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Given a smooth vector field on a closed smooth manifold, does the flow of that vector field exists for all time $t \in R$? This may be a well-known and elementary conclusion, but I just wnat to confirm it to me.

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    If "closed" means compact, then yes.2010-10-23
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    If "closed" does not mean compact, then no. $\dot x=x^2$2010-10-23
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    Usually *closed*, as applied to manifolds, means *compact and without boundary*. (On a compact manifold with boundary it is clear that solutions need tobe defined on all of $\mathbb R$: consider for example on the manifold $[0,1]\times S^2$ the vector field $d/dr$ with hopefully self-evident notation)2010-10-23

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