Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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Every connection on a riemannian vector bundle is metric compatible

Statement

Every connection on a vector bundle equipped with a Riemannian bundle metric is compatible with that metric.

Facts & Assumptions

Given: The asserted automatic compatibility with the supplied metric.

[F1]

Compatibility requires Xh(s,t)=h(Xs,t)+h(s,Xt) for all local sections (Metric compatible connection on a riemannian vector bundle).

[F2]

A smooth matrix of one-forms in a global frame defines a connection (Local connection forms glue exactly when they obey the transformation law).

Refutation

1.1

On E=R×R take the usual fibre metric h(ue,ve)=uv for its constant unit frame e, and the connection (ue)=(du+udx)e. Its coefficient dx is smooth, so [F2] makes this a connection. It has xe=e.

F2given
2.1

With s=t=e and X=x, the left side of [F1] is x1=0 and the right side is h(e,e)+h(e,e)=2. Thus the equality fails at every point, even though the metric is positive definite. A zero section would give no discrepancy; the unit section is an explicit witness. Compatibility is a condition additional to the existence of a metric and a connection.

F1step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources