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 $E$ the structure of a module over the connection algebra over $M$.

Definitions used

Connection

Further information: Connection

Connection algebra

Further information: Connection algebra

Proof

We start with a connection $\mathrm{\nabla }$ on $E$ and show how $\mathrm{\nabla }$ naturally equips $E$ with the structure of a module over $\mathcal{C}(M)$.

Let $D^1(M)$ denote the Lie algebra of first-order differential operators of $M$. For now, we're thinking of $D^1(M)$ as a $C^{\mathrm{\infty }}(M)$-bimodule. Let $B$ be the algebra of all smooth fiber-preserving linear maps from $\mathrm{\Gamma }(E)$ to $\mathrm{\Gamma }(E)$.

A connection gives a map:

$D^1(M)\to B$

as follows:

$X\mapsto (s\mapsto {\mathrm{\nabla }}_X(s))$

First observe that the map sends $C^{\mathrm{\infty }}(M)\subset D^1(M)$ 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 argue that the map is in fact a $C^{\mathrm{\infty }}(M)$-bimodule map. The fact that it is a left $C^{\mathrm{\infty }}(M)$-module map follows from the fact that:

${\mathrm{\nabla }}_{fX}=f{\mathrm{\nabla }}_X$

while the fact that it is a right $C^{\mathrm{\infty }}(M)$-module map follows from the Leibniz rule.

The upshot is that the induced map:

$D^1(M)\to B$

is a $C^{\mathrm{\infty }}(M)$-bimodule map, and hence extends to a $\mathbb{R}$-linear homomorphism from the tensor algebra of $D^1(M)$ over $C^{\mathrm{\infty }}(M)$ to $B$. Clearly, $m(1)-1$ acts trivially, so we obtain a homomorphism:

$\mathcal{C}(M)\to B$

giving $E$ the structure of a module over $\mathcal{C}(M)$.

References

Textbook references

• Book:Globalcalculus^{More info}