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I can easily see a proof that shows its going to be an immersed submanifold . (I am removing the case if the vector field at that point is 0). I am not able to see if it's a embedded submanifold or not? Thank you.

Illuminata
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In general, no, absolutely not! Consider $M = \Bbb R^2/\Bbb Z^2$ and $X= \partial/\partial x + a\partial/\partial y$, where $a$ is irrational. Then an integral curve of this is a line with irrational slope, which is dense in $M$. About as far from embedded you can get! You should expect that the "generic" vector field has non-embedded integral curves.

  • Will you please elaborate the last statement sir ? – Illuminata Feb 06 '16 at 17:55
  • @Illuminata: It's not a rigorous statement, though it could be made to be. I just mean that if I hand you a "random" vector field, you should expect that it has non-embedded integral curves. It's rare for all your integral curves to be embedded. –  Feb 06 '16 at 17:56
  • Okk ...I understood. ..thanks a lot. – Illuminata Feb 06 '16 at 17:58