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 that the map $\nabla \mapsto f_\nabla$ is a $C^\infty(M)$-bimodule map from $D^1(M)$ to $B$, i.e., left and right multiplication by $m(g)$ can be pulled out of the $f_\nabla$:

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

$f_\nabla(m(g) \cdot (X + m(h)))(s) = f_\nabla(gX + m(gh))(s) = \nabla_{gX}(s) + (gh)(s)= g\nabla_X(s) + (gh)(s) = g(\nabla_X(s) + hs) = m(g)f_\nabla(X + m(h))(s)$.

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

$f_\nabla((X + m(h)) \cdot m(g))(s) = f_\nabla(m(Xg) +g\nabla_X + m(gh))(s) = (Xg)(s) + g\nabla_X(s) + (gh)s = \nabla_X(gs) + (gh)(s) = (f_\nabla(X + m(h)) \cdot m(g))(s)$.

Since $\nabla \mapsto f_\nabla$ is a $C^\infty$-bimodule map, it extends to a unique $C^\infty$-bimodule map from the $C^\infty$-tensor algebra over $\mathcal{D}^1(M)$. By definition, the element $m(1) - 1$ induces the zero map on $\Gamma(E)$, so the map descends to a homomorphism $\mathcal{C}(M) \to B$, as desired.

From a module structure to a connection

Given: A manifold $M$, a vector bundle $E$ over $M$. $B$ 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 $M$. A module structure of $\Gamma(E)$ over $\mathcal{C}(M)$.

References

Textbook references

• Book:Globalcalculus^{More info}, Page 64