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Mathematics, 06.06.2021 22:50 SillyEve
A compact, oriented manifold M possesses a no-where
vanishing vector field if and only if its Euler characteristic is zero.
Proof. The idea of the proof is to start off with a vector field with
only finitely many zeros, and use an isotopy to move all the zeros to
single coordinate chart. Then the Hopf Degree theorem and a converse
to the Extension Lemma is used to homotope the vector field to one
without zeros, while leaving it constant outside the coordinate chart.
See Guillemin and Pollack ([2], p.144) for details, as well as the outline
of proofs of the Hopf degree theorem and the converse to the Extension
Lemma. ¤
Since every manifold M2k+1 of odd dimension has zero Euler characteristic (χ(E) = I(M, M) = (−1)2k+1I(M, M) = −χ(E)), this proves
that every odd dimensional manifold has a no-where vanishing vector
field. In fact, this theorem represents a complete answer to the question: When does a manifold posses a no-where vanishing vector field?
References
[1] K. Burns and M. Gidea, Differential Geometry and Topology With a View to Dynamical Systems. Chapman and Hall/CRC, 2005.
[2] V. Guillemin and A. Pollack, Differential Topology. Prentice-Hall, 1974.
[3] R. Bott and L. Tu, Differential Forms in Algebraic Topology. Springer, New
York, 1982.
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A compact, oriented manifold M possesses a no-where
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