Talk:Parallelizable manifold
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Questions
editIs there a relation between parallelizable and orientable?
What is the rational for the name, or rather how does it relate with the elementary concept of "parallel" lines?
What are some (physical?) consequences of a manifold being or not being parallelizable?
Shouldn't this page mention whether TM=MxR^n?
Cesiumfrog (talk) 03:53, 7 August 2009 (UTC)
- A parallelizable manifold is a smooth manifold whose tangent bundle is trivial. The class of parallelizable manifolds is an important class which includes all Lie groups. For instance, although the circle is a parallelizable manifold, the sphere is not by the Hairy ball theorem. Often the algebraic topology of a manifold has a deep connection with its tangent bundle, and thus parallelizability also has such a deep connection. The möbius strip is not a parallelizable manifold due to an overall "twist" in its structue; furthermore, this manifold is also not orientable. The following site, [1] consists of some information on parallelizable manifolds. Hope this answers your question. --PST 06:42, 7 August 2009 (UTC)
- There might be some expansion required, but TM=MxRn up to bundle isomorphism is a consequence of the definition of trivial bundle, so is mentioned. "Physical" consequences assumes there is some physics of the differential topology to which the concept belongs. Starting from the trivial tangent bundle, various characteristic classes will also be trivial, e.g. the Stiefel-Whitney classes. It is proved that M must have Euler characteristic equal to 0 (Milnor-Stasheff book on characteristic classes). One book talks about a "globsl notion of parallel transport" as a differential-geometry way of looking at the issue. Charles Matthews (talk) 07:18, 7 August 2009 (UTC)
Merge into Parallelization (mathematics)?
editIt seems to me the two should be the same article: they say essentially the same thing. This should perhaps be a section of the other. — Quondum 17:02, 2 September 2012 (UTC)
Potential mistake
editThe article claims, in the examples, that every orientable 3-manifold is parallelizable. Is this correct? I know that every orientable 3-manifold has one non-vanishing vector field, but this is not enough to conclude. — Preceding unsigned comment added by 129.236.182.167 (talk) 14:19, 9 February 2022 (UTC)
- The assumption of the manifold being closed was missing. I also added a reference with some proofs. - Saung Tadashi (talk) 16:35, 12 February 2022 (UTC)