Difference between revisions of "Connection is splitting of first-order symbol sequence"

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(New page: ==Statement== Suppose <math>E</math> is a vector bundle over a differential manifold <math>M</math>.)
 
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==Statement==
 
==Statement==
  
Suppose <math>E</math> is a [[vector bundle]] over a [[differential manifold]] <math>M</math>.
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Suppose <math>E</math> is a [[vector bundle]] over a [[differential manifold]] <math>M</math>. Denote by <math>\mathcal{E}</math> the [[sheaf of sections of a vector bundle|sheaf of sections]] of <math>E</math>. Consider the first-order symbol sequence for <math>E</math>, given by:
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<math>0 \to Hom(\mathcal{E},\mathcal{A}) \to \mathcal{D}^1(\mathcal{E},\mathcal{A}) \to Hom(\mathcal{E},\mathcal{T}) \to 0</math>
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Here <math>\mathcal{A}</math> is the [[sheaf of infinitely differentiable functions]] on <math>M</math>, <math>\mathcal{D}^1</math> denotes the space of first-order differential operators from <math>\mathcal{E}</math> to <math>\mathcal{A}</math>, and <math>\mathcal{T}</math> denotes the [[sheaf of derivations]] of <math>M</math>.
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A '''connection''' on <math>E</math> is equivalent to a choice of splitting for this sequence.
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==Proof==
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A splitting of the above sequence is equivalent to a section map:
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<math>Hom(\mathcal{E},\mathcal{T}) \to \mathcal{D}^1(\mathcal{E},A)</math>
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which is equivalent to a map (satisfying some additional conditions):
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<math>\mathcal{T} \to \mathcal{E} \otimes \mathcal{D}^1(\mathcal{E},\mathcal{A})</math>
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The right side is equivalent to <math>\mathcal{D}^1(\mathcal{E},\mathcal{E})</math>, so a splitting of the sequence is equivalent to a map:
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<math>\mathcal{T} \to \mathcal{D}^1(\mathcal{E},\mathcal{E})</math>
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satisfying some additional conditions. Clearly, a '''connection''' is also a map of the above form, so it remains to check that the additional condition that comes from it being a splitting, is equivalent to the Leibniz rule for the connection.

Revision as of 19:43, 6 April 2008

Statement

Suppose E is a vector bundle over a differential manifold M. Denote by \mathcal{E} the sheaf of sections of E. Consider the first-order symbol sequence for E, given by:

0 \to Hom(\mathcal{E},\mathcal{A}) \to \mathcal{D}^1(\mathcal{E},\mathcal{A}) \to Hom(\mathcal{E},\mathcal{T}) \to 0

Here \mathcal{A} is the sheaf of infinitely differentiable functions on M, \mathcal{D}^1 denotes the space of first-order differential operators from \mathcal{E} to \mathcal{A}, and \mathcal{T} denotes the sheaf of derivations of M.

A connection on E is equivalent to a choice of splitting for this sequence.

Proof

A splitting of the above sequence is equivalent to a section map:

Hom(\mathcal{E},\mathcal{T}) \to \mathcal{D}^1(\mathcal{E},A)

which is equivalent to a map (satisfying some additional conditions):

\mathcal{T} \to \mathcal{E} \otimes \mathcal{D}^1(\mathcal{E},\mathcal{A})

The right side is equivalent to \mathcal{D}^1(\mathcal{E},\mathcal{E}), so a splitting of the sequence is equivalent to a map:

\mathcal{T} \to \mathcal{D}^1(\mathcal{E},\mathcal{E})

satisfying some additional conditions. Clearly, a connection is also a map of the above form, so it remains to check that the additional condition that comes from it being a splitting, is equivalent to the Leibniz rule for the connection.