Connection is module structure over connection algebra: Difference between revisions

Revision as of 21:50, 5 April 2008

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 $\nabla$ on $E$ and show how $\nabla$ naturally equips $E$ with the structure of a module over $C (M)$ .

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 $C^{\infty} (M)$ -bimodule. Let $B$ be the algebra of all smooth fiber-preserving linear maps from $Γ (E)$ to $Γ (E)$ .

A connection gives a map:

$D^{1} (M) \to B$

as follows:

$X \mapsto (s \mapsto \nabla_{X} (s))$

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

$\nabla_{f X} = f \nabla_{X}$

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

The upshot is that the induced map:

$D^{1} (M) \to B$

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

$C (M) \to B$

giving $E$ the structure of a module over $C (M)$ .

@@ Line 15: / Line 15: @@
 We start with a connection <math>\nabla</math> on <math>E</math> and show how <math>\nabla</math> naturally equips <math>E</math> with the structure of a module over <math>\mathcal{C}(M)</math>.
-First, observe that a connection gives a rule for the [[Lie algebra of first-order differential operators]] to ''act'' on <math>E</math>, hence the tensor algebra generated by it as a vector space, acts on <math>E</math>. We need to check that under this action <math>m(1) - 1</math> acts trivially on <math>E</math>.
+Let <math>D^1(M)</math> denote the [[Lie algebra of first-order differential operators]] of <math>M</math>. For now, we're thinking of <math>D^1(M)</math> as a <math>C^\infty(M)</math>-bimodule. Let <math>B</math> be the algebra of all smooth fiber-preserving linear maps from <math>\Gamma(E)</math> to <math>\Gamma(E)</math>.
+A connection gives a map:
+<math>D^1(M) \to B</math>
+as follows:
+<math>X \mapsto (s \mapsto \nabla_X(s))</math>
+First observe that the map sends <math>C^\infty(M) \subset D^1(M)</math> to <math>C^\infty(M) \subset B</math>, and is the identity restricted to that subset. In other words, the differential operator of multiplication by a function <math>f</math>, goes to the operator of multiplication by the function <math>f</math>.
+We now argue that the map is in fact a <math>C^\infty(M)</math>-bimodule map. The fact that it is a left <math>C^\infty(M)</math>-module map follows from the fact that:
+<math>\nabla_{fX} = f \nabla_X</math>
+while the fact that it is a right <math>C^\infty(M)</math>-module map follows from the Leibniz rule.
+The upshot is that the induced map:
+<math>D^1(M) \to B</math>
+is a <math>C^\infty(M)</math>-bimodule map, and hence extends to a <math>\R</math>-linear homomorphism from the tensor algebra of <math>D^1(M)</math> over <math>C^\infty(M)</math> to <math>B</math>. Clearly, <math>m(1) - 1</math> acts trivially, so we obtain a homomorphism:
+<math>\mathcal{C}(M) \to B</math>
+giving <math>E</math> the structure of a module over <math>\mathcal{C}(M)</math>.