Connection is module structure over connection algebra: Difference between revisions

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First observe that the map sends <math>C^\infty(M) \subset \mathcal{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>.  
First observe that the map sends <math>C^\infty(M) \subset \mathcal{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 prove some basic results about <math>f_\nabla</math>:
We now prove that the map <math>\nabla \mapsto f_\nabla</math> is a <math>C^\infty(M)</math>-bimodule map from <math>D^1(M)</math> to <math>B</math>, i.e., left and right multiplication by <math>m(g)</math> can be ''pulled out'' of the <math>f_\nabla</math>:


* <math>f_\nabla</math> is <math>\R</math>-bilinear: This is obvious.
* <math>f_\nabla</math> is <math>\R</math>-bilinear: This is obvious.
* For any element <math>X + m(g)</math> in <math>\mathcal{D}^1(M)</math> and any <math>h \in C^\infty(M)</math>, we have <math>f_\nabla(m(g) \dot (X + m(h))(s) = m(g)f_\nabla(X + m(h))(s)</math>. This essentially follows from the fact that a connection is [[tensorial map|tensorial]] in the direction of differentiation:
* Left module map property: For any element <math>X + m(g)</math> in <math>\mathcal{D}^1(M)</math> and any <math>h \in C^\infty(M)</math>, we have <math>f_\nabla(m(g) \cdot (X + m(h))(s) = m(g) \cdot f_\nabla(X + m(h))(s)</math>. This essentially follows from the fact that a connection is [[tensorial map|tensorial]] in the direction of differentiation:


<math>f_\nabla(m(g) \dot (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)</math>.
<math>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)</math>.


* For any element <math>X + m(g)</math> in <math>\mathcal{D}^1(M)</math> and any <math>h \in C^\infty(M)</math>, we have  <math>f_\nabla((X + m(h)) \dot m(g))(s) = f_\nabla(X + m(h))(m(g)s)</math>. This essentially follows from the Leibniz rule property.
* For any element <math>X + m(g)</math> in <math>\mathcal{D}^1(M)</math> and any <math>h \in C^\infty(M)</math>, we have  <math>(f_\nabla((X + m(h)) \cdot m(g))(s) = (f_\nabla(X + m(h)) \circ m(g))(s)</math>. This essentially follows from the Leibniz rule property.


<math>f_\nabla((X + m(h)) \dot 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)</math>.
<math>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)</math>.


==References==
==References==

Revision as of 00:26, 24 July 2009

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))=sX(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 hC(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)=gX(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 hC(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)+gX+m(gh))(s)=(Xg)(s)+gX(s)+(gh)s=X(gs)+(gh)(s)=(f(X+m(h))m(g))(s).

References

Textbook references