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 $\mathbb {R}$-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 $f_{\nabla }(m(g)\cdot (X+m(h))(s)=m(g)\cdot f_{\nabla }(X+m(h))(s)$. 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.

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

References

Textbook references

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