Connection is module structure over connection algebra

From Diffgeom

Statement

Let E be a vector bundle over a differential manifold M. Then, a connection on E is equivalent to giving Γ(E) (the vector space of sections of E) the structure of a module over the connection algebra of M. Equivalently, it gives 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 ∇ on E. B is the algebra of smooth fiber-preserving maps from Γ(E) to Γ(E). D1(M) is the Lie algebra of first-order differential operators on M and C(M) is the connection algebra on M.

To prove: ∇ gives rise to a homomorphism from C(M) to B.

Proof: ∇ gives rise to a map:

f∇:D1(M)→B

as follows:

f∇(X+m(g))=s↦∇X(s)+(gs).

First observe that the map sends C∞(M)⊂D1(M) to C∞(M)⊂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 ∇↦f∇ is a C∞(M)-bimodule map from D1(M) to B, i.e., left and right multiplication by m(g) can be pulled out of the f∇:

  • f∇ is R-bilinear: This is obvious.
  • Left module map property: For any element X+m(g) in D1(M) and any h∈C∞(M), we have f∇(m(g)⋅(X+m(h))(s)=m(g)⋅f∇(X+m(h))(s). This essentially follows from the fact that a connection is tensorial in the direction of differentiation:

f∇(m(g)⋅(X+m(h)))(s)=f∇(gX+m(gh))(s)=∇gX(s)+(gh)(s)=g∇X(s)+(gh)(s)=g(∇X(s)+hs)=m(g)f∇(X+m(h))(s).

  • For any element X+m(g) in D1(M) and any h∈C∞(M), we have (f∇((X+m(h))⋅m(g))(s)=(f∇(X+m(h))∘m(g))(s). This essentially follows from the Leibniz rule property.

f∇((X+m(h))⋅m(g))(s)=f∇(m(Xg)+g∇X+m(gh))(s)=(Xg)(s)+g∇X(s)+(gh)s=∇X(gs)+(gh)(s)=(f∇(X+m(h))⋅m(g))(s).

References

Textbook references