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 $\mathrm{\Gamma }(E)$ (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 $\mathrm{\nabla }$ on $E$. $B$ is the algebra of smooth fiber-preserving maps from $\mathrm{\Gamma }(E)$ to $\mathrm{\Gamma }(E)$. ${\mathcal{D}}^1(M)$ is the Lie algebra of first-order differential operators on $M$ and $\mathcal{C}(M)$ is the connection algebra on $M$.

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

Proof: $\mathrm{\nabla }$ gives rise to a map:

$f_{\mathrm{\nabla }}:D^1(M)\to B$

as follows:

$f_{\mathrm{\nabla }}(X+m(g))=s\mapsto {\mathrm{\nabla }}_X(s)+(gs)$.

First observe that the map sends $C^{\mathrm{\infty }}(M)\subset {\mathcal{D}}^1(M)$ to $C^{\mathrm{\infty }}(M)\subset B$, 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 $f_{\mathrm{\nabla }}$:

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

$f_{\mathrm{\nabla }}(m(g)\dot{(}X+m(h)))(s)=f_{\mathrm{\nabla }}(gX+m(gh))(s)={\mathrm{\nabla }}_{gX}(s)+(gh)(s)=g{\mathrm{\nabla }}_X(s)+(gh)(s)=g({\mathrm{\nabla }}_X(s)+hs)=m(g)f_{\mathrm{\nabla }}(X+m(h))(s)$.

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

$f_{\mathrm{\nabla }}((X+m(h))\dot{m}(g))(s)=f_{\mathrm{\nabla }}(m(Xg)+g{\mathrm{\nabla }}_X+m(gh))(s)=(Xg)(s)+g{\mathrm{\nabla }}_X(s)+(gh)s={\mathrm{\nabla }}_X(gs)+(gh)(s)$.

References

Textbook references

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