# Difference between revisions of "Connection is module structure over connection algebra"

## Statement

Let $E$ be a vector bundle over a differential manifold $M$. Then, a connection on $E$ is equivalent to giving $\Gamma(E)$ (the vector space of sections of $E$) the structure of a module over the connection algebra of $M$. Equivalently, it gives $\mathcal{E}$ (the sheaf of sections of $E$) the structure of a module over the sheaf of connection algebras over $M$.

## Definitions used

### Connection

Further information: Connection

### Connection algebra

Further information: Connection algebra

## Proof

### From a connection to a module structure

The outline of the proof is as follows:

• We first show that a connection gives an action of the first-order differentiable operators on the space of sections.
• Next, we show that the Leibniz rule property of connections allows us to extend this to a well-defined action of the connection algebra.

Given: A manifold $M$, a vector bundle $E$ over $M$, a connection $\nabla$ on $E$. $B$ is the algebra of smooth fiber-preserving maps from $\Gamma(E)$ to $\Gamma(E)$. $\mathcal{D}^1(M)$ is the Lie algebra of first-order differential operators on $M$ and $\mathcal{C}(M)$ is the connection algebra on $M$.

To prove: $\nabla$ gives rise to a homomorphism from $\mathcal{C}(M)$ to $B$.

Proof: $\nabla$ gives rise to a map: $f_\nabla: D^1(M) \to B$

as follows: $f_\nabla(X+m(g)) = s \mapsto \nabla_X(s) + (gs)$.

First observe that the map sends $C^\infty(M) \subset \mathcal{D}^1(M)$ to $C^\infty(M) \subset B$, and is the identity restricted to that subset. In other words, the differential operator of multiplication by a function $f$, goes to the operator of multiplication by the function $f$.

We now prove some basic results about $f_\nabla$:

• $f_\nabla$ is $\R$-bilinear: This is obvious.
• For any element $X + m(g)$ in $\mathcal{D}^1(M)$ and any $h \in C^\infty(M)$, we have $f_\nabla(m(g) \dot (X + m(h))(s) = m(g)f_\nabla(X + m(h))(s)$. This essentially follows from the fact that a connection is tensorial in the direction of differentiation: $f_\nabla(m(g) \dot (X + m(h)))(s) = f_\nabla(gX + m(gh))(s) = \nabla_{gX}(s) + (gh)(s)= g\nabla_X(s) + (gh)(s) = g(\nabla_X(s) + hs)$.

• For any element $X + m(g)$ in $\mathcal{D}^1(M)$ and any $h \in C^\infty(M)$, we have $f_\nabla((X + m(h)) \dot m(g))(s) = f_\nabla(X + m(h))(m(g)s)$. This essentially follows from the Leibniz rule property. $f_\nabla((X + m(h)) \dot m(g))(s) = f_\nabla(m(Xg) +g\nabla_X + m(gh))(s) = (Xg)(s) + g\nabla_X(s) + (gh)s = \nabla_X(gs) + (gh)(s)$.