Difference between revisions of "Connection is module structure over connection algebra"

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==Proof==
 
==Proof==
  
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>.
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===From a connection to a module structure===
  
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>.
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The outline of the proof is as follows:
  
A connection gives a map:
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* We first show that a connection gives an action of the first-order differentiable operators on the space of sections.
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* Next, we show that the Leibniz rule property of connections allows us to extend this to a well-defined action of the connection algebra.
  
<math>D^1(M) \to B</math>
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'''Given''': A manifold <math>M</math>, a vector bundle <math>E</math> over <math>M</math>, a connection <math>\nabla</math> on <math>E</math>. <math>B</math> is the algebra of smooth fiber-preserving maps from <math>\Gamma(E)</math> to <math>\Gamma(E)</math>. <math>\mathcal{D}^1(M)</math> is the Lie algebra of first-order differential operators on <math>M</math> and <math>\mathcal{C}(M)</math> is the connection algebra on <math>M</math>.
  
as follows:
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'''To prove''': <math>\nabla</math> gives rise to a homomorphism from <math>\mathcal{C}(M)</math> to <math>B</math>.
  
<math>X \mapsto (s \mapsto \nabla_X(s))</math>
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'''Proof''': <math>\nabla</math> gives rise to a map:
  
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>.
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<math>f_\nabla: D^1(M) \to B</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:
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as follows:
  
<math>\nabla_{fX} = f \nabla_X</math>
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<math>f_\nabla(X+m(g)) = s \mapsto \nabla_X(s) + (gs)</math>.
  
while the fact that it is a right <math>C^\infty(M)</math>-module map follows from the Leibniz rule.
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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>.  
  
The upshot is that the induced map:
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We now prove some basic results about <math>f_\nabla</math>:
  
<math>D^1(M) \to B</math>
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* <math>f_\nabla</math> is <math>\R</math>-bilinear: This is obvious.
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* 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:
  
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:
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<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)</math>.
  
<math>\mathcal{C}(M) \to B</math>
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* 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.
  
giving <math>\Gamma(E)</math> the structure of a module over the algebra <math>\mathcal{C}(M)</math>.
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<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>.
  
 
==References==
 
==References==

Revision as of 00:18, 24 July 2009

Statement

Let E be a vector bundle over a differential manifold M. Then, a connection on E is equivalent to giving \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 \nabla on E. B is the algebra of smooth fiber-preserving maps from \Gamma(E) to \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: \nabla gives rise to a homomorphism from \mathcal{C}(M) to B.

Proof: \nabla gives rise to a map:

f_\nabla: D^1(M) \to B

as follows:

f_\nabla(X+m(g)) = s \mapsto \nabla_X(s) + (gs).

First observe that the map sends C^\infty(M) \subset \mathcal{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 prove some basic results about f_\nabla:

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

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).

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

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).

References

Textbook references