchored bundle and $V$ a vector bundle over $M$. An $\mathbb R$-bilinear map $\nabla:\Gamma(A)\times \Gamma(V)\to \Gamma(V)$ is called an $A$-connection on $V$, if

  • it is $C^\infty(M)$-linear in the first entry, i.e. $\nabla_{fa}\sigma=f\nabla_a\sigma$ for all $f\in C^\infty(M)$, $a\in\Gamma(A)$, $\sigma\in \Gamma(V)$.
  • it satisfies the Leibniz rule in the second entry: $\nabla_{a}(f\sigma)=f\nabla_a\sigma+ \rho(a)(f)\sigma$ (again for all $a,\sigma,f$).

As usual, having a connection is not very difficult:

Lemma

$A$-connections exist on any vector bundle $V$. Two $A$-connections differ by an element of $\Gamma(A^*\otimes End(V))$.

Proof

The second statement comes from the fact that the difference of two connections is $C^\infty(M)$-linear in both entries. For the first statement one first observes that for a trivial bundle $V=M\times \mathbb R^N$ we can define a connection uniquely determined by the fact that it vanishes on the standard frame (i.e. $\nabla_ae_i=0$ for all $i$). To obtain the general case, one either glues connections using a partition of unity, or embeds $V$ into a trivial bundle $M\times \mathbb R^N$ and uses the inclusion/projection to induce a connection on $V$ from one on $M\times \mathbb R^N$.

Curvature

Of course, we can look at $A$-connections on $V=A$, and obtain a natural notion of curvature:

Definition

$R:\Gamma(A)^3\to \Gamma(A), R(a,b)c := [\nabla_a, \nabla_b]c - \nabla_{(\nabla_ab-\nabla_ba)}c $

Note that this definition of curvature needs no bracket on $A$, we are implicitly using $\nabla_ab-\nabla_ba$ as a bracket (but more on that later). The condition $R\equiv 0$, i.e. the flatness of the connection is exactly the statement that the associator of $\nabla$ is symmetric in the first two arguments, i.e. that $\triangleright=\nabla$ turns $A$ into a Koszul-Vinberg algebroid. Those were originally introduced here and named left-symmetric algebroids here. Sometimes they are also called pre-Lie algebroids (e.g. here). The reason to call them pre-Lie algebroids (in analogy with pre-Lie algebras) is that they induce Lie algebroids

Lemma

If $R\equiv 0$, then $\llbracket a,b\rrbracket:=\nabla_ab-\nabla_ba$ defines a Lie algebroid structure on $A$.

Remark

The converse is not true, for instance if we take $\rho=id:TS^2\to TS^2$ and equip $S^2$ with its standard connection, then $\nabla$ has non-zero curvature but its commutator still gives $TS^2$ its standard (vector field) Lie bracket.

Being a Lie algebroid implies the identity $\rho(\llbracket a,b\rrbracket)=[\rho(a),\rho(b)]$ for all $a,b$ (by the argument from this paper). Hence, the curvature $R$ vanishing implies that the torsion vanishes, too, where torsion is defined as follows:

Definition

For any $A$-connection on $A$, we call $\tau:\Gamma(A)^2\to \Gamma(TM)$,

$$\tau(a,b):=\rho(\llbracket a,b\rrbracket)-[\rho(a),\rho(b)]=\rho(\nabla_ab)-\rho(\nabla_ba) - [\rho(a),\rho(b)] $$

the torsion of $\nabla$.

Let’s just record what we have seen:

Proposition

$R\equiv 0 \implies \tau \equiv 0$

Torsion

The torsion is $C^\infty(M)$-linear in both entries and skew-symmetric, hence can be seen as a section of $\Lambda^2A^*\otimes TM$. Obtaining a torsion-free connection is always possible.

Lemma

For any anchored bundle $A$, there exists a torsion-free connection $\nabla$. Two torsion-free connections differ by a section of $S^2A^*\otimes A$.

Proof

Let $\nabla$ be an initial connection with torsion $\tau$. We now use the involutivity of $\rho_*(\Gamma(A))$, to obtain a lift $\tilde \tau:\Gamma(A)^2\to\Gamma(A)$ of the torsion and define a new connection $\bar\nabla$ by $\bar\nabla_ab=\nabla_ab-\frac{1}{2}\tilde\tau(a,b)$. This new connection is torsion-free. The second statement is a direct verification.

One nice property of torsion-free connections is that their commutator induces almost-Lie algebroid structures.

Definition

An almost-Lie algebroid is an anchored bundle $(A,\rho)$ with a bracket $\llbracket \cdot,\cdot\rrbracket$, which is a biderivation (with respect to the anchor), such that $\rho(\llbracket a,b\rrbracket)=[\rho(a),\rho(b)]$ for all $a,b$.

Hence, we recover the following fact for free (e.g. Proposition 2.2.4 here):

Proposition

For any anchored bundle $A$, there exists a compatible almost-Lie algebroid structure.

Relation between curvature and torsion

But what does the torsion measure? It turns out, it measures exactly the failure of $R$ to be $C^\infty(M)$-linear in its third entry. More precisely:

Lemma

The curvature $R$ is skew-symmetric and $C^\infty(M)$-bilinear in the first two entries. Moreover, $R(a,b)(fc)-fR(a,b)c = \tau(a,b)(f)\cdot c$ for all $f,a,b,c$.

This explains why $R\equiv 0$ implies $\tau\equiv 0$: The zero map is clearly $C^\infty(M)$-linear.

Relation to curvature and torsion on a Lie algebroid

Let us now assume that we start with a Lie algebroid $(A,\rho,[\cdot,\cdot])$ equipped with a connection. Following this article, one can then define different notions of torsion and curvature, this time with respect to the Lie algebroid bracket

$$ \tilde \tau(a,b) = \llbracket a,b\rrbracket - [a,b] $$

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

We then have $\rho_*\circ \tilde \tau = \tau$ and $R(a,b)=\tilde R(a,b) + \nabla_{\tilde\tau(a,b)}$.